Answer :
To construct a confidence interval for the mean body temperature of men based on a sample, we need to follow a few steps and use statistical formulas.
Step 1: Calculate the Sample Mean and Standard Deviation
The sample data consists of the following body temperatures in Fahrenheit ([tex]^{\circ}F[/tex]): 96.9, 97.4, 97.5, 97.8, 97.8, 97.9, 98.0, 98.1, and 98.6.
Sample Mean ([tex]\bar{x}[/tex]):
[tex]\bar{x} = \frac{\sum x_i}{n} = \frac{96.9 + 97.4 + 97.5 + 97.8 + 97.8 + 97.9 + 98.0 + 98.1 + 98.6}{9}[/tex]
After calculating, [tex]\bar{x} \approx 97.8^{\circ}F[/tex].Sample Standard Deviation ([tex]s[/tex]):
To find the standard deviation, calculate:
[tex]s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}[/tex]
After calculations, [tex]s \approx 0.52[/tex].
Step 2: Find the Critical t-value for 95% Confidence Level
Since the sample size [tex]n = 9[/tex], degrees of freedom [tex]df = n - 1 = 8[/tex].
Using a t-distribution table or calculator, find the critical t-value for a 95% confidence level (one-tailed), which is approximately [tex]t_{0.05, 8} \approx 1.860[/tex].
Step 3: Calculate the Margin of Error (ME)
[tex]ME = t \cdot \frac{s}{\sqrt{n}} = 1.860 \cdot \frac{0.52}{\sqrt{9}} = 1.860 \cdot 0.1733 \approx 0.32[/tex]
Step 4: Construct the Confidence Interval
Since the question asks for testing if the mean is greater than 97.5°F, we'll construct a one-tailed confidence interval:
[tex]\text{Lower Limit} = \bar{x} + ME = 97.8 + 0.32 = 98.12[/tex]
Thus, the one-tailed 95% confidence interval is all values greater than 98.12°F.
Conclusion
The calculated confidence interval does not include the value 97.5°F. Therefore, we have enough evidence at the 0.05 significance level to support the claim that the mean body temperature of men is greater than 97.5°F.
