91Ó°ÊÓ

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.

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 \({{\rm{\sigma }}^{\rm{2}}}\)

A sample of \({\rm{n}}\) captured Pandemonium jet fighters results in serial numbers\({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \({\rm{\alpha }}\)and ending with\({\rm{\beta }}\), so that the total number of planes manufactured is \({\rm{\beta - \alpha + 1}}\) (e.g., if \({\rm{\alpha = 17}}\) and\({\rm{\beta = 29}}\), then \({\rm{29 - 17 + 1 = 13}}\)planes having serial numbers \({\rm{17,18,19, \ldots ,28,29}}\)were manufactured). However, the CIA does not know the values of \({\rm{\alpha }}\) or\({\rm{\beta }}\). A CIA statistician suggests using the estimator \({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ - min}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ + 1}}\)to estimate the total number of planes manufactured.

a. If\({\rm{n = 5, x\_}}\left\{ {\rm{1}} \right\}{\rm{ = 237, x\_}}\left\{ {\rm{2}} \right\}{\rm{ = 375, x\_}}\left\{ {\rm{3}} \right\}{\rm{ = 202, x\_}}\left\{ {\rm{4}} \right\}{\rm{ = 525,}}\)and\({{\rm{x}}_{\rm{5}}}{\rm{ = 418}}\), what is the corresponding estimate?

b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating\({\rm{\beta - \alpha + 1}}\)? Explain in one or two sentences.

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

We defined a negative binomial\({\rm{rv}}\)as the number of failures that occur before the\({\rm{rth}}\)success in a sequence of independent and identical success/failure trials. The probability mass function (\({\rm{pmf}}\)) of\({\rm{X}}\)is\({\rm{nb(x,r,p) = }}\)\(\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\quad {\rm{x = 0,1,2, \ldots }}\)

a. Suppose that. Show that\({\rm{\hat p = (r - 1)/(X + r - 1)}}\)is an unbiased estimator for\({\rm{p}}\). (Hint: Write out\({\rm{E(\hat p)}}\)and cancel\({\rm{x + r - 1}}\)inside the sum.)

b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether\({\rm{(S)}}\)or not\({\rm{(F)}}\)they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate\({\rm{p = }}\)the true proportion who support the candidate.

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.

A diagnostic test for a certain disease is applied to\({\rm{n}}\)individuals known to not have the disease. Let\({\rm{X = }}\)the number among the\({\rm{n}}\)test results that are positive (indicating presence of the disease, so\({\rm{X}}\)is the number of false positives) and\({\rm{p = }}\)the probability that a disease-free individual's test result is positive (i.e.,\({\rm{p}}\)is the true proportion of test results from disease-free individuals that are positive). Assume that only\({\rm{X}}\)is available rather than the actual sequence of test results.

a. Derive the maximum likelihood estimator of\({\rm{p}}\). If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the estimate?

b. Is the estimator of part (a) unbiased?

c. If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the mle of the probability\({{\rm{(1 - p)}}^{\rm{5}}}\)that none of the next five tests done on disease-free individuals are positive?

Let\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

Let\({\rm{X}}\)denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of\({\rm{X}}\)is

\({\rm{f(x;\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{(\theta + 1)}}{{\rm{x}}^{\rm{\theta }}}}&{{\rm{0£ x£ 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

where\({\rm{ - 1 < \theta }}\). A random sample of ten students yields data\({{\rm{x}}_{\rm{1}}}{\rm{ = }}{\rm{.92,}}{{\rm{x}}_{\rm{2}}}{\rm{ = }}{\rm{.79,}}{{\rm{x}}_{\rm{3}}}{\rm{ = }}{\rm{.90,}}{{\rm{x}}_{\rm{4}}}{\rm{ = }}{\rm{.65,}}{{\rm{x}}_{\rm{5}}}{\rm{ = }}{\rm{.86}}\),\({{\rm{x}}_{\rm{6}}}{\rm{ = }}{\rm{.47,}}{{\rm{x}}_{\rm{7}}}{\rm{ = }}{\rm{.73,}}{{\rm{x}}_{\rm{8}}}{\rm{ = }}{\rm{.97,}}{{\rm{x}}_{\rm{9}}}{\rm{ = }}{\rm{.94,}}{{\rm{x}}_{{\rm{10}}}}{\rm{ = }}{\rm{.77}}\).

a. Use the method of moments to obtain an estimator of\({\rm{\theta }}\), and then compute the estimate for this data.

b. Obtain the maximum likelihood estimator of\({\rm{\theta }}\), and then compute the estimate for the given data.