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 (□.□).
(round to two decimal places as needed.)

Explanation:

Step1: Find the critical value \( z_{\alpha/2} \) for 90% confidence

For 90% confidence, \( \alpha=1 - 0.90=0.10 \), and \( \alpha/2 = 0.05 \).
From the standard normal distribution table, \( z_{0.05}=1.645 \).

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 \( n = 40 \), \( \sigma=19.50 \), and \( z_{\alpha/2}=1.645 \).
\( E=1.645\times\frac{19.50}{\sqrt{40}}\)
\(=\frac{1.645\times19.50}{6.324555}\)
\(=\frac{32.0775}{6.324555}\approx5.07\)

Step3: Construct the confidence interval

The confidence interval formula is \( \bar{x}-E<\mu <\bar{x} + E \).
Given \( \bar{x}=132 \).
\( 132-5.07<\mu<132 + 5.07\)
\( 126.93<\mu<137.07\)

Answer:

The 90% confidence interval is \((126.93,137.07)\)