Question

A supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal.

Answer 1

Explanation:

Our aim is to determine a 98% confidence interval for the mean repair cost for the dryers

Number of samples. n = 25

Mean, u = $93.36

Standard deviation, s = $19.95

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula,

Confidence interval

= mean ± z × standard deviation/√n

It becomes

93.36 ± 2.33 × 19.95/√25

= 93.36 ± 2.33 × 3.99

= 93.36 ± 9.2967

The lower boundary of the confidence interval is 93.36 - 9.2967 =84.0633

The upper boundary of the confidence interval is 93.36 + 9.2967 = 102.6567

Therefore, with 98% confidence interval, the mean repair costs for the dryers is between $84.0633 and $102.6567

Answer 2

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Explanation:

We have a small sample size n = 25,
`\bar{x} = 93.36` and s = 19.95. The confidence interval is given by
`\bar{x}\pm t_(\alpha/2)((s)/(√(n)))` where
`t_(\alpha/2)` is the
`\alpha/2`th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that
`\alpha = 0.02` and the confidence interval is
`93.36\pm t_(0.01)((19.95)/(√(25)))` where
`t_(0.01)` is the 1st quantile of the t distribution with 24 df, i.e.,
`t_(0.01) = -2.4922`. Then, we have
`93.36\pm (-2.4922)((19.95)/(√(25)))` and the 98% confidence interval is given by (83.4161, 103.3039).