Satellite Data: coolest monthly tropics temperature in over 10 years

From Dr. Roy Spencer:

The Version 6.0 global average lower tropospheric temperature (LT) anomaly for June, 2022 was +0.06 deg. C, down (again) from the May, 2022 value of +0.17 deg. C.

Tropical Coolness

The tropical (20N-20S) anomaly for June was -0.36 deg. C, which is the coolest monthly anomaly in over 10 years, the coolest June in 22 years, and the 9th coolest June in the 44 year satellite record.

The linear warming trend since January, 1979 still stands at +0.13 C/decade (+0.11 C/decade over the global-averaged oceans, and +0.18 C/decade over global-averaged land).

Various regional LT departures from the 30-year (1991-2020) average for the last 18 months are:

YEAR MO GLOBE NHEM. SHEM. TROPIC USA48 ARCTIC AUST 
2021 01 0.12 0.34 -0.09 -0.08 0.36 0.50 -0.52
2021 02 0.20 0.32 0.08 -0.14 -0.66 0.07 -0.27
2021 03 -0.01 0.13 -0.14 -0.29 0.59 -0.78 -0.79
2021 04 -0.05 0.05 -0.15 -0.28 -0.02 0.02 0.29
2021 05 0.08 0.14 0.03 0.06 -0.41 -0.04 0.02
2021 06 -0.01 0.30 -0.32 -0.14 1.44 0.63 -0.76
2021 07 0.20 0.33 0.07 0.13 0.58 0.43 0.80
2021 08 0.17 0.26 0.08 0.07 0.32 0.83 -0.02
2021 09 0.25 0.18 0.33 0.09 0.67 0.02 0.37
2021 10 0.37 0.46 0.27 0.33 0.84 0.63 0.06
2021 11 0.08 0.11 0.06 0.14 0.50 -0.43 -0.29
2021 12 0.21 0.27 0.15 0.03 1.63 0.01 -0.06
2022 01 0.03 0.06 0.00 -0.24 -0.13 0.68 0.09
2022 02 -0.00 0.01 -0.02 -0.24 -0.05 -0.31 -0.50
2022 03 0.15 0.27 0.02 -0.08 0.22 0.74 0.02
2022 04 0.26 0.35 0.18 -0.04 -0.26 0.45 0.60
2022 05 0.17 0.24 0.10 0.01 0.59 0.23 0.19
2022 06 0.06 0.07 0.04 -0.36 0.46 0.33 0.11

The full UAH Global Temperature Report, along with the LT global gridpoint anomaly image for June, 2022 should be available within the next several days here.

The global and regional monthly anomalies for the various atmospheric layers we monitor should be available in the next few days at the following locations:

Lower Troposphere: http://vortex.nsstc.uah.edu/data/msu/v6.0/tlt/uahncdc_lt_6.0.txt
Mid-Troposphere: http://vortex.nsstc.uah.edu/data/msu/v6.0/tmt/uahncdc_mt_6.0.txt
Tropopause: http://vortex.nsstc.uah.edu/data/msu/v6.0/ttp/uahncdc_tp_6.0.txt
Lower Stratosphere: http://vortex.nsstc.uah.edu/data/msu/v6.0/tls/uahncdc_ls_6.0.txt

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fretslider
July 1, 2022 9:43 am

It isn’t that globally heated in Southern England!

John Tillman
Reply to  fretslider
July 1, 2022 1:55 pm

Arctic sea ice extent yesterday was higher than on that date in 9 of the previous 12 years.

Reply to  John Tillman
July 1, 2022 4:09 pm

As these clowns discovered

https://youtu.be/QKUX3OqD7p8

Dave Fair
Reply to  Phil Salmon
July 1, 2022 4:40 pm

I didn’t see Bozo in the shot.

Gerry, England
Reply to  fretslider
July 4, 2022 5:09 am

Certainly isn’t. Seems a lot like last summer where although the sun was shining there was always quite a breeze and if there were clouds around it was very cool. I was outside this weekend at a show and a festival and it was long sleeved shirt with a body warmer and in lengthy cloudy spells it was still chilly.

Coolengineer
July 1, 2022 9:44 am

… anomaly for May, 2022 was +0.06 deg. C, down (again) from the May, 2022 value…

You mean down from the May 2021 value, no?

SMC
Reply to  Coolengineer
July 1, 2022 9:58 am

Typo. I think he meant to say, ‘June’, 2022 was +0.06…

Well, anyway, that’s how I read it.

Mario Lento
Reply to  SMC
July 1, 2022 10:14 am

No, the +0.06 degree above the baseline, is up less than… and that means down.

Johannes Vermeer
Reply to  Coolengineer
July 1, 2022 10:02 am

“Anomaly for June, 2022″… the error has been copied from Roy Spencer’s site.

Editor
Reply to  Coolengineer
July 1, 2022 12:49 pm

Fixed it.

Thanks for pointing it out.

It is June, 2022

Last edited 1 month ago by Sunsettommy
Dave Fair
Reply to  Coolengineer
July 1, 2022 4:51 pm

Actually, no: The quote is: “The Version 6.0 global average lower tropospheric temperature (LT) anomaly for June, 2022 was +0.06 deg. C, down (again) from the May, 2022 value of +0.17 deg. C.”

The June value of +0.06 ℃ was down from May’s +0.17 ℃, which in itself was down from April’s +0.26 ℃. The sentence was grammatically correct and accurately reflected a continuing decline over a three-month period.

JMurphy
July 1, 2022 9:55 am

And yet the global anomalies have been consistently higher than last year since March.

Doonman
Reply to  JMurphy
July 1, 2022 10:53 am

But they are way cooler since 2015. It must be because increasing CO2 in the atmosphere has a cooling effect. Oh wait. The heat is hiding in the ocean. Yeah, thats it.

ResourceGuy
Reply to  Doonman
July 1, 2022 11:02 am

It’s whatever the rap sheep for the week says.

MarkW
Reply to  Doonman
July 1, 2022 11:24 am

Whatever time period that shows warming, is the only one that matters.

Dave Fair
Reply to  MarkW
July 1, 2022 4:57 pm

What about the time period post-2016 that shows dramatic cooling? Oh, I forgot … Leftist thinking is rigidly only one way.

Rhoda R.
Reply to  Dave Fair
July 1, 2022 7:42 pm

If the temps are rising that’s catastrophic climate change, if the temps are falling that’s weather.

Derg
Reply to  JMurphy
July 1, 2022 11:40 am

Why are we so cold then?

gbaikie
Reply to  Derg
July 1, 2022 12:58 pm

That has always been the question.
It started with why do we have glaciation period, or what was generally
called Ice Age.
But then we know we in a 33.9 million year, Ice Age called the Late Cenozoic Ice Age. And within this Ice Age, the last 2.5 million years has been the coldest.And everyone knows we headed for the next glaciation period.
But warming from CO2 is suppose to delay this.
But we don’t know how much warming has been caused by CO2 or how much
will be caused by any increase of CO2 levels.
What we do know, is it has less the projected.
And know our ocean is still cold.

JMurphy
Reply to  Derg
July 1, 2022 2:56 pm

You obviously don’t live in Europe! We are baking hot here so I feel for you being so cold. What temperature have you got?

rah
Reply to  JMurphy
July 1, 2022 3:08 pm

Yea! That CO2 sure gets around.

gbaikie
Reply to  JMurphy
July 1, 2022 4:25 pm

I live fairly near hottest recorded daytime temperature, which occurred:
“Official world record remains 134°F at Furnace Creek in 1913”
Or I live in a desert. Today temperature peaked around 95 F but in last week
it reached 103 F, is suppose reach 102 [39 C} next Friday. But it’s so far been a cool summer, and later in summer we get 110 F or warmer. Or it was about last summer, and was cool summer in general.
But this has nothing to do with global warming. Or because it was 134 F in 1913 does indicate there has been no global warming since then.
Global cooling is drier world, warming is wetter. Or during coldest time in glaciation period, we could have been breaking daytime high temperature- because this desert could manage to become drier.
It terms of average temperature it is about 15 C here, In Europe it averages about 9 C, US about 12 C, China about 8 C, and India around 25 C. It gets hotter here than in India. But India lacks a winter, and has warmer nights.
Or Furance Creek also gets colder than where live. Furnace Creek might get colder than whatever you live. Europe is warmed by the Gulf Stream, otherwise you have temperatures like Canada, which average around -3 C yearly. A bigger factor in terms daytime [and night temperatures] could Urban Heat island effects [if in or near cities].
Or I live in half the world that does not cool down a lot, if in a Glaciation period- which also prevents the gulf stream from warming Europe.

Redge
Reply to  JMurphy
July 1, 2022 11:15 pm

Not in my part of Europe

JMurphy
Reply to  Redge
July 3, 2022 6:38 am

Which part of Europe is that?

Redge
Reply to  JMurphy
July 3, 2022 8:36 am

The UK

JMurphy
Reply to  Redge
July 4, 2022 10:23 am

You need to get up to date: the UK is not a part of Europe either geographically or politically. It is an island off the European continent mostly under the influence of maritime polar air from the north-west, i.e. even though temperatures are gradually generally rising, as in the rest of the world, the UK is still affected by cool, moist air. However, continental Europe (a lot of which is currently baking hot and dry) will try to send you some warmth if you send some cool, moist weather back in return.

Redge
Reply to  JMurphy
July 4, 2022 10:43 am

Say what?

The UK is most definitely part of the European continent and is indeed within the Eurasian tectonic plate.

John Tillman
Reply to  JMurphy
July 4, 2022 1:02 pm

It’s only an island during interglacials. Until two floods about 425 and 225 Ka, it was an island even during glaciations.

comment image

Last edited 1 month ago by John Tillman
Dave Fair
Reply to  JMurphy
July 1, 2022 4:58 pm

But the trend is downward.

JMurphy
Reply to  Dave Fair
July 3, 2022 6:39 am

Which trend?

Josh Scandlen
July 1, 2022 10:14 am

is that pic the fake Blue Marble from Nasa?

Dave Fair
Reply to  Josh Scandlen
July 1, 2022 5:00 pm

I sincerely hope that was sarcasm.

Ron Long
July 1, 2022 10:15 am

What’s with this cold trend? As David Letterman said “it was so cold in Central Park that the squirrels were heating their nuts with hair dryers”. Here in Argentina it is off to an unusually cold start to winter. Miami Beach, anyone?

Eben
Reply to  Ron Long
July 1, 2022 11:07 am

Cooling is the new warming

Fraizer
Reply to  Ron Long
July 1, 2022 7:32 pm

So cold that politicians had their hands in their own pockets.

Julian Flood
Reply to  Fraizer
July 1, 2022 10:50 pm

In wish I’d said that.

JF

Robert Wager
Reply to  Ron Long
July 4, 2022 8:08 am

West Coast of Canada has had a very cold winter and cold/wet spring and summer is cool. La Niña sure cools things off on a grand scale.

Derg
July 1, 2022 10:31 am

Is that why MN experienced the 4th coldest April ever?

Bruce Cobb
July 1, 2022 11:21 am

Here in central New Hampshire as of July 1, I have not yet had to put in the window AC. Cool.

Reply to  Bruce Cobb
July 1, 2022 3:27 pm

Lived down near Rindge for about six years back in the 1980s – that’s only somewhat unusual. Although when I lived there, we had one summer where we had several days that went over 100. (If I hadn’t known better, I would have said the humidity was over 100, too. Bogs and lakes all over the place…).

But – wife and I went through there in the 90s, while back for a friend’s June wedding in MA. We got up there from Boston – and had to run into the Walmart just over the border for thick sweaters.

Bruce Cobb
Reply to  writing observer
July 2, 2022 7:45 am

Yep, we had some quite cool weather last month, with daytime highs sometimes only in the low to mid 50’s. We do get some effect from the ocean, as we are only about 40 miles away.

July 1, 2022 11:24 am

From my updated Blogpost

The End of the UNFCCC /IPCC Global Warming Meme is Confirmed by the Arctic Sea Ice.
1.The Millennial Global Temperature Cycle.
Planetary orbital and solar activity cycles interact and combine to drive global temperatures. Because of the thermal inertia of the oceans there is a 12+/- year delay between these drivers and global temperature. The amount of CO2 in the atmosphere is 0.058% by weight. That is one 1,720th of the whole atmosphere. It is inconceivable thermodynamically that such a tiny tail could wag so big a dog. The Oulu galactic cosmic ray count provides a useful proxy for driver amplitude. 
The statements below are supported by the Data, Links and Reference in parentheses ( ) at   https://climatesense-norpag.blogspot.com/2021/08/c02-solar-activity-and-temperature.html
A Millennial Solar Activity Turning Point (MSATP) was reached in 1991/2.The correlative temperature peak and Millennial Temperature Turning Point (MTTP ) was in 2004 as also reported in Nature Climate Change Zhang, Y., Piao, S., Sun, Y. et al. Future reversal of warming-enhanced vegetation productivity in the Northern Hemisphere. Nat. Clim. Chang. (2022) .(Open Access)
Because of the thermal inertia of the oceans the UAH 6.0 satellite Temperature Lower Troposphere anomaly was seen at 2003/12 (one Schwab cycle delay) and was + 0.26C.(34) The temperature anomaly at 06/2022 was +0.06C (34).There has been no net global warming for the last 18 years. Earth passed the peak of a natural Millennial temperature cycle trend in 2004  and will generally cool until 2680 – 2700…………….
See more at http://climatesense-norpag.blogspot.com/

Macha
Reply to  Norman J Page
July 3, 2022 6:18 pm

As far as correlations go, this appears pretty good…and simple .

https://reality348.wordpress.com/2021/06/14/the-linkage-between-cloud-cover-surface-pressure-and-temperature/

 As albedo falls away, temperature increases. The relationship is watertight. No other influence needs to be invoked other than ENSO which throws a spanner in the works unrelated to the underlying change in the Earths energy budget.

t-and-a-mid-latitudes-2.png
Simonsays
July 1, 2022 12:39 pm

Can I ask, how does these monthly results compare with the other temperature records being kept?

bdgwx
Reply to  Simonsays
July 1, 2022 2:10 pm

Here is the comparison. Notice that the agreement between datasets improved significantly in April and May with closeness being the best it has ever been. I’m currently investigating the anomaly.

comment image

Last edited 1 month ago by bdgwx
Lance Wallace
Reply to  bdgwx
July 1, 2022 4:06 pm

So if CO2 controls temperature rate, and it hardly varies by whether above land or ocean, why then is the rate over land (0.18 oD/decade) 50% higher than over ocean (0.12 oC/decade)?

Maybe UHI?

Richard M
Reply to  Lance Wallace
July 1, 2022 5:13 pm

UHI homogenized over rural areas.

bdgwx
Reply to  Lance Wallace
July 1, 2022 5:24 pm

It is a great question.

There are a few reasons why land warms faster than the ocean.

1) The ocean has a much larger thermal inertia than land. The same amount of energy uptake by land causes a bigger temperature increase than the uptake by the ocean.

2) The ocean has a small lapse rate above it than land. This is due to the fact the atmosphere above the ocean contains more water vapor since there is no shortage of water that can be evaporated into it. This lapse differential promotes faster warming over land as compared to the ocean.

3) In a warming environment evaporation or latent flux is one mechanism in which the surface can shed excess energy. The ocean’s latent flux is higher than that of land due to the abundance of water that can be evaporated. This helps keep the warming pressure muted relative to land.

4) There are many other secondary players as well including winds, how the land is distributed globally, etc. that may be making it more favorable for land to warm faster, but I believe these effects are minor compared to the those mentioned above.

The UHI effect (not be confused with the UHI bias) enhances temperatures in urban areas. Agriculture suppresses temperatures in rural areas. The net effect of all land use changes ends up being mostly a wash with perhaps an ever so slight cooling effect if anything.

Last edited 1 month ago by bdgwx
RickWill
Reply to  Lance Wallace
July 1, 2022 6:25 pm

why then is the rate over land (0.18 oD/decade) 50% higher than over ocean (0.12 oC/decade)?

The precession cycle has been shifting peak surface solar intensity further north for about 500 years and has another 9,000 years to go before perihelion of Earth’s orbit occurs in July. This, combined with the fact that 10% of the Southern Hemisphere is land while 50% of the Northern Hemisphere is land, means the average solar intensity over land is increasing while the average intensity over water is reducing.

Ocean surfaces cannot exceed 30C so more ocean surface is getting to 30C due to warmer land masses reducing the moist air convergence from the oceans. Land runoff back into oceans is in long term decline.

The next observation will be accumulation of snow on northern land masses as the winters get colder – the flip side of warmer summers. This will emerge in the next 1000 years. Earth is already in the early stages of the current cycle of glaciation. Ice accumulation rate will peak in 9,000 years, rate will reduce for 10,000 years but oceans will not recover to current level before they continue to go down.

Climate is always changing and CO2 has no direct influence. The indirect influence is quite profound:
https://earth.nullschool.net/#2021/12/31/0800Z/wind/surface/level/overlay=relative_humidity/orthographic=-81.46,0.62,388/loc=-71.880,-3.303
Rainforests create ocean like conditions over land. Chopping down trees to erect wind turbines is the worst possible environmental vandalism. It will turn the globe into the Sahara Desert. The resultant dust will prevent glaciation though.

Johne Morton
Reply to  RickWill
July 1, 2022 7:37 pm

Also, and I lost the link to the relevant data, so if anyone could provide a link, that would be great, but I’ve seen data showing that Northern Hemisphere aerosols have dropped significantly because we have cleaner energy, plus the last great low-latitude volcanic eruption was Mt. Pinatubo in 1991. Basically, more sunlight is hitting the N. Hemisphere.

Julian Flood
Reply to  Johne Morton
July 1, 2022 11:21 pm

The water surfaces of seas, lakes, rivers are being coared with oil and surfactant which decreases evaporation and lowers albedo. Sewage and farming run-off feed phytoplankton, particularly diatoms which release lipids when blooms die. Both of these effects are greater in the NH.

See TCW Defending Freedom for a guess about mechanisms.

JF
My DIL threatens to buy me a turquoise tracksuit like David Icke, the man who believes we are ruled by shape changing aliens. I tell my theory can be tested.

Julian Flood
Reply to  Lance Wallace
July 1, 2022 11:09 pm

Much more interesting is why are some parts of our water planet warming much more than others?

Examples: Lakes Michigan, Superior, Tanganyika; Seas Mediterranean, Black, Baltic, Marmora. The latter is warming at double the global rate, probably because there’s a huge cloud of CO2 hovering over it. Or it may be because of the surface and subsurface pollution.

I have a post on the UK blog TCW Defending Freedom, Are We Smoothing our Way to a Warmer Planet? where I suggest the mechanisms causing that.

There’s a section on here where Anthony asks for post ideas but no-one looks at it. My suggestion would be to examine the data on those anomalies.

Nearly three quarters of the Earth’s surface is covered in water. Our planet is ruled by Oceana, not Gaia.

