Question

A statistics practitioner took a random sample of 51 observations from a population whose standard deviation is 28 and computed the sample mean to be 108. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.

A. Estimate the population mean with 95% confidence. Confidence Interval =

B. Estimate the population mean with 90% confidence. Confidence Interval =

C. Estimate the population mean with 99% confidence. Confidence Interval =

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 108

Population standard deviation = = 28

Sample size = n = 51

A)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (28 / 51 )

= 7.68

At 95% confidence interval estimate of the population mean is,

- E < < + E

108 - 7.68 < < 108 + 7.68

100.32 < < 115.68

Confidence Interval = (100.32 , 115.68)

B)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (28 / 51 )

= 6.45

At 90% confidence interval estimate of the population mean is,

- E < < + E

108 - 6.45 < < 108 + 6.45

101.55 < < 114.45

Confidence Interval = (101.55 , 114.45)

C)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (28 / 51 )

= 10.10

At 99% confidence interval estimate of the population mean is,

- E < < + E

108 - 10.10 < < 108 + 10.10

97.90 < < 118.10

Confidence Interval = (97.90 , 118.10)