A statistics practitioner took a random sample of 53 observations from a population whose standard deviation is 35 and computed the sample mean to be 95.

Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.

A. Estimate the population mean with 95% confidence.

Confidence Interval =

B. Estimate the population mean with 90% confidence.

Confidence Interval =

C. Estimate the population mean with 99% confidence.

Confidence Interval =

To estimate the population mean with different confidence levels, we will use the formula for the confidence interval of the mean when the population standard deviation (\(\sigma\)) is known. The formula is:

\[ \text{CI} = \left( \bar{x} - Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}, \bar{x} + Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \right) \]

Where:
- \(\bar{x}\) is the sample mean.
- \(Z_{\alpha/2}\) is the critical value from the standard normal distribution for the desired confidence level.
- \(\sigma\) is the population standard deviation.
- \(n\) is the sample size.

Given:
- Sample mean, \(\bar{x} = 95\)
- Population standard deviation, \(\sigma = 35\)
- Sample size, \(n = 53\)

### A. 95% Confidence Interval

For a 95% confidence level, the critical value \(Z_{\alpha/2}\) is approximately 1.96.

The confidence interval is:

\[ \text{CI}_{95} = \left( 95 - 1.96 \cdot \frac{35}{\sqrt{53}}, 95 + 1.96 \cdot \frac{35}{\sqrt{53}} \right) \]

Calculate the margin of error:

\[ \text{Margin of Error} = 1.96 \cdot \frac{35}{\sqrt{53}} \approx 9.42 \]

So the confidence interval is:

\[ \text{CI}_{95} = (95 - 9.42, 95 + 9.42) = (85.58, 104.42) \]

### B. 90% Confidence Interval

For a 90% confidence level, the critical value \(Z_{\alpha/2}\) is approximately 1.645.

The confidence interval is:

\[ \text{CI}_{90} = \left( 95 - 1.645 \cdot \frac{35}{\sqrt{53}}, 95 + 1.645 \cdot \frac{35}{\sqrt{53}} \right) \]

Calculate the margin of error:

\[ \text{Margin of Error} = 1.645 \cdot \frac{35}{\sqrt{53}} \approx 7.90 \]

So the confidence interval is:

\[ \text{CI}_{90} = (95 - 7.90, 95 + 7.90) = (87.10, 102.90) \]

### C. 99% Confidence Interval

For a 99% confidence level, the critical value \(Z_{\alpha/2}\) is approximately 2.576.

The confidence interval is:

\[ \text{CI}_{99} = \left( 95 - 2.576 \cdot \frac{35}{\sqrt{53}}, 95 + 2.576 \cdot \frac{35}{\sqrt{53}} \right) \]

Calculate the margin of error:

\[ \text{Margin of Error} = 2.576 \cdot \frac{35}{\sqrt{53}} \approx 12.38 \]

So the confidence interval is:

\[ \text{CI}_{99} = (95 - 12.38, 95 + 12.38) = (82.62, 107.38) \]

### Summary

A. 95% Confidence Interval:
\[ (85.58, 104.42) \]

B. 90% Confidence Interval:
\[ (87.10, 102.90) \]

C. 99% Confidence Interval:
\[ (82.62, 107.38) \]

Would you like more details or have any questions about this process?

Here are 8 related questions that might expand on this information:
1. What is the significance of the critical value \(Z_{\alpha/2}\) in confidence intervals?
2. How does the sample size affect the width of a confidence interval?
3. Why is the population standard deviation used in these calculations instead of the sample standard deviation?
4. How would the confidence intervals change if the sample size were larger?
5. Can confidence intervals be used for proportions as well as means?
6. How do you interpret a 95% confidence interval in the context of real-world data?
7. What assumptions are made when calculating confidence intervals for the mean?
8. How would you construct a confidence interval if the population standard deviation were unknown?

**Tip:** When calculating confidence intervals, it's essential to use the correct critical value for the desired confidence level to ensure the accuracy of your interval estimates.

Learn how to estimate the population mean with confidence intervals using a statistics example. Understand the calculation process for 95%, 90%, and 99% confidence levels with given sample mean, standard deviation, and sample size.

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