To calculate the 95% confidence interval for the mean serum cholesterol level in this case, we can follow these steps:

1. Identify the given values:

   • Sample mean $(\bar{x}) = 191.5$
   • Sample standard deviation $(s) = 6.4$
   • Sample size $(n) = 16$
   • Confidence level = 95%

2. Determine the t-score for a 95% confidence interval with $n - 1 = 15$ degrees of freedom. For 95% confidence, the t-score for 15 degrees of freedom is approximately $t \approx 2.131$.

3. Calculate the standard error of the mean (SEM): $\text{SEM} = \frac{s}{\sqrt{n}} = \frac{6.4}{\sqrt{16}} = \frac{6.4}{4} = 1.6$

4. Calculate the margin of error (ME): $\text{ME} = t \times \text{SEM} = 2.131 \times 1.6 = 3.4096$

5. Determine the confidence interval:

   • Lower limit = $\bar{x} - \text{ME} = 191.5 - 3.4096 = 188.1$ (rounded to one decimal place)
   • Upper limit = $\bar{x} + \text{ME} = 191.5 + 3.4096 = 194.9$ (rounded to one decimal place)

Answer:

• Lower limit: 188.1
• Upper limit: 194.9

Would you like further details or have any questions about this calculation?

Here are five related questions to explore further:

1. How does the sample size affect the width of the confidence interval?
2. What would change if we used a 99% confidence interval instead of 95%?
3. How is a t-score different from a z-score, and when should each be used?
4. What assumptions are necessary for this confidence interval to be valid?
5. How would the confidence interval change if the standard deviation were larger?

Tip: For small sample sizes, use the t-distribution instead of the z-distribution to account for the increased variability.