Binomial Data, Axioms of Probability, and Bayes Rule. n! Pr(k successes in n draws)' k!(n&k)! Bk (1&B) n&k

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1 Prof. Green Intro Stats // Fall 1998 Binomial Data, Axioms of Probability, and Bayes Rule In this class, we link two sets of topics, sampling and probability. To do so coherently, we must blend ideas from Moore and McCabe s chapters 3, 4, and 5. As you review this material, you may wish to read these chapters in tandem. Let s start with a simple probability problem. Suppose that state legislators in the US came in two varieties, Democrats and Republicans. Suppose Republicans comprise 40% of all legislators. If we draw at random four members of the (large) population of state legislators, what is the probability that all four will be Republicans? To answer this question, we reason that for the outcome to obtain, the first draw we make must be a Republican. The probability of this outcome is.40. And the second draw must come up Republican, too, which also has a probability of.40. Since the first and second draws are independent, the probability that both are Republican is (.40)*(.40). Following this logic to its conclusion implies that the 4 probability that all four draws are Republican = (.40) = Suppose the question were: What is the probability of drawing exactly three Republicans in four draws? One tedious way to answer this question is to list the sample space, that is, the set of all possible outcomes. A more convenient way is the binomial formula: n! Pr(k successes in n draws)' k!(n&k)! Bk (1&B) n&k where π represents the true population proportion of Republicans. In this case, (see Table C, p. T-7) 4! Pr(3 Republicans in 4 draws)' 3!(4&3)! (.4)3 (1&.4) 4&3 '(4)(.064)(.6)'.1536 Note that this formula simplifies nicely when we want to know the probability of at least one Republican being selected in four draws. That is one minus the probability of no Republicans being selected. 4! Pr(at least one Republican in 4 draws)'1& 0!(4&0)! (.4)0 (1&.4) 4&0 '1&(.6) 4 '.8704 This kind of simplification can be very useful for a large class of problems. Suppose we manage an airline that runs 100,000 flights each year. Suppose for the sake of argument that the probability of a crash is independent from flight to flight. Suppose that probability is (read one-in-100,000). Contrary to the way laypeople often look at problems such as this, the probability of at least one crash each year is not equal to 1!

2 Pr(at least one crash in 100,000 draws)'1&(1&.00001) 100,000 '.6321 x [Calculator note: use the y button.] Now suppose that your airline tightens its safety procedures so that the probability of a crash is one-in-a-million per flight. That s more like it! Pr(at least one crash in 100,000 draws)'1&(1& ) 100,000 '.0952 When working with binomial problems for cases where (nπ > 30), assuming that the total population is large or that the cases are drawn with replacement, it is convenient to work with the normal approximation instead of the binomial formula. Let X be the observed number of successes in n draws. Let p be the sample proportion, calculated as X/n. For a population that has a proportion π of successes, the sample proportion p will be distributed normally with mean π and standard deviation equal to: Standard deviation of sampling distribution of a sample proportion' B(1&B) n Notice that the population proportion is used here (if it is known). Thus, if we draw 100 legislators at random, we would expect to find 40/100 Republicans, and we would expect that across hypothetical replications of this sampling procedure, our sample estimates will vary with a standard deviation of.049. Using the +/- 2 standard deviation rule, we would suppose that 95% of our samples would fall between about 30% and 50% Republican. When we do not know the population proportion, two strategies are available. One is to make a conservative assumption, namely that π =.50. Note that this value of π makes the numerator of the standard error as large as possible. Thus, our the actual level of sampling variability will be no greater than the value we obtain under this assumption. An alternative is to assume, for purposes of calculation, that p = π. In keeping with the discussion last time regarding two-sample comparison, we can write down a formula for the standard error of the difference between two proportions (p - p ): 1 2 SE of difference in proportions' B 1 (1&B 1 ) n 1 % B 2 (1&B 2 ) n 2 For example, the Outward bound experiment showed that 39.5% of 86 all-white group subjects scored a 14 on the tolerance scale, as compared to 52.8% of the 178 integrated group subjects. The difference is therefore 13.3 with a standard error of 6.5%. Note that the +/-2 SE rule creates a 95% confidence interval [.3, 26.3] that just barely excludes zero.

