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Chapter 6: Q19E (page 264)

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\({\rm{n}}\)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\({\rm{100}}\)cards, of which\({\rm{50}}\)are of type I and\({\rm{50}}\)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\({\rm{0 , 1 , or 2}}\)(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\({\rm{p}}\)denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let\({\rm{\lambda = P}}\)(yes response). Then\({\rm{\lambda }}\)and\({\rm{p}}\)are related by\({\rm{\lambda = }}{\rm{.5p + (}}{\rm{.5)(}}{\rm{.3)}}\).

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

b. Use the fact that\({\rm{E(Y/n) = \lambda }}\)to show that your estimator\({\rm{\hat p}}\)is unbiased.

c. If there were\({\rm{70}}\)type I and\({\rm{30}}\)type II cards, what would be your estimator for\({\rm{p}}\)?

Short Answer

Expert verified

a) The estimated value is \({\rm{\hat p = 2 \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - 0}}{\rm{.3}}\).

b) \({\rm{Estimator in unbiased}}\).

c) The estimated value for p is \({\rm{\hat p = }}\frac{{{\rm{10}}}}{{\rm{7}}}{\rm{ \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - }}\frac{{\rm{9}}}{{{\rm{70}}}}\).

Step by step solution

01

Introduction

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

02

Explanation

a)

Given unbiased estimator of \({\rm{\lambda }}\)as\(\frac{{\rm{Y}}}{{\rm{n}}}\).

And the fact that\({\rm{Y:Bin(n,\lambda )}}\).

Determine the estimator of probability \({\rm{p}}\) (proportion of honor-code violators).

From relation,

\({\rm{\lambda = 0}}{\rm{.5 \times p + 0}}{\rm{.5 \times 0}}{\rm{.3}}\)

The probability \({\rm{p}}\)is:

\({\rm{p = 2\lambda - 0}}{\rm{.3}}\)

Because\(\frac{{\rm{Y}}}{{\rm{n}}}\)is unbiased estimator of\({\rm{\lambda }}\), the estimator of \({\rm{p}}\)could be,

\({\rm{\hat p = 2 \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - 0}}{\rm{.3}}\)

For \({\rm{n = 80}}\)and\({\rm{y = 20}}\), the estimate is:

\(\begin{aligned} \hat p &= 2 \times \frac{{\rm{y}}}{{\rm{n}}}{\rm{ - 0}}{\rm{.3}}\\ \ &= \frac{{{\rm{2 \times 20}}}}{{{\rm{80}}}}{\rm{ - 0}}{\rm{.3}}\\ &= 0 {\rm{.2}}{\rm{.}}\end{aligned}\)

Therefore, the estimated value is\({\rm{\hat p = 2 \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - 0}}{\rm{.3}}\).

03

Explanation

b)

If the expected value of \({\rm{\hat p}}\)is\({\rm{p}}\), then the estimator is unbiased. Therefore, since the expected value is

\(\begin{aligned} E(\hat p) &= E \left( {{\rm{2 \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - 0}}{\rm{.3}}} \right)\\&= 2 \times E \left( {\frac{{\rm{Y}}}{{\rm{n}}}} \right){\rm{ - 0}}{\rm{.3}}\\&= 2 \times \lambda - 0 {\rm{.3}}\\ &= p\end{aligned}\)

Hence, the estimated value is unbiased.

04

Explanation

c)

The probability \({\rm{\lambda = P}}\)(yes response) would change to,

\({\rm{\lambda = 0}}{\rm{.7 \times p + (1 - 0}}{\rm{.7) \times 0}}{\rm{.3}}\)

Therefore, the probability \({\rm{p}}\)would be,

\({\rm{p = }}\frac{{{\rm{10}}}}{{\rm{7}}}{\rm{\lambda - }}\frac{{\rm{9}}}{{{\rm{70}}}}\)

And the estimator would be, analogously as before,

\({\rm{\hat p = }}\frac{{{\rm{10}}}}{{\rm{7}}}{\rm{ \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - }}\frac{{\rm{9}}}{{{\rm{70}}}}{\rm{.}}\)

Hence, the estimated value for p is \({\rm{\hat p = }}\frac{{{\rm{10}}}}{{\rm{7}}}{\rm{ \times }}\frac{{\rm{Y}}}{{\rm{n}}}{\rm{ - }}\frac{{\rm{9}}}{{{\rm{70}}}}\).

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Most popular questions from this chapter

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

Let\({\rm{X}}\)represent the error in making a measurement of a physical characteristic or property (e.g., the boiling point of a particular liquid). It is often reasonable to assume that\({\rm{E(X) = 0}}\)and that\({\rm{X}}\)has a normal distribution. Thus, the pdf of any particular measurement error is

\({\rm{f(x;\theta ) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \theta }}} }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\theta }}}}\quad {\rm{ - \yen < x < \yen}}\)

(Where we have used\({\rm{\theta }}\)in place of\({{\rm{\sigma }}^{\rm{2}}}\)). Now suppose that\({\rm{n}}\)independent measurements are made, resulting in measurement errors\({{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{.}}\)Obtain the mle of\({\rm{\theta }}\).

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)

Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a pdf that is symmetric about \({\rm{\mu }}\). An estimator for \({\rm{\mu }}\) that has been found to perform well for a variety of underlying distributions is the Hodges鈥揕ehmann estimator. To define it, first compute for each \({\rm{i}} \le {\rm{j}}\) and each \({\rm{j = 1,2, \ldots ,n}}\) the pairwise average \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{ = }}\left( {{{\rm{X}}_{\rm{i}}}{\rm{ + }}{{\rm{X}}_{\rm{j}}}} \right){\rm{/2}}\). Then the estimator is \({\rm{\hat \mu = }}\) the median of the \({{\rm{\bar X}}_{{\rm{i,j}}}}{\rm{'s}}\). Compute the value of this estimate using the data of Exercise \({\rm{44}}\) of Chapter \({\rm{1}}\). (Hint: Construct a square table with the \({{\rm{x}}_{\rm{i}}}{\rm{'s}}\) listed on the left margin and on top. Then compute averages on and above the diagonal.)

Suppose a certain type of fertilizer has an expected yield per acre of \({{\rm{\mu }}_{\rm{2}}}\)with variance \({{\rm{\sigma }}^{\rm{2}}}\)whereas the expected yield for a second type of fertilizer is with the same variance \({{\rm{\sigma }}^{\rm{2}}}\).Let \({\rm{S}}_{\rm{1}}^{\rm{2}}\) and \({\rm{S}}_{\rm{2}}^{\rm{2}}\)denote the sample variances of yields based on sample sizes \({{\rm{n}}_{\rm{1}}}\)and \({{\rm{n}}_{\rm{2}}}\),respectively, of the two fertilizers. Show that the pooled (combined) estimator

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\left( {{{\rm{n}}_{\rm{1}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{1}}^{\rm{2}}{\rm{ + }}\left( {{{\rm{n}}_{\rm{2}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{2}}^{\rm{2}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ - 2}}}}\)

is an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\)
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