Question

Historical delivery times for Jackson Trucking, Inc. have had a mean of 3 hours and a standard deviation of 0.5 hours. A sample of 22 deliveries over the past month provides a sample mean of 3.1 hours and a sample standard deviation of 0.75 hours. Compute 95% confidence interval for the population variance. Select one: A. 0.5770 21.0718 B. 0.3329 21.1487 C. 0.4439 21.5317 D. 0.6663 21.2376

I know the answer is B

(n-1)s2 / x2 mui/2 <= o2 <= (n-1) s2 / X2 1- mui/2

(22-1)0.75sqrt / 35.479 <= o2 <= (22-1)0.75sqrt / 10.283

= 0.3329 <= o2 <= 1.1487

how do you get the 35.479 and 10.283?????

Answer 1

The formula for the confidence interval is as follows:

The values 35.479 and 10.283 are the upper tail critical chi-square value and lower tail critical chi-square value respectively

Alpha = 0.05, Upper tail = 0.975, Lower tail = 0.025

We need to look into the critical chi-square table at df = n - 1 = 22 - 1 = 21:

Upper tail critical value = 35.479 at df = 21 and 1 - alpha/2 = 0.975

Lower tail critical value at df = 21 and alpha/2 = 0.025 is 10.283