To calculate the 95% confidence interval for a population with a known standard deviation, we can use the formula:

$\text{Confidence Interval} = \bar{x} \pm z_{\alpha/2} \times \left(\frac{\sigma}{\sqrt{n}}\right)$

Where:

• $\bar{x}$ = sample mean (98 in this case)
• $z_{\alpha/2}$ = z-score corresponding to the 95% confidence level (1.96 for 95%)
• $\sigma$ = population standard deviation (22)
• $n$ = sample size (2 for part a)

Step-by-step Calculation:

1. Find the standard error (SE): $SE = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{2}} = \frac{22}{1.414} \approx 15.556$

2. Find the margin of error: $\text{Margin of Error} = z_{\alpha/2} \times SE = 1.96 \times 15.556 \approx 30.689$

3. Find the confidence interval: $\text{Lower Limit} = \bar{x} - \text{Margin of Error} = 98 - 30.689 \approx 67.311$ $\text{Upper Limit} = \bar{x} + \text{Margin of Error} = 98 + 30.689 \approx 128.689$

Conclusion:

One can be 95% confident that the true population mean is between 67.31 and 128.69.

Would you like further details on this or the next parts of the question?

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Here are 5 relative questions to expand your understanding:

1. How does increasing the sample size affect the confidence interval?
2. What is the effect of a larger population standard deviation on the confidence interval?
3. How does changing the confidence level (e.g., 99%) change the width of the confidence interval?
4. What happens if the population doesn't follow a normal distribution?
5. How is the confidence interval interpreted in practical research?

Tip: As the sample size increases, the confidence interval becomes narrower, making your estimate more precise.