Question

To determine the efficiency of a batch of production of electric motors, 8 motors are tested. The average efficiency of the sample is calculated to be 88.5% and the standard deviation of the sample is 0.5 percentage point. (a) Determine a 95% confidence interval for the mean efficiency of the given batch of motors. (b) If we want to cut the confidence interval to one third, how many more motors should be tested

Answer 1

The 95% confidence interval for the mean efficiency of the given batch of motors is 0.885 $\pm$ 0.418 and 41 motors should be tested to cut the confidence interval to one-third

The 95% confidence interval

The given parameters are:

• Mean, $\mu$ = 88.5%
• Standard deviation, $\sigma$ = 0.5
• Sample size, n = 8

Here, n is less than 30.

So, we use the t-score

At 95% confidence interval and n = 8, the t-score is:

t = 2.365

So, we have:

$CI = \mu \pm t\frac{\sigma}{\sqrt n}$

So, we have:

$CI = 88.5\% \pm 2.365 * \frac{0.5}{\sqrt {8}}$

$CI = 88.5\% \pm \frac{1.1825}{\sqrt(8)}$

Evaluate the quotient

CI = 88.5% $\pm$ 0.418

Express percentage as decimal

CI = 0.885 $\pm$ 0.418

Hence, the 95% confidence interval for the mean efficiency of the given batch of motors is 0.885 $\pm$ 0.418

The number of motors to be tested

When the interval is cut into one third, we have:

$CI = \frac 13 * (0.885 \pm 0.418)$

Evaluate

$CI = (0.885 \pm \frac 13 *0.418)$

$CI = (0.885 \pm 0.139)$

Recall that:

$CI = \mu \pm t\frac{\sigma}{\sqrt n}$ or $CI = \mu \pm z\frac{\sigma}{\sqrt n}$

Here, we use the z-score i.e.

$CI = \mu \pm z\frac{\sigma}{\sqrt n}$

By comparison, we have:

$z\frac{\sigma}{\sqrt n} = 0.139$

Substitute $\sigma$ = 0.5 and z = 1.960

$1.96 * \frac{0.5}{\sqrt n} = 0.139$

Evaluate the product

$\frac{0.98}{\sqrt n} = 0.139$

Solve for $\sqrt n$

$\sqrt n = \frac{0.98}{0.139}$

Evaluate the quotient

$\sqrt n = 7.1$

Square both sides

n = 49

The initial sample size is 8.

Subtract 8 from 49

Difference = 49 - 8

Difference = 41

Hence, 41 motors should be tested to cut the confidence interval to one-third

Read more about confidence intervals at:

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