Question

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

(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

Define interval

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

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 \({\rm{(57}}{\rm{.1,59}}{\rm{.5)}}\).

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 \({\rm{(57}}{\rm{.7,58}}{\rm{.9)}}\).

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 \({\rm{(57}}{\rm{.5,59}}{\rm{.1)}}\).

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 \({\rm{(57}}{\rm{.9,58}}{\rm{.7)}}\).

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