Question

Fifteen different Thursday evening programs reported that a commercial cost an average of $169,000 with a standard deviation of $81,000. What is the 95% confidence interval for the true mean

Answer 1

The 95% confidence interval for the true mean is between $0 and $342,729

Explanation:

We have the standard deviation for the sample, so we use the 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 = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.1448

The margin of error is:

M = T*s = 2.1448*81000 = 173,729.

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 169,000 - 173,729 = -4,... = $0(cannot be negative)

The upper end of the interval is the sample mean added to M. So it is 169,000 + 173,729 = $342,729

The 95% confidence interval for the true mean is between $0 and $342,729