Answer:
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:
Then, using a calculator, with and 169 df, we have that the two-tailed critical value is .
The margin of error is:
In this problem, , then:
The confidence interval is:
In this problem, , then:
The confidence interval is (3.25, 4.85).
Step-by-step explanation:
It should be A. Normal Distribution.
The factors of a 18 are 1, 2, 3, 6, 9, 18.
we need to get y alone so divide both sides of the equation by -3
y = (-2/3)x + 7/3
You need to set up proportions:
3/8=x/192
(3 cups to 8 ounces = x cups to 192 ounces)
cross multiply:
8x=3×192
8x=576
x=72
72 cups are needed