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Chapter 6: Q11E (page 263)

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Short Answer

Expert verified

(a) It is proved that\({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\)has an unbiased estimator\(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\).

(b) The standard error of the estimator in part (a) is\({\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right){\rm{ = }}\sqrt {\frac{{{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \).

(c) The values\({{\rm{x}}_{\rm{1}}}\)and\({{\rm{x}}_{\rm{2}}}\)are used to estimate the standard error of the estimator as\({\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right){\rm{ = }}\sqrt {\frac{{\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \).

(d) The estimate of\({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\)is\({{\rm{\hat p}}_{\rm{1}}}{\rm{ - }}{{\rm{\hat p}}_{\rm{2}}}{\rm{ = - 0}}{\rm{.245}}\).

(e) The standard error of the estimator is \({\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right) \approx 0.0411\).

Step by step solution

01

Concept Introduction

The average of the given numbers is computed by dividing the total number of numbers by the sum of the given numbers.

The median is the middle number in a list of numbers that has been sorted ascending or descending, and it might be more descriptive of the data set than the average. When there are outliers in the series that could affect the average of the numbers, the median is sometimes utilised instead of the mean.

The standard deviation is a statistic that measures the amount of variation or dispersion in a set of numbers.

02

Unbiased Estimator

(a)

\({{\rm{X}}_{\rm{i}}}\)represents the number of successes (filter cigarettes) among a sample of\({{\rm{n}}_{\rm{i}}}\)individuals with probability of success\({{\rm{p}}_{\rm{i}}}{\rm{(i = 1,2)}}\).

The number of successes among a fixed sample size with a constant probability of success has a binomial distribution with parameters\({\rm{n}}\)and\({\rm{p}}\).

\(\begin{array}{c}{{\rm{X}}_{\rm{1}}}{\rm{\~b}}\left( {{{\rm{n}}_{\rm{1}}}{\rm{,}}{{\rm{p}}_{\rm{1}}}} \right)\\{{\rm{X}}_{\rm{2}}}{\rm{\~b}}\left( {{{\rm{n}}_{\rm{2}}}{\rm{,}}{{\rm{p}}_{\rm{2}}}} \right)\end{array}\)

The mean of a binomial distribution is the product of the sample size\({\rm{n}}\)and the probability\({\rm{p}}\).

\(\begin{array}{c}{\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = }}{{\rm{\mu }}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{1}}}{{\rm{p}}_{\rm{1}}}\\{\rm{E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ = }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{{\rm{p}}_{\rm{2}}}\end{array}\)

For the linear combination\({\rm{W = a}}{{\rm{X}}_1} + b{X_2}\), the following properties hold for the mean and variance 鈥�

\(\begin{array}{c}{{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}\\{\rm{\sigma }}_{\rm{W}}^{\rm{2}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}{\rm{ (If }}{{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ are independent)}}\end{array}\)

Then determine the expected value of\(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\)鈥�

\(\begin{array}{c}{\rm{E}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right){\rm{ = }}\frac{{\rm{1}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{{\rm{n}}_{\rm{2}}}}}{\rm{E}}\left( {{{\rm{X}}_{\rm{2}}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{n}}_{\rm{1}}}}}{{\rm{n}}_{\rm{1}}}{{\rm{p}}_{\rm{1}}}{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{n}}_{\rm{2}}}}}{{\rm{n}}_{\rm{2}}}{{\rm{p}}_{\rm{2}}}\\{\rm{ = }}{{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\end{array}\)

Since the expected value of\(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\)is equal to\({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

Therefore, \(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\) is an unbiases estimator of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

03

Standard Error of the Estimator

(b)

The variance of a binomial distribution is the product of the sample size \({\rm{n}}\) and the probabilities \({\rm{p}}\) and \(q\) 鈥�

\(\begin{array}{c}{\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ = }}{{\rm{n}}_{\rm{1}}}{{\rm{p}}_{\rm{1}}}{{\rm{q}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{1}}}{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)\\{\rm{V}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ = \sigma }}_{\rm{2}}^{\rm{2}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{{\rm{p}}_{\rm{2}}}{{\rm{q}}_{\rm{2}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)\end{array}\)

Determine the variance of \(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\)鈥�

