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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%?

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.

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article 鈥淭esting Practices for the AC Breakdown Voltage Testing of Insulation Liquids'' (IEEE)gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.

\(\begin{array}{l}{\rm{62\;50\;53\;57\;41\;53\;55\;61\;59\;64\;50\;53\;64\;62}}\\{\rm{\;50\;68 54\;55\;57\;50\;55\;50\;56\;55\;46\;55\;53\;54\;}}\\{\rm{52\;47\;47\;55 57\;48\;63\;57\;57\;55\;53\;59\;53\;52\;}}\\{\rm{50\;55\;60\;50\;56\;58}}\end{array}\)

a. Construct a boxplot of the data and comment on interesting features.

b. Calculate and interpret a \({\rm{95\% }}\)CI for true average breakdown voltage m. Does it appear that m has been precisely estimated? Explain.

c. Suppose the investigator believes that virtually all values of breakdown voltage are between \({\rm{40 and 70}}\). What sample size would be appropriate for the \({\rm{95\% }}\)CI to have a width of 2 kV (so that m is estimated to within 1 kV with \({\rm{95\% }}\)confidence)

a.Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter 位 of an exponential distribution, and calculate the bound based on the data given in the example.

b.If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P(X>t) = e-位迟. Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

The Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?
See all solutions

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