Respuesta :
Using the t-distribution, it is found that the 99% confidence interval for the population mean is: (100.61, 111.07).
We suppose that:
- The sample mean is of [tex]\overline{x} = 105.84[/tex].
- The sample standard deviation is of [tex]s = 14.27[/tex].
- We have the standard deviation for the sample, thus, the t-distribution is used to solve this question.
The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, thus [tex]df = 32 - 1 = 31[/tex].
Then, we find the critical value for a 99% confidence interval with 31 df, which, looking at the t-table, is t = 2.0395.
The margin of error is:
[tex]M = t\frac{s}{\sqrt{n}} = 2.0395\frac{14.27}{\sqrt{31}} = 5.23[/tex]
The confidence interval is:
[tex]\overline{x} \pm M[/tex]
Then
[tex]\overline{x} - M = 105.84 - 5.23 = 100.61[/tex]
[tex]\overline{x} + M = 105.84 + 5.23 = 111.07[/tex]
The 99% confidence interval for the population mean is (100.61, 111.07).
A similar problem is given at https://brainly.com/question/14885491
