/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 An investigator wishes to estima... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 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 \(\lambda=P\) (yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim \operatorname{Bin}(n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) 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 \(p\) ?

Short Answer

Expert verified
Estimate for p is 0.2, and the estimator is unbiased.

Step by step solution

01

Derive the estimator for p

The relationship between \( \lambda \) and \( p \) is given by \( \lambda = 0.5p + 0.15 \). To find an estimator for \( p \), solve this equation for \( p \):\[\lambda = 0.5p + 0.15 \p = 2\lambda - 0.3\]Now substitute \( Y/n \) for \( \lambda \):\[\hat{p} = 2 \left( \frac{Y}{n} \right) - 0.3\]This gives us the estimator \( \hat{p} = 2\frac{Y}{n} - 0.3 \).
02

Calculate the estimate for p with n=80 and y=20

Substitute \( n = 80 \) and \( y = 20 \) into the estimator:\[\hat{p} = 2 \left( \frac{20}{80} \right) - 0.3\]This simplifies to:\[\hat{p} = 0.5 - 0.3 = 0.2\]Thus, the estimate for \( p \) is 0.2.
03

Show the estimator is unbiased using E(Y/n)

The expectation \( E(Y/n) = \lambda \), where \( \lambda = 0.5p + 0.15 \). Substituting the expression for \( \lambda \), we have:\[p = 2 \lambda - 0.3\]Substituting \( \lambda = E(Y/n) \),\[E(\hat{p}) = E(2(Y/n) - 0.3) = 2E(Y/n) - 0.3 = 2\lambda - 0.3 = p\]Thus, \( \hat{p} \) is an unbiased estimator of \( p \).
04

Adjust the estimator for a different card distribution

With 70 type I cards and 30 type II cards, the equation changes to account for the new distribution.The new relationship is:\[\lambda = 0.7p + 0.3 \times 0.3 = 0.7p + 0.09\]Solving for \( p \),\[p = \frac{\lambda - 0.09}{0.7}\]Substituting \( Y/n \) for \( \lambda \), the estimator becomes:\[\hat{p} = \frac{Y/n - 0.09}{0.7}\]This is the altered estimator for \( p \) with the new card distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Randomized Response Technique
The randomized response technique is an ingenious method used to collect sensitive data while keeping individual responses confidential. This technique helps in reducing bias that might occur due to respondents giving socially desirable answers. When individuals are asked about sensitive topics like violating an honor code, they may not respond truthfully.
  • To counteract this, a deck of cards is used with two types of questions.
  • The first type (Type I) directly asks if they have violated the honor code.
  • The second type (Type II) asks a totally irrelevant question such as whether the last digit of their telephone number is a 0, 1, or 2.
Each participant picks one card at random. By mixing these cards, the randomness ensures that aggregate data retains privacy for respondents. They answer the question on the card drawn, ensuring higher truthfulness in responses.
Binomial Distribution
A binomial distribution is a statistical term that describes the number of 'successes' in a fixed number of experiments or trials, each having the same probability of success. In the randomized response technique scenario, each student鈥檚 response can be seen as a trial, where 'success' corresponds to a 'yes' response.
  • Since there are two types of cards that a student could answer 'yes' to, this strategy introduces a binomial model.
  • For our investigation, the number of 'yes' responses among the surveyed students follows a binomial distribution.
  • This is captured by the notation: \[Y \sim \text{Bin}(n, \lambda)\]
Here, \( n \) is the total number of students surveyed, and \( \lambda \) represents the probability of a 'yes' response.
Unbiased Estimator
An unbiased estimator is a key topic in statistics. It is essentially a statistical measure that predicts the actual parameter value in a population without systematic error. In this exercise, our goal is to determine the true proportion \( p \) of students violating the honor code.
To achieve this, we derived an estimator for \( p \) using the observations.
  • The formula for \( \hat{p} \), as derived from the randomized response model, is:\[\hat{p} = 2\frac{Y}{n} - 0.3\]
  • This formula rearranges the expression relating \( \lambda \) and \( p \), ensuring that the expected value equals \( p \).
  • By leveraging the expression \( E(Y/n) = \lambda \), we showed that \( \hat{p} \) is unbiased since its expected value matches the actual proportion \( p \).
Probability Estimation
Probability estimation is about quantifying the likelihood of a particular outcome. In the context of this exercise, it concerns estimating how many students have violated the honor code based on survey responses.
  • With a sample of students responding through randomized response, our task was to estimate the probability \( p \).
  • The expression \( \lambda = 0.5p + 0.15 \) provides a derived scenario for \( \lambda \), i.e., the likelihood of a 'yes' response.Subsequently, estimated \( \lambda \) is observed through formula:\[\hat{p} = 2 \left(\frac{Y}{n}\right) - 0.3 \]
  • Using this formula, we can substitute observed data and calculate our best estimate for \( p \).
This approach illustrates how probability estimation builds from observed data to make educated guesses about population parameters.

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Most popular questions from this chapter

Suppose waiting time for delivery of an item is uniform on the interval from \(\theta_{1}\) to \(\theta_{2}\) (so \(f\left(x ; \theta_{1}, \theta_{2}\right.\) ) \(=1 /\left(\theta_{2}-\theta_{1}\right)\) for \(\theta_{1}

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The principle of unbiasedness (prefer an unbiased estimator to any other) has been criticized on the grounds that in some situations the only unbiased estimator is patently ridiculous. Here is one such example. Suppose that the number of major defects \(X\) on a randomly selected vehicle has a Poisson distribution with parameter \(\lambda\). You are going to purchase two such vehicles and wish to estimate \(\theta=P\left(X_{1}=0, \quad X_{2}=0\right)=e^{-2 \lambda}\), the probability that neither of these vehicles has any major defects. Your estimate is based on observing the value of \(X\) for a single vehicle. Denote this estimator by \(\hat{\theta}=\delta(X)\). Write the equation implied by the condition of unbiasedness, \(E[\delta(X)]=e^{-2 \lambda}\), cancel \(e^{-\lambda}\) from both sides, then expand what remains on the right-hand side in an infinite series,and compare the two sides to determine \(\delta(X)\). If \(X=200\), what is the estimate? Does this seem reasonable? What is the estimate if \(X=199 ?\) Is this reasonable?

The probability that any particular component of a certain type works in a satisfactory manner is \(p\). If \(n\) of these components are independently selected, then the statistic \(X\), the number among the selected components that perform in a satisfactory manner, is sufficient for \(p\). You must purchase two of these components for a particular system. Obtain an unbiased statistic for the probability that exactly one of your purchased components will perform in a satisfactory manner. [Hint: Start with the statistic \(U\), the indicator function of the event that exactly one of the first two components in the sample of size \(n\) performs as desired, and improve on it by conditioning on the sufficient statistic.]

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