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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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Most popular questions from this chapter

a. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a uniform distribution on \({\rm{(0,\theta )}}\). Then the mle of \({\rm{\theta }}\) is \({\rm{\hat \theta = Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\). Use the fact that \({\rm{Y}} \le {\rm{y}}\) if each \({{\rm{X}}_{\rm{i}}} \le {\rm{y}}\) to derive the cdf of Y. Then show that the pdf of \({\rm{Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\) is \({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}}&{{\rm{0}} \le {\rm{y}} \le {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

b. Use the result of part (a) to show that the mle is biased but that \({\rm{(n + 1)}}\)\({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{/n}}\) is unbiased.

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 103, 156, 118, 89, 125, 147, 122, 109, 138, 99. Let m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

At time \({\rm{t = 0}}\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \({\rm{\lambda }}\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \({\rm{\lambda }}\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (\({\rm{\lambda }}\)) variables, which is exponential with parameter \({\rm{2\lambda }}\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential \({\rm{rv}}\) with parameter \({\rm{3\lambda }}\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \({\rm{25}}{\rm{.2,41}}{\rm{.7,51}}{\rm{.2,55}}{\rm{.5,59}}{\rm{.5,61}}{\rm{.8}}\) (from which you should calculate the times between successive births). Derive the mle of l. (Hint: The likelihood is a product of exponential terms.)

A vehicle with a particular defect in its emission control system is taken to a succession of randomly selected mechanics until\({\rm{r = 3}}\)of them have correctly diagnosed the problem. Suppose that this requires diagnoses by\({\rm{20}}\)different mechanics (so there were\({\rm{17}}\)incorrect diagnoses). Let\({\rm{p = P}}\)(correct diagnosis), so\({\rm{p}}\)is the proportion of all mechanics who would correctly diagnose the problem. What is the mle of\({\rm{p}}\)? Is it the same as the mle if a random sample of\({\rm{20}}\)mechanics results in\({\rm{3}}\)correct diagnoses? Explain. How does the mle compare to the estimate resulting from the use of the unbiased estimator?

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\({\rm{\theta }}\).
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