WSU STAT 360
Class Session 11 Summary and Notes: November 10, 2000
Back on November 3, 2000, Jessie expressed utter disbelief in one statement that I made reagarding the interpretation of Chi Squared. Just to remind all of you, we were examining how to test a die as being fair or not. We looked at the size of Chi Squared as determined by the outcome of 60 tosses. It was obvious to everyone that a large value of Chi Squared would indicate a loaded die. However, I imagine that nearly everyone thought I was crazy to claim that an extremely small value of Chi Squared is suspicious also. Jessie was the only one who dared raise his concern publically. He said, in effect, that if 10 on each face has the highest probability of occurence, and this outcome also gives a Chi Square of zero, then obviously a zero value for (I'm so sick of writing Chi Squared that I'm going to write instead) C2 is entirely expected. Now I'm going to show you that what is obvious is not always true.
The reason that things don't work out as expected is as follows. Indeed, 10 of each face is our most probable outcome of 60 rolls. But, we are not interested in the most probable outcome of rolls, but the most probable value of C2 as we calculate it. It is C2 that we are using for our test statistic.
The distribution that the die actually follows is multinomial. We can calculate the probability of 10 rolls on each face as...
P(10,10,10,...,10)=60!*(1/6)60/(10!)6 This is 8.4x10-5
In other words, 10 of each face value may be our most likely outcome, but it is pretty unusual, occuring only 1/10,000 tries. Now this is also the only case that provides C2 = 0.0. Therefore a zero value of our test statistics is quite unusual. Let's calculate the value of C2 for several other hypothetical cases.
P(11,11,11,9,9,9)=60!*(1/6)60/(11!39!3) = 6.3x10-5 P(12,8,10,10,8,12)=60!*(1/6)60/(10!212!28!2) = 3.9x10-5 P(13,12,11,9,8,7)=60!*(1/6)60/(13!12!11!9!8!7!) = 0.77x10-5
Each of these successive outcomes is less probable than P(10,10,...,10) but there are more ways to arrive at them. For example, p(11,9,...11,9) provides a C2 of 0.6, there are a total of 6!/3!3!=20 ways to arrive at C2 = 0.6 (Can you show this?), so the probability of C2 = 0.6 is 15 times that of 0.0. Likewise, there are 6!/2!2!2! = 60 ways to arrive at C2 = 1.6, p(12,10,8,8,10,12) being only one of them. Therefore, this value of C2 is about 42 times as likely as 0.0. Finally p(13,12,11,9,8,7) produces a C2 of 2.8 and there are 6! ways of arriving at the same value. It is 66 times as likely to occur as C2 = 0.0.
You might now ask, what is the most likely outcome in terms of C2? You might think that this is also a difficult question to answer, but it is not. Our problem, or our statistics, has 5 degrees of freedom you recall. The pdf of C2 for 5 degrees of freedom is...
pdf(C2) = (2n/2-1*G(n/2))-1*(C2)3/2*exp(-C2/2);
By differentiating this expression with respect to C2 and setting the result equal to zero to find the maximum, we will arrive at the equation (C2 - 3)=0; which places the maximum of C2 at a value of 3.0.
This was a long winded problem, but I think it is fair to say that it shows that C2 = 0.0 is suspicious. Now there is another consideration which Jesse brought up, and which is quite valid. He said that an outcome like (12,8,10,10,8,12) is as likely as (10,...,10). We have shown here that it is actually a little less likely, but that it leads to a more likely value of C2. This is one of those areas in inference where a person can get caught in a terrible tangle, in which I view the most likely outcome as suspicious and a less likely outcome as more reasonable. Here I can offer only the following guidance. My suspicion is raised because I expect there to be some variation and there is none. That can only happen one way in 10,000. Certainly (11,9,11,9,11,9) would only happen once in 20,000. But all 10's is a singular event. If I were to run the experiment a second time and get (10,...,10) again I'd be dead set against that die. Yet, if I obtained (11,9,9,9,11,11) I would have gotten a less likely result again, but I'd be even more convinced of the fairness of the die, because this result would show once again expected variation, but of a different form.
The lesson is that a person cannot place blind faith in statistical measures. He has to exhibit some judgement as well. I expressed this in last week's notes when I said that your rejection/acceptance region ought to be based on everything you know about the problem. Confirmation, truth, and knowledge, are very slippery things.
Link forward to the next set of class notes for Friday, November 17, 2000