Answer :
To test the hypothesis that the population mean body temperature is not 98.6°F, we will perform a two-tailed t-test for a single sample.
Step-by-step Solution:
Define the Hypotheses:
- Null Hypothesis [tex]H_0[/tex]: The mean temperature [tex]\mu = 98.6°F[/tex].
- Alternative Hypothesis [tex]H_a[/tex]: The mean temperature [tex]\mu \neq 98.6°F[/tex].
Gather the Sample Data:
The sample of body temperatures is: 98.2, 98.8, 99.0, 96.4, 98.3, 98.6, 97.4, 99.1, 98.9, 97.2.Calculate the Sample Mean ([tex]\bar{x}[/tex]):
[tex]\bar{x} = \frac{98.2 + 98.8 + 99.0 + 96.4 + 98.3 + 98.6 + 97.4 + 99.1 + 98.9 + 97.2}{10} = 98.29[/tex]Calculate the Sample Standard Deviation ([tex]s[/tex]):
[tex]s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}[/tex]First, calculate each [tex](x_i - \bar{x})^2[/tex]:
- (98.2 - 98.29)² = 0.0081
- (98.8 - 98.29)² = 0.2601
- (99.0 - 98.29)² = 0.5041
- (96.4 - 98.29)² = 3.5521
- (98.3 - 98.29)² = 0.0001
- (98.6 - 98.29)² = 0.0961
- (97.4 - 98.29)² = 0.7921
- (99.1 - 98.29)² = 0.6561
- (98.9 - 98.29)² = 0.3721
- (97.2 - 98.29)² = 1.1881
Sum of squares = 7.429
[tex]s = \sqrt{\frac{7.429}{9}} \approx 0.898[/tex]
Calculate the Test Statistic ([tex]t[/tex]):
[tex]t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{98.29 - 98.6}{0.898/\sqrt{10}} \approx -1.232[/tex]Determine the Degrees of Freedom ([tex]df[/tex]):
[tex]df = n - 1 = 10 - 1 = 9[/tex].Determine the Critical t-value:
For a significance level of 0.05 and two-tailed test, lookup a t-table or use a calculator for [tex]t_{\text{critical}}[/tex] for df=9. [tex]t_{\text{critical}} \approx ±2.262[/tex].Make a Decision:
Compare [tex]|t| = 1.232[/tex] with [tex]t_{\text{critical}} = 2.262[/tex]. Since [tex]1.232 < 2.262[/tex], we fail to reject the null hypothesis.
Conclusion
At a 0.05 significance level, there is not enough evidence to reject the null hypothesis. Therefore, we do not have sufficient evidence to say that the population mean body temperature is different from 98.6°F.
