Question

A simple random sample of 50 items from a population with = 6 resulted in a sample mean of 39. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean. to

b. Provide a 95% confidence interval for the population mean. to

c. Provide a 99% confidence interval for the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 39

Population standard deviation =    = 6

Sample size = n = 50

a) 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 * (6 /  50)

= 1.40

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

  ± E

39 ± 1.40

( 37.60, 40.40)  

b) 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 * ( 6 /  50 )

= 1.66

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

  ± E

39 ± 1.66

( 37.34, 40.66)  

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 * ( 6 /  50 )

= 2.19

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

  ± E

39 ± 2.19

( 36.81, 41.19)