Tangentially though, a big question here is how wise it is for the US education system to be based so heavily on multiple choice questions. There are other countries with decent education systems that do not do this.
Tangentially though, a big question here is how wise it is for the US education system to be based so heavily on multiple choice questions. There are other countries with decent education systems that do not do this.
Instead, I suspect there is a particular concept or perspective they have been trying to impress on their students recently, and in this case they are expecting the answer that most demonstrates they have paid attention to the recent lessons.
I could be wrong of course.
I've never seen a test like that for school work.
Those exams got easier after a couple years. I imagine there were some words to be had with the dean.
There was a great blog post I read a while back on redesigning multiple choice tests to allow the student to indicate the "confidence" of their response, with a more confident answer being rewarded/penalized more heavily than a response with low confidence. This allowed for a statistically better sample of how well the student learned the material.
I thought for sure the post was written by Scott Aaronson, but I haven't been able to find it despite extensively searching his blog, so maybe it was someone else.
Found with the following Google query: "multiple choice" "school" "confidence" "blog"
In the case of taking a test, let’s say you’re answering a true/false question, true represented by 1 and false represented by 0. Let’s also assume you have no idea which one is correct, it’s a coin flip to you.
If you choose True, 50% of the time, the correct answer is true and you’ll have 0 loss, because (1-1)^2 is 0. The other 50% you’ll have (1-0)^2 is 1.
So your expected loss is 0.5(1)+0.5(0)=0.5
On the other hand, if you guess 0.5 (true with a confidence level of 50%), then 100% of the time your error is 0.5, and your mean squared error is 0.25.
In other words, you minimize your expected loss by guessing your true confidence level. This can be mathematically proven to work for any confidence level.
This could be adapted to multiple choice questions by treating each option as a true/false question.
Sorry was that’s very wordy but hopefully you can get the point.
An easier to understand, but perhaps less sensible example would be to do the same thing in a quiz about arithmetic, so 5+5=9 and 6+2=7 is less wrong than 5+5=10 and 6+2=1.
But you can't survey non-respondents--because they don't respond!
What option A will actually result in is having a mixture of people who responded on the first try, and people who responded on the second try. In principle, neither of those will be representative of non-respondents. Whether this creates a significant problem in practice will depend on lots of other assumptions.
My own first choice would be either E or D, depending on (a) whether you have enough resources available to send 30 more emails, and (b) how much statistical power you are sacrificing with a sample of only 90 instead of 120. (If you were doing things properly, the number of emails you send out initially would be larger than the actual number you needed to get enough statistical power, by some factor that would depend on what fraction of people you expected to respond. In which case D would be the obvious option.)
Perhaps. But you are also creating a potential confounder, since you now have two categories of responders instead of one.
The decennial census is not a random sample, though the Census Bureau does sruveys that are random samples separately.
Yes, and you're clawing back a non-random portion of that non-random set, which can be expected to improve the sample quality overall.
I would go for A’: email all of them, reminding them to fill in the questionnaire, if they haven’t already done so.
Who said anything about "US education system"?
Where in the original article does it say "US"?