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 40 home theater systems has a mean price of $132.00. assume the population standard deviation is $19.50.
construct a 90% confidence interval for the population mean.
the 90% confidence interval is (126.93, 137.07).
(round to two decimal places as needed.)
construct a 95% confidence interval for the population mean.
the 95% confidence interval is (□,□).
(round to two decimal places as needed.)

Explanation:

Step1: Find the z - value for 95% confidence interval

For a 95% confidence interval, the significance level \(\alpha=1 - 0.95=0.05\). Then \(\alpha/2=0.025\). Looking up in the standard normal distribution table, \(z_{\alpha/2}=z_{0.025} = 1.96\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(n = 40\), \(\sigma=19.50\), \(z_{\alpha/2}=1.96\)

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x} + E\). Given \(\bar{x}=132\)

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Answer:

The 95% confidence interval is \((125.96,138.04)\)