1a The waiting times in minutes of a random sample of 22 peo

1a) The waiting times (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 4.6 minutes. Construct a confidence interval for the population variance ² and the population standard deviation . Use a 90% level of confidence. Assume the sample is from a normally distributed population.

What is the confidence interval for the population variance ²?

(        ,      ) Round to one decimal as needed

What is the confidence interval for the population standard deviation ?

(          ,             ) Round to one decimal as needed

1b) Use technology to construct the confidence intervals for the population variance ² and the population standard deviation . Assume the sample is taken from a normally distributed population.

c=0.99, s=35, n-16

The confidence interval for the population variance is

(           ,          ) Round to two decimal places as needed

The confidence interval for the population standard deviation is

(           ,          ) Round to two decimal places as needed

Solution

1a)

CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.1
^2 right = (1 - Confidence Level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 21 df are 32.6706 , 11.591
S.D( S^2 )=4.6
Sample Size(n)=22
Confidence Interval for Variance= [ 21 * 21.16/32.6706 < ^2 < 21 * 21.16/11.591 ]
= [ 444.36/32.6706 < ^2 < 444.36/11.5913 ]
= [ 13.6012 < ^2 < 38.3356 ]       

Confidence Interval for S.D = [ Sqrt(13.6012) < < Sqrt(38.3356) ] = [ 3.687 < < 6.1916 ]