To calculate the 98% confidence interval for the population mean, we will use the following formula for a confidence interval when the population standard deviation is unknown and the sample size is greater than 30 (which allows us to use the normal distribution approximation):

$CI = \bar{x} \pm z^* \cdot \frac{s}{\sqrt{n}}$

Where:

• $\bar{x} = 38.2$ (sample mean),
• $s = 6.7$ (sample standard deviation),
• $n = 72$ (sample size),
• $z^*$ is the critical value corresponding to the 98% confidence level.

The critical value $z^*$ for a 98% confidence level is approximately 2.33 (from the standard normal distribution).

Now, let's compute the margin of error (ME):

$ME = z^* \cdot \frac{s}{\sqrt{n}} = 2.33 \cdot \frac{6.7}{\sqrt{72}} = 2.33 \cdot \frac{6.7}{8.485} \approx 1.84$

So, the confidence interval is:

$CI = 38.2 \pm 1.84$

The lower bound of the interval is:

$38.2 - 1.84 = 36.36$

The upper bound of the interval is:

$38.2 + 1.84 = 40.04$

Thus, the 98% confidence interval is approximately $(36.4, 40.0)$.

Would you like further details on how this interval was calculated or explanations on any part?

Here are 5 related questions to expand on the topic:

1. How does the sample size affect the width of a confidence interval?
2. Why do we use a $z^*$-value for large samples and a $t^*$-value for small samples?
3. How would the confidence interval change if the confidence level was reduced to 95%?
4. What assumptions must be met for this confidence interval formula to be valid?
5. How does increasing the confidence level affect the critical value $z^*$?

Tip: When calculating confidence intervals, always double-check whether you should use the $z^*$-value or $t^*$-value depending on your sample size and the population standard deviation information.