Toke-Home-Ouiz Question 3 of 10 (i) point) I Question Attempt 1 of Unilmited Suppose that the population of the number of hours slept by all college students the night before finals is approximately normally distributed. A report claimed that the mean of this population is 6.75 hours. As a student wellness advocate, you want to test this claim, so you select a random sample of 19 college students and record the number of hours each slept the night before finals. Follow the steps below to construct a $99 \%$ confidence interval for the population mean of all the numbers of hours slept by college students the night before finals. Then state whether the confidence interval you construct contradicts the report's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample \begin{tabular}{|c|c|c|} \hline Number of students & Sample mean & \begin{tabular}{c} Sample standard \\ deviation \end{tabular} \\ \hline 19 & 6.504 & 1.421 \\ \hline \end{tabular} Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your $99 \%$ confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". \begin{tabular}{|l|l|} \hline \begin{tabular}{l} Sample size: \\ \hline \end{tabular} & Standard error: \\ \hline \begin{tabular}{l} Point estimate: \\ $\square$ \end{tabular} & \\ \hline \begin{tabular}{l} Sample standard deviation: \\ $\square$ \end{tabular} & Margin of error: \\ \hline \begin{tabular}{l} Critical value: \\ $\square$ \end{tabular} & $99 \%$ confidence interval: \\ \hline Compute \\ \hline \end{tabular} \begin{tabular}{|l|} \hline Critical values \\ \hline$t_{0.005}=2.878$ \\ \hline$t_{0.010}=2.552$ \\ \hline$t_{0.025}=2.101$ \\ \hline$t_{0.050}=1.734$ \\ \hline$t_{0.100}=1.330$ \\ \hline \end{tabular} (h) Raced on vour camole aranh the $99 \%$ confidence interval for the nonulation mean of all the numbers of hours elent hu collene students the ninht Check Save For Later Submit Assignment
Solution
Step 1 ：Calculate the standard error using the formula \( SE = \frac{s}{\sqrt{n}} \) where \( s \) is the sample standard deviation and \( n \) is the sample size.
Step 2 ：Calculate the margin of error using the formula \( ME = t \times SE \) where \( t \) is the critical value.
Step 3 ：Find the lower bound of the confidence interval using the formula \( \text{lower bound} = \bar{x} - ME \) where \( \bar{x} \) is the sample mean.
Step 4 ：Find the upper bound of the confidence interval using the formula \( \text{upper bound} = \bar{x} + ME \).
Step 5 ：The \( 99\% \) confidence interval for the population mean of all the numbers of hours slept by college students the night before finals is \( \boxed{(5.566, 7.442)} \).
Step 6 ：This confidence interval does not include the reported mean of 6.75 hours, which suggests that the report's claim may not be accurate.
