A binomial problem involving 2 different random variables.

[Try Hard0]

In a recent federal appeals court case, a special 11-judge panel sat to decide on a certain particular legal issue under certain particular facts. Of the 11 judges, 3 were appointed by political party A, and 8 were appointed by political party B. Of the party-A judges, 2 of 3 sided with the Plaintiff on the issue, and of the party-B judges 6 of 8 sided with the Plaintiff. Now in a later appeal of a different case involving the same legal issue and same exact facts, but where a regular 3-judge panel sat to decide the appeal, these 3 judges decided unanimously (3 of 3) against the Plaintiff. The 3-judge panel was composed of two (2) party-A type judges, and one (1) party-B type judge. The Plaintiff claimed that the 3 judge panel was biased and prejudiced against him. Is there any merit to what the Plaintiff is claiming?

I see this as a binomial type problem, but with different probabilities involved that need to be accounted for – a bivariate type situation.

But can I simplify this problem by averaging the probabilities of the judges voting for or against the Plaintiff? If I can, then this becomes a single-variable type problem and I can just use the good-old binomial formula to determine a z-score and confidence interval.(?)

If my averaging method is okay to do, this is what I come up with:
The probability of a decision against the Plaintiff by a party-A judge is 0.33, and the probability of a decision against the Plaintiff by a party-B judge is 0.25. There were two party-A judges and one party-B judge on the panel, so the probability average of the judges voting against the Plaintiff is (0.33+0.33+0.22)/3 = 0.30.

Thus, I can plug the binomial formula like so --> p=0.30, q=0.70, n=3, and x=3. Really, in such a situation, the binomial formula boils down to 0.30^3 = 0.027 <-- call it 0.03

Now, I can also find the s.d. with these numbers, and I find s.d.= 0.794

Also, the expected (mean) number of votes against the Plaintiff would be 0.33+0.33+0.25 = 0.91

Plugging the z score formula I get z = (x – u)/(s.d.) = (3 – 0.91)/0.794 = 2.63

That is 2.63 s.d. from the mean (of 0.91 total average votes against the Plaintiff by a randomly selected and unbiased 3-judge appellate panel), which translates to a 0.0043 probability that such an occurrence would happen from random chance alone. This would indicate that the judges were, indeed, biased and prejudiced against the Plaintiff. There is “merit” in what the Plaintiff claims.

Okay, so what is the right way to do this problem, and with a confidence interval? (Will be interesting to see the difference between doing it the right way and the way I did it.)

 

[Stephen Tashi]

This is a type of problem that can be approached in many ways and the one that is counted as correct can only be guessed by knowing what material was in the chapter where the problem appeared! An alternate approach to yours would be to assume that if the judges remain "unbiased" then they would vote the same way in the second case as the first. So if the panel of 3 is unbiased then each judge would vote the same way as he or she did in the first case. How the vote goes depends on which judges are selected (not on a binomial random variable representing the decision of each judge).

That probability problem would be like this: There are 3 apricot colored balls. Two have "For" written on them and one has "Against" written on it. There are 8 blue colored balls. Six of them have "For" written on them and two have "Against" written on them. Three balls are randomly selected from the 11 balls. The selection contains 2 apricot colored balls and 1 blue ball. What is the probability that all the balls have "Against" written on them?

As to confidence intervals, you worry about confidence intervals when you are trying to estimate the value of a parameter. You might be confusing "confidence intervals" with "acceptance regions".

 

[Try Hard0]

I will think about what you said, Stephen.

I think I am supposed to assume something in the problem, like that the decision by the 11 judges was "unbiased," and that on the average the probability of a vote against the plaintiff is only 0.27, and that the probability of a vote against the Plaintiff is 0.73. I am not to worry about the political party affiliations, because such minor differences that show up between the statistics on how the judges in the 11 judge panel could obviously be attributed to random chance (i.e. Party-A type judges vote against the Plaintiff 33% of the time, and for the Plaintiff 67% of the time, and Party-B judges vote against the Plaintiff 25% of the time, and for the Plaintiff 75% of the time -- too close of percentages to make any blanket statements about judges here based on such a small sample, and so I assume they all see things pretty much the same, and politics has little to do with their decision making).

Also, I realize this is a discrete random variable problem.