JF

Dave Fair
Reply to  bdgwx
July 1, 2022 5:13 pm

Just from memory, the linear trend of the CMIP5 average of all ensembles was much greater (double?) than the average of each of the radiosonde and satellite datasets. Without explanation of your graph, I’ll go with my memory, bdgwx.

bdgwx
Reply to  Dave Fair
July 1, 2022 6:09 pm

Here is what the various radiosonde datasets show. The average of RATPAC, RAOBCORE, and RICH is +0.20 C/decade. CMIP5 is deviates by +0.03 C/decade. UAH deviates by -0.07 C/decade. That means the CMIP5 prediction of the trend is a better match to radiosondes than the UAH trend observation.
.
comment image

[Christy et al. 2020]

Jim Gorman
Reply to  bdgwx
July 2, 2022 4:58 am

What is the variance or standard deviation of the absolute temps used to calculate the anomalies and what is the variance or standard deviation of the anomalies?

An average of a distribution has no meaning unless you quote the other statistical descriptors that define the distribution. Skewness and kurtosis for each along with the variance of the above would be even better.

Simonsays
Reply to  bdgwx
July 2, 2022 12:49 am

Thanks for that, the differences between most of them looks like splitting hairs. So what is the bottom line? Are we living in the fastest warming period in recent history or has the data pulled the bottom out of that argument?

AGW is Not Science
Reply to  Simonsays
July 2, 2022 12:15 pm

The fact that you need to qualify your last question with ‘recently’ is telling, since anything NOT “recent” can only be determined from proxy measurements that average changes over longer periods of time and hence cannot provide the ‘resolution’ of today’s instrument records.

Yet we can STILL find more rapid warming in the paleoclimate record.

So the answer is “It doesn’t matter.” It isn’t “unprecedented” in any way nor is there ant empirical evidence that CO2 is the driver – just like always.

Tom Abbott
Reply to  Simonsays
July 2, 2022 2:56 pm

“So what is the bottom line? Are we living in the fastest warming period in recent history or has the data pulled the bottom out of that argument?”

There are several periods in the recent past that had the same magnitude of warming as we have had in the satellite era (1979 to present).

comment image

bigoilbob
Reply to  Tom Abbott
July 4, 2022 7:53 am

Look at the post 1980 trend, (including the pause) and see if you can find another 42 year period with anything approaching it.

Mary Brown
July 1, 2022 12:45 pm

June should be a touch higher than May… but not significant

Dave Fair
Reply to  Mary Brown
July 3, 2022 10:21 am

June’s anomaly was lower than May’s.

Tom Abbott
Reply to  Dave Fair
July 3, 2022 11:29 am

More CO2 in the atmosphere, yet the temperatures continue to cool.

How do alarmists explain this? To hear them tell it, the more CO2 in the atmosphere, the hotter it should be. But then reality intrudes.

Dave Fair
Reply to  Tom Abbott
July 3, 2022 1:07 pm

The coincidence of rising temperatures and increasing atmospheric CO2 concentrations during the late 20th Century allowed CliSciFi practitioners to gin up climate models that predicted rising future temperatures with expected increases in atmospheric CO2 concentrations. Playing with historic aerosol levels to come up with unrealistically high values for ECSs, combined with wildly inaccurate CO2 emissions scenarios, enabled them to comply with their political paymasters’ demands for catastrophic anthropogenic global warming (CAGW) projections to justify radical socialists’ takeover of Western societies, economies and energy systems.

Leftist control of governments and media has prevented widespread publication of the failures of CliSciFi models to predict the approximately 20-year pause in global warming beginning in the late 1990s nor the fact that it took a Super El Niño to temporarily increase global temperatures, which are now falling.

Sadly, it appears that it will take the collapse of Western economies to wake up the low-information voters. Even then Leftist will try to blame it on those evil fossil fuels.

Robber
July 1, 2022 2:49 pm

But, but but, it’s a climate emergency/catastrophe/calamity/crash/debacle/fiasco.

Carlo, Monte
July 1, 2022 3:30 pm

Here are the UAH monthly data points replotted with generous uncertainty intervals of U(T) = ±1.4K. Notice that the standard deviation of the slope is 26% of the slope itself, and the correlation coefficient is just 0.47.

Also plotted is the histogram of the regression residuals, the shape of which indicates much of the month-to-month variation is random noise.

UAH LT globe 2023-06.jpg
Last edited 1 month ago by Carlo, Monte
Carlo, Monte
Reply to  Carlo, Monte
July 1, 2022 3:38 pm

The UAH 20-year baseline data, which are subtracted from the monthly averages, show some very unusual features. Here are the baseline data for June, shown as a global intensity map, along with the histogram of all the values. Notice they never get more than a few degrees above freezing, where there is a large spike of values.

Screen Shot 2022-06-30 at 6.58.51 PM.jpg
Carlo, Monte
Reply to  Carlo, Monte
July 1, 2022 4:06 pm

Another experiment with an unexpected result: I took the global anomaly map data for June 2021 (June 2022 has not been uploaded to the UAH archive yet), subtracted from it the anomaly map data for June 1980, and made another intensity map of the difference. The most surprising result to me was the appearance of hot or cold “blobs” at latitudes higher than ±50°. I’m guessing these represent near-polar weather patterns that persisted for significant periods of either month.

The histogram of the difference has a Gaussian shape, with a peak at about -0.3°C. The full-width-at-half-maximum is only ±1°C.

When I looked at other pairs of months (selected semi-randomly), they all showed similar hot/cold spots, but in different positions and polarities.

The tropics are remarkable in that I did not see any spots, the differences were mostly uniform.

Screen Shot 2022-06-30 at 7.00.04 PM.jpg
Richard Page
Reply to  Carlo, Monte
July 1, 2022 5:13 pm

Do these line up with high or low pressure areas?

Carlo, Monte
Reply to  Richard Page
July 1, 2022 6:04 pm

Possibly, it would require searching back through weather histories to verify. For this example, the Antarctic was either unusually warm in 1980, or unusually cold in 2021.

Last edited 1 month ago by Carlo, Monte
RickWill
Reply to  Carlo, Monte
July 1, 2022 5:31 pm

Another experiment with an unexpected result:

It is exactly what is expected. Orbital mechanics are slowly shifting peak solar intensity north so Southern Hemisphere is cooling and Northern Hemisphere warming. However tropical oceans are limited to 30C so they remain almost constant at that value.

What I find unexpected is that a large proportion of informed adults believe that there is a Greenhouse Effect influencing Earth’s temperature and CO2 is the demon molecule.

SSTvSolarEMR.png
Carlo, Monte
Reply to  RickWill
July 1, 2022 6:05 pm

Unexpected for me as I am not a climate scientist, but this makes sense.

RickWill
Reply to  Carlo, Monte
July 1, 2022 7:13 pm

I am not a climate scientist, 

That is obvious!. If it was not produced by a climate model then climate scientists treat observations as fantasy. Only curious observers point out such results as yours.

All climate models of the 2000 vintage predicted warm ocean surfaces would be regularly sustaining a temperature higher than 30C by now. All provably wrong and yet there are still people who believe these dullards.

Later vintage models were tuned to the temperature of the day so have these regions cooler than they are now but with steeper trends so they get the 2.8C warming by 2100; unless of course we change our ways and stop burning carbon.

Nino34_CSIRO_CIMP3.png
Carlo, Monte
Reply to  RickWill
July 1, 2022 8:44 pm

Another important point that is often overlooked—the UAH temperatures are from altitudes above the surface. For LT lower tropopause series, they are a convolution of the ~0-10km exponential temperature profile with the response function of the satellites’ microwave sounding units.

Reply to  RickWill
July 2, 2022 12:19 am

Rick
If you are referring to obliquity change, and/or precession, these Milankovitch orbital parameters change very slowly over tens of thousands of years so I doubt would cause changes that rise above noise over just a decade or two.

RickWill
Reply to  Phil Salmon
July 2, 2022 2:00 am

Precession is a 23,000 year cycle and is the dominant cycle behind the annual cycle of course. Earth is a bit more than 500 years past the last time perihelion occurred before the austral summer solstice. Perihelion is now advanced to early January so the changes are accelerating.

The changes eventually become enormous. In April 2020 had 2020 the northern land masses averaged 2.1W/m^2 more than in 1500. September was down by 2W/m^2. Over the year the northern land masses averaged 1.1W/m^2 in 2020 than 1500. This aligns quite well with what CO2 has purportedly done.

Jim Gorman
Reply to  Carlo, Monte
July 1, 2022 6:38 pm

Nice analysis’. You ever wonder why some of this doesn’t show up in climate scientists analysis of the trends? It seems like OLS and averaging is about all you ever see.

Carlo, Monte
Reply to  Jim Gorman
July 2, 2022 8:00 am

Which leads into more analysis:

The UAH process can be summarized as:

1) divide the globe into a spherical 144×66 grid, so that 360° / 144 longitude (= 2.5°) and 80° / 66 latitude (= 2.71°) (latitudes higher than ±85° are not reported)

2) note that because the lengths of latitude lines vary as the cosine, a grid at 5° is almost 6x larger in area than a grid at 80°

3) the satellites are in polar orbits with periods of about 90 minutes, so the scans are not continuous

4) collect all the satellite scan data and sort them into the grid points, and calculate the corresponding temperatures

5) collect all the temperatures during a single calendar month for each grid point

6) calculate the mean temperature for each grid point (the statistics from the mean are not kept)

7) collect the mean temperatures for each month over a twenty-year period (currently 1990-2010), and calculate the means at each grid point (statistics are not kept from these means); this produces a baseline temperature map for each month of the year

8) for the current month, subtract the corresponding baseline month from the means calculated in #6; this produces the monthly “anomaly”

9) calculate the global mean temperature anomaly for each month from the 66×144 grid points—these numbers are the points on the UAH global LT graph

Thus, each point on the graph is the result of three separate average calculations, but only the result of the last one is reported.

Carlo, Monte
Reply to  Carlo, Monte
July 2, 2022 8:36 am

What are magnitudes of the standard deviations of the means? Answering these questions requires digging through the UAH raw data archives. Starting from last one and working backward:

The anomaly map data are contained in these yearly files, here is the one for 2022 (does not yet hold the June 2022 data):

https://www.nsstc.uah.edu/data/msu/v6.0/tlt/tmtmonamg.2022_6.0

This is the anomaly map for May 2022; notice that the temperature range for the entire globe is quite small, less than ±4K, and the corresponding histogram very narrow. The peak is at +0.25K, while the standard deviation is 0.89K, 3x larger than the peak. Large weather patterns for the month are easily seen, however.

Screen Shot 2022-07-02 at 9.26.30 AM.jpg
Carlo, Monte
Reply to  Carlo, Monte
July 2, 2022 8:51 am

I posted the anomaly map for May above:

(https://wattsupwiththat.com/2022/07/01/satellite-data-coolest-monthly-global-temperature-in-over-10-years/#comment-3545869)

The standard deviation is quite large, 13K, and the distribution is completely non-normal.

Carlo, Monte
Reply to  Carlo, Monte
July 2, 2022 9:36 am

What about the grid point averaging? This information is not contained in the archive files, but some hints can be found in the satellite temperature publications. Figure 3 inside an RSS paper:

CA Mears, FJ Wentz, P Thorne, D Bernie, “Assessing uncertainty in estimates of atmospheric temperature changes from MSU and AMSU using a Monte-Carlo technique”, J of Geo Res, 116, D08112, doi:10.1029/2010JD014954, 2011

This is a plot of sampling by two satellites in 1980 (NOAA-11 & NOAA-12) at 40°N, 170°E in the Pacific Ocean, against a “mean of hourly data” generated by “the CCM3 climate model”. The important point to see here is that the sampling is discontinuous even at this mid-latitude location, and there are entire days without any points.

The authors point out that at high latitudes, grid points are sampled several times each day, but in the tropic as many as three days can lapse between samples.

From Fig. 3 here, at mid-latitudes the number of points in this month was 29 for NOAA-11 and 26 for NOAA-12. By redigitizing the points I was able to calculate for the two sets:

NOAA-11: T = 242.33K, sd(T) = 3.6K
NOAA-12: T = 242.17K, sd(T) = 3.6K

There is a lot of information in this paper (a lot more than I have time to study), which was an attempt to apply Monte Carlo methods to the issue of satellite temperature uncertainty; unfortunately the authors continued the common problem of thinking uncertainty is error, and ended up with comparing trends (see Fig. 13).

jgrd16826 fig 3.jpg
Carlo, Monte
Reply to  Carlo, Monte
July 2, 2022 9:50 am

One more to look at, a UAH paper:

JR Christy, RW Spencer, WB Norris, WD Braswell, DE Parker, “Error Estimates of Version 5.0 of MSU–AMSU Bulk Atmospheric Temperatures”, J of Atmos and Oceanic Tech, 20, 613-629, May 2003

From Sec. 3 (part of which is copied here) on comparison with radiosonde data; notice that sigmas are being divided by the sqr root of N, number of satellite points (26 in this case). This is completely inappropriate because these are (sparsely) sampled time series, so that multiple measurements of the same quantity are not being averaged. Also note the assumption of normal distributions and random errors, without any justifications being provided.

Screen Shot 2022-07-02 at 10.43.42 AM.jpg
Jim Gorman
Reply to  Carlo, Monte
July 2, 2022 11:28 am

Is it any wonder that scientists think their calculations are so accurate?

Dividing a sample standard deviation of samples by the √N tells you nothing

The SEM is the standard deviation of the samples and only tells how closely to the population mean you are. It is an interval around the sample mean, not how precision/resolution of the sample mean.

Here is a pertinent document

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2959222/#

Carlo, Monte
Reply to  Jim Gorman
July 3, 2022 5:10 am

Succinct and right to the point—notice the statement “and for data not following the normal distribution, it is median and range“—climate science certainly ignores this also.

Tim Gorman
Reply to  Carlo, Monte
July 3, 2022 8:13 am

They should provide the 5-number description:
minimum
1st quartile
median
3rd quartile
maximum

They totally ignore this.

Bellman
Reply to  Jim Gorman
July 3, 2022 1:16 pm

I’ve said before I agree with that letter. If you are using SEM to describe the variability in the sample, you are making a huge mistake.

But I don’t know how that leads you to the conclusion that therefore SEM tells you nothing. As the letter says, it’s an estimate of the precision of the sample mean, it gives an indication of the range of uncertainty for your sample mean compared with the true mean.

The SD won’t tell you that. It will tell you how much variation there is amongst the population, and that’s an important value to know in many cases, but it tells you nothing about how precise your sample mean is.

Jim Gorman
Reply to  Bellman
July 3, 2022 1:57 pm

For the millionth time.

From:

https://byjus.com/maths/standard-error/

“In statistics, the standard error is the standard deviation of the sample distribution. The sample mean of a data is generally varied from the actual population mean. It is represented as SE. It is used to measure the amount of accuracy by which the given sample represents its population.”

If you are dealing with a sample, the SD of that sample distribution is also the Standard error. An SD gives you an interval within which the population mean may lay. The precision of the mean is still determined by significant digits, not the spread of where the population may may actually be.

Bellman
Reply to  Jim Gorman
July 3, 2022 3:15 pm

Sorry. Can’t speak for Byjus as a whole, but that page and your quote is incoherent.

If you are dealing with a sample, the SD of that sample distribution is also the Standard error.

It really isn’t.

An SD gives you an interval within which the population mean may lay.

Of course the population mean lies within an interval centered on the population mean. And it probably lies within a sample SD of the sample mean, but only because it probably lies within the SEM of the mean.

The precision of the mean is still determined by significant digits, not the spread of where the population may may actually be.

Nonsense. Precision is defined by how closely different measurements agree. Applying that to sample means, the smaller the SEM the closer the sample means will be. Both links you’ve given state this. Byjust says

The standard error is calculated by dividing the standard deviation by the sample size’s square root. It gives the precision of a sample mean by including the sample-to-sample variability of the sample means.

Bellman
Reply to  Bellman
July 3, 2022 3:49 pm

If you are dealing with a sample, the SD of that sample distribution is also the Standard error

I think this might be where the problem lies. The standard error is the standard deviation of the sample distribution.

But you are talking here about a single sample, and I think assuming the SD of the sample distribution is the SD of that sample.

Tim Gorman
Reply to  Bellman
July 4, 2022 6:04 am

 Applying that to sample means, the smaller the SEM the closer the sample means will be.”

Neither of which addresses the accuracy of either the sample mean or the population mean.

It gives the precision of a sample mean by including the sample-to-sample variability of the sample means.”

Precision is not accuracy. That’s why I eschew the use of the term SEM. It is the standard deviation of the sample means. Nothing more. Even if all the sample means are the same, i.e. their standard deviation is zero, that doesn’t mean they are accurate. Accuracy can only be determined by propagating the uncertainty of the individual data elements into any value calculated from them.

Bellman
Reply to  Tim Gorman
July 4, 2022 11:16 am

I said nothing about accuracy, just the precision.

Tim Gorman
Reply to  Bellman
July 5, 2022 6:34 am

I said nothing about accuracy, just the precision.”

Precision without accuracy is meaningless. Something you just can’t seem to get through your head. If the true value is X and you very precisely calculate a value of Y, then of what possible use is Y?

Bellman
Reply to  Tim Gorman
July 5, 2022 9:56 am

Correct. Also if you correctly measure X, but your result is very imprecise, what use is it?

Tim Gorman
Reply to  Bellman
July 6, 2022 4:45 pm

Correct. Also if you correctly measure X, but your result is very imprecise, what use is it?”

You are still showing your lack of knowledge of the real world. Infinite precision is impossible in the real world.

If you need more precision then get a measuring device with more precision. That’s why I have both a 3 1/2 digit frequency counter and an 8 digit frequency counter. And you can get counters with even more digits of resolution if you are willing to pay the price.

It’s why I have an analog caliper marked in tenths of an inch as well as a micrometer. Different tools for different uses.

If you can reduce systematic uncertainty to a level lower than the precision of the measuring device then you can get a “true value” by taking multiple measurements of the same thing. You will still only be as precise as the measuring device allows but if that level of precision is all you need then that true value is perfectly usable. Take a crankshaft journal. You can only buy journal bushings in specified steps. You don’t need any more precision than is needed to determine which step to buy.

Bellman
Reply to  Tim Gorman
July 6, 2022 5:53 pm

You are still showing your lack of knowledge of the real world. Infinite precision is impossible in the real world.

Strawman time. I said nothing about infinite precision. In fact I was specifically talking about imprecise measurements.

If you need more precision then get a measuring device with more precision.

I think you’re talking about instrument resolution, not measurement precision here.

That’s why I have both a 3 1/2 digit frequency counter and an 8 digit frequency counter.”

That’s no guarantee of precision.

If you can reduce systematic uncertainty to a level lower than the precision of the measuring device then you can get a “true value” by taking multiple measurements of the same thing.

You can’t. Surely you are the ones who keep insisting it’s impossible to ever get a measurement with no uncertainty. All you are really saying here, I think, is that of the resolution of your instrument is worse than its precision or trueness, you will always get the same value, but this value will not be correct. You will have a systematic error caused by the the rounding to the nearest unit of resolution.

