An investigator wishes to estimate the proportion of studentsat a certain university who

Chapter 6, Problem 19

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An investigator wishes to estimate the proportion of studentsat a certain university who have violated the honorcode. Having obtained a random sample of n students, sherealizes that asking each, Have you violated the honorcode? will probably result in some untruthful responses.Consider the following scheme, called a randomizedresponse technique. The investigator makes up a deck of100 cards, of which 50 are of type I and 50 are of type II.Type I: Have you violated the honor code (yes or no)?Type II: Is the last digit of your telephone number a 0,1, or 2 (yes or no)?Each student in the random sample is asked to mixthedeck, draw a card, and answer the resulting questiontruthfully. Because of the irrelevant question on type IIcards, a yes response no longer stigmatizes the respondent,so weassume that responses are truthful. Let p denotethe proportion of honorcode violators (i.e., the probabilityof a randomly selected student being a violator),and let l 5 P(yes response).Then l and p are related byl 5 .5p 1 (.5)(.3).a. Let Y denote the number of yes responses, so Y , Bin(n,l). Thus Yyn is an unbiased estimator of l. Derivean estimator for p based on Y. If n 5 80 and y 5 20,what is your estimate? [Hint: Solve l 5 .5p 1 .15 forp and then substitute Yyn for l.]b. Use the fact that E(Yyn) 5 l to show that your estimatorp is unbiased.c. If there were 70 type I and 30 type II cards, whatwould be your estimator for p?

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