I also realize that any other outcome than a "unanimous" decision would be in the "normal" range. But that a unanimous decision came down against the plaintiff does show that there probably was bias and prejudice against the plaintiff in the later appeal involving just 3 judges -- because that outcome is outside the "normal" range of likelihood, and, indeed, it was a rare event -- only a 0.019 probability of happening.

Thanks also for straightening me out on the confidence interval too. But, what is an "acceptance region" for this situation where there is a discrete random variable with only 4 possible outcomes (i.e. 3 judges for the plaintiff, 2 for and 1 against, 1 for and 2 against, all 3 against). I take it that an event with only a 0.019 probability of happening is outside the "acceptance region?" Is there any way to do a "hypothesis test" here?

 

[Stephen Tashi]

I take it that an event with only a 0.019 probability of happening is outside the "acceptance region?"

The acceptance region is a set of values of the statistic (or statistics) that a person is using to test a hypothesis. (These values have associated probabilities, but it isn't the probabilities that make them part of the acceptance region.) Assuming the "null hypothesis" is true there is a certain total probability (e.g. 0.95) that the statistic being used will fall in the acceptance region. If the statistic does then the person doing the hypothesis test "accepts" the null hypothesis. Otherwise he "rejects" it.

Hypothesis testing is a procedure that is both systematic and completely arbitrary. In your problem, the statistic could be the total number of votes against the plaintiff. The acceptance region could be (one or fewer votes) or (two or fewer votes) etc.. depending on how much total probability you want to put in the acceptance region.

 

[Try Hard0]

Thanks for the explanation Stephen;
I would want to use a 95% type probability/acceptance region. I take it that because the chances of all 3 judges rendering a decision against the Plaintiff falls outside the 95% region that the result is to be considered "suspect."

But then that "confidence interval". . . seems difficult to say what the confidence interval should be here! (If not impossible.)

Or am I wrong about that?

If there is a confidence interval, would that entail something like a fraction of a vote?

Here I am thinking of the mean for the decision, and the mean would be either 0.91 votes for the Defendant (or 2.09 votes for the Plaintiff) OR just 1 vote for the Defendant, and 2 votes for the Plaintiff. (Which one would it be?)

I am thinking that if the confidence interval can entail something like a fraction of a vote, like, say, 0.49 votes or less one way or the other, then that would still mean the decision was to be suspect at the 95% confidence level. Likewise, if the confidence interval was, say, 0.50 votes or more one way or the other, then there is a chance that such an event as all 3 judges deciding against the Plaintiff would be a random event, and not to be suspected as a matter of unfair bias and prejudice on the part of the judges.

Am I on the right track here?

 

[Stephen Tashi]

I would want to use a 95% type probability/acceptance region. I take it that because the chances of all 3 judges rendering a decision against the Plaintiff falls outside the 95% region that the result is to be considered "suspect."
...
Am I on the right track here?

You might be on the right track if you understand that nothing about those acceptance intervals or confidence intervals will tell you "There is a 0.95 probability that the panel was biased" (nor will they quantify the probability the panel was not biased). The probabilities that can be calculated are being calculated on the assumption that the panel was not biased.

This is the typical situation in "frequentist" statistics. A person's question is "Given the data, what is the probability that this idea is correct" and all the statistics can answer is "Given this idea is correct, here is the probability of the data". The two probabilities do not have to be equal and without additional information,one of them does not determine the other. So when you say that the result "would be suspect", this is simply your subjective judgment. It might be as good as anyone elses, but realize that there is no mathematics that says you have a 95% chance of being correct.

For a statistic with only discrete values, you may as well consider the possible acceptance regions to be set of discrete values since you can't compute the probability of a non-discrete outcome. Defining an acceptance region that is "equal or greater than 0 and less than or equal 2.5 votes against the plaintiff" amounts to defining an acceptance region that is "zero or 1 or 2 votes against the plaintiff".

Confidence intervals are relevant to quantifying how good a statistical estimate is and deciding how many samples ought to be taken. As far as I can see, you aren't doing either of those things.

 

[Try Hard0]

Okay Stephen, I am lost, so I will get to studying more with your statements in mind. When I come to understand what you are talking about, I will probably know what I am doing.

Thanks.