Question

# You are given the sample mean and the population standard deviation. Use this information to construct...

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 ​\$135.00 . Assume the population standard deviation is ​\$15.90 .

A.) the 90% confidence interval is

B.) the 95% confidence interval is

C.) Which interval is wider?

Solution :

Given that,

= \$135

= \$15.90

n = 60

A)At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * (15.90 / 60)

= 0.436

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

- E < < + E

135 - 0.436 < < 135 + 0.436

134.564 < < 135.436

(134.564,135.436 )

The 90% confidence interval is \$134.564 to \$135.436

B)At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (15.90 / 60)

= 0.519

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

- E < < + E

135 - 0.519 < < 135 + 0.519

134.481 < < 135.519

(134.481,135.519 )

The 95% confidence interval is \$134.481 to \$135.519

C) The 95% interval is wider.

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