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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 o 100 cards, of which 50 are of type I and 50 are of type II. Type I: Type II: Have you violated the honor code (yes or no)? 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 denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and le and p are related by λ = 0.5p+ (0.5)(0.3). λ = P(yes response). Then (a) Let Y denote the number of yes responses, so Y~ Bin(n, A). Thus Y/n is an unbiased estimator of λ. Derive an estimator for p based on Y. [Hint: Solve λ = 0.5p+ 0.15 for p and then substitute Y/n for λ.] (Enter your answer in terms of Y and n.) p= If n = 80 and y = 20, what is your estimate? (b) Use the fact that E(Y/n) = λ to show that your estimator p is unbiased. (c) If there were 70 type I and 30 type II cards, what would be your estimator for p? (Enter your answer in terms of Y and n.) p =

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 o 100 cards, of which 50 are of type I and 50 are of type II. Type I: Type II: Have you violated the honor code (yes or no)? 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 denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and le and p are related by λ = 0.5p+ (0.5)(0.3). λ = P(yes response). Then (a) Let Y denote the number of yes responses, so Y~ Bin(n, A). Thus Y/n is an unbiased estimator of λ. Derive an estimator for p based on Y. [Hint: Solve λ = 0.5p+ 0.15 for p and then substitute Y/n for λ.] (Enter your answer in terms of Y and n.) p= If n = 80 and y = 20, what is your estimate? (b) Use the fact that E(Y/n) = λ to show that your estimator p is unbiased. (c) If there were 70 type I and 30 type II cards, what would be your estimator for p? (Enter your answer in terms of Y and n.) p =

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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: Type II: Have you violated the honor code (yes or no)? 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 λ = P(yes response). Then and p are related by λ = 0.5p+ (0.5) (0.3). = (a) Let Y denote the number of yes responses, so Y~ Bin(n, λ). Thus Y/n is an unbiased estimator of λ. Derive an estimator for p based on Y. [Hint: Solve > substitute Y/n for λ.] (Enter your answer in terms of Y and n.) p= If n = 80 and y = 20, what is your estimate? (b) Use the fact that E(Y/n) = A to show that your estimator p is unbiased. (c) If there were 70 type I and 30 type II cards, what would be your estimator for p? (Enter your answer in terms of Y and n.) p= 0.5p+ 0.15 for p and then
Transcribed Image Text: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: Type II: Have you violated the honor code (yes or no)? 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 λ = P(yes response). Then and p are related by λ = 0.5p+ (0.5) (0.3). = (a) Let Y denote the number of yes responses, so Y~ Bin(n, λ). Thus Y/n is an unbiased estimator of λ. Derive an estimator for p based on Y. [Hint: Solve > substitute Y/n for λ.] (Enter your answer in terms of Y and n.) p= If n = 80 and y = 20, what is your estimate? (b) Use the fact that E(Y/n) = A to show that your estimator p is unbiased. (c) If there were 70 type I and 30 type II cards, what would be your estimator for p? (Enter your answer in terms of Y and n.) p= 0.5p+ 0.15 for p and then
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