ARCUS 2012 PAN-ARCTIC sea ice forecast submitted

Sea ice, ice berg and fog.

Sea ice, ice berg and fog. (Photo credit: Derek Keats)

PAN-ARCTIC OUTLOOK Submitted by WUWT today, June 4th to Helen Wiggins of ARCUS, details here

About the Sea Ice Outlook

The SEARCH Sea Ice Outlook is an international effort to provide a community-wide summary of the expected September arctic sea ice minimum. Monthly reports released throughout the summer synthesize community estimates of the current state and expected minimum of sea ice—at both a pan-arctic and regional scale.

The intent of the SEARCH Sea Ice Outlook effort is not to issue predictions, but rather to summarize all available data and observations to provide the scientific community, stakeholders, and the public the best available information on the evolution of arctic sea ice.

Sea Ice Outlook activities are supported by the National Science Foundation (NSF), the National Oceanic and Atmospheric Administration (NOAA), and through the volunteer efforts of contributors. The Outlook is organized by the SEARCH Project Office at the Arctic Research Consortium of the U.S. (ARCUS). The pan-arctic monthly reports are synthesized by James Overland, NOAA; the regional monthly reports are synthesized by Hajo Eicken and Adrienne Tivy, University of Alaska Fairbanks.

The WUWT Submission:

  1. Extent Projection: 4.9 million square kilometers.

Readers polled, responded with 37.37% of responses in the range of 4.8 to 5.1 million square kilometers.

5.0 to 5.1 Million km2  13.67%  (79 votes)

4.9 to 5.0 Million km2  10.03%  (58 votes)

4.8 to 4.9 Million km2  13.67%  (79 votes)

Since there was a tie between the 4.8-4.9 and the 5.0-5.1, values, the median value of 4.9 million square kilometers was chosen for submission.

Techniques: web poll of readers

DATA:

Greater than 6.0 Million km2  2.77%  (16 votes)

Greater than 5.5 Million km2  5.71%  (33 votes)

5.4 to 5.5 Million km2  3.46%  (20 votes)

5.3 to 5.4 Million km2  4.84%  (28 votes)

5.2 to 5.3 Million km2  4.5%  (26 votes)

5.1 to 5.2 Million km2  5.02%  (29 votes)

5.0 to 5.1 Million km2  13.67%  (79 votes)

4.9 to 5.0 Million km2  10.03%  (58 votes)

4.8 to 4.9 Million km2  13.67%  (79 votes)

4.7 to 4.8 Million km2  12.63%  (73 votes)

4.6 to 4.7 Million km2  7.61%  (44 votes)

4.5 to 4.6 Million km2  6.92%  (40 votes)

Less than 4.5 Million km2  5.54%  (32 votes)

Less than 4.0 Million km2  1.9%  (11 votes)

Less than 3.5 Million km2  0.87%  (5 votes)

Less than 1.0 Million km2 (Zwally’s “ice free” threshold)  0.87%  (5 votes)

  1. Rationale: Composite of projections by readers, projection bracket with the highest response is the one submitted.
  2. Executive Summary: Website devoted to climate and weather polled its readers for the best estimate of 2012 sea ice extent minimum by choosing bracketed values from a web poll which can be seen at:   http://wattsupwiththat.com/2012/06/01/sea-ice-news-volume-3-number-6-sea-ice-outlook-forecasting-contest-for-2012-is-online/
  3. Estimate of Forecast Skill: none

Submitted by Anthony Watts June 4, 2012

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19 thoughts on “ARCUS 2012 PAN-ARCTIC sea ice forecast submitted

  1. Really!?! It WAS a poll based on what we thought it would be, not what we hoped it would be… Right?
    4.6 Tops.
    4.5 seems more likely.
    4.4 Low end.

    REPLY: You are welcome to submit this to ARCUS today, as it is open to everyone. Whining about what we have submitted won’t change anything – Anthony

  2. Oh, I think it could go lower than 4.4M, but surely everyone understands that predicting sea ice in early June is 90% entertainment purposes!

  3. I don’t have a clue what to outcome might be, and a guess would be above my pay grade.

    I applaud Anthony for his efforts and offering in detail how he arrived to any possible outcome. “Kudos’ to you!

  4. geo, you say “but surely everyone understands that predicting sea ice in early June is 90% entertainment purposes“

    Maybe you should look at last year`s estimates. Environment Canada predicted 4.7 million sq kms in June, 4.6 in July and August. The correct number was 4.6. Let us see how they do this year. Mind you, for us here in Canada, this is not a useless exercise. The Canadian Coast Guard must escort the supply vessels which carry the heavy loads to our Arctic communities. If they are going to get to Eureka in September, as well as carrying out all the other vital convoys, it is essential that their forecasts be as accurate as possible. The planning for this process is already ongoing.

