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 ​$120.00. Assume the population standard deviation is ​$15.40. Construct a​ 90% confidence interval for the population mean.

Answer 1

Solution :

Given that,

= 120

= 15.40

n = 60

(a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (15.40 / 60)

= 3.27

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

- E < < + E

120 - 3.27 < < 120 + 3.27

116.73 < < 123.27

($116.73 < $123.27)

b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96.

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

= 1.96 * (15.40 / 60)

= 3.90

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

- E < < + E

120 - 3.90 < < 120 + 3.90

116.1 < < 123.90

($116.10 < $123.90)