/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q1E The accompanying data on flexura... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 \(20.39\% \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)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:

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

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:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

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, 鈥� ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi 鈥檚 constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi 鈥檚 form a random sample (independent of the Xi 鈥檚) 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.

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?

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.