Using the t-distribution, the 99% confidence interval for the true average number of alcoholic drinks all UF female students (over 21) have in a one week period is (3.25, 4.85).
We have the standard deviation for the sample, thus, the t-distribution is used to solve this question.
First, we find the number of degrees of freedom, which is the sample size subtracted by 1, thus:
[tex]df = 170 - 1 = 169[/tex]
Then, using a calculator, with [tex]\alpha = 1 - 0.01 = 0.99[/tex] and 169 df, we have that the two-tailed critical value is [tex]t = 2.605[/tex].
The margin of error is:
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
In this problem, [tex]s = 4, n = 170[/tex], then:
[tex]M = 2.605\frac{4}{\sqrt{170}} = 0.8[/tex]
The confidence interval is:
[tex]\overline{x} \pm M[/tex]
In this problem, [tex]\overline{x} = 4.05[/tex], then:
[tex]\overline{x} - M = 4.05 - 0.8 = 3.25[/tex]
[tex]\overline{x} + M = 4.05 + 0.8 = 4.85[/tex]
The confidence interval is (3.25, 4.85).
A similar problem is given at https://brainly.com/question/22596713
