Question

A person measures the contents of 25 pop cans and finds the mean content to be 12.1 fluid ounces with a standard deviation of 0.2 ounces. Construct a 99% confidence interval for the average fluid content of a can.

Answer 1

The 99% confidence interval for the average fluid content of a can is between 11.54 and 12.66 fluid ounces.

Explanation:

We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995([tex]t_(995)`). So we have T = 2.797

The margin of error is:

M = T*s = 2.797*0.2 = 0.56

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.56 = 11.54 fluid ounces.

The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.56 = 12.66 fluid ounces.

The 99% confidence interval for the average fluid content of a can is between 11.54 and 12.66 fluid ounces.