To solve this problem and construct the 95% confidence interval for the population mean, we will follow these steps:

Given Information:

• Sample size (n) = 99
• Population variance (σ²) = 3636
• Sample values:
  • 54, 46, 54, 45, 43, 51, 46, 61, 48
• Confidence level = 95% (0.95)

Step 1: Calculate the sample mean ($\bar{x}$)

$\bar{x} = \frac{54 + 46 + 54 + 45 + 43 + 51 + 46 + 61 + 48}{9}$

$\bar{x} = \frac{448}{9} \approx 49.78$

Step 2: Standard Deviation ($\sigma$) from the population variance

The population variance ($\sigma^2$) is given as 3636, so the standard deviation is: $\sigma = \sqrt{3636} = 60.29$

Step 3: Calculate the standard error of the mean (SE)

Since we know the population variance, we use the standard deviation and the sample size to calculate the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}} = \frac{60.29}{\sqrt{99}} \approx \frac{60.29}{9.95} \approx 6.06$

Step 4: Find the z-value for a 95% confidence interval

For a 95% confidence interval, the z-value (z*) is 1.96.

Step 5: Calculate the margin of error (ME)

$ME = z^* \times SE = 1.96 \times 6.06 \approx 11.88$

Step 6: Compute the confidence interval

The confidence interval is given by: $\left( \bar{x} - ME, \bar{x} + ME \right)$ $CI = \left( 49.78 - 11.88, 49.78 + 11.88 \right) = (37.90, 61.66)$

Conclusion:

The 95% confidence interval for the population mean is approximately:

$(37.90, 61.66)$

Would you like further clarification or a deeper explanation of any of the steps?

Follow-up Questions:

1. How does population variance affect the width of the confidence interval?
2. What changes if we use a t-distribution instead of a z-distribution?
3. How would the confidence interval change if the sample size were increased?
4. Why is it important to know the population is normally distributed?
5. What does a 95% confidence interval mean in practical terms?

Tip: Larger sample sizes reduce the margin of error, leading to narrower confidence intervals.