3 Unconditional and Conditional Probabilities Central to the study of probability is the conceptual distinction between conditional and unconditional probabilities. Consider, by way of illustration, the following crosstabulation: Study hard Goof off Total Get good grades Get bad grades Total The probability that one gets good grades is 175/300 =.583. Call this Pr(G). The probability that one gets good grades given that one studies hard is 75/100 =.75. Call this Pr(G S). If we want to know whether studying pays off, we would compare Pr(G S) to the probability call it Pr(G ~S) that one gets good grades given that one does not study hard. That probability is.50. It should be clear that Pr(G) is not equal to Pr(G S) unless getting good grades and studying are statistically independent events. A different set of questions is answered if we percentage the table across the rows. The probability that one studies hard is 100/300 =.333 = P(S). The probability that one studies hard given that one gets good grades is 75/175 =.429. The fact that most people who get good grades goof off is quite compatible with the observation made above that those who study have a higher probability of getting good grades. People with a weak understanding of statistics often run these two points together. They should have studied harder. Note the following axioms of probability: 1. Pr(G) + Pr(~G) = 1. In this example, 175/ /300 = Pr(G and S) = Pr(G)Pr(S G) = Pr(S)Pr(G S). Here, 75/300 = (175/300)(75/175) = (100/300)(75/100). 3. Pr(S) = Pr(S G)Pr(G) + Pr(S ~G)Pr(~G). Here, (100/300) = (75/175)(175/300)+(25/125)(125/300). 4. Pr(G S) = Pr(G)P(S G) / Pr(S). Obtained by rearranging terms in axiom 2. Putting lines 3 and 4 together generates Bayes rule, an alternative expression for Pr(G S)...

4 Bayes Rule Terminology and Assumptions: Pr(H) =probability that an hypothesis is true (a prior) Pr(E) =probability that one observed a given form of evidence Pr(H E ) =probability that the hypothesis is true given that one observed this evidence (posterior) Pr(E H) =probability that this evidence is observed given that the hypothesis is true (likelihood) Pr(E ~H)=probability that this evidence is observed given that the hypothesis is false (likelihood) Pr(~H) = 1 - Pr(H) = probability that the hypothesis is false Bayes Rule: Pr(E H)Pr(H) Pr(H E)' Pr(E H)Pr(H)%Pr(E -H)Pr(-H) Consequences of observing a study that supports the deterrent effect of the death penalty (E), for two observers with different priors (H) OBSERVER 1 Input: Pr(H) = prior probability that death penalty deters Pr(E H) =0.800 prob of coming up with evidence of deterrent effect, given deterrence Pr(E ~H)=0.300 prob of coming up with evidence of deterrence, given no effect Output: Pr(H E)=0.123 posterior prob that death penalty deters OBSERVER 2 Input: Pr(H) = prior probability that death penalty deters Pr(E H) =0.800 prob of coming up with evidence of deterrent effect, given deterrence Pr(E ~H)=0.300 prob of coming up with evidence of deterrence, given no effect Output: Pr(H E)=0.981 posterior prob that death penalty deters

5 Notice the following features of the foregoing illustration: < If Pr(E H) > Pr(E ~H) for both people, then a given piece of evidence will push them both in the same direction; that is, both become more convinced of the hypothesis. In this case, for two people with the same likelihoods, the percentage-point gap between their posteriors (here.858) will be smaller than the initial gap between their priors (here.900). < If a person believes that Pr(E H) = Pr(E ~H), then Pr(H) = Pr(H E); new information has no effect on someone who believes this evidence to be nondiagnostic. This fact follows directly from Bayes rule. < Bayes rule has nothing to say about where priors come from; it focuses solely on the process by which these priors are updated in light of new information. Note also that Bayes rule can be applied to a succession of learning experiences. Consider, for example, what happens when Observer 1 is exposed to another piece of information. The prior is now.123. Plugging the numbers into the formula gives a new posterior of.272. If the process is repeated again, this time with a prior of.272, the posterior is.499. And so on, until eventually the two observers beliefs converge. Relevance? To real world? A normative or descriptive model? Can it be both? Prospective Economic Evaluations of the Parties, by Party Identification in the Initial Wave of Each Panel Study (Entries are percentage of each partisan group saying that the Democratic Party does a better job of handling the economy.) Panel Study Party Identification in 1990 Economic evaluations Economic evaluations Economic evaluations N of cases (405) (319) (261) Panel Study Party Identification in 1992 Economic evaluations Economic evaluations Economic evaluations N of cases (201) (233) (179) Panel Study Party Identification in 1994 Economic evaluations Economic evaluations N of cases (232) (215) (182)