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Chapter 6: Q1E (page 261)

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.

Short Answer

Expert verified

a. \(8.1407,\)Sample mean.

b. \({\rm{7}}{\rm{.7,}}\) Sample mean.

c. \({\rm{1}}{\rm{.6595,}}\) Sample standard deviation.

d. \({\rm{0}}{\rm{.1481 }}\)of \({\rm{14}}{\rm{.81 \% ,}}\)Sample Proportion.

e. \({\rm{0}}{\rm{.2039 }}\)or \(20.39{\rm{ }}\% ,\)Sample coefficient of variation.

Step by step solution

01

Concept Introduction

We can compute probabilities with regard to means using Z scores because the sample means are normally distributed.

We used Z scores to compute probabilities for values among individuals in a population.

02

Step 2:Estimating of the mean value

(a)

Given,

\(\begin{array}{c}\sum {{{\rm{x}}_{\rm{i}}}} {\rm{ = 219}}{\rm{.8}}\\{\rm{n = 27}}\end{array}\)

The sample mean, which is the total of all values divided by the number of values, is the point estimate of the mean value:

\(\begin{array}{c}\bar x = \frac{{\sum {{x_i}} }}{n}\\ = \frac{{219.8}}{{27}}\\ \approx 8.1407\end{array}\)

Therefore, the estimate of the mean value is \(8.1407\)

03

Finding the Median Point

(b)

The median is the point estimate that divides the weakest \(50\% \)from the greatest \(50\% \).Sort the data values as follows:

\(\begin{array}{l}{\rm{5}}{\rm{.9,6}}{\rm{.3,6}}{\rm{.3,6}}{\rm{.5,6}}{\rm{.8,6}}{\rm{.8,7,7,7}}{\rm{.2,7}}{\rm{.3,7}}{\rm{.4,7}}{\rm{.6,7}}{\rm{.7,7}}{\rm{.7,7}}{\rm{.8,7}}{\rm{.8,7}}{\rm{.9,8}}{\rm{.1,8}}{\rm{.2,8}}{\rm{.7,9,9}}{\rm{.7,9}}{\rm{.7,10}}{\rm{.7,11}}{\rm{.3,11}}{\rm{.6,11}}{\rm{.8}}\\\end{array}\)

The median is the middle (14th data value) of the sorted data set, because there are 27 data values in the data set.

\({\rm{M = 7}}{\rm{.7}}\)

Hence, the Median point is \({\rm{M = 7}}{\rm{.7}}\)

04

Estimating of the population standard deviation.

(c)

Given:

\(\begin{array}{c}\sum {x_i^2} = 1860.94\\\sum {{x_i}} = 219.8\\n = 27\end{array}\)

The sample standard deviation is the point estimate of the population standard deviation.

\(\begin{aligned} s &= \sqrt {\frac{{\sum {{x^2}} - {{\left( {\sum x } \right)}^2}/n}}{{n - 1}}} \\ &= \sqrt {\frac{{1860.94 - {{(219.8)}^2}/27}}{{27 - 1}}} \\ \approx 1.6595\end{aligned}\)

As a result, the population standard deviation is \(1.6595\)

05

Calculating the sample proportion

(d)

The sample proportion is the point estimate of the proportion. The sample proportion is calculated by dividing the number of successes (values greater than 10) by the sample size \(n = 27:\)

\(\begin{array}{c}\hat p = \frac{4}{{27}}\\ \approx 0.1481\\ = 14.81\% \end{array}\)

Such that, the sample proportion is \(14.81\% \).

06

Estimating the population coefficient

(e)

The sample coefficient of variation \(s/\bar x\)is the point estimate of the population coefficient of variation \(\sigma /\mu \)

\(\begin{aligned}CV &= \frac{s}{{\bar x}}\\ &= \frac{{1.6595}}{{8.1407}}\\ \approx 0.2039\\ &= 20.39\% \end{aligned}\)

Therefore, the sample coefficient variation is \(20.39\% \).

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Most popular questions from this chapter

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) from the pdf

\({\rm{f(x;\theta ) = }}{\rm{.5(1 + \theta x)}}\quad {\rm{ - 1£ x£ 1}}\)

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\({\rm{\mu = E(X) = E(\bar X)}}\).)

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.

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

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

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.

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 \({\rm{n = 20}}\)?

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:

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