A data set includes 106 body temperatures of healthy adult humans having a mean of and a standard deviation
of . Construct a confidence interval estimate of the mean body temperature of all healthy humans. What
does the sample suggest about the use of as the mean body temperature?
Click here to view a t distribution table.
Click here to view page 1 of the standard normal distribution table.
Click here to view page 2 of the standard normal distribution table.
What is the confidence interval estimate of the population mean ?

(Round to three decimal places as needed.)
What does this suggest about the use of as the mean body temperature?
A. This suggests that the mean body temperature is higher than .
B. This suggests that the mean body temperature is lower than .
C. This suggests that the mean body temperature could very possibly be .

Question

A data set includes 106 body temperatures of healthy adult humans having a mean of and a standard deviation
of . Construct a confidence interval estimate of the mean body temperature of all healthy humans. What
does the sample suggest about the use of as the mean body temperature?
Click here to view a t distribution table.
Click here to view page 1 of the standard normal distribution table.
Click here to view page 2 of the standard normal distribution table.
What is the confidence interval estimate of the population mean ?

(Round to three decimal places as needed.)
What does this suggest about the use of as the mean body temperature?
A. This suggests that the mean body temperature is higher than .
B. This suggests that the mean body temperature is lower than .
C. This suggests that the mean body temperature could very possibly be .

Ask by Fitzgerald Barrett. in the United States

Dec 09,2024

UpStudy AI · Accepted Answer

To construct a $$ 99\% $$ confidence interval for the mean body temperature, we can use the formula for the confidence interval:

$$
\bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
$$

where:
- $$\bar{x}$$ is the sample mean,
- $$z^*$$ is the z-value corresponding to the desired confidence level,
- $$s$$ is the sample standard deviation,
- $$n$$ is the sample size.

### Step 1: Identify the values
- Sample mean ($$\bar{x}$$) = $$98.9^{\circ} \mathrm{F}$$
- Sample standard deviation ($$s$$) = $$0.63^{\circ} \mathrm{F}$$
- Sample size ($$n$$) = $$106$$

### Step 2: Find the z-value for a $$ 99\% $$ confidence level
For a $$ 99\% $$ confidence level, the z-value (from the standard normal distribution table) is approximately $$ 2.576 $$.

### Step 3: Calculate the standard error (SE)
$$
SE = \frac{s}{\sqrt{n}} = \frac{0.63}{\sqrt{106}} \approx \frac{0.63}{10.2956} \approx 0.0612
$$

### Step 4: Calculate the margin of error (ME)
$$
ME = z^* \cdot SE = 2.576 \cdot 0.0612 \approx 0.1575
$$

### Step 5: Construct the confidence interval
$$
\text{Lower limit} = \bar{x} - ME = 98.9 - 0.1575 \approx 98.7425
$$
$$
\text{Upper limit} = \bar{x} + ME = 98.9 + 0.1575 \approx 99.0575
$$

### Step 6: Round to three decimal places
$$
98.743^{\circ} \mathrm{F} < \mu < 99.058^{\circ} \mathrm{F}
$$

### Conclusion
The confidence interval estimate of the population mean $$ \mu $$ is:
$$
98.743^{\circ} \mathrm{F} < \mu < 99.058^{\circ} \mathrm{F}
$$

### Interpretation of the results
Now, regarding the use of $$ 98.6^{\circ} \mathrm{F} $$ as the mean body temperature:

- The entire confidence interval $$ (98.743, 99.058) $$ is above $$ 98.6^{\circ} \mathrm{F} $$.
- This suggests that the mean body temperature is likely higher than $$ 98.6^{\circ} \mathrm{F} $$.

Thus, the correct answer is:
**A. This suggests that the mean body temperature is higher than $$ 98.6^{\circ} \mathrm{F} $$.**
The confidence interval estimate of the population mean $$ \mu $$ is $$ 98.743^{\circ} \mathrm{F} < \mu < 99.058^{\circ} \mathrm{F} $$. This suggests that the mean body temperature is likely higher than $$ 98.6^{\circ} \mathrm{F} $$.

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