Question

A CI is desired for the true average stray-load loss ? (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with

? = 2.8. (Round your answers to two decimal places.)(a) Compute a 95% CI for ? when n = 25 and x = 51.4.

 

(b) Compute a 95% CI for ? when n = 100 and x = 51.4.

(c) Compute a 99% CI for ? when n = 100 and x = 51.4.

(d) Compute an 82% CI for ? when n = 100 and x = 51.4.

(e) How large must n be if the width of the 99% interval for ? is to be 1.0? (Round your answer up to the nearest whole number.)

Answer 1

Concepts and reason

Confidence interval:

A range of values such that the population parameter can expected to contain for the given confidence level is termed as the confidence interval. In other words, it can be defined as an interval estimate of the population parameter which is calculated for the given data based on a point estimate and for the given confidence level.

Moreover, the confidence level indicates the possibility that the confidence interval can contain the population parameter. Usually, the confidence level is denoted by . The value is chosen by the researcher. Some of the most common confidence levels are 90%, 95%, and 99%.

Confidence interval interpretation:

It is $100\left( {1 - \alpha } \right)\%$ confident that the population means lies within the confidence interval.

Mean: Sum of observations is divided by number of observations in the sample is named as mean, which is denoted by . Moreover, the sample mean signifies the measure of center of the data.

Fundamentals

The formula for the sample mean is as follows:

$\bar x = \frac{{\sum x }}{n}$

Margin of error:

$E = {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right)$

Confidence interval:

$\bar x \pm E = \bar x \pm {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right)$

The general formula for the sample size is given below:

$n = {\left( {\frac{{\sigma {Z_{\frac{\alpha }{2}}}}}{E}} \right)^2}$

(a)

From the given information, $\bar x = 51.4$ , $\sigma = 2.8$ , $n = 25$ and the level of significance, $\alpha = 0.05$ .

From the “Standard normal distribution table”, the value of ${z_{\frac{\alpha }{2}}}$ that is ${z_{\left( {\frac{{0.05}}{2}} \right)}}$ is $\pm 1.96$ . The 95% confidence interval is,

$\begin{array}{c}\\\bar x \pm {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right) = 51.4 \pm 1.96\left( {\frac{{2.8}}{{\sqrt {25} }}} \right)\\\\ = 51.4 \pm 1.96\left( {0.56} \right)\\\\ = \left( {51.4 - 1.0976,{\rm{ }}51.4 + 1.0976} \right)\\\\ = \left( {50.3024,{\rm{ }}52.4976} \right)\\\\ \approx \left( {50.30,{\rm{ }}52.50} \right)\\\end{array}$

(b)

From the given information, $\bar x = 51.4$ , $\sigma = 2.8$ , $n = 100$ and the level of significance, $\alpha = 0.05$ .

From the “Standard normal distribution table”, the value of ${z_{\frac{\alpha }{2}}}$ that is ${z_{\left( {\frac{{0.05}}{2}} \right)}}$ is $\pm 1.96$ . The 95% confidence interval is,

$\begin{array}{c}\\\bar x \pm {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right) = 51.4 \pm 1.96\left( {\frac{{2.8}}{{\sqrt {100} }}} \right)\\\\ = 51.4 \pm 1.96\left( {0.28} \right)\\\\ = \left( {51.4 - 0.5488,{\rm{ }}51.4 + 0.5488} \right)\\\\ = \left( {50.8512,{\rm{ }}51.9488} \right)\\\\ \approx \left( {50.85,{\rm{ }}51.95} \right)\\\end{array}$

(c)

From the given information, $\bar x = 51.4$ , $\sigma = 2.8$ , $n = 100$ and the level of significance, $\alpha = 0.01$ .

From the “Standard normal distribution table”, the value of ${z_{\frac{\alpha }{2}}}$ that is ${z_{\left( {\frac{{0.01}}{2}} \right)}}$ is $\pm 2.576$ . The 99% confidence interval is,

$\begin{array}{c}\\\bar x \pm {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right) = 51.4 \pm 2.576\left( {\frac{{2.8}}{{\sqrt {100} }}} \right)\\\\ = 51.4 \pm 2.576\left( {0.28} \right)\\\\ = \left( {51.4 - 0.7213,{\rm{ }}51.4 + 0.7213} \right)\\\\ = \left( {50.67872,{\rm{ }}52.12128} \right)\\\\ \approx \left( {50.68,{\rm{ }}52.12} \right)\\\end{array}$

(d)

From the given information, $\bar x = 51.4$ , $\sigma = 2.8$ , $n = 100$ and the level of significance, $\alpha = 0.18$ .

From the “Standard normal distribution table”, the value of ${z_{\frac{\alpha }{2}}}$ that is ${z_{\left( {\frac{{0.18}}{2}} \right)}}$

$\begin{array}{c}\\{z_{\left( {\frac{{0.18}}{2}} \right)}} = Norm\sin v\left( {0.18/2} \right)\\\\ = 1.341\\\end{array}$

The 82% confidence interval is,

$\begin{array}{c}\\\bar x \pm {z_{\frac{\alpha }{2}}}\left( {\frac{s}{{\sqrt n }}} \right) = 51.4 \pm 1.341\left( {\frac{{2.8}}{{\sqrt {100} }}} \right)\\\\ = 51.4 \pm 1.341\left( {0.28} \right)\\\\ = \left( {51.4 - 0.37548,{\rm{ }}51.4 + 0.37548} \right)\\\\ = \left( {51.02452,{\rm{ }}51.77548} \right)\\\\ \approx \left( {51.02,{\rm{ }}51.78} \right)\\\end{array}$

(e)

From the given information, $\sigma = 2.8$ , ${Z_{\frac{\alpha }{2}}} = 2.576$ and the width of the 99% interval for population mean is to be 1.0. This implies the difference of $\left( {\bar x - E,{\rm{ }}\bar x + E} \right)$ is 1.

Therefore, $E = 0.5$

From the Standard Normal Table, the required ${Z_{0.01}}$ value for 99% confidence level is 2.576.

The numbers of stray-load is obtained below:

$\begin{array}{c}\\n = {\left( {\frac{{\sigma {Z_{\frac{\alpha }{2}}}}}{E}} \right)^2}\\\\ = {\left( {\frac{{2.8 \times 2.576}}{{0.5}}} \right)^2}\\\\ = 208.0656\\\\ \approx 209\\\end{array}$

Ans: Part a

The 95% confidence interval for true average stray-load loss is $\left( {50.30,{\rm{ }}52.50} \right)$ .

Part b

The true average stray-load loss at 95% confidence level is $\left( {50.85,{\rm{ }}51.95} \right)$

Part c

The true average stray-load loss at 99% confidence level is $\left( {50.68,{\rm{ }}52.12} \right)$ .

Part d

The true average stray-load loss at 82% confidence level is $\left( {51.02,{\rm{ }}51.78} \right)$

Part e

The sample size is 209.