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Chapter 7: Q4E (page 285)

A CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

Short Answer

Expert verified

(a) The value is \({\rm{(57}}{\rm{.1,59}}{\rm{.5)}}\).

(b) The value is \({\rm{(57}}{\rm{.7,58}}{\rm{.9)}}\).

(c) The value is \({\rm{(57}}{\rm{.5,59}}{\rm{.1)}}\).

(d) The value is \({\rm{(57}}{\rm{.9,58}}{\rm{.7)}}\).

(e) The n must be \({\rm{240}}\).

Step by step solution

01

Define interval

An interval is a set of numbers that includes all the real numbers between the two endpoints of the interval.

02

Explanation


(a) When a normal population is given,

\({\rm{100(1 - \alpha )\% confidence interval}}\)

the mean is calculated using,

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

when it is known what the value\({{\rm{\sigma }}^{\rm{2}}}\)is.

For given values, the\({\rm{95\% }}\)percent confidence interval is,

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{58}}{\rm{.3 - 1}}{\rm{.96 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{25}}} }}{\rm{,58}}{\rm{.3 + 1}}{\rm{.96 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{25}}} }}} \right)\\{\rm{ = (57}}{\rm{.1,59}}{\rm{.5)}}\end{array}\)

Where,

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 95}}\\{\rm{\alpha = 0}}{\rm{.05}}\end{array}\)

and

\(\begin{array}{c}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\end{array}\)

(1) : this is a result of

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ = 0}}{\rm{.025}}\)

and from the appendix's normal probability table a software can also be used to calculate the probability.

Therefore, the value is \({\rm{(57}}{\rm{.1,59}}{\rm{.5)}}\).

03

Explanation


(b) For given values, the \({\rm{95\% }}\) percent confidence interval is,

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{58}}{\rm{.3 - 1}}{\rm{.96 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}{\rm{,58}}{\rm{.3 + 1}}{\rm{.96 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}} \right)\\{\rm{ = (57}}{\rm{.7,58}}{\rm{.9)}}\end{array}\)

Where,

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 95}}\\{\rm{\alpha = 0}}{\rm{.05}}\end{array}\)

and

\(\begin{array}{c}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.96}}\end{array}\)

(1) : this is a result of

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ = 0}}{\rm{.025}}\)

and from the appendix's normal probability table a software can also be used to calculate the probability.

Therefore, the value is \({\rm{(57}}{\rm{.7,58}}{\rm{.9)}}\).

04

Explanation


(c) For given values, the \({\rm{99\% }}\) percent confidence interval is,

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{58}}{\rm{.3 - 2}}{\rm{.58 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}{\rm{,58}}{\rm{.3 + 2}}{\rm{.58 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}} \right)\\{\rm{ = (57}}{\rm{.5,59}}{\rm{.1)}}\end{array}\)

Where,

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 99}}\\{\rm{\alpha = 0}}{\rm{.01}}\end{array}\)

and

\(\begin{array}{c}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.01/2}}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{2}}{\rm{.58}}\end{array}\)

(1) : this is a result of

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}} \right){\rm{ = 0}}{\rm{.005}}\)

and from the appendix's normal probability table a software can also be used to calculate the probability.

Therefore, the value is \({\rm{(57}}{\rm{.5,59}}{\rm{.1)}}\).

05

Explanation


(d) For given values, the \({\rm{82\% }}\) percent confidence interval is,

\(\begin{array}{l}\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\\{\rm{ = }}\left( {{\rm{58}}{\rm{.3 - 1}}{\rm{.34 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}{\rm{,58}}{\rm{.3 + 1}}{\rm{.34 \times }}\frac{{\rm{3}}}{{\sqrt {{\rm{100}}} }}} \right)\\{\rm{ = (57}}{\rm{.9,58}}{\rm{.7)}}\end{array}\)

Where,

\(\begin{array}{c}{\rm{100(1 - \alpha ) = 82}}\\{\rm{\alpha = 0}}{\rm{.18}}\end{array}\)

and

\(\begin{array}{c}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.18/2}}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.34}}\end{array}\)

(1) : this is a result of

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.09}}}}} \right){\rm{ = 0}}{\rm{.09}}\)

and from the appendix's normal probability table a software can also be used to calculate the probability.

Therefore, the value is \({\rm{(57}}{\rm{.9,58}}{\rm{.7)}}\).

06

Explanation


(e) In order to calculate the confidence interval,

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

having width\({\rm{w}}\), the

\({\rm{the necessarysample size n}}\)

Is

\({\rm{n = }}{\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\rm{w}}}} \right)^{\rm{2}}}\)

The larger the \({\rm{n}}\), the smaller the width \({\rm{w}}\) must be.

It is given\({\rm{w = 1,}}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}{\rm{ = 2}}{\rm{.58(see(c)),n}}\)will be,

\(\begin{aligned}n &= {\left( {{\rm{2}}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\rm{w}}}} \right)^{\rm{2}}}\\ &= {\left( {{\rm{2 \times 2}}{\rm{.58 \times }}\frac{{\rm{3}}}{{\rm{1}}}} \right)^{\rm{2}}}\\ &= 239 {\rm{.62}}\end{aligned}\)

Because an integer number is required, round it up to,

\({\rm{n = 240}}\).

Therefore, the \({\rm{n = 240}}\).

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Most popular questions from this chapter

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p 藛?

Determine the values of the following quantities

\(\begin{array}{l}{\rm{a}}{\rm{.}}{{\rm{x}}^{\rm{2}}}{\rm{,1,15}}\\{\rm{b}}{\rm{.}}{{\rm{X}}^{\rm{3}}}{\rm{,125}}\\{\rm{c}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{01,25}}\\{\rm{d}}{\rm{.}}{{\rm{X}}^{\rm{2}}}{\rm{00525}}\\{\rm{e}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{9925}}\\{\rm{f}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{995,25}}\end{array}\)

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

High concentration of the toxic element arsenic is all too common in groundwater. The article 鈥淓valuation of Treatment Systems for the Removal of Arsenic from Groundwater鈥� (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152鈥�157) reported that for a sample of n = 5 water specimens selected for treatment by coagulation, the sample mean arsenic concentration was 24.3 碌g/L, and the sample standard deviation was 4.1. The authors of the cited article used t-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal.

a.Calculate and interpret a 95% CI for true average arsenic concentration in all such water specimens.

b.Calculate a 90% upper confidence bound for the standard deviation of the arsenic concentration distribution.

c.Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.
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