Bernoulli TrialsThe Binomial DistributionComputational DifficultiesThe Normal Curve Approximation to the BinomialCalculating Binomial Probabilities When p is Unknown Runs in Dichotomous TrialsThe Runs TestThe Test Statistics V and WSummaryPractice ProblemsSolutions to Practice ProblemsHomework ProblemsChapter 11Bernoulli Trials11.1 The Binomial DistributionIn the previous chapter, we learned about i.i. d. trials. Recall that there are three ways we can havei.i.d. trials:1. Our units are trials and we have decided to assume that they are i.i.d.2. We have a finite population and we will select our sample of its members at random withreplacement—the dumb form of random sampling. The resu lt is that we have i.i . d. randomvariables which m eans the sam e thing as having i.i.d. trials.3. We h ave a finite p opulation and we h ave selected our sample of its members at randomwithout replacement—the smart form of random sampli ng. If n/N—the ratio of samp lesize to population size—is 0.05 or smaller, then we get a good approximation if we treat ourrandom variables as i. i.d.In this chapter, we study a very important special case of i.i.d. trials, called Bernoulli trials. If eachtrial has exactly two possibl e outcomes, then we have Bernoulli trials. For convenient reference, Iwill now explicitly state the assumptions of Bernoulli trials.Definition 11.1 (The assumptions of Bernoulli trials.) If we have a collection of trials that sat-isfy the three condit ions below, then we say that we have Bernoul li tria ls.1. Each trial results in on e of two poss ible outcomes, deno ted success (S) or f ailure (F ).2. The probability of a success remains constant from tria l-to-trial and is denoted by p. Writeq = 1 − p for the constan t probability of a failure.3. The trials are independ ent.We wi ll use the method described on page 168 of Chapter 8 to assign the labels s uccess and failure.When w e are involved in mathematical arguments, it will be convenient to represent a success bythe number 1 and a failure by the number 0. Finally,255We are not interested in either of th e trivial cases in which p = 0 or p = 1. Thus, werestrict attention to situations in which 0 < p < 1.One reason that Bernoulli trials are so important, is that if we have Bernoulli trials, we can calculateprobabilities of a great m any events. Our first tool for calculation is the multiplication rule that w elearned in Chapter 10. For example, suppose that we have n = 5 Bernoulli trials with p = 0.70.The probability that the Bernoulli trials y ield four successes followed by a failure is:P (SSSSF ) = ppppq = (0.70)4(0.30) = 0.0720.Our next tool is extremely powerful and very useful in science. It is the binomial probabilitydistribution. Suppose that we plan to perform/observe n Bernoulli trials. Let X denote the totalnumber of successes in th e n t ri al s. The probability distribution of X is given by the followingequation.P (X = x) =n!x!(n − x)!pxqn−x, for x = 0, 1, . . . , n. (11.1)Equation 11.1 is called the binomial probability distribution with parameters n and p; it isdenoted by the Bin(n, p) distri bution. I will illustrate the use of this equation below, a compilationof the Bin(5,0.60) distribution. I replace n with 5, p with 0.60 and q with (1 − p) = 0.40. Below,I will evaluate Equation 11 . 1 six times, for x = 0, 1, . . . , 5: You shoul d check a couple of thefollowing computations to make sure you are comfortable using Equ at ion 11.1 , but you don’t needto verify all of them.P (X = 0) =5!0!5!(0.60)0(0.40)5= 1(1)(0.01024) = 0.01024.P (X = 1) =5!1!4!(0.60)1(0.40)4= 5(0.60)(0.0256) = 0.07680.P (X = 2) =5!2!3!(0.60)2(0.40)3= 10(0.36)(0.064) = 0.23040.P (X = 3) =5!3!2!(0.60)3(0.40)2= 10(0.216)(0.16) = 0.34560.P (X = 4) =5!4!1!(0.60)4(0.40)1= 5(0.1296)(0.40) = 0.25920.P (X = 5) =5!5!0!(0.60)5(0.40)0= 1(0.07776)(1) = 0.07776.Whenever probabili ties for a random variable X are given by Equation 11.1 we say that Xhas a binomial probability (sampling) distribution with parameters n and p and write this as X ∼Bin(n, p).There are a number of di fficulties that arise when one attempts to use the binom ial probabilitydistribution. The most obvious i s that each trial n eeds to give a dichot omous response. Sometimesit is obvious that we have a dichotomy: For example, if my trials are shooting free throws orattempting golf putts, then the natural response is that a trial results in a make or miss. Other256times, the nat u ral response migh t not be a dichotomy, but the response of interest is. For example,in my example of t he American rou lette wheel in Chapter 10, the natural response is the winningnumber, but if I like to bet on red, t hen the response of interest has possible values red and notred. Similarly, in the game of craps, I might be primarily interested in whether or not my come outresults in a pass line win, a dichotomy.Thus, let’s su ppose th at we have a dichotomou s response. The next di fficulty is that in orderto calculate probabili ties, we need to kn ow the numerical values of n and p. Almost always, n isknown to t he researcher and if it is unknown, we might be able to salvage something by using thePoisson d istribution, which you will learn about in Chapter 13. There are situations in which p isknown, including the following:• Mendelian inheritance; my roulette example above, as suming the wheel is fair; and my crapsexample, assuming both dice are fair and they beha ve independently.In other words, sometimes I feel that the phenomenon under st udy is sufficiently well understoodthat I feel comfortable in my belief that I know the nu merical value of p. Obvious ly, in manysituations I won’t know the numerical value of p. For shooting free th rows or attempting golfputts I usually won’t know exactly how skilled the player/golfer is . When my in terest is in a finitepopulation, typically I won’t know the composition of the population; hence, I won’t know thevalue of p.Before I explore how to deal with the difficulty of p being unknown, let’s perform a few com-putations when p is known.Example 11.1 (Mendelian inheritance with the 3:1 ratio.) If you are unfamilia r with Mendelianinheritance, you can find an explanation of th e 1:2:1, 3:1 and 9:3:3:1 ratios at:http://en.wikipedia.org/wiki/Mendelian_inheritance.Each trial—offspring—will possess either the dominant (success) or recessive (failure) phenotype.Mendelian inheri tance tells us that these will occur in th e ratio
View Full Document
Unlocking...