To construct a 99% confidence interval for the mean body temperature of all healthy humans, we can use the formula for a confidence interval for the mean when the standard deviation is known:

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

where:

• $\bar{x}$ is the sample mean (98.198°F),
• $z$ is the z-score corresponding to a 99% confidence level,
• $\sigma$ is the standard deviation of the sample (0.56°F),
• $n$ is the sample size (103).

1. Find the z-score for a 99% confidence interval: For a 99% confidence interval, the z-score is approximately $2.576$.

2. Calculate the margin of error (ME): $ME = z \cdot \frac{\sigma}{\sqrt{n}} = 2.576 \cdot \frac{0.56}{\sqrt{103}}$

3. Compute the confidence interval: $CI = \bar{x} \pm ME$

Let's calculate this step-by-step.The 99% confidence interval for the mean body temperature of all healthy humans is:

$98.056^\circ F < \mu < 98.340^\circ F$

Interpretation

This confidence interval does not contain 98.6°F, suggesting that the commonly used value of 98.6°F may not accurately represent the mean body temperature of healthy humans. The sample mean of 98.198°F lies lower, implying that the actual mean body temperature might be slightly less than 98.6°F.

Would you like more details on this calculation or have any questions?

Here are five related questions you might consider:

1. How would the confidence interval change with a 95% confidence level?
2. What is the impact of a larger sample size on the confidence interval width?
3. How is the margin of error affected if the standard deviation is larger?
4. What happens to the confidence interval if the sample mean were higher than 98.6°F?
5. How does a t-distribution compare to a z-distribution in constructing confidence intervals?

Tip: Confidence intervals offer a range within which the true mean likely falls, but they don't guarantee that the population mean is within that range; they reflect statistical confidence, not absolute certainty.