To calculate a 95% confidence interval for the population mean based on the sample values, we can follow these steps:

Given information:

• Sample size $n = 9$ (since 9 values are provided)
• Population variance $\sigma^2 = 3636$
• The sample values are: $54, 46, 54, 45, 43, 51, 46, 61, 48$
• Confidence level = 95%

Steps:

1. Calculate the sample mean $\bar{x}$: $\bar{x} = \frac{54 + 46 + 54 + 45 + 43 + 51 + 46 + 61 + 48}{9}$

2. Calculate the population standard deviation $\sigma$: $\sigma = \sqrt{3636}$

3. Find the critical value $Z_{\alpha/2}$ for a 95% confidence interval:
   For a 95% confidence level, the critical value $Z_{\alpha/2}$ from the standard normal distribution is approximately 1.96.

4. Calculate the margin of error $E$: $E = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

5. Construct the confidence interval:
   The 95% confidence interval is given by: $\bar{x} - E \quad \text{to} \quad \bar{x} + E$

Now let's compute the result step-by-step.

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

$\bar{x} = \frac{54 + 46 + 54 + 45 + 43 + 51 + 46 + 61 + 48}{9} = \frac{448}{9} \approx 49.78$

Step 2: Calculate the population standard deviation $\sigma$:

$\sigma = \sqrt{3636} \approx 60.30$

Step 3: Critical value $Z_{\alpha/2}$:

For a 95% confidence level, the critical value $Z_{\alpha/2} \approx 1.96$.

Step 4: Calculate the margin of error $E$:

$E = 1.96 \times \frac{60.30}{\sqrt{9}} = 1.96 \times \frac{60.30}{3} = 1.96 \times 20.10 \approx 39.39$

Step 5: Construct the confidence interval:

$\text{Confidence interval} = (49.78 - 39.39, 49.78 + 39.39) = (10.39, 89.17)$

Conclusion:

The 95% confidence interval estimate for the population mean is approximately: $(10.39, 89.17)$

Would you like any more details or have questions?

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Here are 5 related questions to further your understanding:

1. How does increasing the confidence level affect the width of the confidence interval?
2. What is the effect of a larger sample size on the confidence interval?
3. Can you explain the role of the population variance in determining the margin of error?
4. Why do we use the $Z$-score for a normal population when the population variance is known?
5. How would the confidence interval change if the population variance were smaller?

Tip: A larger sample size decreases the margin of error, resulting in a narrower confidence interval.