19. An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code.

Having obtained a random sample of n students, she realizes that asking each, “Have you violated the honor code?” will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 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 mix the deck, draw a card, and answer the resulting question truthfully.

Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let p denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let 
P(yes response). Then  and p are related by 
.5p (.5)(.3).

a. Let Y denote the number of yes responses, so Y Bin (n, ). Thus Y/n is an unbiased estimator of . Derive an estimator for p based on Y. If n
80 and y
20, what is your estimate? [Hint: Solve 
.5p .15 for p and then substitute Y/n for .]

b. Use the fact that E(Y/n)
 to show that your estimator pˆ is unbiased.

c. If there were 70 type I and 30 type II cards, what would be your estimator for p?