Question

A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 15 greeting cards indicates an average value of $3.00 and a s of $0.40. The critical value for a 99% confidence interval is

Answer 1

The critical value for a 99% confidence interval is T = 2.9768

Explanation:

We have the standard deviation of the sample, which means that the t-distribution is used 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

Critical value

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 - 0.99)/(2) = 0.995`. So we have T = 2.9768

The critical value for a 99% confidence interval is T = 2.9768