You will still only be as precise as the measuring device allows but if that level of precision is all you need then that true value is perfectly usable.

It’s not the true value. But you are right that it might be perfectly usable, just as a mean temperature can be perfectly usable even though it will never be the true value.

Tim Gorman
Reply to  Bellman
July 7, 2022 3:29 pm

Strawman time. I said nothing about infinite precision. In fact I was specifically talking about imprecise measurements.”

What is an “imprecise” measurement? Are you trying to say that a measurement of the form “stated value +/- uncertainty” is an imprecise measurement?

“I think you’re talking about instrument resolution, not measurement precision here.”

Resolution defines precision. It does not define uncertainty. You seem to be going down the path of trying to equate the two!

“That’s no guarantee of precision.”

And here we go again. Precision and uncertainty is the same thing! Go away troll!

“You can’t. Surely you are the ones who keep insisting it’s impossible to ever get a measurement with no uncertainty.”

It’s a basic truism. What you must do is make the uncertainty as small as possible. E.g. reduce systematic error to less than the precision of the measuring device and take multiple measurements of the same thing. In this manner you can reduce the standard deviation of the distribution of random measurements. You’ll never make that standard deviation zero but you can reduce to a level below the needed tolerance level.

All you are really saying here, I think, is that of the resolution of your instrument is worse than its precision or trueness, you will always get the same value, but this value will not be correct. “

Precision is *NOT* accuracy. If I calibrate my micrometer using a gauge block that is itself inaccurate I can get very precise measurements but they won’t be accurate!

You will have a systematic error caused by the the rounding to the nearest unit of resolution.”

NO! Accuracy is not precision. Error is not uncertainty. Two rules you continually refuse to internalize.

“It’s not the true value.”

When I said: “You are still showing your lack of knowledge of the real world. Infinite precision is impossible in the real world.” you claimed that was a strawman argument I made up.

But here you are saying it again!

Bellman
Reply to  Tim Gorman
July 7, 2022 3:50 pm

What is an “imprecise” measurement? Are you trying to say that a measurement of the form “stated value +/- uncertainty” is an imprecise measurement?

I mean a measurement that lacked precision. I wasn’t trying to give an exact definition, just pointing out that no measurement has an infinite precision.

Resolution defines precision. It does not define uncertainty. You seem to be going down the path of trying to equate the two!

Maybe we need to define these often imprecise terms. I keep trying to use the definitions in VIM, as I assume that’s what people who do this in the real world use.

Under those definitions resolution does not define precision. They’re different but related concepts. I am not equating precision and uncertainty.

“That’s no guarantee of precision.”

And here we go again. Precision and uncertainty is the same thing! Go away troll!”

Precision and uncertainty are not the same thing. I’ve no idea why you jump to that conclusion from my comment. And if you want me to go away, I’d suggest not responding to my every comment with so much nonsense in need of correcting.

Precision is *NOT* accuracy.

Why do you keep shouting out these mantra’s in response to comments where I’ve suggested no such thing.

NO! Accuracy is not precision. Error is not uncertainty. Two rules you continually refuse to internalize.

And again, mantras that have nothing to do with what I just said.

In the same way

You: “You will still only be as precise as the measuring device allows but if that level of precision is all you need then that true value is perfectly usable.

Me: “It’s not the true value.

You: “When I said: “You are still showing your lack of knowledge of the real world. Infinite precision is impossible in the real world.” you claimed that was a strawman argument I made up.

But here you are saying it again!

Where am I talking about infinite precision. You claimed it was possible to get a perfectly usable measurement within the need level of precision you required. I agreed but pointed out it wouldn’t be a true value as you claimed, and from that you deduce I’m claiming infinite precision is possible.

Tim Gorman
Reply to  Bellman
July 8, 2022 9:02 am

Precision and uncertainty are not the same thing. I’ve no idea why you jump to that conclusion from my comment.”

because you continue to conflate the precision of the mean with the uncertainty of the mean! Your words are meaningless until you internalize the difference.

Where am I talking about infinite precision. You claimed it was possible to get a perfectly usable measurement within the need level of precision you required. I agreed but pointed out it wouldn’t be a true value as you claimed, and from that you deduce I’m claiming infinite precision is possible.”

You claim that an 8-digt frequency counter doesn’t provide a precise measurement.

Do you remember saying: “That’s no guarantee of precision.”

You are either claiming that precision must be infinite or you are confusing precision with uncertainty as you usually do.



Carlo, Monte
Reply to  Tim Gorman
July 7, 2022 4:11 pm

He should be assigned a homework problem: develop an uncertainty interval for a commercial digital voltmeter from the manufacturer’s error specifications.

This would be interesting!

Tim Gorman
Reply to  Carlo, Monte
July 8, 2022 9:50 am

He won’t do it. He won’t do for either an analog or digital voltmeter.

Tim Gorman
Reply to  Bellman
July 4, 2022 5:59 am

“As the letter says, it’s an estimate of the precision of the sample mean, it gives an indication of the range of uncertainty for your sample mean compared with the true mean.”

Precision is *NOT* accuracy.

Precision doesn’t determine accuracy.

Accuracy is described by uncertainty, precision is not.

Bellman
Reply to  Tim Gorman
July 4, 2022 11:28 am

That’s why I said precision, not accuracy. It was in response to Jim saying that SEM was “…not how precision/resolution of the sample mean.”

Tim Gorman
Reply to  Bellman
July 5, 2022 6:35 am

That’s why I said precision, not accuracy. It was in response to Jim saying that SEM was “…not how precision/resolution of the sample mean.””

Again, of what use is precision without accuracy? You just get a very precise wrong number!

Bellman
Reply to  Tim Gorman
July 5, 2022 9:55 am

Stop deflecting. I said the SEM was a measure of precision in response to Jim saying it wasn’t. This wasn’t a question about how useful precision is.

But in answer to your question, for accuracy you need both precision and trueness. A precise but wrong figure is not accurate, a true but imprecise figure is not accurate. Increasing sample size is a way of increasing precision, there isn’t an easy statistical method for ensuring trueness; you just have to try to improve your methods and correct biases.

Tim Gorman
Reply to  Bellman
July 6, 2022 4:21 pm

Two different definitions of precision. Your sample means should not have more significant digits than the data used to calculate the means. That’s one definition of precision.

Standard deviation of the sample means gives you a measure of how data (the sample means) is dispersed around the mean calculated from the sample means. That is a different measure of precision. A standard deviation of zero means you have calculated the mean of the sample means very precisely – but they must all meet the significant digits precision as well.

But in answer to your question, for accuracy you need both precision and trueness.”

“Trueness”? In the physical world you will *NEVER* reach more accuracy than the measurement uncertainty allows.

“A precise but wrong figure is not accurate, a true but imprecise figure is not accurate.”

Malarky! Accuracy can be no more precise than the device uncertainty used to measure an object. That’s why measurements are always “stated value +/- uncertainty”. Based on your definition you can *never* have a true value because infinite precision is impossible with real world measuring devices. That’s just one more indication of how little you actually know about the real world. The real world exists outside of statistics textbooks but apparently you just can’t get that into your head.

Bellman
Reply to  Tim Gorman
July 6, 2022 5:43 pm

Sorry I thought we were meant to be using the correct terms as defined in the VIM. No matter, you are still wrong. SEM is a measure of precision, the more precise the measurement the average the more significant figures you can use. But we’ve been through all this too many times to have to repeat the arguments at this late stage.

Based on your definition you can *never* have a true value because infinite precision is impossible with real world measuring devices.

It’s not my definition. It’s the definition of metrology as specified in the GUM and VIM.

Tim Gorman
Reply to  Bellman
July 7, 2022 3:18 pm

SEM is a measure of precision, the more precise the measurement the average the more significant figures you can use.”

Malarky! Using more significant figures than the data provides is saying that you can somehow divine unknown values, I assume via messages from God.

As I pointed out to you, precision that is sufficient for the purpose can define a true value. You do *NOT* have to have infinite precision in order to know a true value. I can use a micrometer to measure a crankshaft journal and get a true value useful enough to order crankshaft bushings. I don’t need a micrometer with precision out to a million decimal points!

This is just one more delusion you have – just like the one that the standard deviation of sample means defines the uncertainty of the mean calculated from those sample means!

Carlo, Monte
Reply to  Tim Gorman
July 7, 2022 4:08 pm

I recall that when confronted with the absurdity that U(x_bar)—>0 as N—> infinity, all he could do is invoke some technobabble about “autocorrelation”.

Bellman
Reply to  Carlo, Monte
July 7, 2022 4:37 pm

Your recollection is as bad as your arguments. You kept insisting that the logic of reducing uncertainty with sample size implied that with an infinite sample size you could get zero uncertainty, and somehow that disproved the idea that sample size reduces uncertainty.

I pointed out it wasn’t possible in the real world as,

a) you couldn’t have an infinite sample, and as uncertainty only reduces with the square root of sample size, even trying to get close to zero was likely to be more effort than it was worth.

And b) this is only talking about uncertainty from random errors, and there would always be some systematic error, which would always mean uncertainty would be greater than zero.

I’ve no idea why I’d have mentioned autocorrelation in this regard. I expect you are misremembering as you kept ignoring all my points, but if you have a reference I’ll check what I said.

Tim Gorman
Reply to  Bellman
July 8, 2022 9:48 am

 You kept insisting that the logic of reducing uncertainty with sample size implied that with an infinite sample size you could get zero uncertainty, and somehow that disproved the idea that sample size reduces uncertainty.”

You simply can *NOT* reduce uncertainty with sample size. All you can do is increase precision, i.e. get closer and closer to the mean of the population. And precision is not uncertainty! Something you just can’t seem to get into your head!

If the mean of the population is uncertain due to propagation of uncertainty from the data elements onto the mean then getting closer to the value of that mean does *NOT* decrease the uncertainty of the mean you are trying to get closer to!

Once again you fall back into your old “uncertainty always cancels” meme even though you deny that you do!

 and as uncertainty only reduces with the square root of sample size, even trying to get close to zero was likely to be more effort than it was worth.”

Once again – conflating precision and uncertainty. You say you don’t but you DO IT EVERY TIME!

“And b) this is only talking about uncertainty from random errors, and there would always be some systematic error, which would always mean uncertainty would be greater than zero.”

Which means that larger samples do *NOT* decrease uncertainty of the mean!

Bellman
Reply to  Tim Gorman
July 8, 2022 3:41 pm

You simply can *NOT* reduce uncertainty with sample size. All you can do is increase precision, i.e. get closer and closer to the mean of the population.

If you are increasing precision you are reducing uncertainty. If you are getting closer and closer to the mean of the population you are reducing uncertainty.

If you don’t agree then you have to explain what definition of uncertainty you are using and why you think it is useful.

Once again – conflating precision and uncertainty. You say you don’t but you DO IT EVERY TIME!

Only if you ignore all the words I was using. You and Carlo seem to have a list of trigger words, and have to respond to respond with some variation of uncertainty is not error, or precision is not accuracy, regardless of what I’ve actually said..

Which means that larger samples do *NOT* decrease uncertainty of the mean!

No. It means they reduce it to the systematic part of the uncertainty.

Bellman
Reply to  Tim Gorman
July 7, 2022 4:28 pm

Malarky! Using more significant figures than the data provides is saying that you can somehow divine unknown values, I assume via messages from God.

What’s 2167 divided by 100? 2.167. Same number of significant figures.

I think what you want to say, this is the usual argument, it that you can’t use more decimal places that the original measurements. So would have to say that if any of the measurements making up the sum were integers the average would have to be given as an integer, and 2.167 is rounded to 2.

But all the data in the first average is from measured values, you don’t need to be God to know what it is.

But as I say the best way to do this is to calculate the uncertainty and get the appropriate number of figures from that.

As I pointed out to you, precision that is sufficient for the purpose can define a true value. You do *NOT* have to have infinite precision in order to know a true value.

You’re using the words “true value” in a way I don’t recognize. But as I still don’t know what relevance this and the rest of your comment have to my original point, I’ll leave it there.

Tim Gorman
Reply to  Bellman
July 8, 2022 10:17 am

What’s 2167 divided by 100? 2.167. Same number of significant figures.”

Really? 100 has 4 significant figures?

“I think what you want to say, this is the usual argument, it that you can’t use more decimal places that the original measurements. “

Nope! Not only do you misunderstand uncertainty you don’t understand significant digits.

In addition, your training in statistics is showing again since you don’t include uncertainty intervals with your numbers. E.g. 2167 +/- 2 and 100 +/- 1. Thus you are not quoting measurements which is the subject of the thread.

The last significant digit in a number calculated from uncertain numbers should usually be in the same digit as the uncertainty. In the measurements I’ve provided here that would be the units digit. So your answer would be 2, not 2.167.

If the measurements were 2167 +/- 0.2 and 100 +/- 0.1 then your calculated answer should be stated as 2.2. And your uncertainty would be +/- 0.3.

In fact, your answer should be limited to the quantity of significant digits in the least precise number. That would be 3 significant digits in the stated value of 100.

Thus when you divide the stated values your answer should be 2.17. That would then be further limited by the magnitude of the last significant digit in the uncertainty –> giving a stated value of 2.2.

But all the data in the first average is from measured values, you don’t need to be God to know what it is.”

If you are going to properly state the answer you *do* need to know the uncertainty because it determines the placement of the last significant digit in the stated value!

But as I say the best way to do this is to calculate the uncertainty and get the appropriate number of figures from that.”

But you didn’t quote the uncertainty. So I guess you are left to some how “divine” what it is using your magic “uncertainty divining rod”!

“You’re using the words “true value” in a way I don’t recognize”

Because you have no actual understanding of the real world. I even tried to educate you using the example of how you measure crankshaft journals and determine the size of bushings you need to fit properly. But you just ignore it all and go right on down your merry way never looking left or right to expand your horizons.

Bellman
Reply to  Tim Gorman
July 8, 2022 3:00 pm

Really? 100 has 4 significant figures?

No it has infinitely many significant figures. (To clear, I don’t like these sf rules, I don’t think they make much sense compared to proper uncertainty rules, but you’re the one who brought them up.)

Nope! Not only do you misunderstand uncertainty you don’t understand significant digits.

OK. So to be clear you don;t agree with the rule I keep being told that you can’t have more decimal places in a mean than in the measurements that made up that mean. That’s good to know.

In addition, your training in statistics is showing again since you don’t include uncertainty intervals with your numbers.

I really don’t have much training in statistics. But I thought the whole point of significant figure rules was to avoid explicit uncertainties. The uncertainty is assumed to be ±0.5 of what ever the last digit is.

The last significant digit in a number calculated from uncertain numbers should usually be in the same digit as the uncertainty. In the measurements I’ve provided here that would be the units digit. So your answer would be 2, not 2.167.

That’s what I said two paragraphs ago, and you said I was wrong.

If the measurements were 2167 +/- 0.2 and 100 +/- 0.1 then your calculated answer should be stated as 2.2. And your uncertainty would be +/- 0.3.”

How can there be a ±0.1 in the sample size? And doesn’t this just demonstrate the nonsense of your no division in the mean arguments. You know the sum to within ±0.2. The true sum could be anywhere between 2166.8 and 2167.2. When I divide that by 100 the result can be anywhere between 21.668 and 21.672 (Just noticed my mistake above). But you would say the act of dividing by 100 means the actual result is 21.7 ± 0.3, suggesting the true value could be as small as 21.4 or as large as 22.0.

In fact, your answer should be limited to the quantity of significant digits in the least precise number. That would be 3 significant digits in the stated value of 100.

Again, 100 is an exact value it has infinite significant figures.

If you are going to properly state the answer you *do* need to know the uncertainty because it determines the placement of the last significant digit in the stated value!

As I said the implied uncertainty of the sum was ±0.5, as indicated by the fact I’m only showing the result in integers. Following the sf rules for addition, the number of decimal places should be equal to the figure with the smallest number of places, so you could assume that all values were stated as integers with an implied uncertainty of ±0.5, which should make the sum much less certain, but as I say, I don’t like using these simplistic rules.

Tim Gorman
Reply to  Bellman
July 8, 2022 4:01 pm

No it has infinitely many significant figures. “

Final or trailing zeros are not counted as significant. 100 has one significant digit. That’s all! If no decimal point exists then the rightmost non-zero digit determines the number of significant digits. It is the *least* significant digit. Thus 1 is the least significant digit meaning 100 has one significant digit.

100.00 would have five significant digits.

” (To clear, I don’t like these sf rules, I don’t think they make much sense compared to proper uncertainty rules, but you’re the one who brought them up.)”

And you are the expert that decides how to use significant digits? Good to be you I guess.

“OK. So to be clear you don;t agree with the rule I keep being told that you can’t have more decimal places in a mean than in the measurements that made up that mean. That’s good to know.”

Nope! That *IS* the exact rule I gave you and just repeated above! You can’t have more significant digits than the elements used to find the mean! Doing so means you have somehow “divined” more precision in the mean than the elements in the mean provide!

“I really don’t have much training in statistics. But I thought the whole point of significant figure rules was to avoid explicit uncertainties. “

You don’t seem to have much training in anything used in the real world.

“The uncertainty is assumed to be ±0.5 of what ever the last digit is.”

That was the *OLD* way of determining part of the uncertainty. Like in 1900 when using a thermometer marked only in degrees. It was even more true back then because of having to consistently read the meniscus of the liquid in the thermometer which lead to parallax errors. And this was only true for READING errors. You still had to contend with systematic uncertainty in the device.

That’s what I said two paragraphs ago, and you said I was wrong.”

you said: “What’s 2167 divided by 100? 2.167. Same number of significant figures.”

And that is what I said was wrong. Along with not specifying the uncertainty of the measurements.

“How can there be a ±0.1 in the sample size?”

So you are saying 100 is a CONSTANT? You didn’t specify that! If it was a constant you really should have said 100. (with a period).

Again, 100 is an exact value it has infinite significant figures.”

But you didn’t state that! How was I to know?

You should have *still* specified an uncertainty for 2167 and that would determine last significant figure in the answer. If your uncertainty was in the tenth digit then your answer would still be 2.2.

As I said the implied uncertainty of the sum was ±0.5,”

That is an estimate to be used if you don’t know anything more about the uncertainty. It is determined by the resolution of your measurement device. Analog and digital measurement devices used different methods to determine resolution uncertainty.

This subject is covered pretty well here: https://www.isobudgets.com/calculate-resolution-uncertainty/

I’m not going to try and repeat all this here.

Bellman
Reply to  Tim Gorman
July 8, 2022 4:57 pm

100 has one significant digit.

http://www.ruf.rice.edu/~kekule/SignificantFigureRules1.pdf

Exact numbers, such as the number of people in a room, have an infinite number of significant figures

As I say, I don’t care for these rules. But if you do, I would hope you at least understood them.

And you are the expert that decides how to use significant digits? Good to be you I guess.

No, I just like to think for myself.

That was the *OLD* way of determining part of the uncertainty. Like in 1900 when using a thermometer marked only in degrees.”

That’s my point. These rules are outdated.

And that is what I said was wrong. Along with not specifying the uncertainty of the measurements.

