Question

online courses: a sample of 261 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the most favorable impression. the average score was 5.45, and the standard deviation was 0.90. use the ti-84 plus calculator as needed. part: 0 / 2 part 1 of 2 (a) construct a 99.5% confidence interval for the mean score. round the answers to at least two decimal places. a 99.5% confidence interval for the mean score is \\( \square < \mu < \square \\).

Explanation:

Step1: Determine the critical value

For a \(99.5\%\) confidence interval, the significance level \(\alpha = 1 - 0.995=0.005\). The critical value \(z_{\alpha/2}\) is such that the area in the two - tails is \(\alpha = 0.005\). So the area in each tail is \(\frac{\alpha}{2}=0.0025\). Looking up in the standard normal table or using a calculator (e.g., on TI - 84 Plus: invNorm(0.0025, 0, 1)), we get \(z_{\alpha/2}=2.807\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) for a confidence interval for the population mean when the population standard deviation \(\sigma\) is known (we assume the sample standard deviation \(s\) is a good estimate of \(\sigma\) since \(n = 261\) is large, \(n\geq30\)) is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 0.90\), \(n = 261\), and \(z_{\alpha/2}=2.807\). Then \(E=2.807\times\frac{0.90}{\sqrt{261}}\).
First, \(\sqrt{261}\approx16.16\), and \(\frac{0.90}{16.16}\approx0.0557\). Then \(E = 2.807\times0.0557\approx0.16\).

Step3: Construct the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu <\bar{x} + E\). Given \(\bar{x}=5.45\).
So \(5.45- 0.16<\mu<5.45 + 0.16\).

Answer:

\(5.29<\mu<5.61\)