  5. Due to (probably too many) blockers in my browser and box, I never actually fought through to the poll this year. OTOH, my vote for 4.9-5.0 wouldn’t have changed much.

    National Geographic will presumably be doing the decent thing and sending their estimate in soon. :)

  6. “Since there was a tie between the 4.8-4.9 and the 5.0-5.1, values, the median value of 4.9 million square kilometers was chosen for submission.”

    The median of those values is 4.95 – though I appreciate your need to try and crank the forecast down a bit so your readership don’t look quite as out of touch.

  7. hmmm. I see three numbers Anthony listed when deciding which one to send it, 4.8, 4.9, and 5.0. The median of those three numbers (not the mean, and not the mode), would be 4.9. So Peter, where do you see 4.95?

  8. It is very likely that the Arctic will be ice free, not only that but the ice maker in my freezer will fail too.
    Sadly, no amount of recovery of Arctic ice will be reported in the media or stop the march to cap and tax in the US.
    I’m not even sure a new president would curtail the EPA.

  9. i guessed 5-5.1….i confess this is on the lines of what i would like to see tempered by a look at reality….is the median of a few hundred guesses more accurate than one?

  10. Pamela Gray writes: hmmm. I see three numbers Anthony listed when deciding which one to send it, 4.8, 4.9, and 5.0. The median of those three numbers (not the mean, and not the mode), would be 4.9. So Peter, where do you see 4.95?

    Er, no. Peter Ellis is correct. The median of 4.8, 4.9, 5.0, and 5.1 is … 4.95. I won’t speculate about why Anthony chose to round it down.

  11. Pamela: You missed a bit!
    4.8-4.9
    4.9-5.0
    5.0-5.1 <– there it is!

    You've calculated the median based solely on the lower end of each bracket – the upper end of the top bracket goes up to 5.1

  12. Gopal Panicker asks, “is the median of a few hundred guesses more accurate than one?”

    Gopal, there was a programme on BBC TV a few months ago which touched on this question, using the old summer fête standby, “guess the number of sweets in the jar”. Near the start of the programme, they showed you a few people making wild guesses at the number, then near the end they came back with the mean of several hundred guesses. Amazingly, it was very close – well within 1%, if I remember it right. So a tentative answer to your query is – yes, it appears that there is some sort of “wisdom of crowds”, and that the accuracy of a “crowd guess” seems to tend towards being a lot more accurate than a single guess.

  13. Kent Davies says:
    June 5, 2012 at 5:26 am
    Pamela Gray writes: hmmm. I see three numbers Anthony listed when deciding which one to send it, 4.8, 4.9, and 5.0. The median of those three numbers (not the mean, and not the mode), would be 4.9. So Peter, where do you see 4.95?

    Er, no. Peter Ellis is correct. The median of 4.8, 4.9, 5.0, and 5.1 is … 4.95. I won’t speculate about why Anthony chose to round it down.

    Er, no, you’re wrong. The median value is not calculated; it is based on the bins used unless the data are recorded on a continuous spectrum. There is no bin identified as “4.95”. In fact, since the bins are identified as a ranges, it would have been more correct to have reported the value as 4.8 – 4.9

    There are a total of 578 votes. Half the votes (289) fall at or below bin 4.8 – 4.9, and half above. The median bin might be taken as 4.8 – 4.9 or 4.9 – 5.0.

    The mode value is a tie between bin 4.8 – 4.9 and bin 5.0 – 5.1 with 79 votes each.

    I calculated a weighted average, multiplying the center value of each bin by the number of votes for that bin and dividing that by the total number of votes. For single value bins, I used that value (eg “6” for >6, “3” for <3). The number I calculated for the weighted average is 4.96 million km2. Since I'm running out of lunch hour, others can feel free to tweak as they see fit.

  14. The median is usually defined as the mean of the two middle values, in this case 4.9 and 5.0. The median is, therefore, 4.95. However, this is a “usual definition” and can certainly vary.

  15. There are only three button choices Anthony had in front of him labeled 4.8 to 4.9, 4.9 to 5.0, and 5.0 to 5.1. You could say they are just the labels. If there were four choices I don’t know what Anthony would have done, since a calculated value would not have been one of the choices. Besides, how do you “average” labels. You don’t. You can’t average pictures or names either. I think we are confusing data with labels.

  16. “Folks, you’re nit picking an estimate to death”
    Considering the greater distribution of remaining votes on the low side, I would go for 4,936119208033451… There! Dead as a doornail!

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