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A data set includes 110 body temperatures of healthy adult humans with a mean of [tex]$98.0^{\circ} F$[/tex] and a standard deviation of [tex]$0.74^{\circ} F$[/tex]. Construct a [tex][tex]$99 \%$[/tex][/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?

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What is the confidence interval estimate of the population mean [tex]\mu[/tex]?
[tex]\square \, \square^{\circ} F[/tex]
(Round to three decimal places as needed.)

Sagot :

To construct a [tex]\(99 \%\)[/tex] confidence interval estimate of the mean body temperature of all healthy humans, we follow these steps:

1. Gather Given Data:
- Mean ([tex]\(\bar{x}\)[/tex]) of the sample: [tex]\(98.0^{\circ} F\)[/tex]
- Standard deviation ([tex]\(\sigma\)[/tex]) of the sample: [tex]\(0.74^{\circ} F\)[/tex]
- Sample size ([tex]\(n\)[/tex]): 110
- Confidence level: [tex]\(99 \%\)[/tex]

2. Calculate the Standard Error of the Mean:
Standard Error (SE) is given by:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the given values:
[tex]\[ SE = \frac{0.74}{\sqrt{110}} \approx 0.0704 \][/tex]

3. Determine the Z-Score for a 99% Confidence Level:
A 99% confidence level corresponds to [tex]\( \alpha = 0.01 \)[/tex], and we split this into two tails, so each tail has an area of [tex]\( \frac{\alpha}{2} = 0.005 \)[/tex]. We need to find the Z-score that leaves [tex]\(0.005\)[/tex] in the tail (which corresponds to [tex]\(99.5\% \)[/tex] in the cumulative distribution).

Using the standard normal distribution table, the Z-score for [tex]\(99.5\%\)[/tex] is approximately [tex]\(2.576\)[/tex].

4. Calculate the Margin of Error:
The Margin of Error (ME) is given by:
[tex]\[ ME = Z \times SE \][/tex]
Substituting the values:
[tex]\[ ME = 2.576 \times 0.0704 \approx 0.1817 \][/tex]

5. Calculate the Confidence Interval:
The confidence interval is given by:
[tex]\[ \text{Lower bound} = \bar{x} - ME = 98.0 - 0.1817 \approx 97.818 \][/tex]
[tex]\[ \text{Upper bound} = \bar{x} + ME = 98.0 + 0.1817 \approx 98.182 \][/tex]

Therefore, the 99% confidence interval estimate of the population mean [tex]\(\mu\)[/tex] is:
[tex]\[ (97.818^{\circ} F, 98.182^{\circ} F) \][/tex]

### Conclusion About the Use of [tex]\(98.6^{\circ} F\)[/tex]:
This sample suggests that the commonly accepted mean body temperature of [tex]\(98.6^{\circ} F\)[/tex] might be an overestimate since the [tex]\(99\%\)[/tex] confidence interval [tex]\((97.818^{\circ} F, 98.182^{\circ} F)\)[/tex] does not include [tex]\(98.6^{\circ} F\)[/tex]. This indicates that the true mean body temperature for healthy adults is likely lower than [tex]\(98.6^{\circ} F\)[/tex].
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