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 35 home theater systems has a mean price of ​$144.00. Assume the population standard deviation is $15.90.

A. Construct a 90% confidence interval for the population mean.

The 90% confidence interval is (__,__)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 144.00

Population standard deviation =    = 15.90

Sample size = n = 35

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 * ( 15.90 /  35 )

= 4.42

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

  ± E

144.00 ± 4.42

( 139.58, 148.42)  

We are 90% confident that the true mean of home theater systems has a mean price between $ 139.58 and $ 148.42.

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 * ( 15.90 /  35 )

= 5.27

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

  ± E

144.00 ± 5.27

( 138.73, 149.27 )  

We are 95% confident that the true mean of home theater systems has a mean price between $ 138.73 and $ 149.27.

confidence level is increases, margin of error is increases