Question

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.

A random sample of 60 home theater systems has a mean price of $111.00. Assume the population standard deviation is $16.20.

The 90% confidence interval is(--,--)

The 99% confidence interval is(--,--)

Which interval is wider?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 111.00

Population standard deviation =    = 16.20

Sample size = n = 60

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 * ( 16.20 /  60)

= 3.44

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

  ± E

111.00 ± 3.44   

( 107.56, 114.44 )  

b) 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 * ( 16.20 /  60)

= 5.39

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

  ± E

111.00 ± 5.39   

( 105.61, 116.39 )  

A 99% confidence interval is wider,