No, the quote you said was wrong was:

“I think what you want to say, this is the usual argument, it that you can’t use more decimal places that the original measurements. “

So you are saying 100 is a CONSTANT? You didn’t specify that!

I was talking about an average/ I thought you could work that out. Is this period thing yet another “rule” that seems to vary from place to place? The document I quoted doesn’t show that.

That is an estimate to be used if you don’t know anything more about the uncertainty.

I’ve looked at numerous documents regarding these precious rules, mainly because of Jim insisting on them. None of them I can recall mentioned showing uncertainty intervals. My impression is you have two options, don;t show uncertainty and use the rules, or work out the uncertainty and use that as an indication as to how many digits you write.

I’m not going to try and repeat all this here.

You could at least point to where it mentions significant figure rules, because I can’t see it anywhere.

Carlo, Monte
Reply to  Bellman
July 8, 2022 6:30 pm

As I say, I don’t care for these rules.

Another indication you are a pseudoscientist.

Tim Gorman
Reply to  Bellman
July 9, 2022 6:21 am

I was talking about an average/ I thought you could work that out.”

And here we go again. An average has an uncertainty propagated from the individual elements. As such you should have shown that uncertainty. You are hopping around from saying 100 is a constant (like the number of elements) to saying it is an average. Which is it? Pick one and stick with it!

 Is this period thing yet another “rule” that seems to vary from place to place? The document I quoted doesn’t show that.”

You just said 100 is an average and not a constant. Which is it? If it is a constant then you have to indicate that! Using a period follow the value is a traditional way of showing that. If it is an average of measurements then it should have an uncertainty!

Again, pick one and stick with it!

“I’ve looked at numerous documents regarding these precious rules, mainly because of Jim insisting on them. None of them I can recall mentioned showing uncertainty intervals.”

Because you’ve never bothered to learn anything about how to handle measurements! Measurements follow significant digit rules but have their own rules as well. For instance, what does 3.756 +/- 0.1 tell you about the measurement? You have included more decimal places in the stated value than you can actually know based on the uncertainty! It should be shown as 3.8 +/- 0.1. That’s not necessarily a significant digit rule, it’s a measurement rule. “Don’t give the stated value more precision that you actually know!”

My impression is you have two options, don;t show uncertainty and use the rules, or work out the uncertainty and use that as an indication as to how many digits you write.”

You are stuck in your usual little box and unable to see outside of it. Expand your horizons for Pete’s sake! They both go together! It isn’t one or the other!

For instance, say you are determining the area of a table top. It measures 3.5″ +/- .1″ by 3.5″+/- .1″. That gives you an area of 3.5″ * 3.5″ = 12.25 sq in. Significant digit rules say that should be rounded to 12 sq in (two significant digits in each measurement). The uncertainties would add as 0.1/3.5 + 0.1/3.5 = 2/3.5 = 0.06. 0.06 * 12 = .7 using significant digit rules. So you *could* actually show the area as 12.3 sq in +/- 0.7 sq in using uncertainty rules but the significant digit rules apply in this situation so stick with 12 sq in +/- 0.7 sq in.

Now say you had the side measurements as 3.125″ +/- 0.1″. The area would work out to 9.765625sq in. Using significant figure rules we would round this to 9.765 sq in. The uncertainty would be 0.1/3.125 + 0.1/3.125 = 0.06 or 6%. 6% of .9765 sq in = .6 sq in. In this case the uncertainty rules would apply and the area should be stated as 9.8 sq in +/- 0.6 sq in.

You could at least point to where it mentions significant figure rules, because I can’t see it anywhere.”

OMG! That’s because the link is about resolution uncertainty and not measurement uncertainty. Resolution uncertainty is just one piece of the measurement uncertainty.

“Go away kid, you bother me!”

Bellman
Reply to  Tim Gorman
July 9, 2022 7:47 am

100 is not the average, it’s the coint. It’s the number I divided the sum by to get the average.

Reply to  Carlo, Monte
July 2, 2022 12:11 am

Inter-hemispheric heat piracy

Javier
Reply to  Carlo, Monte
July 2, 2022 12:55 am

What you are seeing in that temperature pattern map is the solar warming pattern. Particularly telling is the 45-65ºN intense warming (+0.6K in the solar cycle) and the Southern Ocean wave pattern with the warm areas separated about 4000 km.

This figure is from Lean 2017 Solar review, comparing the effect on surface temperature of the solar cycle with paleoclimatic effects of solar activity changes.

comment image

Carlo, Monte
Reply to  Javier
July 2, 2022 6:26 am

This is likely different from my graph above which represents only two months, 06/1980 and 06/2021. Shifting to different years the patterns change a lot.

Dave Fair
Reply to  Carlo, Monte
July 3, 2022 1:19 pm

Monte, what are the conclusions one may draw from this? CAGW or meh?

Jim Gorman
Reply to  Dave Fair
July 3, 2022 1:44 pm

Not sure what Monte makes of that but when I look at the attached I see something that always returns to a baseline. That is, no growth in warming or cooling. Recently, there has been some warming but the last few years we have been on a cooling trend back to base line.

202204_Bar.png
Last edited 1 month ago by Jim Gorman
Carlo, Monte
Reply to  Dave Fair
July 3, 2022 5:39 pm

From the difference map? Because the 1990-2010 baseline has been subtracted from both June 2021 and June 1980, in the additional subtraction the baseline drops out, so this map is the same as subtracting the absolute temperatures:

D = (A – C) – (B – C) = A – C

Here is another histogram, this is March 2021 – March 1979: the standard deviation is 1.7K, the mean is 0.4K, but notice the peak of the histogram is just slightly above 0K. The total difference across the globe is only ±6K. Granted that these are just single months and I have only sampled a few, but the temperature rise from these is 0.1-0.4K across 40 years, which just 3-10 mK. Hardly catastrophic, and well within the statistical noise. If this is the effect of CO2, then I have to go with meh.

Screen Shot 2022-07-03 at 6.25.46 PM.jpg
RickWill
Reply to  Carlo, Monte
July 1, 2022 5:49 pm

The peak corresponds to 30C at the surface. That is where deep convection runs out of steam because the atmospheric ice forming over the warm pools limits the surface sunlight. It is very precise just over 30C surface temperature where the convection limits.

If you looked at the surface temperature (only over oceans) you will find your peak aligns with 30C on the surface. Just keep in mind that there is a lag of about 20 days between the surface and the ice cloud. That is how long it takes to build the head of steam to get to 30C from 29C. The warm pools move around so the surface will not align exactly with the atmosphere. If you time shift the surface temperature later or the atmospheric earlier by 20 to 25 days you should find close agreement. If you do not have daily time resolution then one month time shift will be closer than the same month.

Tim Gorman
Reply to  Carlo, Monte
July 1, 2022 4:43 pm

Nice! The uncertainty interval is *far* larger than the actual plotted data. So who knows what the h*ll is going on?

Carlo, Monte
Reply to  Tim Gorman
July 1, 2022 6:13 pm

Thanks, Tim; and I just plucked a number out of the air, which could easily be considerably larger.

I have one more little tidbit to post, coming soon…

Dave Fair
Reply to  Carlo, Monte
July 1, 2022 5:24 pm

Monte, why not take out the effects of large volcanoes and ENSO on the series? With those in I doubt the meaningfulness of any statistical analyses. Autocorrelation would also seem to be an issue.

Carlo, Monte
Reply to  Dave Fair
July 1, 2022 6:09 pm

You are probably quite right, but removing these is beyond my capabilities.

Dave Fair
Reply to  Carlo, Monte
July 3, 2022 1:39 pm

IIRC, Dr. Roy Spencer (and/or others) did some work. I don’t care enough to look it all up.

Bellman
Reply to  Carlo, Monte
July 2, 2022 3:01 pm

Here are the UAH monthly data points replotted with generous uncertainty intervals of U(T) = ±1.4K.”

Are you claiming that it’s plausible that say June 2022, could have a temperature that is more than 1.4K hotter or colder than the 1991-2020 average?

Jim Gorman
Reply to  Bellman
July 2, 2022 7:16 pm

You have never figured out uncertainty have you? It is an interval where you don’t know where the real value lays. It doesn’t mean that value could be at the max or min, it means that you can’t KNOW where the value is within that interval. The interval, when stated with a value, shows how uncertain you are about the value.

Do you consider 0.6 ± 1.4 a good accurate number?

Carlo, Monte
Reply to  Jim Gorman
July 2, 2022 8:51 pm

He has not figured it out, and likely never will.

Bellman
Reply to  Jim Gorman
July 3, 2022 8:25 am

The 1.4K figure was quoted as standard uncertainty. That should mean it’s plausible the true value could be outside that range.

What Carlo means by it is anyone’s guess. That’s why I was asking the question. But as always he just jokes it off.

As you say a standard uncertainty of 1.4 isn’t good when monthly temperature changes are being measured in tenths of a degree, but I would like to see some evidence that supports that uncertainty value. I’ve long argued that UAH shouldn’t be regarded as the only reliable data set, but that doesn’t mean I think it’s fair to traduce Spencer’s work just by plucking an insane figure out of the air.

Jim Gorman
Reply to  Bellman
July 3, 2022 9:06 am

The point is that as you claim more and more resolution from averaging, you can’t ignore that there is uncertainty that follows thru from the original temperatures.

I’ll be honest I am no satellite measurement expert. I do know that when you are using samples dividing the SD by √N is not the correct way to get a standard deviation for the sample distribution.

Bellman
Reply to  Jim Gorman
July 3, 2022 1:02 pm

I’m not making any direct claims about how to calculate the uncertainty of the UAH data. I just fond the 1.4K uncertainty figure difficult to justify given the actual data. I have doubts about the accuracy of satellite data, at least compared to the sorts of claims made for them a few years ago, but can’t conceive how such a large uncertainty could be correct given the coherence of the data, both with itself and other data sets.

Bellman
Reply to  Jim Gorman
July 3, 2022 1:04 pm

I do know that when you are using samples dividing the SD by √N is not the correct way to get a standard deviation for the sample distribution.”

Of course it’s not. You don’t need to divide SD by anything to get the sampling distribution, because that’s what the SD is. At least, that’s what I assume you mean by sampling distribution.

Jim Gorman
Reply to  Bellman
July 3, 2022 2:01 pm

Yet that Is what is being done to justify small, small uncertainty values.

Bellman
Reply to  Jim Gorman
July 3, 2022 3:22 pm

Because the SD is not the uncertainty of the mean. Or at least not the value you need if you want to compare one months mean to another.

The SD is useful if you need to know who close any random place on the earth might be to the mean, but that’s not what I’m interested in when looking at the trend of global averages.

Carlo, Monte
Reply to  Jim Gorman
July 3, 2022 6:03 pm

It’s right there in the paper!

A question he should ask of himself—why does he care so much about me putting uncertainty limits on the graph? UAH has never displayed either U limits or even error bars.

Bellman
Reply to  Carlo, Monte
July 4, 2022 3:13 am

I don’t care about putting uncertainty intervals on the graph. I think it would be a good idea if UAH did this. I just think those uncertainty intervals should reflect the actual uncertainty rather than your fantasy ones.

If you don’t think UAH is trustworthy because they don’t estimate uncertainty limits, and if you really believe the true uncertainty is at least 1.4°C, you should be asking why WUWT gives it so much publicity, including these and Monckton’s monthly updates.

Tim Gorman
Reply to  Bellman
July 3, 2022 9:25 am

The 1.4K figure was quoted as standard uncertainty. That should mean it’s plausible the true value could be outside that range.”

he merely stated the uncertainty interval. It is *you* that are extending that, not him.

That uncertainty interval is certainly wide enough to dwarf the differences trying to be identified.

+/- 1.4K is the same as +/- 1.4C. That is a *very* reasonable assumption for the uncertainty interval. If you don’t like it then put +/- 0.5K uncertainty lines on the graph. All of the data will fit inside that uncertainty interval meaning you simply don’t know what the trend line will come out to be. We’ve had this argument before and I gave you several internet links saying the same thing and even showing graphs.

Bellman
Reply to  Tim Gorman
July 3, 2022 12:56 pm

My mistake. I thought the 1.4K figure was for the standard uncertainty, but it seems GUM uses capital U to mean expanded uncertainty. But he never states the coverage factor, or what the level of confidence is, so what that actually means is still a mystery.

I don;t believe this value, whatever the confidence is reasonable. But at the moment I’m just trying to understand where you think the uncertainty comes from.

At the very least I think it’s a disservice to Spencer and Christy to allege their data is so inaccurate and claims made for it are almost fraudulent, without directing your concerns to them directly.

Tim Gorman
Reply to  Bellman
July 4, 2022 5:54 am

Do you *really* not understand where uncertainty comes into play when measuring the radiance at different locations?

You simply can’t directly measure all the actual atmospheric conditions that can affect the radiance. All you can get is the radiance itself. Therefore there will *always* be some uncertainty in what you measure. Couple that with the uncertainty associated with the measuring devices themselves plus uncertainty contributed by the conversion algorithm from radiance to temperature and you wind up with a measure of uncertainty that is as significant as that associated with land based thermometers.

Because all of the satellite measurements are of different things at different times you simply cannot assume that the uncertainty profile cancels as you are wont to do. Therefore the uncertainty contributed to the “average” adds as you add measurements.

UAH is not a “Global Average Temperature”. It is a metric that is consistent and useful in measuring differences – as long as the uncertainty in those differences are recognized.

If the climate models were forecasting actual absolute temperatures or if the observations themselves weren’t “annual average anomalies” then you couldn’t even directly compare those with the UAH.

At the very least I think it’s a disservice to Spencer and Christy to allege their data is so inaccurate and claims made for it are almost fraudulent, without directing your concerns to them directly.”

Do you have even a glimmer of understanding as to what you are saying here? I know Spencer reads WUWT so we *are* directly passing our concerns along. The climate model authors and the IPCC? They wouldn’t accept our criticisms even if they were to read WUWT!

*ALL* of the climate stuff is questionable. No one seems to address uncertainty correctly and all of it assumes they can identify differences that are far smaller than the uncertainties of the inputs. You can’t average away uncertainty. Average uncertainty is not the same thing as uncertainty of the average. And you can’t just assume that multiple measurements of different things generates an uncertainty profile such that all uncertainty cancels out.

As a summer intern for a power company I once got to help an EE professor measure RF signal levels along the route of a new high voltage line from a new power plant in order to help address any complaints that might be made after the line was installed. It was my first introduction to uncertainty and that professor absolutely knew his stuff. I learned:

  1. You can’t identify differences smaller than the resolution of the measuring device. Unknowns remain unknowns.
  2. It took us two weeks to cover the entire route. Propagation was different each day and generated an uncertainty in the actual measurement at each location. You can’t just ignore that uncertainty.
  3. Since we were measuring different things you couldn’t just generate an average uncertainty and apply it to all of the measurements. That would never pass muster in a legal process resolving an interference complaint at a specific location.
  4. The same logic applies to generating an average signal strength from all the individual measurements of different things. It wouldn’t hold up in a legal proceeding. Since the signal strength contribution to total signal strength along the route was distance dependent to high powered radio stations an average value for the whole route was useless in resolving individual complaints.

We ran the route during the day. The professor was going to run it at night at a later time. You couldn’t just average the day and night measurements because it would give you another useless number. Individual complaints had to be time resolved as well as location.

I write all this in the faint hope that maybe it will get across that uncertainty has a *real* application in the *real* world. You can’t just assume things away that are inconvenient and you can’t identify unknowns that are beyond detection. Statistics just don’t help, they are descriptions of the data you have, they are not measurements themselves. The data you have includes the uncertainty associated with the individual measurements as well as their total.

Carlo, Monte
Reply to  Tim Gorman
July 4, 2022 7:57 am

There is also the uncertainty associated with the satellite sampling: it is not continuous in time, varies with latitude, and done from spherical grid points that are not equal in area!

Bellman
Reply to  Tim Gorman
July 4, 2022 11:40 am

I’ve been arguing that satellite data has large uncertainties, back in the days when to say such things was considered heresy. None of that means you can just make up improbably large uncertainty ranges, that imply UAH data is effectively worthless.

If the uncertainty really was 1.4K or has a 95% confidence interval of around 1K, it would have to be an incredible coincidence that it agrees so closely with other data sets using completely different methods.

Tim Gorman
Reply to  Bellman
July 4, 2022 2:04 pm

Why would it have to be an incredible coincidence? Is it an incredible coincidence that the newest land based temp measuring stations have just about the same uncertainty as the Argo floats? Or do all of these uncertainties stem from our engineering capability for field measurement devices at this time?

Bellman
Reply to  Tim Gorman
July 4, 2022 11:58 am

Therefore the uncertainty contributed to the “average” adds as you add measurements.

I’m pretty sure Carlo does nothing of the sort. If he did the uncertainty range would be much much bigger.

Tim Gorman
Reply to  Bellman
July 4, 2022 2:07 pm

So what’s your point? The number he came up with was more than sufficient to question the actual slope of any trend line let alone the actual values. What good would it do to use a higher number?

Bellman
Reply to  Tim Gorman
July 4, 2022 3:06 pm

The point is, not even Carlo accepts your nonsense that the uncertainty increases as you increase the samples. If that was the case you could improve the uncertainty bound just by removing random observations.

Carlo, Monte
Reply to  Bellman
July 4, 2022 9:24 pm

Do you need another quarter?

Bellman
Reply to  Carlo, Monte
July 5, 2022 9:57 am

No mercy, no quarter.

Carlo, Monte
Reply to  Bellman
July 5, 2022 11:14 am

Ouch this hurts.

Tim Gorman
Reply to  Bellman
July 5, 2022 6:26 am

The point is, not even Carlo accepts your nonsense that the uncertainty increases as you increase the samples. If that was the case you could improve the uncertainty bound just by removing random observations.”

Again, SO WHAT? You would wind up in the final removal just having one measurement! Which is what I tried to point out in my message above on the survey of signal strengths!

When you are measuring different things finding their average is pretty much a useless exercise. It doesn’t matter whether you are averaging temperature minimums and maximums or temperatures from different locations. The average simply doesn’t provide any expectation of what the next measurement of a different thing at a different location will actually be. If the average can’t help identify an expected value then of what use is it? As I said, you couldn’t use it in a court of law to resolve a complaint since it wouldn’t be site specific, nor would it identify the actual temperature profile at question since many different min/max temps give the same mid-range value.

As you add independent, random measurements of different things the variance increases. You can find this in *any* statistical textbook you wish to study. As variance increases the possible value of the next measurement increases also. This is exactly how uncertainty works. It’s why the same techniques work on both variance and uncertainty. Why you simply can’t accept this is beyond me.

And MC merely came up with an uncertainty that puts the trend line laid out in question. It is only one of the possible trend lines that will fit inside the uncertainty interval used. So who cares if the actual uncertainty is even wider than the one he used? All that does in increase the number of possible trend lines. YOU COME TO THE SAME CONCLUSION!