\(\begin{array}{c}{\rm{V}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{n}}_{\rm{1}}^{\rm{2}}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + }}\frac{{\rm{1}}}{{{\rm{n}}_{\rm{2}}^{\rm{2}}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{2}}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n}}_{\rm{1}}^{\rm{2}}}}{{\rm{n}}_{\rm{1}}}{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right){\rm{ + }}\frac{{\rm{1}}}{{{\rm{n}}_{\rm{2}}^{\rm{2}}}}{{\rm{n}}_{\rm{2}}}{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)\\{\rm{ = }}\frac{{{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}\end{array}\)

The standard error is the square root of the variance 鈥�

\(\begin{array}{c}{\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right){\rm{ = }}\sqrt {{\rm{V}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)} \\{\rm{ = }}\sqrt {\frac{{{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \end{array}\)

Therefore, the standard error is \(\sqrt {\frac{{{{\rm{p}}_{\rm{1}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{1}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{{{\rm{p}}_{\rm{2}}}\left( {{\rm{1 - }}{{\rm{p}}_{\rm{2}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \).

04

Standard Error of the Estimator

(c)

Let the observed values be\({{\rm{x}}_{\rm{1}}}\) and\({{\rm{x}}_{\rm{2}}}\). The estimates of the proportions are the number of successes\({{\rm{x}}_i}\) divided by the sample size\({n_i}\)鈥�

\(\begin{array}{c}{{{\rm{\hat p}}}_{\rm{1}}}{\rm{ = }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}\\{{{\rm{\hat p}}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\end{array}\)

The standard error can then be estimated by replacing the proportions\({p_i}\)by their estimates\({\hat p_i}\).

\(\begin{array}{c}{\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right) \approx \sqrt {\frac{{{{{\rm{\hat p}}}_{\rm{1}}}\left( {{\rm{1 - }}{{{\rm{\hat p}}}_{\rm{1}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{{{{\rm{\hat p}}}_{\rm{2}}}\left( {{\rm{1 - }}{{{\rm{\hat p}}}_{\rm{2}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \\{\rm{ = }}\sqrt {\frac{{\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \end{array}\)

Therefore, the standard error is \(\sqrt {\frac{{\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}} \right)}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ + }}\frac{{\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\left( {{\rm{1 - }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)}}{{{{\rm{n}}_{\rm{2}}}}}} \).

05

Estimator of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\)

(d)

It is given that 鈥�

\(\begin{array}{c}{{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200}}\\{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\\{{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\end{array}\)

\(\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\)is estimator of\({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\)-

\(\begin{array}{c}{{{\rm{\hat p}}}_{\rm{1}}}{\rm{ - }}{{{\rm{\hat p}}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{x}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{x}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}\\{\rm{ = }}\frac{{{\rm{127}}}}{{{\rm{200}}}}{\rm{ - }}\frac{{{\rm{176}}}}{{{\rm{200}}}}\\{\rm{ = - }}\frac{{{\rm{49}}}}{{{\rm{200}}}}\\{\rm{ = - 0}}{\rm{.245}}\end{array}\)

Therefore, the value is obtained as \({\rm{ - 0}}{\rm{.245}}\).

06

Standard Error of the Estimator

(e)

It is given that 鈥�

\(\begin{array}{c}{{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200}}\\{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\\{{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\end{array}\)

Use the formula found in part (c) to estimate the standard error 鈥�

\(\begin{array}{c}{\rm{SE}}\left( {\frac{{{{\rm{X}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ - }}\frac{{{{\rm{X}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right) \approx \sqrt {\frac{{\frac{{{\rm{127}}}}{{{\rm{200}}}}\left( {{\rm{1 - }}\frac{{{\rm{127}}}}{{{\rm{200}}}}} \right)}}{{{\rm{200}}}}{\rm{ + }}\frac{{\frac{{{\rm{176}}}}{{{\rm{200}}}}\left( {{\rm{1 - }}\frac{{{\rm{176}}}}{{{\rm{200}}}}} \right)}}{{{\rm{200}}}}} \\{\rm{ = }}\frac{{\sqrt {{\rm{26990}}} }}{{{\rm{4000}}}} \approx {\rm{0}}{\rm{.0411}}\end{array}\)

Therefore, the value is obtained as \(0.0411\).

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