MATH SOLVE

3 months ago

Q:
# A study on students drinking habits asks a random sample of 170 female UF students (over 21) how many alcoholic beverages they have consumed in the past week. The sample reveals an average of 4.05 alcoholic drinks, with a standard deviation of 4.66. Construct a 99% confidence interval for the true average number of alcoholic drinks all UF female students (over 21) have in a one week period.

Accepted Solution

A:

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