A data set includes 110 body temperatures of healthy adult humans having a mean of [tex]$98.1^{\circ} F$[/tex] and a standard deviation of [tex]$0.64^{\circ} F$[/tex]. Construct a [tex]$99 \%$[/tex] confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of [tex]$98.6^{\circ} F$[/tex] as the mean body temperature?

What is the confidence interval estimate of the population mean [tex]$\mu$[/tex]?

[tex]$\square$[/tex] [tex]${ }^{\circ} F \ \textless \ \mu \ \textless \ $[/tex] [tex]$\square$[/tex] [tex]${ }^{\circ} F$[/tex]

(Round to three decimal places as needed.)



Answer :

To find a \(99\%\) confidence interval estimate of the mean body temperature of all healthy humans given the data, we can follow these steps:

### 1. Understanding the given data:
- Sample size (\(n\)) = 110
- Sample mean (\(\bar{x}\)) = \(98.1^{\circ} F\)
- Sample standard deviation (\(s\)) = \(0.64^{\circ} F\)
- Confidence level = \(99\%\)

### 2. Determine the critical value for the given confidence level:
For a \(99\%\) confidence level, the significance level (\(\alpha\)) is calculated as follows:
[tex]\[ \alpha = 1 - \text{confidence level} = 1 - 0.99 = 0.01 \][/tex]
Since it's a two-tailed test, we divide \(\alpha\) by 2:
[tex]\[ \alpha/2 = 0.01/2 = 0.005 \][/tex]
We then use the standard normal distribution (Z-distribution) to find the critical value \(z_{\alpha/2}\):
[tex]\[ z_{\alpha/2} \approx 2.576 \][/tex]
(This value is obtained from the standard normal distribution table corresponding to \(\alpha/2 = 0.005\)).

### 3. Calculate the standard error of the mean (SE):
The standard error of the mean is given by:
[tex]\[ SE = \frac{s}{\sqrt{n}} = \frac{0.64}{\sqrt{110}} \approx 0.061 \][/tex]

### 4. Calculate the margin of error (E):
The margin of error is calculated using:
[tex]\[ E = z_{\alpha/2} \times SE = 2.576 \times 0.061 \approx 0.157 \][/tex]

### 5. Construct the confidence interval:
The confidence interval is determined by:
[tex]\[ \text{Lower bound} = \bar{x} - E = 98.1 - 0.157 \approx 97.943 \][/tex]
[tex]\[ \text{Upper bound} = \bar{x} + E = 98.1 + 0.157 \approx 98.257 \][/tex]

So, the \(99\%\) confidence interval estimate of the population mean body temperature \(\mu\) is:
[tex]\[ 97.943 \,^{\circ}\text{F} < \mu < 98.257 \,^{\circ}\text{F} \][/tex]

### Interpretation:
This confidence interval suggests that the mean body temperature of all healthy humans is likely to lie between \(97.943^{\circ}F\) and \(98.257^{\circ}F\) with \(99\%\) confidence. Since \(98.6^{\circ}F\) is not within this range, it suggests that the commonly accepted mean body temperature of \(98.6^{\circ}F\) may be higher than the actual mean body temperature of the population.

Therefore, given this confidence interval, the use of [tex]\(98.6^{\circ}F\)[/tex] as the mean body temperature might need to be revisited, as it appears to be an overestimate based on the sample data provided.