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, [tex]\bar{x} = 93.36[/tex] and s = 19.95. The confidence interval is given by [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]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 [tex]\alpha = 0.02[/tex] and the confidence interval is [tex]93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})[/tex] where [tex]t_{0.01}[/tex] is the 1st quantile of the t distribution with 24 df, i.e., [tex]t_{0.01} = -2.4922[/tex]. Then, we have [tex]93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})[/tex] and the 98% confidence interval is given by (83.4161, 103.3039).
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