Question

A statistics practitioner took a random sample of 51 observations from a population whose standard deviation is 30 and computed the sample mean to be 99.

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 = = 99

Population standard deviation =    = 30

Sample size = n = 51

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 * ( 30 /  51 )

= 8.23

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

- E < < + E

99 - 8.23 <   < 99 + 8.23

90.77 <   < 107.23

( 90.77 , 107.23 )

The 95% Lower confidence interval is 90.77

The 95% Upper confidence interval is 107.23

B.

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 * ( 30 /  51 )

= 6.91

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

- E < < + E

99 - 6.91 <   < 99 + 6.91

92.09 <   < 105.91

( 92.09 , 105.91 )

The 90% Lower confidence interval is 92.09

The 90% Upper confidence interval is 105.91

C.

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z_{0.005 = 2.576}

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

= 2.576 * ( 30 /  51 )

= 10.82

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

- E < < + E

99 - 10.82 <   < 99 + 10.82

88.18 <   < 109.82

( 88.18 , 109.82 )

The 99% Lower confidence interval is 88.18

The 99% Upper confidence interval is 109.82