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 45 45 home theater systems has a mean price of $ 137.00 137.00. Assume the population standard deviation is $ 16.60 16.60. Construct a 90% confidence interval for the population mean. The 90% confidence interval is ( nothing , nothing ). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is ( nothing , nothing ). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. B. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%. C. With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Answer #1

Solution :

Given that,

Point estimate = sample mean =
= 137.00

Population standard deviation =
= 16.60

Sample size = n = 45

1) 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 * ( 16.60 /
45)

= 4.07

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

± E

137.00 ± 4.07

( 132.93, 141.07 )

2) 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 * ( 16.60 /
45)

= 4.85

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

± E

137.00 ± 4.85

( 132.15, 141.85)

correct option is =A

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