$$ 1 \leftarrow \quad \begin{array}{l} ext { A data set includes } 106 ext { body temperatures of healthy adult humans having a mean of } 98.9^{\circ} \mathrm{F} ext { and a standard deviation of } 0.62^{\circ} \mathrm{F} ext {. Construct a } 99 \% ext { confidence interval estimate of the mea } \ ext { healthy humans. What does the sample suggest about the use of } 98.6^{\circ} \mathrm{F} ext { as the mean body temperature? } \ ext { Click here to view a t distribution table. } \ ext { Click here to view page } 1 ext { of the standard normal distribution table. } \ ext { Click here to view page } 2 ext { of the standard normal distribution table. } \ ext { What is the confidence interval estimate of the population mean } \mu ext { ? } \ \square^{\circ} \mathrm{F}

Question

$$ 1 \leftarrow \quad \begin{array}{l}	ext { A data set includes } 106 	ext { body temperatures of healthy adult humans having a mean of } 98.9^{\circ} \mathrm{F} 	ext { and a standard deviation of } 0.62^{\circ} \mathrm{F} 	ext {. Construct a } 99 \% 	ext { confidence interval estimate of the mea } \ 	ext { healthy humans. What does the sample suggest about the use of } 98.6^{\circ} \mathrm{F} 	ext { as the mean body temperature? } \ 	ext { Click here to view a t distribution table. } \ 	ext { Click here to view page } 1 	ext { of the standard normal distribution table. } \ 	ext { Click here to view page } 2 	ext { of the standard normal distribution table. } \ 	ext { What is the confidence interval estimate of the population mean } \mu 	ext { ? } \ \square^{\circ} \mathrm{F}<\mu<\square^{\circ} \mathrm{F} \ 	ext { (Round to three decimal places as needed.) }\end{array} $$ ?

UpStudy AI · Accepted Answer

We are given a sample of $$ n = 106 $$ healthy adult body temperatures with a sample mean of  
$$
\bar{x} = 98.9^\circ \mathrm{F}
$$
and a sample standard deviation of  
$$
s = 0.62^\circ \mathrm{F}.
$$

Since the population standard deviation is unknown and the sample size is moderate, we use the $$ t $$-distribution to construct a $$ 99\% $$ confidence interval for the population mean $$ \mu $$.

1. Define the confidence interval formula using the $$ t $$-distribution:
   $$
   \bar{x} \pm t_{\alpha/2,\,n-1} \cdot \frac{s}{\sqrt{n}}.
   $$

2. For a $$ 99\% $$ confidence level, the significance level is 
   $$
   \alpha = 0.01 \quad \text{and} \quad \frac{\alpha}{2} = 0.005.
   $$
   With degrees of freedom $$ n-1 = 105 $$, from a $$ t $$-table we find an approximate critical value
   $$
   t_{0.005,\,105} \approx 2.63.
   $$

3. Compute the standard error (SE):
   $$
   \text{SE} = \frac{s}{\sqrt{n}} = \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602.
   $$

4. Compute the margin of error:
   $$
   \text{Margin} = t_{\alpha/2,\,n-1} \times \text{SE} \approx 2.63 \times 0.0602 \approx 0.158.
   $$

5. Construct the $$ 99\% $$ confidence interval:
   $$
   \bar{x} \pm 0.158 \quad \Rightarrow \quad 98.9 - 0.158 < \mu < 98.9 + 0.158,
   $$
   which gives
   $$
   98.742^\circ \mathrm{F} < \mu < 99.058^\circ \mathrm{F}.
   $$

6. Interpretation regarding $$ 98.6^\circ \mathrm{F} $$:

Notice that the entire $$ 99\% $$ confidence interval is above $$ 98.6^\circ \mathrm{F} $$. This suggests that the true mean body temperature for healthy adults is very likely higher than $$ 98.6^\circ \mathrm{F} $$, calling into question the use of $$ 98.6^\circ \mathrm{F} $$ as the average body temperature.

Thus, the $$ 99\% $$ confidence interval for the population mean is 
$$
98.742^\circ \mathrm{F} < \mu < 99.058^\circ \mathrm{F},
$$
and the sample suggests that the typical body temperature of healthy adults is greater than $$ 98.6^\circ \mathrm{F} $$.
The $$ 99\% $$ confidence interval for the mean body temperature is $$ 98.742^\circ \mathrm{F} < \mu < 99.058^\circ \mathrm{F} $$. This suggests that the average body temperature of healthy adults is likely higher than $$ 98.6^\circ \mathrm{F} $$.

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