The image presents a problem where you are asked to construct a 99% confidence interval for the population mean based on the given information:

• Sample size $n = 40$
• Sample mean $\bar{x} = 121.1$
• Sample standard deviation $s = 13.2$

Steps to calculate the 99% confidence interval:

1. Find the critical value: Since it's a 99% confidence level, the critical value $z_{\alpha/2}$ for a two-tailed test with 99% confidence can be obtained from the standard normal distribution table. For 99%, $z_{\alpha/2} \approx 2.576$.

2. Calculate the standard error of the mean (SE): $SE = \frac{s}{\sqrt{n}} = \frac{13.2}{\sqrt{40}} \approx 2.087$

3. Calculate the margin of error (ME): $ME = z_{\alpha/2} \times SE = 2.576 \times 2.087 \approx 5.376$

4. Find the confidence interval: The confidence interval is given by: $\bar{x} \pm ME = 121.1 \pm 5.376$

   • Lower bound: $121.1 - 5.376 = 115.724$, rounded to two decimal places: 115.72
   • Upper bound: $121.1 + 5.376 = 126.476$, rounded to two decimal places: 126.48

Final Answer:

• Lower bound: 115.72
• Upper bound: 126.48

This differs slightly from the answer shown in the image (115.45 and 123.71), possibly due to different critical values or rounding at intermediate steps.

Would you like further clarification, or have any questions?

Here are 5 related questions for further exploration:

1. What is the significance of a 99% confidence interval?
2. How does the sample size affect the width of the confidence interval?
3. Why is the critical value $z_{\alpha/2}$ different for varying confidence levels?
4. How does using the sample standard deviation instead of the population standard deviation affect the result?
5. What is the difference between a confidence interval and a prediction interval?

Tip: When calculating confidence intervals, remember that larger sample sizes reduce the standard error, leading to narrower intervals and more precise estimates.