1a The monthly incomes of 13 randomly selected individuals w

1a) The monthly incomes of 13 randomly selected individuals who have recently graduated with a bachelor\'s degree in economics have a sample standard deviation of $347. Construct a confidence interval for the population variance ² and the population standard deviation Use a 99% level of confidence. Assume the sample is from a normally distributed population.

What is the confidence interval for the population variance ²?

(        ,      ) Round to the nearest interger as needed.

What is the confidence interval for the population standard deviation ?

(          ,             ) Round to the nearest interger as needed.

1b) The monthly incomes of 13 randomly selected individuals who have recently graduated with a bachelor\'s degree in economics have a sample standard deviation of $348. Construct a confidence interval for the population variance ² and the population standard deviation . Use a 95% level of confidence. Assume the sample is from a normally distributed population.

What is the confidence interval for the population variance ²?

(        ,      ) Round to the nearest interger as needed.

What is the confidence interval for the population standard deviation ?

(          ,             ) Round to the nearest interger as needed.

Solution

Confidence Interval
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 =0.01
^2 right = (1 - Confidence Level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
^2 left = 1 - ^2 right = 1 - 0.005 = 0.995
the two critical values ^2 left, ^2 right at 12 df are 28.2995 , 3.074
S.D( S^2 )=347
Sample Size(n)=13
Confidence Interval = [ 12 * 120409/28.2995 < ^2 < 12 * 120409/3.074 ]
= [ 1444908/28.2995 < ^2 < 1444908/3.0738 ]
Confidence Interval Variance= [ 51057.7219 , 470072.2233 ]       

C.I for S.D = [ Sqrt(51057.7219) < < Sqrt(470072.2233) ] = [ 225.95 < < 685.618 ]