Hello,

Let me preface this by saying Daniel Kahneman asked this question to a series of graduate students. Here is the question ---


The mean IQ of the population of eight graders in a city is known to be 100. You have selected a random sample of 50 children for a study of educational achievements. The first tested has an IQ of 150. What do you expect the mean IQ to be for the whole sample?


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Would anyone like to venture on the correct answer and how they arrived at that.

answer:


further insight:
A surprisingly large number of people believe that the expected IQ for the sample is still 100. This expectation can be justified only be the belief that a random process is self correcting. Idioms such "errors cancel each other out" reflect the image of an active self-correcting process. Some familiar processes in nature obey such laws: a deviation from a stable equilibrium produces a force that restores the equilibrium. The laws of chance, in contrast do not work that way: deviations are not cancelled as a sampling proceeds, they are merely diluted.
-- Daniel Kahneman.

Also why would anyone expect a sample of 50 be a representative sample of the population?

Hello checkcheckfold,

Interesting question, one that could use further thought, one may suggest that a common heuristic the law of small of numbers can be a potential explanation. The law of small numbers is defined as, 'a judgmental bias which occurs when it is assumed that the characteristics of a sample population can be estimated from a small number of observations or data points'.

(100*49 + 150) / 50 = 101

Because that is how sampling theory works, and as grad students, they might know something about it. Their error is in not treating the remaining 49 as a separate random sample, and adjusting their calculation. A sample of any size, if randomly selected from a population, is representative of the population. And for each sample size there is an associated sampling error.

Would be slightly lower than 101 depending how big the city is

I'm pretty sure the correct answer is dependent on the total population size.

Population size = population
1. Average IQ of the population excluding Mr 150 = pop_average
2. Total IQ of the 50 = 150 + ( pop_average x 49 )
3. Total IQ of the "rest" = ( pop_average x ( population - 50 ) ) / 50

Combining 2 and 3 we get:
4. Average IQ of the population = ( (150 + ( pop_average x 49 ) ) + ( pop_average x ( population - 50 ) ) / population = 100

Solving 4:
pop_average = ( ( 100 x population ) - 150 ) / ( population - 1 )

So for example for a population of 100,000 total
pop_average = 9,999,850 / 99,999 = 99.9986 IQ

So for example for a population of 51 total
pop_average = 99 IQ

All the info was given. The question asked about the mean of the sample, not of the population. We already know the sample size to be 50.

The mean of the sample is dependent on the mean of the "other" 49 in the sample. The mean of the other 49 in the sample has been assumed to be 100 but it is not, it is a little less than 100 (as one large outlier in the population is not being counted).

Assume the population is 500 people, and all of them have exactly the same IQ except for Mr 150. And the mean IQ for the population is 100.

What IQ does each of the 499 have? Easy to calculate;
Define: ( 150 + IQ x 499 ) / 500 = 100
Therefore: IQ = 99.90

So in this case for our sample the average would be:
( 150 + ( 49 x 99.90 ) ) / 50 = 100.90 (not 101)

I think you are making the flawed assumption that the "mean of everyone else other than Mr 150" is the same as "mean of the population". When it cannot be, as Mr 150's absence changes it.

You're right, good catch!

The sample of 50 has been randomly selected, and therefore the expected value of the mean would be 100 (no reason to think it is specifically higher or lower). If we further divided that sample of 50 into two samples, say 25 and 25, each sub-sample would have an expected mean of 100. If we calculate the mean of group 2 and find it to be 105, that does nothing to change the expected mean of sample 1.

Similarly, if we divide them into sub-samples of 49 and 1, each one has an expected mean of 100, regardless of the mean of the other one. So the expected sum of the values in the group of 49 = 4900, which we add to the empirical value of 150 from the other, and get 5050 / 50 = 101.

Mr. 150's absence or presence does change the expected mean of the entire sample of 50, but not of the 49. That is what the entire exercise is about.

In your mathematical example, David, the 49 are not a random sample of the population. In a random sample, it is possible to get all of the values that are in it, and in your example it is not possible for the 49 to have a value higher than 100. The sample is biased, because you have constructed it that way. In the problem, there is almost as much chance of there being a value of 150 in the sample of 49 as there was of there being a value of 150 in the sample of 50 (slightly lower because of the slightly lower sample size).

Two people are playing hold'em heads-up and are dealt preflop hands. The expected number of red cards in each hand is 1. If we expose one person's hand and both cards are red, that certainly changes the other hand's expected # of reds.

Suppose the population is 51 and that the mean IQ of the population is 100 (ie the sum of all 51 IQ's is 5100). Suppose instead of knowing the 1 person's IQ, we know 49 IQ's and that they're all 101, for a total sum of 4949. Does that info not change the expected IQ of the 50th person in the sample? Or better yet, suppose an IQ of 0 and non-integer IQ's are possible and suppose the 49 IQ's are all 5100/49. What is then the expected IQ of the 50th person?

