Question

You may need to use the appropriate appendix table or technology to answer this question.

A sample of 22 items provides a sample standard deviation of 5.

(a)Compute the 90% confidence interval estimate of the population variance. (Round your answers to two decimal places.)

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(b)Compute the 95% confidence interval estimate of the population variance. (Round your answers to two decimal places.)

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(c) Compute the 95% confidence interval estimate of the population standard deviation. (Round your answers to one decimal place.)

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Answer 1

Confidence interval for population variance

Given,

Sample size : n=22

sample standard deviation : s = 5

(a) 90% confidence interval estimate of the population variance.

for 90% confidence level = (100-90)/100 =0.1

/2 = 0.1/2 =0.05

1-0.05=0.95

Degrees of freedom = n-1=22-1=21

90% confidence interval estimate of the population variance.

90% confidence interval estimate of the population variance. = (3.21) to (9.06)

(b) 95% confidence interval estimate of the population variance.

for 95% confidence level = (100-95)/100 =0.05

/2 = 0.05/2 =0.025

1-0.025=0.975

Degrees of freedom = n-1=22-1=21

95% confidence interval estimate of the population variance.

95% confidence interval estimate of the population variance. = (2.96) to (10.21)

(c) 95% confidence interval estimate of the population standard deviation.

95% confidence interval estimate of the population standard deviation = (1,72) to (3.20)