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Chapter 6: Q8E (page 262)

In a random sample of 80 components of a certain type, 12 are found to be defective.

a. Give a point estimate of the proportion of all such components that are not defective.

b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here.

The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. (Hint: If p denotes the probability that a component works properly, how can P (system works) be expressed in terms of p ?)

Short Answer

Expert verified

(a) The required point estimate value is\({\rm{0}}{\rm{.85}}\).

(b) The required probability of interest value is\(0.723\).

Step by step solution

01

Concept introduction

In statistics, point estimation is the process of estimating an estimated value of a population's parameter—such as the mean (average)—from random samples of the population. The exact accuracy of any one approximation is unknown, but probabilistic claims about the accuracy of numbers determined over a large number of experiments can be built.

02

Calculating using a point estimate

(a)

There are 80 components in all, 12 of which are faulty. As a result, there are\({\rm{80 - 12 = 68}}\)components that are in good working order. The percentage of all such components that are not faulty is calculated using a point estimate.

\(\begin{array}{c}{\rm{\hat p = }}\frac{{{\rm{68}}}}{{{\rm{80}}}}\\{\rm{ = 0}}{\rm{.85}}\end{array}\)

Hence, the required point estimate value is \({\rm{0}}{\rm{.85}}\).

03

Calculating probability of interest

(b)

The probability of interest is,

\(\begin{array}{c}{\rm{P( system works )}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{p \times p}}\\{\rm{ = }}{{\rm{p}}^{\rm{2}}}\end{array}\)

1) Both components must function properly.

Therefore, to calculate this probability, replace \(p\)with \(\hat p\)and get,

\(P({\rm{ system works }}) \approx {0.85^2} = 0.723.\)

Hence, the required probability of interest value is \(0.723\).

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Most popular questions from this chapter

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

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

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where \({\rm{ - 1£ \theta £ 1}}\) (this distribution arises in particle physics). Show that \({\rm{\hat \theta = 3\bar X}}\) is an unbiased estimator of\({\rm{\theta }}\). (Hint: First determine\({\rm{\mu = E(X) = E(\bar X)}}\).)

When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?

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\({\rm{\delta }}{\rm{.)}}\)

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

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Prior to obtaining data, denote the beam strengths by X1, … ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi ’s constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi ’s form a random sample (independent of the Xi ’s) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). Calculate the estimate for the given data.

b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error.

c. Calculate a point estimate of the ratio \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)of the two standard deviations.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \({\rm{X - Y}}\) between beam strength and cylinder strength.
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