Question

(1 point) A statistics practitioner took a random sample of 52 observations from a population whose standard deviation is 35 and computed the sample mean to be 96.

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 = = 96

Population standard deviation =    = 35

Sample size = n = 52

A) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 35 /  52 )

= 9.51

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

  ± E

96 ± 9.51   

( 86.49, 105.51)  

B) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * ( 35 /  52 )

= 7.98

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

  ± E

96 ± 7.98

( 88.02, 103.98)

C) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * ( 35 /  52 )

= 12.50

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

  ± E

96 ± 12.50  

( 83.50, 108.50)