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
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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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:** 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 \( \lambda = P(\text{yes response}) \). Then \( \lambda \) and \( p \) are related by \( \lambda = 0.5p + (0.5)(0.3) \).
### (a)
Let \( Y \) denote the number of yes responses, so \( Y \sim \text{Bin}(n, \lambda) \). Thus \( Y/n \) is an unbiased estimator of \( \lambda \). Derive an estimator for \( p \) based on \( Y \). [Hint: Solve \( \lambda = 0.5p + 0.15 \) for \( p \) and then substitute \( Y/n \) for \( \lambda. \) (Enter your answer in terms of \( Y \) and \( n \).)]
\[ \hat{p} = \]
If \( n = 80 \) and \( y = 20 \), what is your estimate?
\[ \hat{p} = \]
### (b)
Use the fact that \( E(Y/n) = \lambda \) to show that your estimator \( \hat{p} \) is unbiased.
\[ \]
### (c)
If there were 70 type I and 30 type II cards, what would be your estimator for \( p \)? (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1665bdf4-5411-4ca1-a903-736ef5c9444c%2F95c8c39a-0f41-4157-9001-39b5d973f73b%2F3osqhm_processed.png&w=3840&q=75)
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:** 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 \( \lambda = P(\text{yes response}) \). Then \( \lambda \) and \( p \) are related by \( \lambda = 0.5p + (0.5)(0.3) \).
### (a)
Let \( Y \) denote the number of yes responses, so \( Y \sim \text{Bin}(n, \lambda) \). Thus \( Y/n \) is an unbiased estimator of \( \lambda \). Derive an estimator for \( p \) based on \( Y \). [Hint: Solve \( \lambda = 0.5p + 0.15 \) for \( p \) and then substitute \( Y/n \) for \( \lambda. \) (Enter your answer in terms of \( Y \) and \( n \).)]
\[ \hat{p} = \]
If \( n = 80 \) and \( y = 20 \), what is your estimate?
\[ \hat{p} = \]
### (b)
Use the fact that \( E(Y/n) = \lambda \) to show that your estimator \( \hat{p} \) is unbiased.
\[ \]
### (c)
If there were 70 type I and 30 type II cards, what would be your estimator for \( p \)? (
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