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To solve this problem, let's construct a 99% confidence interval for the mean body temperature using the provided dataTo solve this problem, let's construct a 99% confidence interval for the mean body temperature using the provided 

**Given dataTo solve this problem, let's construct a 99% confidence interval for the mean body temperature using the provided 

**Given **
- Sample mean (\(\bar{x}\)) = 98.3°F
- Sample standard deviation (\(s\)) = 0.69°F
- Sample size (\(n\)) = 108
- Confidence level = 99%

### Step 1: Find the Z-score for a 99% confidence level
For a 99% confidence interval, the Z-score corresponding to the two-tailed distribution is approximately 2.576.

### Step 2: Calculate the standard error (SE)
The standard error of the mean is calculated as:
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.69}{\sqrt{108}}
\]

Calculating the SE:
\[
SE \approx \frac{0.69}{10.39} \approx 0.0664
\]

### Step 3: Construct the confidence interval
The confidence interval is given by:
\[
\text{CI} = \bar{x} \pm (Z \times SE)
\]
Substitute the values:
\[
\text{CI} = 98.3 \pm (2.576 \times 0.0664)
\]

Calculating the margin of error:
\[
\text{Margin of error} = 2.576 \times 0.0664 \approx 0.171
\]

### Step 4: Determine the confidence interval
\[
\text{Lower limit} = 98.3 - 0.171 = 98.129
\]
\[
\text{Upper limit} = 98.3 + 0.171 = 98.471
\]

### Final Answer
The 99% confidence interval for the population mean body temperature is approximately:
\[
98.129^\circ\text{F} < \mu < 98.471^\circ\text{F}
\]

---

Would you like further explanation on any steps, or do you have any questions?

### Additional Related Questions:
1. How would the confidence interval change with a 95% confidence level?
2. What if the standard deviation were higher? How would that affect the interval?
3. Why is a Z-score used instead of a t-score in this case?
4. How does increasing the sample size affect the standard error?
5. What does this interval suggest about using 98.6°F as a "normal" body temperature?

**Tip:** Increasing the confidence level will widen the interval, as it requires a larger Z-score for greater certainty.

A data set includes 108 body temperatures of healthy adult humans having a mean of 98.3°F and a standard deviation of 0.69°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What is the confidence interval estimate of the population mean μ?

Calculate a 99% confidence interval for the mean body temperature of healthy adults using a sample mean of 98.3°F, standard deviation of 0.69°F, and sample size of 108. This problem demonstrates confidence interval estimation for large samples, with explanations of standard error and Z-scores.

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