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 (31.5, 42.0). (round to one decimal place as needed.) the 99% confidence interval is (28.6, 44.9). (round to one decimal place as needed.) which interval is wider? the 99% confidence interval the 90% confidence interval

Explanation:

Step1: Recall confidence interval formula

The formula for a confidence interval for the population mean when the population standard deviation \(\sigma\) is known is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Here, \(n = 32\), \(\sigma=18.0\).

Step2: Calculate the sample mean \(\bar{x}\)

\(\bar{x}=\frac{32 + 46+34 + 46+24 + 28+28 + 40+18 + 42+16 + 38+28 + 37+34 + 14+71 + 16+5 + 57+25 + 53+32 + 71+41 + 32+18 + 44+35 + 26+60 + 85}{32}=\frac{1176}{32}=36.75\)

Step3: Find \(z\) - values

For a \(90\%\) confidence interval, \(\alpha=1 - 0.90 = 0.10\), \(\alpha/2=0.05\), and \(z_{\alpha/2}=z_{0.05}\approx1.645\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\), and \(z_{\alpha/2}=z_{0.005}\approx2.576\)

Step4: Calculate the margin of error for \(90\%\) confidence interval

\(E_{90}=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.645\times\frac{18}{\sqrt{32}}\approx1.645\times3.182\approx5.24\)
The \(90\%\) confidence interval is \(\bar{x}\pm E_{90}=36.75\pm5.24=(31.5,42.0)\)

Step5: Calculate the margin of error for \(99\%\) confidence interval

\(E_{99}=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{18}{\sqrt{32}}\approx2.576\times3.182\approx8.15\)
The \(99\%\) confidence interval is \(\bar{x}\pm E_{99}=36.75\pm8.15=(28.6,44.9)\)

Step6: Compare the widths

The width of a confidence interval is \(2E\). For the \(90\%\) confidence interval, width \(=2\times5.24 = 10.48\). For the \(99\%\) confidence interval, width \(=2\times8.15=16.3\)

Answer:

The \(99\%\) confidence interval is wider.