Question

use the information to construct 90% and 99% confidence intervals for the population mean. interpret the results and compare the widths of the confidence intervals. researchers surveyed a random sample of 32 u.s. workers ages 16 years and over who did not work from home and asked how long (in minutes) it takes them to get from home to work. the responses are listed. 32 46 34 46 24 28 28 40 18 42 16 38 28 37 34 14 71 16 5 57 25 53 32 71 41 32 18 44 35 26 60 85 from past studies, the researchers assume that σ is 18.0 minutes. construct the 90% and 99% confidence intervals for the population mean. which interval is wider? if convenient, use technology to construct the confidence intervals. the 90% confidence interval is (□,□). (round to one decimal place as needed.)

Explanation:

Step1: Calculate the sample mean

First, sum up all the data values:

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The sample size \(n = 32\). The sample mean \(\bar{x}=\frac{1216}{32}=38\)

Step2: Find the critical values

For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2=0.05\). The critical value \(z_{\alpha/2}=z_{0.05}\approx1.645\) (from the standard normal distribution table)
For a \(99\%\) confidence interval, the significance level \(\alpha = 1- 0.99=0.01\), and \(\alpha/2=0.005\). The critical value \(z_{\alpha/2}=z_{0.005}\approx2.576\)

Step3: Calculate the margin of error for \(90\%\) confidence interval

The formula for the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 18\), \(n = 32\)
\(E_{90}=1.645\times\frac{18}{\sqrt{32}}\approx1.645\times\frac{18}{5.6568}\approx1.645\times3.182\approx5.2\)

Step4: Calculate the \(90\%\) confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
\(38-5.2<\mu<38 + 5.2\)
\(32.8<\mu<43.2\)

Answer:

The \(90\%\) confidence interval is \((32.8,43.2)\)