Problem
https://virtualback... MATH1127: Introduction to Statistics (60022) Lesson 8.2 A Single Population Mean using the Student t-Distribution Unit 3 Chapter 8 : Uuestion $\angle U$ Suppose you are a researcher in a hospital. You are experimenting with a new sedative. You collect data from a random sample of 9 patients. The period of effectiveness of the sedative for each patient (in hours) is as follows: \begin{tabular}{|c|} \hline Hours \\ \hline 2.4 \\ \hline 2.1 \\ \hline 2.9 \\ \hline 2.6 \\ \hline 2.7 \\ \hline 2 \\ \hline 2.4 \\ \hline 2 \\ \hline 2.1 \\ \hline \end{tabular} a) What is a point estimate for the population mean length of time? (Round answer to 4 decimal places) b) What must be true in order to construct a confidence interval in this situation? Select an answer c) Construct a $99 \%$ confidence interval for the population mean length of time. Enter your answer as an open-interval (i.e., parentheses example (5.2314,8.1245)) Round upper and lower bounds to 4 decimal places.
Solution
Step 1 ：Given data is [2.4, 2.1, 2.9, 2.6, 2.7, 2, 2.4, 2, 2.1].
Step 2 ：The point estimate for the population mean length of time can be calculated by finding the average of the given data.
Step 3 ：The formula to calculate the mean is \(\frac{\sum x}{n}\), where \(\sum x\) is the sum of all the values and \(n\) is the number of values.
Step 4 ：Sum of all the values is \(2.4 + 2.1 + 2.9 + 2.6 + 2.7 + 2 + 2.4 + 2 + 2.1 = 20.2\).
Step 5 ：The number of values is 9.
Step 6 ：Substituting these values into the formula, we get \(\frac{20.2}{9} = 2.355555555555556\).
Step 7 ：Rounding this to 4 decimal places, we get 2.3556.
Step 8 ：So, the point estimate for the population mean length of time is \(\boxed{2.3556}\).
