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 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%.

Homework Answers

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 = Z0.05 = 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 = Z0.025 = 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

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%

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