As usual with a troll you are nit-picking. An argumentative fallacy known as “Logic Chopping” – Focusing on trivial details of an argument, rather than the main point of the argumentation. It’s a form of the Red Herring argumentative fallacy.

You don’t want to admit that what MC shows puts the trend line in question so you throw up the red herring that he didn’t get the uncertainty interval right.

Pathetic.

Carlo, Monte
Reply to  Tim Gorman
July 5, 2022 7:02 am

The 1.4 number should have been a huge clue for him, but he missed it entirely.

He also doesn’t seem to realize that the regression line can fall anywhere inside the confidence interval that he puts on his own UAH plots!

Bellman
Reply to  Carlo, Monte
July 5, 2022 9:49 am

How can it be a clue. I asked you about it and you refused to answer. All you’ve said is you plucked it out the air. If you are claiming you derived it by adding all the uncertainties together, than I think it should be much bigger, and wronger.

He also doesn’t seem to realize that the regression line can fall anywhere inside the confidence interval that he puts on his own UAH plots!

Of course I realize that. That’s the whole point of me displaying confidence intervals when relevant. Now, what do you think the confidence interval is for the pause period?

Carlo, Monte
Reply to  Bellman
July 5, 2022 11:16 am

You are still unable to comprehend that I will not participate in your little trendology games, mr condescending expert.

Bellman
Reply to  Carlo, Monte
July 5, 2022 12:35 pm

No. I expect you not to answer my questions. That’s why asking them is useful. It shows me the gulf between what you claim to know and the reality.

Carlo, Monte
Reply to  Bellman
July 5, 2022 3:32 pm

Reality must be strange in the world you inhabit.

That I refuse to play ring-around-the-merry-go-round with you lot does not imply any answers to the questions you try to goad me with.

But I suppose this to be expected with climastrology.

Tim Gorman
Reply to  Bellman
July 5, 2022 3:48 pm

How can it be a clue. I asked you about it and you refused to answer. All you’ve said is you plucked it out the air. If you are claiming you derived it by adding all the uncertainties together, than I think it should be much bigger, and wronger.”

You continue to miss the entire point! If the uncertainty is larger then what does that prove that MC didn’t already show? You are *STILL* nit-picking! It’s like saying you didn’t correctly calculate how many angels can fit on the head of a pin!

Bellman
Reply to  Tim Gorman
July 5, 2022 4:08 pm

I think at least one of us is missing the entire point.

  1. You say you add the uncertainties when calculating the uncertainty of the mean.
  2. I say you don’t and suggest that Carlo doesn’t do this.
  3. My evidence for this is that if he was adding the uncertainties I would expect the quoted standard uncertainty to be much larger than 0.5K.
  4. Carlo indicates he may or may not agree with you, and pleads the 5th.
  5. You seem to now be claiming you don’t care how the uncertainty is calculated, or what it shows, as long as the value plucked out of the air is large enough to prove the pause is meaningless, or something.
Carlo, Monte
Reply to  Bellman
July 5, 2022 5:19 pm

It is not my job to correct your ignorance about propagating uncertainty, regardless of how much you whine.

CMoB had you pegged to a tee right off the bat.

Tim Gorman
Reply to  Bellman
July 7, 2022 6:46 am

“You say you add the uncertainties when calculating the uncertainty of the mean.”

Absolutely. The data elements are “stated value +/- uncertainty”. You can’t just simply dismiss the uncertainties of the data elements they way you do.

“My evidence for this is that if he was adding the uncertainties I would expect the quoted standard uncertainty to be much larger than 0.5K.”

Again, SO WHAT? Remember that when you are graphing a trend line you are using each individual data element separately, not added together. The uncertainty of each individual element should also be shown on the graph. Doing so determines the area within which the true trend line might lie. What does summing uncertainties have to do with this process?

“Carlo indicates he may or may not agree with you, and pleads the 5th.”

No, he pleads that since you will never understand it is a waste of time trying to educate you.

“You seem to now be claiming you don’t care how the uncertainty is calculated, or what it shows, as long as the value plucked out of the air is large enough to prove the pause is meaningless, or something.”

The issue is that you just simply can’t seem to get basic rules right. The uncertainty intervals shown on a graph of individual data elements don’t ADD as you move along the x-axis! And the value that MC picked *is* reasonable enough to show the invalidity of a trend line developed solely based on stated values while ignoring the uncertainty interval that goes along with the stated values – which you, for some unknown reason, just can’t quite figure out!

Carlo, Monte
Reply to  Tim Gorman
July 7, 2022 12:42 pm

It is absolutely incredible the degree to which he misunderstands plain English!

Carlo, Monte
Reply to  Tim Gorman
July 5, 2022 7:21 am

“The point is, not even Carlo accepts your nonsense that the uncertainty increases as you increase the samples. …”

And HTH did he get this ridiculous notion?

Bellman
Reply to  Carlo, Monte
July 5, 2022 9:43 am

If you do believe that nonsense then I apologize. But whenever I ask a direct question you get evasive, and I can’t figure out how you get a standard uncertainty as low as 0.5K if you do believe that.

So for once, could you give a straight answer. Do you think that as you increase sample size the uncertainty of the mean increases? Specifically if you take 100 measurements the measurement uncertainty will be the sum of all the individual uncertainties?

Carlo, Monte
Reply to  Bellman
July 5, 2022 11:17 am

You’ve been educated about this about five hundred thousand times, 500,000 + 1 is a waste of time.

Bellman
Reply to  Carlo, Monte
July 5, 2022 12:31 pm

Lame. It’s a simple question, and one I’ve never seen you answer once, let alone 500000 times. Tim Gorman is clear, and wrong, but at least argues his case. You hide behind ambiguity and weak jokes.

Do you think that as you increase sample size the uncertainty of the mean increases? Yes, no or it’s complicated. I just want you to commit to an answer, give some indication you are not just making it up as you go along.

Carlo, Monte
Reply to  Bellman
July 5, 2022 3:34 pm

/yawn/

Find another victim.

Bellman
Reply to  Carlo, Monte
July 5, 2022 3:54 pm

Do you think that as you increase sample size the uncertainty of the mean increases?

1) Yes
2) No
3) Stop making me think.

Carlo, Monte
Reply to  Bellman
July 5, 2022 4:01 pm

You are in no position to put such demands on anyone, lastwordbellcurveman.

Bellman
Reply to  Carlo, Monte
July 5, 2022 4:16 pm

I’m not making any demands. You can answer the question, or you can keep making up dumb nicknames. Any independent reader can decide for themselves why you refuse to answer.

Carlo, Monte
Reply to  Bellman
July 5, 2022 5:21 pm

Of course you are, don’t lie. And because I refuse to participate in your little trendology jihad, you whine. A lot.

Grow up.

Tim Gorman
Reply to  Bellman
July 7, 2022 6:37 am

The uncertainty of the mean remains that uncertainty propagated from the individual elements. If you add more uncertainty into the data set then the uncertainty increases.

When you increase the sample size you are adding elements that are “stated value +/- uncertainty”. The mean you calculate comes from the stated values. *YOU* continually want to ignore the “+/- uncertainty” part of the data elements, however.

If you add elements and the mean changes all that shows is that your first, smaller sample was not representative of the entire population. A changing mean, however, does *NOT* change the uncertainty of the mean you calculated. The uncertainty of the mean must be recalculated using the uncertainties of the original elements plus the uncertainties of the added elements.

Bellman
Reply to  Tim Gorman
July 7, 2022 11:40 am

You keep using that word “adding”.

Do you mean as in summing or as in increasing elements in a sample? In this case it seems to be the later.

When you increase the sample size you are adding elements that are “stated value +/- uncertainty”. The mean you calculate comes from the stated values. *YOU* continually want to ignore the “+/- uncertainty” part of the data elements, however.

You keep making this lie. We are discussing the measurement uncertainties. I am not ignoring them.

If you add elements and the mean changes all that shows is that your first, smaller sample was not representative of the entire population.

Yes. That’s regression toward the mean.

A changing mean, however, does *NOT* change the uncertainty of the mean you calculated.

Obviously it does.If my small sample was unrepresentative of the mean, and the larger sample is likely to be closer to the mean I’ve reduced the uncertainty.

The uncertainty of the mean must be recalculated using the uncertainties of the original elements plus the uncertainties of the added elements.

And then divided by the new sample size.

You can keep rewording your mistake all you like, we still end up back to the same point. You think the uncertainty of the mean is the uncertainty of the sum of all elements, when it should be the uncertainty of all elements divided by sample size. And please don’t say that’s the same as the average uncertainty.

Tim Gorman
Reply to  Bellman
July 8, 2022 4:33 am

You keep making this lie. We are discussing the measurement uncertainties. I am not ignoring them.”

You keep saying this but EVERY SINGLE TIME that it comes down to actually using them you just wind up totally ignoring them!

Words don’t matter, it’s the actions that matter. Your action is to always ignore uncertainty!

“Yes. That’s regression toward the mean.”

It also implies that the mean you calculate from the sample is inaccurate!

“Obviously it does.If my small sample was unrepresentative of the mean, and the larger sample is likely to be closer to the mean I’ve reduced the uncertainty.”

And you STILL demonstrate that you simply don’t understand uncertainty! Standard deviation of the sample means is a measure of how precise you have calculated the resultant mean.

That is *NOT* the accuracy of the sample means or of the average you calculate from the sample means.

The population mean has an uncertainty propagated from each individual element. It has a stated value +/- uncertainty.

What you are claiming is that you are getting closer to the stated value of the population mean – WHILE IGNORING THE UNCERTAINTY part of the mean!

You said “I am not ignoring them” when it comes to uncertainty of the data elements. But you turn right around and IGNORE THEM!

Like I said, your actions belie your words!

Tim Gorman
Reply to  Bellman
July 5, 2022 3:53 pm

Every time you say that variances don’t add when combining independent, random variables you are violating the precepts of statistics.

And you simply can’t admit that for some reason. In your world, every single statistics textbook in existence is wrong – variances don’t add when combining independent, random variables.

And you are questioning others?

Carlo, Monte
Reply to  Tim Gorman
July 5, 2022 4:03 pm

Don’t ever forget, he is the world’s expert on the GUM.

/snort/

Bellman
Reply to  Carlo, Monte
July 5, 2022 4:13 pm

Billhooks! Just because I was able to use an equation you insisted had to be used, does not make me any sort of expert on it.

Carlo, Monte
Reply to  Bellman
July 5, 2022 5:22 pm

“WHAAAA! MOMMY!”

Bellman
Reply to  Tim Gorman
July 5, 2022 4:26 pm

Every time you say that variances don’t add when combining independent, random variables you are violating the precepts of statistics.”

You keep equivocating on the word combine. If you combine them by adding, you add the variances. If you combine them by taking the mean you have to add the variances and divide them by the square of the number of elements.

This is not violating any precept of statistics, it’s absolutely central to combining variances, it’s how the formula for the SEM is derived, and it is easy to establish for yourself.

Here’s a reference for you

https://online.stat.psu.edu/stat414/lesson/24/24.4

Screenshot 2022-07-06 002535.png
Bellman
Reply to  Tim Gorman
July 5, 2022 10:20 am

When you are measuring different things finding their average is pretty much a useless exercise.

How a big a bonfire do we need for all the text books over the last century or so? Is it time to cancel Dr Spencer?

The average simply doesn’t provide any expectation of what the next measurement of a different thing at a different location will actually be.

Firstly, it does.

Secondly, that’s not the point of this exercise. I don;t want to predict what the next days temperature will be in a random part of the planet. The point of the monthly global averages is to compare it with previous values, to see if there has been any significant change.

As you add independent, random measurements of different things the variance increases.

And you still haven’t grasped this simple point – As you add random variables the variance increases, but when you take their mean the variance decreases.

You can find this in *any* statistical textbook you wish to study.

Exactly.

As variance increases the possible value of the next measurement increases also.

Wut? How do previous measurements increase the variance of the next? What part of independent are you not understanding?

It’s why the same techniques work on both variance and uncertainty.

Yes. And that technique shows you are wrong in both cases.

It is only one of the possible trend lines that will fit inside the uncertainty interval used.

Which uncertainty interval are you talking about here? I think there’s been some confusion because he shows what he claims is the uncertainty interval for monthly values, but that is not the same as the confidence interval for the trend. Carlo’s stated uncertainty for the trend shows that the warming trend over the last 40 or so years is statistically significant, and the trend line could be anywhere from 0.6 to 2.0°C / decade.

You don’t want to admit that what MC shows puts the trend line in question…

Firstly, he doesn’t show anything. He admits he plucked the uncertainty interval from the air.

Secondly, as I explained above it doesn’t put the trend line in question. His standard error is about 50% bigger than the one provided by Skeptical Science, but not enough suggest the trend could be non existent.

… the red herring that he didn’t get the uncertainty interval right.”

You’re missing my point. I’m not saying his uncertainty should be bigger. On the contrary I think his current claim is probably far too big. What I’m saying is if he had used you technique of adding the uncertainty of each measurement, his monthly uncertainty would have been much bigger.

Carlo, Monte
Reply to  Bellman
July 5, 2022 11:20 am

On the contrary I think his current claim is probably far too big.

Evidence today, dr trendologist?

What I’m saying is if he had used you technique of adding the uncertainty of each measurement, his monthly uncertainty would have been much bigger.

A big red flag that you don’t read/understand what other people write.

Bellman
Reply to  Carlo, Monte
July 5, 2022 12:24 pm

If you want me to understand you, maybe you could answer some of my questions rather than hide behind lame name calling.

Carlo, Monte
Reply to  Bellman
July 5, 2022 3:36 pm

stomp yer feet and yell please-please-please-please-please-please-please-please-please-please-please-please-please-please-please-please-please-please

Do you really think I care if you understand or not?

Tim Gorman
Reply to  Bellman
July 5, 2022 4:55 pm

How a big a bonfire do we need for all the text books over the last century or so? Is it time to cancel Dr Spencer?”

*YOU* are the only one with the bonfire with your claim that variances don’t add when combining random, independent variables.

“Firstly, it does.”

No, it doesn’t, except in your bizzaro world. If you pick up 5 random boards from the ditches around your house and they measure 2′, 5′, 8′, 10′, and 3″ (avg = 5.05′) then what does the average tell you about what length to expect for the next random board you find?

Ans: NOTHING!

It could be 20′ long, 4′ long, and it might even be cut at a 45deg angle on one end so it’s longer on one side than on the other! The only expectation you could possible have is that its length is greater than zero! It’s hard to see a board of length 0!

Variance(2,5) = 2.3
Variance(2,5,8 = 6
Variance(2,5,8,10) = 9.2
Variance(2,5,8,10,0.25) = 13.1

The variance goes up with each random, independent addition to the series, just as I and all the textbooks say. Uncertainty does the same thing. If each of the board measurements have an uncertainty of +/- .5″, +/- .6″, +/- .75″, +/- 1″, and +/- .1″ then all five added together will have an uncertainty of +/- 3″ if added directly.

It’s no different with independent, random boards than it is with independent, random temperature measurements. They are all measurements of different things. The variances add and the uncertainties add. If you want to add the uncertainties directly or by root-sum-square they *still* add. The growth is just not as fast with root-sum-square.

Most of us live in the real world, not in your bizzaro world. In the real world variances don’t cancel out and neither do uncertainties. And averages of independent, random things don’t give you an expectation of what the next value will be.

Bellman
Reply to  Tim Gorman
July 5, 2022 5:51 pm

It could be 20′ long, 4′ long, and it might even be cut at a 45deg angle on one end so it’s longer on one side than on the other!

Could be, but it’s more likely to be closer to the average of the previous boards. Those who can;t learn from the past and all that.

Ans: NOTHING!

So why keep picking things out of the ditch if you don;t want to learn anything about what’s in the ditch? For some reason people keep dumping planks of wood in your ditch. Don’t you think it might be possible that a random sample will tell you something about what sort of planks they are?

The variance goes up with each random, independent addition to the series, just as I and all the textbooks say.

As usual you are talking about the sum not the average.

It’s no different with independent, random boards than it is with independent, random temperature measurements.

The difference being it makes some sense to want to add the lengths of various boards to see what it would come to if they were laid end to end. It makes no sense to care about the sum of various temperature readings, you are only interested in the average, not the sum.

Most of us live in the real world

You are going to need to provide some evidence for that claim.

And averages of independent, random things don’t give you an expectation of what the next value will be.

Say you were an alien with no knowledge of human anatomy. You select 20 or so people at random and measure their height. The average say is 1.8m. Are you not capable of inferring anything from that knowledge. Do you assume that there’s no way of telling if the net person you abduct might be more likely to be closer to 1.8m than 100m, or 0.01m? Of course, more information, such as the standard deviation, range or any other parameters will be useful, but it’s absurd to simply state that the average tells you nothing.

Carlo, Monte
Reply to  Bellman
July 5, 2022 8:54 pm

Could be, but it’s more likely to be closer to the average of the previous boards.

Handwaving, you don’t know this.

Tim Gorman
Reply to  Bellman
July 7, 2022 7:04 am

Could be, but it’s more likely to be closer to the average of the previous boards. Those who can;t learn from the past and all that.”

Wrong! If the variance increases then the possible values increase also. Graph those lengths. That histogram will tell you that it does *NOT* resemble a Gaussian distribution which would be necessary for the next value to probably be close to the mean!

So why keep picking things out of the ditch if you don;t want to learn anything about what’s in the ditch? For some reason people keep dumping planks of wood in your ditch. Don’t you think it might be possible that a random sample will tell you something about what sort of planks they are?”

Maybe I am building bird houses using the boards. Maybe I am using them to build a small bridge across a drainage ditch in my back yard. There are a myriad of reasons why I might be collecting them.

A random sample of multiple independent things is not likely to give you anything in the way of determining an expected value for the next board you find. These aren’t multiple measurements of the same thing but multiple measurements of different things!

“As usual you are talking about the sum not the average.”

Is the average not found by taking a sum? When you divide by the number of elements you are merely spreading an equal length to each element. You are finding an “average length”. It’s the same thing that you do with your uncertainties. You find an “average uncertainty” which you can spread across all elements. That is *NOT* the uncertainty of the average. It is an average uncertainty!

“The difference being it makes some sense to want to add the lengths of various boards to see what it would come to if they were laid end to end. It makes no sense to care about the sum of various temperature readings, you are only interested in the average, not the sum.”

And, once again, how do you calculate an average? Do you not sum up all the individual elements? So how is laying boards end-to-end to get the sum of their length any different than laying temperatures end-to-end to get the sum of their values?

The sum of those temperatures will wind up with an uncertainty in the sum, just like the sum of the lengths of the boards will have an uncertainty. And those uncertainties translate to the mean you calculate.

The uncertainty of (x1 + x2 + … +xn)/n is:

u_x1 + u_x2 + … + u_xn + u_n.

The uncertainty of n is zero so the uncertainty of the average is the uncertainty of the sum!

I *still* don’t understand how you can’t get this through your skull into your brain.

“You are going to need to provide some evidence for that claim.”

You said yourself that you have little knowledge of the real world. You said you’ve never built anything. I have! I grew up doing all kinds of things in the real world. There are all kinds of us that live in the real world that keep telling you that your delusions about how uncertainty works are just that – delusions!

