Let's go through the questions provided in the image step-by-step.

Part (a) - Determine the Margin of Error, $E$

The confidence interval given is from 19.6 months to 47.2 months. The margin of error $E$ is half the width of this confidence interval.

$E = \frac{47.2 - 19.6}{2} = \frac{27.6}{2} = 13.8 \text{ months}$

So, the margin of error is 13.8 months.

Part (b) - Interpret the Meaning of $E$

In this context, the margin of error $E$ represents the maximum amount that the sample mean may differ from the true mean of the duration of imprisonment for political prisoners, with a 95% confidence level. It provides an estimate of how accurate the sample mean is in estimating the population mean.

The correct answer is:

• C: One can be 95% confident that the maximum error made in estimating $\mu$ by $\bar{x}$ is as found in part (a).

Part (c) - Calculate the Required Sample Size for a Margin of Error of 12 Months at a 99% Confidence Level

To find the required sample size, we use the formula for the sample size based on a desired margin of error:

$n = \left( \frac{z \cdot \sigma}{E} \right)^2$

where:

• $z$ is the z-score for a 99% confidence level,
• $\sigma$ is the standard deviation (given as 38 months),
• $E$ is the desired margin of error (12 months).

For a 99% confidence level, the z-score $z \approx 2.576$.

$n = \left( \frac{2.576 \cdot 38}{12} \right)^2$

Calculating the inner part first:

$\frac{2.576 \cdot 38}{12} \approx 8.16$

Then, squaring this result:

$n \approx 8.16^2 \approx 66.57$

Rounding up to the nearest whole number, the required sample size is 67 prisoners.

Part (d) - Find a 99% Confidence Interval for the Mean Duration of Imprisonment

This part depends on the sample mean provided (36.5 months) and the sample size calculated in part (c) (67 prisoners). The formula for the confidence interval is:

$\bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}$

Substituting in the values:

• $\bar{x} = 36.5$,
• $z = 2.576$,
• $\sigma = 38$,
• $n = 67$.

Calculating the margin part:

$z \cdot \frac{\sigma}{\sqrt{n}} = 2.576 \cdot \frac{38}{\sqrt{67}}$

First, find $\sqrt{67} \approx 8.19$.

Then,

$\frac{38}{8.19} \approx 4.64$

Now,

$2.576 \cdot 4.64 \approx 11.95$

Thus, the 99% confidence interval is:

$36.5 \pm 11.95$

So, the interval is approximately:

$(24.55, 48.45) \text{ months}$

Let me know if you have any questions or want further clarification on any part.

Here are 5 related questions to explore further:

1. What is the significance of choosing a 95% vs. 99% confidence level?
2. How does increasing the sample size affect the margin of error?
3. Why is the z-score higher for a 99% confidence level compared to a 95% confidence level?
4. How would the confidence interval change if the standard deviation were smaller?
5. What assumptions are we making about the data by using this confidence interval method?

Tip: Remember, as the sample size increases, the margin of error decreases, resulting in a more precise estimate of the population parameter.