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

Question

3 years ago

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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] · Answer 1 · 2022-07-24T22:54:58+03:00

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 $\alpha/2$ 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 $93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})$ 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).