Question

The waiting times​ (in minutes) of a random sample of 21 people at a bank have a sample standard deviation of 4.7 minutes. Construct a confidence interval for the population variance σ2 and the population standard deviation σ  Use a 99%

level of confidence. Assume the sample is from a normally distributed population.

What is the confidence interval for the population variance σ2​? (__,__)  ​(Round to one decimal place as​ needed.)

What is the confidence interval for the population standard deviation σ​? (__, __) ​(Round to one decimal place as​ needed.)

Answer 1

Solution :

Given that n = 21 , standard deviation s = 4.7

=> df = n - 1 = 20

=> for 99% level of confidence , c = 0.99 then p = 0.005

XR^2 = 40 , XL^2 = 7.434

=> The 99% confidence interval of population variance is

=> ((n - 1)*s^2)/XR^2 < σ^2 < ((n - 1)*s^2)/XL^2

=> (20*(4.7^2))/40 < σ^2 < (20*(4.7^2))/7.434

=> 11.0450 < σ^2 < 59.4296

=> 11.0 < σ^2 < 59.4 (rounded)

=> (11.0 , 59.4)

=> The 99% confidence interval of population standard deviation is

=> sqrt(((n - 1)*s^2)/XR^2) < σ < sqrt(((n - 1)*s^2)/XL^2)

=> sqrt((20*(4.7^2))/40) < σ < sqrt((20*(4.7^2))/7.434)

=> sqrt(11.0450) < σ < sqrt(59.4296)

=> 3.3234 < σ < 7.7091

=> 3.3 < σ < 7.7 (rounded)

=> (3.3 , 7.7)