Bellman
Reply to  Tim Gorman
July 7, 2022 11:53 am

Wrong! If the variance increases then the possible values increase also.

More goal posts being shifted. We were talking about pulling planks out of the ones dumped in your ditch. Why is it’s variance increasing as I pull more planks out?

That histogram will tell you that it does *NOT* resemble a Gaussian distribution which would be necessary for the next value to probably be close to the mean!

Does not matter.

There are a myriad of reasons why I might be collecting them.

I’m not interested in what you plan to do with them. I want to know why you are taking the average of the first five and then ignoring what they might tell you about the next length.

A random sample of multiple independent things is not likely to give you anything in the way of determining an expected value for the next board you find.

That’s exactly what it will give you. The mean of your sample is the best estimate of the mean of the population, and the mean of the population is the expected value for the next board.

These aren’t multiple measurements of the same thing but multiple measurements of different things!

Measurements of different things is one of the main uses of averaging. There isn’t much point of getting an average when everything’s the same.

Is the average not found by taking a sum?

The logic you live by! If an average is derived from a sum then an average is a sum.

You find an “average uncertainty” which you can spread across all elements.

Wrong. Is it really worth my while to keep pointing out every time you make mistake? Let’s go back to your original 100 thermometers with measurement uncertainty of ±0.5°C and all uncertainties independent and random.

The average uncertainty is not surprisingly ±0.5°C. The uncertainty if the mean is ±0.05°C. THEY ARE NOT THE SAME.

Bellman
Reply to  Bellman
July 7, 2022 3:20 pm

Part 2:

And, once again, how do you calculate an average? Do you not sum up all the individual elements?”

No you sum up the elements and then divide by sample size.

“So how is laying boards end-to-end to get the sum of their length any different than laying temperatures end-to-end to get the sum of their values?

How do you lay temperatures end to end? Seriously, yes you calculate the mean in the same way. The point I was trying to make is that there is an intrinsic reason why you might just want the sum of the length of boards, but no reason to want to know the sum of temperatures except as way to get to the mean.

And those uncertainties translate to the mean you calculate.

Yes, but you never want to do the translation. It’s a very simple translation, just divide the uncertainty of the sum by the sample size.

The uncertainty of (x1 + x2 + … +xn)/n is:
u_x1 + u_x2 + … + u_xn + u_n.

It isn’t. I’ve explained to you repeatedly why it isn’t. You claim to have read Taylor and done all the exercises but you seemed to have missed the part where he explains you cannot mix add/subtraction with multiplication/division.

The uncertainty of n is zero so the uncertainty of the average is the uncertainty of the sum!

I’m really getting to the stage where I think this is some clever piece of performance art, or worry that you might have some cognitive issues.

You cannot add the uncertainty of n to the sum of the xs because they are being added, and n is being divided. You have to first calculate the uncertainty of the sum, then use the multiplication/division equation to calculate the uncertainty of the sum divided by n. This requires writing the sum and the result as relative uncertainties. Hence

u(mean)/mean = u(sum)/sum + 0

Then you have to convert these back to absolute uncertainties, at which point anyone studying elementary algebra, or who understands how proportions work, will see that as mean < sum, u(mean) has to be less than u(sum), and specifically

u(mean) = u(sum) / n

I *still* don’t understand how you can’t get this through your skull into your brain.

It’s a mystery. Maybe I don’t to being endlessly told something I can see is false, and cannot possibly be true.

There are all kinds of us that live in the real world that keep telling you that your delusions about how uncertainty works are just that

Out of interest, could you point to a time in all your real world experience, where it was necessary to determine the uncertainty of a mean of a large number of values, and it was possible to see if your calculation was correct?

Carlo, Monte
Reply to  Bellman
July 7, 2022 4:02 pm

1. Calculate the uncertainty of the sum = u(sum)
2. Calculate the uncertainty of u(sum)/N

You get a different answer than your hallowed NIST calculator.

Oops.

Bellman
Reply to  Carlo, Monte
July 7, 2022 4:41 pm

It’s not my NIST calculator. I’d forgotten all about it until you brought it up just now. I only saw it in the first place because you kept invoking NIST on uncertainty and I thought it interesting that their calculator disagreed with your claims. At which point you seemed to lose all interest in NIST on uncertainty.

You may well get different results using NIST as it’s a monte carlo simulation IIRC. Could you show me what you did in the calculator, and how it differs from u(sum)/N? Or will you just throw another tantrum?

Tim Gorman
Reply to  Bellman
July 8, 2022 10:24 am

The NIST calculator does not determine uncertainty. There is no place to enter the uncertainty of your stated values.

It determines the best fit of the stated values to the trend line while ignoring the uncertainty of the stated values.

Best fit is *NOT* uncertainty. It is purely a measure of how well your trend line fits the stated values, nothing more.

Bellman
Reply to  Tim Gorman
July 8, 2022 3:04 pm

Then take it up with Carlo, he was the one saying it gave different results. This double act is becoming quite tiresome.

It determines the best fit of the stated values to the trend line while ignoring the uncertainty of the stated values.

I’m not sure if you know what you are talking about here. There is nothing about a trend line.

I suspect you’re confusing this with the trend calculator we were talking about on another thread.

Bellman
Reply to  Carlo, Monte
July 7, 2022 4:50 pm

So I ran a test with the calculator. 9 values. Each with a normal distribution and a standard deviation of 0.5.

The sum has a standard deviation of 1.5, exactly what you’d expect. sqrt(9) times 0.5.

The mean had a standard deviation of 0.167. Which is 1.5 / 9.

Repeated using rectangular distributions, got the same result.

Tim Gorman
Reply to  Bellman
July 8, 2022 10:31 am

There you go again with a “normal” distribution. Why won’t you admit that the temperatures used to create a “GAT” aren’t a “normal” distribution?

Rectangular distributions are independent, identically distributed data just like Gaussian distributions.

Again, why won’t you admit that jamming NH and SH temps together don’t create an iid distribution?

Tim Gorman
Reply to  Bellman
July 8, 2022 8:06 am

No you sum up the elements and then divide by sample size.”

And what is the uncertainty of the sample size?

“How do you lay temperatures end to end? Seriously, yes you calculate the mean in the same way. The point I was trying to make is that there is an intrinsic reason why you might just want the sum of the length of boards, but no reason to want to know the sum of temperatures except as way to get to the mean.”

And what is the uncertainty of the value of that mean if it isn’t made up of the propagation of uncertainty from the individual elements used to create the sum?

And what is the usefulness of the mean if the distribution is not iid? The mean doesn’t appear in the 5-number statistical description.

“Yes, but you never want to do the translation. It’s a very simple translation, just divide the uncertainty of the sum by the sample size.”

The uncertainty of the sample size is zero. It can’t add to the uncertainty. You divide the sum by the sample size to get the mean. You ADD the uncertainties of all the elements to get the total uncertainty. u(sum) + u(sample size) = u(sum)

“It isn’t. I’ve explained to you repeatedly why it isn’t. You claim to have read Taylor and done all the exercises but you seemed to have missed the part where he explains you cannot mix add/subtraction with multiplication/division.”

I haven’t mixed anything! It truly is not my problem that you can’t do simple algebra.

ẟq/q = ẟx/x + ẟn/n is how you do multiplication/divide in Taylor’s Rule 3.8 as an example. Just define u(q) = ẟq/q, ẟx/x as u(x), and ẟn/n as u(n) and you get u(q) = u(x) + u(n). A simple algebraic substitution. Since ẟn = 0, both ẟn/n and u(n) = 0!

I don’t know why I waste my time with you. It’s just useless! You can’t even get simple algebra correct!

Out of interest, could you point to a time in all your real world experience, where it was necessary to determine the uncertainty of a mean of a large number of values, and it was possible to see if your calculation was correct?”

Absolutely! I have designed a number of beams to span house foundations. I have designed several small bridges spanning fixed points across a stream.

Each beam and each strut were made up of combinations of shorter lengths of boards. If you didn’t allow for the uncertainty of the boards making up the beam and bridge struts you couldn’t guarantee that the beam and struts would meet at the support points. For instance, a support beam is made up of overlapping boards that are nailed and glued together. You offset each joint to maximize the strength of the overall beam. Once you have glued and nailed the beam together it is difficult to extend its length since it is impossible to un-nail and un-glue it! Same with the bridge struts!

Of course YOU wouldn’t care. You’d just go out and buy a new set of boards as many times as needed in order to finally get one beam or strut that is of sufficient length to span the needed distance – AND THEN CHARGE THE CUSTOMER FOR THE USELESS INVENTORY YOU COULDN’T USE! Your reputation would be quickly shot and your business would go broke! But who cares? You could always fall back on your knowledge of statistics and uncertainty to get a job somewhere, right?

Bellman
Reply to  Tim Gorman
July 8, 2022 3:11 pm

And what is the uncertainty of the sample size?

Assuming you’ve figured out how to count by now, close to zero.

The uncertainty of the sample size is zero. It can’t add to the uncertainty.

And no surprise you still have grasped that it is not the contribution of the uncertainty of the count that changes the uncertainty, it’s the difference in size between the mean and the sum. It’s just that this difference happens to be 1/N.

ẟq/q = ẟx/x + ẟn/n is how you do multiplication/divide in Taylor’s Rule 3.8 as an example. Just define u(q) = ẟq/q, ẟx/x as u(x), and ẟn/n as u(n) and you get u(q) = u(x) + u(n). A simple algebraic substitution. Since ẟn = 0, both ẟn/n and u(n) = 0!

Good. Now all you need to remember is how you defined u(q) and u(x), and what that means for ẟq and ẟx. As you say, it’s simple algebra, but for some reason you can never complete it to conclusion.

Tim Gorman
Reply to  Bellman
July 8, 2022 4:54 am

More goal posts being shifted. We were talking about pulling planks out of the ones dumped in your ditch. Why is it’s variance increasing as I pull more planks out?”

The only limits on the length of the boards you collect are zero (i.e. no boards) and the length of board you can carry in your vehicle. So why wouldn’t the variance (i.e. the range) of your collection increase (i.e. longer boards) until you’ve maxed out the length (determined by your vehicle)?

“Does not matter.”

Of course it matters. You don’t have to necessarily have a Gaussian distribution but you *must* have, at the very least, an independent and identically distributed set of values in order to have a possibility of ignoring the uncertainty! And you don’t have that!

You keep saying you don’t ignore uncertainty but you do ignore it in everything you post.

Measurements of different things is one of the main uses of averaging. There isn’t much point of getting an average when everything’s the same.”

The average is only useful if you have an independent, identically distributed distribution – typically a Gaussian distribution.

If you don’t have that then you should use the 5-number statistical description of the distribution. And the 5-number statistical description doesn’t include the average!

“The logic you live by! If an average is derived from a sum then an average is a sum.”

You STILL don’t get it! The uncertainty of the average derives from the sum of the values plus the uncertainty of a constant. Since the uncertainty of a constant is ZERO it contributes nothing to the uncertainty of the average, only the sum does! That doesn’t mean an average is a sum!

The average uncertainty is not surprisingly ±0.5°C. The uncertainty if the mean is ±0.05°C. THEY ARE NOT THE SAME.”

And you STILL can’t get it right. The average uncertainty IS 0.5. The uncertainty of the average is sqrt(0.5^2 * 100) = 0.5*10 = 5.

I give up. You will NEVER understand the subject of uncertainty. There are none so blind as those who will not see. Willful ignorance is *not* a survival trait!

Carlo, Monte
Reply to  Tim Gorman
July 8, 2022 6:49 am

And he still invokes the same ignorant argument the climate scientists in academia used against Pat Frank’s paper: “the error can’t be this large, therefore the analysis is wrong.”

He STILL doesn’t understand that uncertainty is not error.

Bellman
Reply to  Carlo, Monte
July 8, 2022 6:11 pm

It’s a pretty useful test of your analysis. If it produces impossible results it’s probably wrong.

Carlo, Monte
Reply to  Bellman
July 8, 2022 6:32 pm

It’s bullshit that shows you just another clueless climastrologer using huffing and bluffing to cover your abject ignorance.

Bellman
Reply to  Tim Gorman
July 8, 2022 3:13 pm

And you STILL can’t get it right. The average uncertainty IS 0.5. The uncertainty of the average is sqrt(0.5^2 * 100) = 0.5*10 = 5.

Sorry, I’ve given up banging my head against this wall for now. Believe whatever nonsense you like.

Carlo, Monte
Reply to  Bellman
July 8, 2022 6:33 pm

Yet 3 hours later, here you are again pushing your garbage pseudoscience.

Carlo, Monte
Reply to  Tim Gorman
July 7, 2022 12:48 pm

The NIST uncertainty calculator tells him what he wants to hear.

Tim Gorman
Reply to  Carlo, Monte
July 8, 2022 4:55 am

It’s confirmation bias all the way. That seems to be a common malady among CAGW advocates.

Tim Gorman
Reply to  Bellman
July 5, 2022 5:14 pm

Secondly, that’s not the point of this exercise. I don;t want to predict what the next days temperature will be in a random part of the planet.”

That is EXACTLY what you are doing when you use the Global Average Temperature. If the GAT can’t give you an expectation of what the next temperature measurement somewhere on the earth will be then of what use it?

“The point of the monthly global averages is to compare it with previous values, to see if there has been any significant change.”

When you do this the uncertainties will outweigh any “difference” you might see! So what is the purpose? You either find a way to lower the uncertainty by using better measurement devices or you wait until the measurement changes exceed the uncertainty intervals in order to determine what is happening! If your two measurements are 51C +/- 0.5C and 52C +/- 0.5C you can’t even be sure that a change has happened because the true value of the first measurement could be 51.5C (51 + .05) and of the second 51.5 (52 – 0.5) => THE SAME VALUE!

And it gets worse when you add temperatures in the vain attempt to develop an average that actually tells you something. They are independent, random temperatures, i.e. measurements of different things, and their uncertainties ADD, just like variances add when combining random, independent variables. Add T1 and T2 together and divide by two and the uncertainty of the average is u(T1) + u(T2) + u(2) = u(T1) + u(T2) = +/- 1.0C.

Your average value has more uncertainty that either of the two temperature measurements alone. So if you use the average value to compare against another temperature or another average temp you will need a difference great enough to overcome the uncertainty in order to tell if there has bee a difference. And the more temperatures you include in your average the wider the uncertainty interval will become whether you do it by direct addition or root-sum-square.

You can run and hide from this by saying that all uncertainty cancels when using temperatures from different locations taken at different times but you are only fooling yourself.

Random, independent variables:

  1. Variances add.
  2. Uncertainties add

Fundamental truisms in the real world.

Carlo, Monte
Reply to  Tim Gorman
July 5, 2022 5:28 pm

And he still can’t figure out that a sampled temperature time series is NOT random sampling of a distribution!

Bellman
Reply to  Carlo, Monte
July 5, 2022 6:21 pm

Never said it was. We are not talking about a time series, we are talking about random independent variables, and in particular what happens to the variance when you take their average. This should avoid all the distractions of how you define uncertainty, what a temperature is etc. It’s a basic mathematical operation, well defined and easily tested, yet for some reason Tim, and possibly you, still manage to get it wrong.

Carlo, Monte
Reply to  Bellman
July 5, 2022 8:53 pm

What is this “we” business?

WTH do you think the UAH is? Dice?

Scheesh.

And you still have no clues what uncertainty is—”Unskilled and Unaware”.

Bellman
Reply to  Carlo, Monte
July 6, 2022 2:59 pm

“we” being myself and Tim, who were discussing variance in random variables.

Bellman
Reply to  Tim Gorman
July 5, 2022 6:09 pm

That is EXACTLY what you are doing when you use the Global Average Temperature. If the GAT can’t give you an expectation of what the next temperature measurement somewhere on the earth will be then of what use it?

I answer that in the comment you quoted.

When you do this the uncertainties will outweigh any “difference” you might see!

That’s the point of doing significance testing. It’s just that you don’t know how to do it correctly. But as always your incorrect uncertainty calculations are arguments of convenience. You have no objection to the claims that this month was cooler than the previous month, or that there has been no trend over the last 7.75 years. You only kick off about uncertainty when it appears there is a significant positive trend.

They are independent, random temperatures, i.e. measurements of different things, and their uncertainties ADD, just like variances add when combining random, independent variables.

Do you still fail to understand why this is wrong, or do you believe that the act of endlessly repeating it will somehow make it true?

Add T1 and T2 together and divide by two and the uncertainty of the average is u(T1) + u(T2) + u(2) = u(T1) + u(T2) = +/- 1.0C.”

You asked me to explain why this was wrong last time. I did, so you ignore it and just keep repeating the same mistakes.

You do not add uncertainties like that when you are combining adding and dividing. They have to be treated as two separate operations because adding and dividing use two different equations. You claim to have read Taylor, I’ve pointed you to the part where Taylor explains this, but for some reason you cannot or will not understand it.

u(T1 + T2) = u(T1) + u(T2), or if they are independent sqrt(u(T1)^2 + u(T2)^2).

u((T1 + T2) / 2) / (T1 + T2) / 2) = u(T1 + T2) / (T1 + T2) + u(2) = u(T1 + T2) / (T1 + T2)

Which implies

u((T1 + T2) / 2) = u(T1 + T2) / 2

So, if u(T1) = u(T2) = 0.5, the uncertainty of the average is either 0.5, or 0.5 / sqrt(2), depending on independence.

Tim Gorman
Reply to  Bellman
July 7, 2022 7:17 am

That’s the point of doing significance testing. It’s just that you don’t know how to do it correctly. “

You can’t just do significance testing on stated values! When you ignore the uncertainties that go along with those stated values your significance testing is useless!

“You have no objection to the claims that this month was cooler than the previous month, or that there has been no trend over the last 7.75 years.”

Really? I have told you *MULTIPLE* times that I have no more confidence in UAH than any other data set — because they all ignore the rules of propagating uncertainty! Like most things, you just ignore reality!

“Do you still fail to understand why this is wrong, or do you believe that the act of endlessly repeating it will somehow make it true?”

It’s not wrong. I’ve given you quote after quote, several from university textbooks, which you just dismiss as wrong without ever showing how they are wrong.

You asked me to explain why this was wrong last time. I did, so you ignore it and just keep repeating the same mistakes.”

Malarky! You never showed anything except how you can calculate an average uncertainty! Which is *NOT* the same thing as the uncertainty of the average!

Which implies

u((T1 + T2) / 2) = u(T1 + T2) / 2
So, if u(T1) = u(T2) = 0.5, the uncertainty of the average is either 0.5, or 0.5 / sqrt(2), depending on independence.”

And here you are again, trying to show that the average uncertainty is the uncertainty of the average!

The average uncertainty is just an artifact you can use to equally distribute uncertainty among all the data elements. It is *NOT* the uncertainty of the sum of the data elements – which you must do in order to calculate the average uncertainty.

If you take 100 boards, each with a different uncertainty interval for its length, and calculate the average uncertainty, that average uncertainty tells you absolutely ZERO about each individual board. When you extract a sample of those boards that average uncertainty will tell you nothing about the actual uncertainty of the sample!

