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Allushta [10]
3 years ago
15

A supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 a

re subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal.
Mathematics
2 answers:
Feliz [49]3 years ago
3 0

Answer:

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

Step-by-step 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}(\frac{s}{\sqrt{n}}) where t_{\alpha/2} is the \alpha/2th 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}(\frac{19.95}{\sqrt{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)(\frac{19.95}{\sqrt{25}}) and the 98% confidence interval is given by (83.4161, 103.3039).

horsena [70]3 years ago
3 0

Answer:

Step-by-step 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

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