Question

A random sample of 42 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 74.5 and 5.6, respectively. Assume that the population is normally distributed.

a. Construct the 90% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

b. Construct the 99% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

c. Use your answers to discuss the impact of the confidence level on the width of the interval.

• The width of the interval decreases with the confidence level.

• The width of the interval increases with the confidence level.

Answer 1

a)

90% confidence interval for is

(n-1 ) S² / /2 < < (n-1 ) S² / 1-/2

(42-1) * 5.6² / 56.942 < < (42-1) * 5.6² / 27.326

22.58 < < 47.05

90% CI is ( 22.58 , 47.05)

b)

90% confidence interval for is

(n-1 ) S² / /2 < < (n-1 ) S² / 1-/2

(42-1) * 5.6² / 68.053< < (42-1) * 5.6² / 21.421

18.89 < < 60.02

99% CI is (18.89, 60.02 )

c)

The width of confidence interval increases with confidence level.

This happens due to 99% confidence interval has more accuracy than 90% confidence interval.