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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

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

b. Suppose the investigators had made a rough guess of \({\rm{320}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of \({\rm{50}}\)ml for a confidence level of \({\rm{95\% }}\)?

Let \({\mathbf{0}} \leqslant {\mathbf{\gamma }} \leqslant {\mathbf{\alpha }}\). Then a 100(1 - 伪)% CI for 碌 when n is large is

\(\left( {\overline {\bf{x}} {\bf{ - }}{{\bf{z}}_{\bf{\gamma }}}{\bf{.}}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}{\bf{,}}\overline {\bf{x}} {\bf{ + }}{{\bf{z}}_{{\bf{\alpha - \gamma }}}}{\bf{ \times }}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}} \right)\)

The choice 蠏 =伪/2 yields the usual interval derived in Section 7.2; if 蠏 鈮� 伪/2, this interval is not symmetric about \(\overline x \). The width of this interval is \({\mathbf{w = s(}}{{\mathbf{z}}_{\mathbf{\gamma }}}{\mathbf{ + }}{{\mathbf{z}}_{{\mathbf{\alpha - \gamma }}}}{\mathbf{)/}}\sqrt {\mathbf{n}} \) .Show that w is minimized for the choice 蠏 =伪/2, so that the symmetric interval is the shortest. (Hints: (a) By definition of za, 蠒 (za) = 1 - 伪, so that za =蠒-1 (1 -伪 );

(b) the relationship between the derivative of a function y= f(x) and the inverse function x= f-1(y) is (dydy) f-1(y) = 1/f鈥�(x).)

Each of the following is a confidence interval for \({\rm{\mu = }}\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \({\rm{(114}}{\rm{.4,115}}{\rm{.6)(114}}{\rm{.1,115}}{\rm{.9)}}\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \({\rm{90\% }}\) and for the other is \({\rm{99\% }}\). Which of the intervals has the \({\rm{90\% }}\) confidence level, and why?

The article 鈥淒istributions of Compressive Strength Obtained from Various Diameter Cores鈥� (ACI Materials J., 2012: 597鈥�606) described a study in which compressive strengths were determined for concrete specimens of various types, core diameters, and length -to-diameter ratios. For one particular type, diameter, and l/d ratio, the 18 tested specimens resulted in a sample mean compressive strength of 64.41 MPa and a sample standard deviation of 10.32 MPa. Normality of the compressive strength distribution was judged to be quite plausible.

a.Calculate a confidence interval with confidence level 98% for the true average compressive strength under these circumstances.

b.Calculate a 98% lower prediction bound for the compressive strength of a single future specimen tested under the given circumstances. (Hint: t.02,17 = 2.224.)

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