Keyshira Thompson
$11/26/238:25$ PM
Question 3, 9.3.9
HW Score: $43.18\mathrm{\%},4.32$ of 10
Part 1 of 2
points
Points: 0 of 1
Save
A simple random sample of size $n=40$ is drawn from a population. The sample mean is found to be $\overline{x}=121.7$ and the sample standard deviation is found to be $s=12.5$. Construct a $99\mathrm{\%}$ confidence interval for the population mean.

The lower bound is $◻$. (Round to two decimal places as needed.)

Step 1 ：We are given a simple random sample of size $n=40$ drawn from a population. The sample mean is found to be $\overline{x}=121.7$ and the sample standard deviation is found to be $s=12.5$. We are asked to construct a $99\mathrm{\%}$ confidence interval for the population mean.

Step 2 ：The formula for a confidence interval is $\overline{x}\pm Z\frac{s}{\sqrt{n}}$, where $\overline{x}$ is the sample mean, $Z$ is the Z-score corresponding to the desired level of confidence, $s$ is the sample standard deviation, and $n$ is the sample size.

Step 3 ：For a $99\mathrm{\%}$ confidence interval, the Z-score is $Z=2.576$.

Step 4 ：Substituting the given values into the formula, we get the lower bound of the confidence interval as $\overline{x}-Z\frac{s}{\sqrt{n}}$.

Step 5 ：Substituting the given values, we get $121.7-2.576\frac{12.5}{\sqrt{40}}$.

Step 6 ：Solving the above expression, we get the lower bound of the confidence interval as approximately 116.61.

Step 7 ：Final Answer: The lower bound of the 99% confidence interval for the population mean is approximately $\boxed{{\displaystyle 116.61}}$.