It really *IS* true that you know little about reality!

Bellman
Reply to  Tim Gorman
July 8, 2022 6:32 pm

You can’t just do significance testing on stated values! When you ignore the uncertainties that go along with those stated values your significance testing is useless!

There’s a lot of competition but that has to be one of the dumbest claims made yet. Read any book on statistics to see how to do significance testing. Most will just use stated values, because it’s still very easy to see if a result is significant or not.

I know. I know. You’ll say that just shows that all statistics over the last century or so is wrong, because they don’t “live in the real world”.

Really? I have told you *MULTIPLE* times that I have no more confidence in UAH than any other data set

And yet you still think you can use it to detect a precise pause.

It’s not wrong. I’ve given you quote after quote, several from university textbooks, which you just dismiss as wrong without ever showing how they are wrong.”

I’m sure that’s what you think you have done, but I’m not sure about your memory at this moment. Still, prove me wrong. Point to one of these the books which claims variances add when taking a mean, or that uncertainties add when taking a mean of different things. Then we can examine it and see if it actually says what you think it says.

Malarky! You never showed anything except how you can calculate an average uncertainty! Which is *NOT* the same thing as the uncertainty of the average!

Do you know you have a tell? Often, when you say something that is completely untrue you yell “Malarky!” just before.

No I did not “show you how to calculate an average uncertainty”. I showed you that you couldn’t mix propagation of uncertainty for adding with that for dividing, how you had to do it in two stages, switching between absolute and fractional uncertainties, and if you did it correctly you got the uncertainty of the average rather than the uncertainty of the sum. Do you remember now?

And here you are again, trying to show that the average uncertainty is the uncertainty of the average!

No it isn’t. This shouldn’t be hard even for you. u(T1 + T2) / 2 is not the average uncertainty. How do I know? Because u(T1 + T2) is not (necessarily) the sum of the uncertainties, it’s the uncertainty of the sum. Hence, dividing by two is not dividing the sum of uncertainties between the two values, it’s dividing the uncertainty of the sum between the two.

Bellman
Reply to  Tim Gorman
July 5, 2022 6:18 pm

And the more temperatures you include in your average the wider the uncertainty interval will become whether you do it by direct addition or root-sum-square.

Unless, by some strange chance you are wrong and every other expert on statistics and uncertainty are right.

You can run and hide from this by saying that all uncertainty cancels when using temperatures…

This strawman is getting quite pathetic. Nobody says all uncertainties cancel..

Variances add.

And variances scale.

If you have a single random variable X, with variance var(X), what do you think the variance of X/2 will be? If you have two random independent variables X and Y, what will be the variance of (X + Y) / 2, or (1/2)X + (1/2)Y?

Tim Gorman
Reply to  Bellman
July 7, 2022 7:35 am

Unless, by some strange chance you are wrong and every other expert on statistics and uncertainty are right.”

I suggest you take another look at Taylor, Rule 3.8. It agrees with me, not you.

If q = x/w then ẟq/q = ẟx/x + ẟw/w

I have merely defined ẟx/x = u(x)

Since w (or in our case “n”) is a constant, u(w) = 0.

EXACTLY as I have done!

Just extrapolate that out: q = (x1+x2+…+xn)/n

where x1+x2+…+xn is the sum of the data elements and is used to calculate the mean. Then the uncertainty is:

u(q) = u(x1) + u(x2) + … +u(xn) + u(n) =
u(x1) + u(x2) + … + u(xn)

since u(n) = 0!

I simply don’t understand why this is so damned hard for you to understand.

*This strawman is getting quite pathetic. Nobody says all uncertainties cancel..”

You do. Every time you claim the accuracy of a mean calculated from sample means is the standard deviation of those sample means. Those sample means are calculated ONLY using stated values. So the mean you calculate from them uses ONLY stated values.

YOU JUST TOTALLY IGNORE THE UNCERTAINTIES OF THE STATED VALUES!!! You do it every single time. And the only justification for doing so is a belief that all uncertainties cancel!

It’s the same thing almost all climate scientists do. They have been trained by statisticians using statistics textbooks that never consider the uncertainty associated with the stated values they use as data sets. So they (AND YOU) get an inbuilt bias that uncertainties don’t matter, they all cancel!

If you have two random independent variables X and Y, what will be the variance of (X + Y) / 2, or (1/2)X + (1/2)Y?”

And we are back to AVERAGE VARIANCE is total variance!

You are stuck in a rut you just can’t seem to climb out of!

Bellman
Reply to  Tim Gorman
July 7, 2022 8:45 am

I have merely defined ẟx/x = u(x)

OK, so you are defining u(x) as relative uncertainty.

Then the uncertainty is: u(q) = u(x1) + u(x2) + … +u(xn) + u(n) =
u(x1) + u(x2) + … + u(xn)

Wrong, wrong and even more wrong.

You are adding values of x1 … xn. You do not add their relative uncertainties, you add the absolute uncertainties.

And you do not combine the uncertainty prolongations for adding/subtracting with those for multiplication/division like this, so you can not simply add the uncertainty of n like this. They have to be done in two separate stages.

You keep trying to do the same thing over and over, come up with some new arrangement of equations to convince yourself that you must be right. And they never work because you are wrong. And a moment’s though should convince you that you are wrong becasue the result you want is clearly nonsense.

I simply don’t understand why this is so damned hard for you to understand.

And again, you can nether accept that the reason it might be so hard to understand is because it’s wrong.

Every time you claim the accuracy of a mean calculated from sample means is the standard deviation of those sample means.

Firstly, I don;t say that. What I say is that the standard error of the means is akin to precision. It’s an indicator of the closeness of the sample means to each other, it does not necessarily mean that it is close to the true value.

Secondly, how does this equate to saying all uncertainties cancel? Random uncertainties whether from measurement or sampling will decrease as sample size increases – they do not all cancel.

YOU JUST TOTALLY IGNORE THE UNCERTAINTIES OF THE STATED VALUES!!! You do it every single time. And the only justification for doing so is a belief that all uncertainties cancel!

No it isn’t.

Firstly, I do not say you can’t take into account measurement uncertainties. This entire past two years have been about me discussing how to calculate the measurement uncertainties in a mean.

Secondly, if I do think it’s often reasonable to ignore measurement uncertainties when calculating a mean, it’s not because I think all uncertainties cancel, it’s because I think that they are largely irrelevant compared with the uncertainties inherent in the sampling process. If the things I’m measuring vary by meters the effect of an uncertainty of a cm in the measurement will be insignificant to the result. I also think that random measurement errors do not have to be accounted for normally, because they are already present in the data.

And we are back to AVERAGE VARIANCE is total variance!

NO WE ARE NOT. You keep thinking that dividing anything by anything is only done to get an average. I am not talking about the average variance but the variance of the average. The fact that after all these weeks you still can’t understand this makes me a little concerned for you.

I asked what the variance of

(X + Y) / 2

was, and the answer is not the average of the variance. The answer is

(Var(X) + Var(Y)) / 4

How is that the average of the variances?

Tim Gorman
Reply to  Bellman
July 7, 2022 5:56 pm

Wrong, wrong and even more wrong.”

Nope. It is *EXACTLY* what Taylor 3.8 shows. EXACTLY.

“You are adding values of x1 … xn. You do not add their relative uncertainties, you add the absolute uncertainties.”

NOPE! Again, look at Taylor 3.8!

If q = x/w then u(q)/q = u(x)/x + u(w)/w

Since u(w) = 0 then u(q)/q = u(x)/x

“And you do not combine the uncertainty prolongations for adding/subtracting with those for multiplication/division like this, so you can not simply add the uncertainty of n like this. They have to be done in two separate stages.”

Nope. You do it EXACTLY as I have explained!

There is no “two separate stages”!

“You keep trying to do the same thing over and over, come up with some new arrangement of equations to convince yourself that you must be right.”

I tried to simplify it so you could understand. As usual you completely failed at understanding.

“Firstly, I don;t say that. What I say is that the standard error of the means is akin to precision. It’s an indicator of the closeness of the sample means to each other, it does not necessarily mean that it is close to the true value.”

You say you don’t but then you turn around and do it every time. When you conflate confidence intervals with uncertainty for temperatures you are totally ignoring any uncertainties in the measurements. It’s just baked into *everything* you do!

“Secondly, if I do think it’s often reasonable to ignore measurement uncertainties when calculating a mean,”

You can’t even get this simple thing straight. The mean is calculated from the stated values not from their uncertainty intervals. The UNCERTAINTY OF THE MEAN is calculated from the uncertainty of the measurements, not the mean!

” it’s not because I think all uncertainties cancel, it’s because I think that they are largely irrelevant compared with the uncertainties inherent in the sampling process.”

The uncertainties in the sampling process have to do with the precision of the mean you calculate from the sample(s). NOT with the uncertainties associated with the sample mean!

“f the things I’m measuring vary by meters the effect of an uncertainty of a cm in the measurement will be insignificant to the result.”

And here we are, back to your absolute lack of knowledge about the real world. How do you fix a cooling pipe in a nuclear power plant that is 20m long but winds up1cm short because you forgot to allow for uncertainty in your measurement of the 20m pipe? Do you just let it hang loose? Do you throw it away and get a new chunk of pipe? Do you cut it and splice in another chunk hoping it will pass inspection?

“I also think that random measurement errors do not have to be accounted for normally, because they are already present in the data.”

The DATA IS “STATED VALUE +/- UNCERTAINTY”. How is that uncertainty able to be discarded?

“You keep thinking that dividing anything by anything is only done to get an average. I am not talking about the average variance but the variance of the average.”

The variance is a NUMBER. It is a number calculated from the stated values, the mean, and the number of elements. You’ve already calculated the AVERAGE (i.e. the mean). How do you get a variance of the average value which is a number, not a distribution?

That’s like saying the mean of a distribution has a standard deviation. It doesn’t. The POPULATION has a standard deviation, not the mean. The mean is a number, not a distribution!

“I asked what the variance of

(X + Y) / 2

was, and the answer is not the average of the variance. The answer is

(Var(X) + Var(Y)) / 4″

(X + Y)/2 = (X/2 + Y/2)

Var(aX + bY) = a^2X + b^2Y = (X + Y)/4

So what? The variance of (X + Y) is Var(X) + Var(Y).

When you are working with measurements you don’t all of a sudden divide them by 2!

X ≠ X/2

Bellman
Reply to  Tim Gorman
July 7, 2022 6:25 pm

Taylor 3.8 – Uncertainty in Products and Quotients (Provisional Rule).

It deals with Products and Quotients, not Adding..

Nope. You do it EXACTLY as I have explained!
There is no “two separate stages”!

Taylor Section 3.8 – Propagation Step by Step

(My emphasis)

Any calculation can be broken down into a sequence of steps, each involving just one of the following types of operation: (1) sums and differences; (2) products and quotients; and (3) computation of a function of one variable

Before I discuss some examples of this step-by-step calculation of errors, let me emphasize three general points. First, because uncertainties in sums or differences involve absolute uncertainties whereas those in products or quotients involve fractional uncertainties the calculations will require some facility in passing from absolute to fractional uncertainties and vice versa…

Last edited 1 month ago by Bellman
Tim Gorman
Reply to  Bellman
July 8, 2022 1:05 pm

So what? I USED fractional uncertainties!

Can you not understand that a function can be defined for uncertainty?

Maybe if I call it f() instead of u you can figure it out?

if f(x) = ẟx/x and f(q) = ẟq/q and f(n) = ẟn/n

Then f(q) = f(x) + f(n). Since f(n) = 0 then we get

f(q) = f(x) –> ẟq/q = ẟx/x

QED!

You *really* don’t know algebraic math at all, do you?

Bellman
Reply to  Tim Gorman
July 8, 2022 3:16 pm

ẟq/q = ẟx/x

You spend all this time to produce exactly what I’m trying to tell you, and then will ignore the consequence of that result.

Let’s try this. What if q = 20 and x = 2000 and ẟx = 5. What is ẟq?

Bellman
Reply to  Tim Gorman
July 7, 2022 6:37 pm

You can’t even get this simple thing straight. The mean is calculated from the stated values not from their uncertainty intervals. The UNCERTAINTY OF THE MEAN is calculated from the uncertainty of the measurements, not the mean!

Yes, my mistake. I meant to say “Secondly, if I do think it’s often reasonable to ignore measurement uncertainties when calculating [the uncertainty of] a mean,”

How do you fix a cooling pipe in a nuclear power plant that is 20m long but winds up1cm short because you forgot to allow for uncertainty in your measurement of the 20m pipe?

No idea, it’s got nothing to do with the point I was making. I’m talking about calculating the mean and how much of an impact 1cm uncertainties will make when what you are measuring varies by meters.

If you need your pipes to be an exact length, there’s little point having pipes that vary by meters and knowing the precise average length.

The DATA IS “STATED VALUE +/- UNCERTAINTY”. How is that uncertainty able to be discarded?

The point is that if the uncertainty is from random errors, those errors will be present in the data. If you measure with low precision there will be more variability in the sample.

Tim Gorman
Reply to  Bellman
July 8, 2022 1:16 pm

I meant to say “Secondly, if I do think it’s often reasonable to ignore measurement uncertainties when calculating [the uncertainty of] a mean,””

“often reasonable”? You do it ALL THE TIME!

“No idea, it’s got nothing to do with the point I was making.”

OMG! Uncertainty has nothing to do with the point? That’s because you *always* ignore uncertainty!

 I’m talking about calculating the mean and how much of an impact 1cm uncertainties will make when what you are measuring varies by meters.”

SO AM I! Coming up 1cm short on a 20m pipe MAKES ONE HELL OF BIG DIFFERENCE!

“If you need your pipes to be an exact length, there’s little point having pipes that vary by meters and knowing the precise average length.”

Unfreakingbelievable! The pipes don’t vary by meters, they vary by centimeters and millimeters. When they have to reach a connector every single millimeter matters!

You are right, however. There is little use in knowing a precise average length! That’s what I keep telling you and you keep refusing to believe. The average length of dissimilar things IS USELESS!

The point is that if the uncertainty is from random errors, those errors will be present in the data. If you measure with low precision there will be more variability in the sample.”

Temperature measurements are *NOT* random errors. They are stated values +/- uncertainty. The true value lies between stated value minus the uncertainty and stated value plus the uncertainty.

Precision and uncertainty ARE NOT THE SAME THING. Uncertainty is an estimate of accuracy. Accuracy is not precision!

It’s like beating your head against the wall trying to educate you!

Bellman
Reply to  Tim Gorman
July 8, 2022 6:00 pm

SO AM I! Coming up 1cm short on a 20m pipe MAKES ONE HELL OF BIG DIFFERENCE!”

Which has nothing to do with the mean.

Unfreakingbelievable! The pipes don’t vary by meters, they vary by centimeters and millimeters.

Then you are not addressing my point.

The average length of dissimilar things IS USELESS!

Then stop using it as an example. If your example is of something where it’s not useful to know the mean, then it’s not an example of something where it is useful to know the mean. This is the problem with people who only live in “the real world”. They can’t conceive things can be different outside their own little real world.

Temperature measurements are *NOT* random errors.

Of course not. But the measurement contains random errors.

Precision and uncertainty ARE NOT THE SAME THING.

Why do you keep repeating things I agree with?

It’s like beating your head against the wall trying to educate you!

Maybe if you read what I write you wouldn’t have to.

Carlo, Monte
Reply to  Bellman
July 8, 2022 6:34 pm

Maybe if you read what I write you wouldn’t have to.

What you write is ignorant word salad.

Bellman
Reply to  Tim Gorman
July 7, 2022 6:48 pm

The variance is a NUMBER. It is a number calculated from the stated values, the mean, and the number of elements. You’ve already calculated the AVERAGE (i.e. the mean). How do you get a variance of the average value which is a number, not a distribution?

By using all those equations for combining random variables.

That’s like saying the mean of a distribution has a standard deviation. It doesn’t. The POPULATION has a standard deviation, not the mean. The mean is a number, not a distribution!

The distribution of means is what you would get if you take multiple samples. The variance of the means of that sample distribution is the variance of the mean.

But you don’t need to take multiple samples, becasue you know how random variables work so you calculate the variance, and hence standard error using the rules for combining random variables.

“(X + Y)/2 = (X/2 + Y/2)
Var(aX + bY) = a^2X + b^2Y = (X + Y)/4”

That last line should be
Var(aX + bY) = a^2var(X) + b^2var(Y) = (var(X) + var(Y))/4.

“So what? The variance of (X + Y) is Var(X) + Var(Y).”

The so what is I don;t want to know the variance of the sum of X and Y, I want to know the variance of the mean of X and Y, which is (var(X) + var(Y))/4.

When you are working with measurements you don’t all of a sudden divide them by 2!

You do if you want to know their average.

“X ≠ X/2”

I see your education wasn’t a complete waste.

Tim Gorman
Reply to  Bellman
July 8, 2022 1:19 pm

The distribution of means is what you would get if you take multiple samples. The variance of the means of that sample distribution is the variance of the mean.”

That’s the standard deviation of the sample means. It is not uncertainty!

“The so what is I don;t want to know the variance of the sum of X and Y, I want to know the variance of the mean of X and Y, which is (var(X) + var(Y))/4.”

Why do you want to know the variance of the mean? You just said in the prior message that knowing the precise mean is useless!

b: “If you need your pipes to be an exact length, there’s little point having pipes that vary by meters and knowing the precise average length.”

Bellman
Reply to  Tim Gorman
July 8, 2022 5:36 pm

Why do you want to know the variance of the mean?

I don’t particularly want to know the variance, I want to take the square root to get the standard error.

You just said in the prior message that knowing the precise mean is useless!”

I’m not sure if your reading comprehension isn’t even worse than your maths.

Carlo, Monte
Reply to  Bellman
July 8, 2022 6:34 pm

I see your education wasn’t a complete waste.

So speaketh the professional trendologist.

Bellman
Reply to  Carlo, Monte
July 8, 2022 7:23 pm

I really don’t get why you think an interest in trends is an insult. The same with bellcurve. It’s like you’ve studied the Monckton book of ad hominems, but just don’t have his level of shining wit.

Tim Gorman
Reply to  Bellman
July 5, 2022 5:31 pm

And you still haven’t grasped this simple point – As you add random variables the variance increases, but when you take their mean the variance decreases.”

How can the variance of the population decrease? The variance is the sum of the differences between the mean and each value squared which is then divided by the number of the elements. You calculate the mean FIRST, before finding the variance!

The mean is an average of a group of numbers. The variance is an average of how far each number is from the mean. You can’t change the variance by calculating the mean!

You can find the mean of the population but that has nothing to do with the variance of the population! Finding the mean, therefore, cannot decrease the variance of the population! And it is the variance of the population that is a measure of the accuracy of your mean!

If you sample the population then you can calculate the mean more precisely by taking more samples or by taking larger samples but neither of these decrease the variance of the population.