1) Let's suppose that you allow a computer to randomly assign the values of red or black to two cards (which we will call hand 1). We look at it and see they are both red. We now get to predict how many red cards will come up if we do this a second time (hand 2). Your holdem situation is different because the probability of a red card changes once the values have been revealed, because now you have probability of a red card being 48% instead of 50%.

2) Let's take a look at the scenario you created. You have constructed a population that has 51 people - 49 have a single value (101) and 2 have values that combine to equal 151. Somehow you have randomly selected 2 cases to not be in your large group, and against all odds, the only two that you picked happen to be the weird 2. You are then using basically a degrees of freedom problem showing that if we know the sum of all values and the sum of all the values except 1, we can figure out what that 1 is.

This is just not how sampling theory works - it is a theory based on what happens when you randomly select members of a population to be in a sample. The important word here is "randomly" - not "biased in such a way as to demonstrate a possible outcome".

Again, using the computer selection - imagine you tell the computer to generate a single number at random, from a population with a mean of 100 and a standard deviation of 15. That number is 150. Now, have it do the same thing again and generate 49 additional numbers using the same parameters. The expected value of the mean for that group of 49 is going to be 100.

So, as long as the population size isn't very small with respect to the 50 in the sample, knowing 1 value isn't going to impact what you think about the other 49. Since the population size was not mentioned in the problem, there is no adjustment that can be made to account for it. If there were an adjustment to be made for population size, the answer would vary depending on that size, and therefore there would be no exact correct answer to the problem.

My hold'em situation is different because mine is a valid analogy to the problem of this thread: the "deck" is not infinite. The mean of the population is known, and there can be no negative IQ's, therefore there is only a finite number of IQ points to be distributed (just like a finite number of red cards in the deck). If the population is 51 then the "deck" only has 5100 points. Once you know that one person has hogged 150 of them, that lowers another person's expected number. It's not like all 50 people in the sample can have IQ's of 150, because that would make it impossible for the population mean to be 100 (the 51st person in the population can't have a negative score to average it out). So the peoples' scores are not independent.

Agreed.

Wrong. We've taken some smarter people out of the population, so the others on average are going to be slightly dumber. We have more information, we are allowed use it. That information is not very useful if the total population is huge (e.g. 5 million) but is extremely useful if the total population is small (e.g. 300).

Agreed.

Nope. We are observing the IQ of people we have removed from the possibility of being in a sample. Sample is no longer random.

No, see points above. We know the average of EVERYONE is 100. When we take out the possibility of choosing Mr Super Smart, the average of the rest is known to be lower than 100. So if you randomly choose from this group your expected average is lower than 100.

I think I agree, but the problem here is that by us saying "the average is 100 for the population" is NOT the same as saying "the expected IQ for a person in the population is 100".


Another example to illustrate. The average height of me, my brother, and my two cousins, is 6 foot 2. One of my cousins is 6 foot 8. That means that the rest of us have an average height of 6 foot. If you were to take a sample of two of us and say "one of the lads in the sample of two is 6 foot 8, what is the average height of the sample" I would say 6 foot 4 (calculated my way, assuming the other guy was the 'other guy' average) NOT 6 foot 5 (calculated your way, assuming the other guy was the population average).

I'm pretty certain Kahneman's answer of 101 is correct, from the information given, regardless of population size.

Try a population size of exactly 50.

Then we don't have a sample. The math is not equivalent. With a sample we use probability rules. With the whole population we simply use arithmetic, as the mean is precisely known.

The problem statement was about a sample.

Then please explain the difference between this problem and my hold'em example, or why the "deck" here is infinite despite the population being finite.

It obviously doesn't jump from 100 at 50 to 101 at 51. C'mon man. That's literally saying the unselected guy is always a 50 whenever the first revealed is a 150, which makes no sense.

Nope.

Try a population size of 200.

Known: Average of population of 200 is 100
Known: Sample of 50 is random other than it includes Mr. 150.

So we know that population is (a) 199 guys with an average IQ of X and (b) one guy with an average IQ of 150 and (c) the average of all 200 guys is 100.

Spoiler: the average of the 199 guys (represented above as X) must be 99.74874.

To prove it just calculate the population average like this: ( (199 x 99.74874) + 150 ) ) / 200 = 100.00

So the sample average is this: ( ( 49 x 99.74874 ) + 150 ) / 50 = 100.7538

LOL

my iq is a hair under 150, might be 150 on a test now and then-on the high.


what's my point???


i find it humorous that i cant follow this thread. 101 or 100.9


I am enjoying it tho.

respectfully and best

eric

I concede.

The answer approaches 101 as the population gets large. For any realistic "city" we could assume that a sample of 50 is a small enough percentage that the answer is ~101.

With a population of 1000, the mean sample IQ is 100.95. If we take the population to 5000, the sample IQ is 100.99. If we take the population to 50,000, the sample IQ is 100.999.

I had to run a spreadsheet with all possibilities to see the light.