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 50 home theater systems has a mean price of $118.00. assume the population standard deviation is $19.70. construct a 90% confidence interval for the population mean. the 90% confidence interval is ( , ) (round to two decimal places as needed.)

Explanation:

Step1: Identify the formula

The formula for the confidence - interval for the population mean when the population standard deviation $\sigma$ is known is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Determine the values of $\bar{x}$, $\sigma$, and $n$

We are given that $\bar{x} = 118$, $\sigma=19.70$, and $n = 50$.

Step3: Find the $z$ - score for a 90% confidence interval

For a 90% confidence interval, $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z$ - score $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step4: Calculate the margin of error $E$

$E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.645\times\frac{19.70}{\sqrt{50}}\approx1.645\times\frac{19.70}{7.071}\approx1.645\times2.786\approx4.58$.

Step5: Calculate the confidence interval

The lower limit is $\bar{x}-E=118 - 4.58 = 113.42$.
The upper limit is $\bar{x}+E=118 + 4.58 = 122.58$.

Answer:

$(113.42,122.58)$