Question

A small business owner has two fast food restaurants. The owner wants to determine if there is any difference in the variability of service times at the​ drive-thru window of each restaurant. A sample of size nequals9 is taken from each​ restaurant's drive-thru window. To perform a hypothesis test using the 0.05 level of​ significance, what is the critical​ value? Round to three decimal places as needed.

Answer 1

Answer: The critical value using the t-table at 0.05 level of significance is 2.306.

Explanation:

Since we have given that

n = 9

We will use t- test as n < 30.

so, the degree of freedom v = n-1 = 9-1 = 8

Now, α = 0.05

According to t-table, we get that

`t_(\alpha ,v)=t_(0.05,8)=2.306`

Hence, the critical value using the t-table at 0.05 level of significance is 2.306.