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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

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data:

Assume that X has a Poisson distribution with parameter \({\bf{\mu }}.\)and that X represents the number of scratches on a randomly picked item.

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