Step-by-step solution:
1. State the hypotheses:
- The null hypothesis, [tex]\(H_0\)[/tex]: [tex]\(\mu = 98.6^{\circ} F\)[/tex]
- The alternative hypothesis, [tex]\(H_1\)[/tex]: [tex]\(\mu < 98.6^{\circ} F\)[/tex]
So, we have:
[tex]\[
H_0 : \mu = 98.6^{\circ} F
\][/tex]
[tex]\[
H_1 : \mu < 98.6^{\circ} F
\][/tex]
2. Find the test statistic:
- Given:
- Sample size ([tex]\(n\)[/tex]) = 200
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 98.2
- Population mean ([tex]\(\mu\)[/tex]) = 98.6
- Sample standard deviation ([tex]\(s\)[/tex]) = 0.9
- Significance level ([tex]\(\alpha\)[/tex]) = 0.01
The t-score (test statistic) is calculated using the formula:
[tex]\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\][/tex]
Plugging in the values:
[tex]\[
t = \frac{98.2 - 98.6}{0.9 / \sqrt{200}}
\][/tex]
The calculated t-score is:
[tex]\[
t_0 = -6.29
\][/tex]
(Rounded to two decimal places as needed.)
3. Determine the P-value:
- The calculated t-score is -6.29.
- Degrees of freedom = [tex]\( n - 1 = 200 - 1 = 199 \)[/tex]
Using a t-distribution table or software, the P-value corresponding to [tex]\( t_0 = -6.29 \)[/tex] for 199 degrees of freedom is:
[tex]\[
\text{P-value} = 0.000
\][/tex]
(Rounded to three decimal places as needed.)
4. Conclusion:
Given the information:
- The significance level ([tex]\(\alpha\)[/tex]) = 0.01
- The P-value = 0.000
Since the P-value (0.000) is less than the significance level (0.01), we reject the null hypothesis.
Therefore, the correct conclusion is:
[tex]\[
\text{A. Reject \(H_0\) since the P-value is less than the significance level.}
\][/tex]