You are STILL confusing the standard deviation of the sample means with standard deviation/variance of the population. The standard deviation of the sample means is only a measure of how precisely you have calculated that mean. It has nothing to do with the uncertainty of the mean you have precisely calculated! The uncertainty of that mean will still be the uncertainty of the mean of the entire population.

Precision is not accuracy. How often does this need to be repeated in order for it to finally sink into your skull?

Bellman
Reply to  Tim Gorman
July 5, 2022 6:29 pm

How can the variance of the population decrease?

Are you ever capable of holding on to one idea for more than a second? We are not talking about the variance of the population, we are talking about the variance of the mean of multiple random variables.

And it is the variance of the population that is a measure of the accuracy of your mean!

What are you on about now? How is the variance of the population a measure of the accuracy of the mean? You keep making these assertions as if you saying it makes it true. And you keep failing to see how you are contradicting yourself. First you want the uncertainty of the sample mean to grow with sample size, now you are claiming it will be the same as the population variance which will be the same regardless of the sample size.

Tim Gorman
Reply to  Bellman
July 7, 2022 8:47 am

Are you ever capable of holding on to one idea for more than a second? We are not talking about the variance of the population, we are talking about the variance of the mean of multiple random variables.”

No, you are talking about an AVERAGE VARIANCE! The variance associated with the mean is the variance of the population and not the average variance!

You keep making these assertions as if you saying it makes it true.”

It *is* true! The higher the variance the wider range of numbers are in the distribution.

 First you want the uncertainty of the sample mean to grow with sample size, now you are claiming it will be the same as the population variance which will be the same regardless of the sample size.”

The uncertainty of the sample mean *does* grow as you add more uncertainty elements.

Until you can figure out that average uncertainty is not uncertainty of the mean and that average variance is not variance (how does a mean calculated from stated values have a variance anyway?) of the population you will *never* understand physical reality and the use of uncertainty!

Bellman
Reply to  Tim Gorman
July 7, 2022 9:25 am

No, you are talking about an AVERAGE VARIANCE!

No I am not. I’d don’t know why you think this, I don;t know why you ignore every effort to explain why it isn’t true. You seem to have got this tick in your brain and you can’t let go of it.

I am not talking about the average variance but the variance of the average. If you add multiple independent random variables together the variance will be the sum of the variances. If you average a number of independent random variables the variance of the mean will be the sum of the variances divided by the square of the number of variables. You are not finding the average variance.

It *is* true! The higher the variance the wider range of numbers are in the distribution.

Yes that’s true, but it’s not what you were saying. What you were saying is “it is the variance of the population that is a measure of the accuracy of your mean!

The uncertainty of the sample mean *does* grow as you add more uncertainty elements.

No idea what you mean by “more uncertainty elements”. I think this disussion keeps getting confused because you keep changing the terms. There are two scenarios here.

1) you are taking a sample of elements from the same population. In that case each element is a random variable with the same mean and variance as the population. As you increase sample size the variance of the sample will tend to the variance of the population.

2) you are taking the average of a number of different random variables not from the same population. In that case there is no knowing what will happen to the variance as it depends entirely on what random variables you keep adding.

Until you can figure out that average uncertainty is not uncertainty of the mean

I keep telling you it’s not.

and that average variance is not variance

Not sure what you mean there. You need to define your terms better. The variance of a random variable is by definition an average. The average variance of a sample will be the same as the variance of the population. However the variance of the average of a sample will not be the same as the variance of the population. In fact it will be (assuming independence) equal to the “average” variance, i.e. the variance of the population divided by the sample size.

Tim Gorman
Reply to  Bellman
July 7, 2022 6:21 pm

I am not talking about the average variance but the variance of the average.”

How can a VALUE (a single number), i.e. the average have a variance? It inherits the variance of the population but that *is* the variance of the population and not the variance of the average!

If you average a number of independent random variables the variance of the mean will be the sum of the variances divided by the square of the number of variables. You are not finding the average variance.”

If you have X and Y and you convert that to X/2 and Y/2 you don’t have the same distribution. The variance of X+Y is *NOT* the variance of (X/2 + Y/2).

The value of Var(X+Y) /2 *IS* the average variance and is meaningless! You are throwing out a red herring trying to divert the discussion down a different path!

Yes that’s true, but it’s not what you were saying. What you were saying is “it is the variance of the population that is a measure of the accuracy of your mean!””

So what? The wider the variance the more possible values you can possibly have. That *IS* a measure of the accuracy of your mean!

“No idea what you mean by “more uncertainty elements”. I think this disussion keeps getting confused because you keep changing the terms. There are two scenarios here.”

Malarky! Each data element is a “stated value +/- uncertainty”. How many times have you been told this? When you add data elements you also add uncertainty elements!

 you are taking a sample of elements from the same population. In that case each element is a random variable with the same mean and variance as the population. As you increase sample size the variance of the sample will tend to the variance of the population.”

Single values don’t have a mean or variance. Each data element has a stated value and an uncertainty – neither of which is a variance or mean!

But you *are* correct that the variance of the sample will approach the variance of the population as the sample size grows. So what?

In that case there is no knowing what will happen to the variance as it depends entirely on what random variables you keep adding.”

Nope. If the sample is representative of the population the variance should be close to the variance of the population. If that wasn’t true then your claim that larger sample sizes approach the population variance is incorrect!

I keep telling you it’s not.”

You keep saying that but you have *never* internalized it. If you had you wouldn’t keep ignoring uncertainty – even to trying to rationalize why you ignore it!

“Not sure what you mean there. You need to define your terms better.”

And now you are claiming you don’t know what an average is! ROFL!!

“The variance of a random variable is by definition an average.”

Then why do you want to divide the variance to get a supposed average variance?

The average variance of a sample will be the same as the variance of the population.”

ROFL!! You say you don’t use the average variance and then you turn right around and make this kind of claim?

A sample has a variance. It doesn’t have an AVERAGE variance.

However the variance of the average of a sample “

Again, an average is a single VALUE, it doesn’t have a variance! The population, be it a sample population or the total population, has a variance, not the mean. A mean doesn’t have a standard deviation – it is just a NUMBER. The distribution has a standard deviation around that mean (assuming a Gaussian distribution).

And you chastise me for not being specific with my terms?

 In fact it will be (assuming independence) equal to the “average” variance, i.e. the variance of the population divided by the sample size.”

And now we circle back again! You don’t use the “average variance” but here you are using it!

Tim Gorman
Reply to  Bellman
July 5, 2022 5:50 pm

Wut? How do previous measurements increase the variance of the next? What part of independent are you not understanding?”

That is not what I said. Stop putting words in my mouth. I said: “As variance increases the possible value of the next measurement increases also”

I didn’t say the variance of the next measurement increases.

Variance correlates to the size of the overall range of numbers. The variance is greater when there is a wider range of numbers in the set. When there is a wider range of numbers in the set there is also a wider range of possible expected values for the next measurement.

“Yes. And that technique shows you are wrong in both cases.”

I gave you a very simple example showing how variance grows. Uncertainty does the same. You just keep on proving that you have absolutely no understanding of the real world.

Which uncertainty interval are you talking about here? I think there’s been some confusion because he shows what he claims is the uncertainty interval for monthly values, but that is not the same as the confidence interval for the trend. Carlo’s stated uncertainty for the trend shows that the warming trend over the last 40 or so years is statistically significant, and the trend line could be anywhere from 0.6 to 2.0°C / decade.”

You are *STILL* nit-picking!

If the monthly values have an uncertainty then so does the trend developed from those monthly values!

The monthly values are “stated value +/- uncertainty”. the trend line *must* consider those uncertainties. Like usual, you just want to ignore them. Since the trend line is made up of values with uncertainty then the confidence interval for the trend also has uncertainty!

When you consider the uncertainty intervals the trend line could be positive or negative! It could even change slope in the middle of the interval. How in blue blazes would you know?

is statistically significant, and the trend line could be anywhere from 0.6 to 2.0°C / decade” This is based solely from the stated values with no consideration of the uncertainty interval. Just wipe out EVERY SINGLE STATED VALUE BETWEEN THE UNCERTAINTY LINES. Print the graph out and color the area between the lines black. Because every single point between those uncertainty lines might be the true value for that point on the x-axis. YOU DON’T KNOW.

Now, tell me what the trend line is!

Bellman
Reply to  Tim Gorman
July 5, 2022 6:41 pm

“That is not what I said. Stop putting words in my mouth. I said: “As variance increases the possible value of the next measurement increases also””

If the possible value of the next value increases that implies it has a greater variance. If you are taking random values from a population each should have the same variance.

When there is a wider range of numbers in the set there is also a wider range of possible expected values for the next measurement.

Maybe I’m just being confused by you use of language and you have a different concept in mind. What you said was:

As you add independent, random measurements of different things the variance increases. … As variance increases the possible value of the next measurement increases also. This is exactly how uncertainty works.

I assumed, by adding independent random measurements, you were on your usual practice of adding random variables together, but maybe you mean making a population by mixing sub populations together. We had this confusion before with the ambiguity of the word “add”.

Tim Gorman
Reply to  Bellman
July 7, 2022 8:56 am

If the possible value of the next value increases that implies it has a greater variance.”

Thanks for repeating what I’ve been telling you!

“If you are taking random values from a population each should have the same variance.””

How does a STATED VALUE have a variance? 1 = 1. 2 = 2. 3=3.

The value of 1 doesn’t have a variance. The value of 2 doesn’t have a variance. The value of 3 doesn’t have a variance.

Samples do not necessarily have the same variance as the population. The same thing applies for uncertainty. It depends on how well the sample represents the population. If the population has a wide variance it is more likely that a sample of size “n” will *not* be as good of a representation of the population as a population with a narrow variance.

tg: When there is a wider range of numbers in the set there is also a wider range of possible expected values for the next measurement. ”

tg: “As variance increases the possible value of the next measurement increases also.”

Maybe I’m just being confused by you use of language “

Those quotes are saying the exact same thing! Where’s the confusion?

“I assumed, by adding independent random measurements, you were on your usual practice of adding random variables together, but maybe you mean making a population by mixing sub populations together.”

Each independent, random temperature measurement you try to cram together in a data set so you can calculate an average represents a sub-population of size 1. You are trying to create some kind of red herring argument. Stop it right now!

Bellman
Reply to  Tim Gorman
July 7, 2022 11:23 am

How does a STATED VALUE have a variance? 1 = 1. 2 = 2. 3=3.

You keep getting confused over what the subject is. I’m not talking about specific values but random variables.

Samples do not necessarily have the same variance as the population.

That’s why I said the sample variance will tend to the population variance.

If the population has a wide variance it is more likely that a sample of size “n” will *not* be as good of a representation of the population as a population with a narrow variance.

Yes, that’s why you divide the variance by N, or the SD by root N. The larger the variance in the population the larger the variance in the sample mean.

Those quotes are saying the exact same thing! Where’s the confusion?

They don’t. In the first you are just making the obvious case that the larger the variance in a population the larger range of values a single item can take.

In the second you are talking about increasing the variance. What you actually said was

As you add independent, random measurements of different things the variance increases. … As variance increases the possible value of the next measurement increases also. This is exactly how uncertainty works.

It’s the first part of that quote I’m confused about. As I said, you aren;t clear in what sense you mean “adding”, is it summing or mixing?

Each independent, random temperature measurement you try to cram together in a data set so you can calculate an average represents a sub-population of size 1.

I ask you for some clarity in what you are saying, and I get this. Either think of all possible values as a single random variable representing the population, take your sample from that and the variance of the mean is the variance of the population divided by N, or treat each temperature as it’s own random variable with a distinct mean and variance, in which case the variance of the mean is the sum of all the variances divided by N squared.

Tim Gorman
Reply to  Bellman
July 8, 2022 4:24 am

You keep getting confused over what the subject is. I’m not talking about specific values but random variables.”

So what? A random variable has a mean and a variance. The mean of that random variable does *NOT* have a variance, the random variable does.

You are arguing black is white in order to gain a reply TROLL!

“Yes, that’s why you divide the variance by N, or the SD by root N. The larger the variance in the population the larger the variance in the sample mean.”

You’ve just highlighted another problem with the whole concept of using mid-range temperature values to calculate a global average temp! If you consider the daily max and the daily min as samples which also define the variance of the temperature profile then that mid-range value has a variance of of the temperature profile variance divided by 2.

If your high and low temps are 90F and 70F (about what they were here yesterday) then you get a mid-range value of 80 and a variance of 2.5. [(90-80)^2 + (80-70)^2]/80 Divide that 2.5 by the size of the sample and you get 1.3 for variance. That is +/- 1.1 for the standard deviation.

YOU ALREADY HAVE A VARIANCE OF THE MEAN THAT IS WIDER THAN THE TEMPERATURE DIFFERENTIAL YOU ARE TRYING TO FIND!

It’s greater than adding the uncertainties using root-sum-square (+/- 0.7) and even larger than direct addition of the uncertainties (+/- 1.0)

It’s the first part of that quote I’m confused about. As I said, you aren;t clear in what sense you mean “adding”, is it summing or mixing?”

How are temperatures combined in order to get a global average temperature?

You *know* the answer! Don’t play dumb.

take your sample from that and the variance of the mean is the variance of the population divided by N, or treat each temperature as it’s own random variable with a distinct mean and variance, in which case the variance of the mean is the sum of all the variances divided by N squared.”

And you *totally* miss the impact of this! It makes the mid-range values even more questionable for use in calculating a global average temperature!

You are hoist on your own petard!

Bellman
Reply to  Tim Gorman
July 8, 2022 3:28 pm

If your high and low temps are 90F and 70F (about what they were here yesterday) then you get a mid-range value of 80 and a variance of 2.5. [(90-80)^2 + (80-70)^2]/80 Divide that 2.5 by the size of the sample and you get 1.3 for variance. That is +/- 1.1 for the standard deviation.

Thanks for illustrating you don;t understand variance.

Firstly, there isn’t much point in taking the variance of max and min, they are not a random sample of the daily temperatures.

Secondly, why on earth are you dividing by 80? The variance is (10^2 + 10^2) / 2 = 100. (Or if this is a sample you should divide by 1, so it should be 200).

I presume you want to rescale the temperatures so they are ratio of the daily mean. I don;t know why, but go ahead. But in that case you should convert them to K first. Think about what would happen if the mean temperature was 0°F.

YOU ALREADY HAVE A VARIANCE OF THE MEAN THAT IS WIDER THAN THE TEMPERATURE DIFFERENTIAL YOU ARE TRYING TO FIND!

That’s what I was trying to tell you about variance last time. It’s not a value you can easily interpret in relation to the actual measurements. The variance can be much larger than the range of all your values, because it’s the square of the distances. That’s why it’s more convenient to convert variance to standard deviations.

Bellman
Reply to  Tim Gorman
July 5, 2022 6:50 pm

I gave you a very simple example showing how variance grows.

And it was wrong. Not sure which specific example you mean, but if it was any of the ones involving adding and claiming it’s the mean, it’s wrong.

The monthly values are “stated value +/- uncertainty”. the trend line *must* consider those uncertainties.”

I’m assuming that Carlo’s stated value for sigma was taking into account his fantasy monthly uncertainties. But as he refuses to say how it was calculated, who knows?

When you consider the uncertainty intervals the trend line could be positive or negative!

If Carlo’s values were correct there’s close to zero probability the trend is negative.

This is based solely from the stated values with no consideration of the uncertainty interval.

Again, you’ll have to take that up with Carlo. If his stated value is not taking into account the uncertainty as you wish, then tell him to include it or calculate it yourself. You might also want to calculate the uncertainty over the last 7.75 years.

Tim Gorman
Reply to  Bellman
July 7, 2022 9:09 am

And it was wrong. Not sure which specific example you mean, but if it was any of the ones involving adding and claiming it’s the mean, it’s wrong.”

How do you calculate variance without calculating the mean? You are losing your mind!

“I’m assuming that Carlo’s stated value for sigma was taking into account his fantasy monthly uncertainties. But as he refuses to say how it was calculated, who knows?”

“fantasy monthly uncertainties”. I thought you tried to convince us that you believe in uncertainties. You are back to claiming that all uncertainties cancel and the average is 100% accurate! You just can’t stop yourself, can you?

“But as he refuses to say how it was calculated, who knows?”

What difference does it make as to how it was calculated? Do you even know what the value of 1.4 represents? It’s the value of sqrt(2)! E.g. Something like a daily minimum and a daily maximum with an uncertainty of 0.5C being combined in a mid-range value and the uncertainty of them being calculated using root-sum-square.

That means that even using the uncertainty of a daily mid-range value as the monthly uncertainty is certainly reasonable.

You just *REFUSE* to learn about the real world. Why is that? Is it that scary because it contradicts so much of your fantasy world?

“If Carlo’s values were correct there’s close to zero probability the trend is negative.”

Malarky! The values can take on ANY value within the uncertainty interval. The stated values are *NOT* 100% accurate even though you believe they are! That means the trend could certainly be negative! How do you know it isn’t?

Again, you’ll have to take that up with Carlo.”

No, I understand what he did. *YOU* are the one that seems to have a problem with it!

” If his stated value is not taking into account the uncertainty as you wish, then tell him to include it or calculate it yourself. You might also want to calculate the uncertainty over the last 7.75 years.”

He *DID* take the uncertainty of the stated value into consideration! What do you think the uncertainty lines on the graph are!

If the graph showed a single stated value then adding up all the uncertainties would be appropriate. You can’t even understand this simple concept!

Bellman
Reply to  Tim Gorman
July 7, 2022 9:58 am

How do you calculate variance without calculating the mean? You are losing your mind!

It often feels like it talking to you. And the fact I persist in these arguments when you show zero ability to comprehend anything I say might be another sign.

You keep confusing related things and putting them into some random order with no logical sense.

You need to calculate the mean in order to calculate variance. Correct. But somehow this ends up in your mind as the variance of the sum is the same as the variance of the mean.

“fantasy monthly uncertainties”. I thought you tried to convince us that you believe in uncertainties.

I believe in uncertainties, but not the fantasy ones. Not too difficult to understand.

You are back to claiming that all uncertainties cancel and the average is 100% accurate!?

I absolutely do not think that any monthly average is 100% accurate, especially not UAH.

What difference does it make as to how it was calculated?

Because I’d like to see if his calculations are correct. Empirically they seem wrong, but maybe he knows something I don’t.

Honestly, you and Carlo spent ages last year insisting I couldn’t mention any uncertainty, even in a toy example, without completing a full uncertainty report as specified in the GUM. But now it doesn’t matter which bit of the air the figures were plucked out of, as long as they are large enough for you dismiss UAH as worthless.

Do you even know what the value of 1.4 represents?

No I don’t, that’s why I keep asking him to explain his coverage factor. He says the standard uncertainty is 0.5K. You multiply this by a coverage factor to get a range that represents a reasonable confidence interval. Typically this is around 2 to give a 95% confidence. For some reason he uses a coverage factor of 2.8, and I’d like to know why. The implication is he wants a very high level of confidence, around 99.5%, but I don;t know why he wants such a high value, or how he come to use the specific value of 2.8.

Of course, it’s entirely possible