To address the problem, let's follow the steps and provide a detailed solution.
### Part (a): Formulate the Hypotheses.
The dean claims that the mean number of classroom hours per week for full-time faculty is 11.0.
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean number of classroom hours per week ([tex]\(\mu\)[/tex]) is equal to 11.0.
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The mean number of classroom hours per week ([tex]\(\mu\)[/tex]) is not equal to 11.0.
Mathematically, the hypotheses can be formulated as:
- [tex]\(H_0: \mu = 11.0\)[/tex]
- [tex]\(H_a: \mu \neq 11.0\)[/tex]
Among the given options, the correct choice is:
- C. [tex]\(H_0: \mu = 11.0\)[/tex], [tex]\(H_a: \mu \neq 11.0\)[/tex]
### Part (b): Calculate the P-value from the Sample Data.
Given sample data on weekly classroom hours for full-time faculty:
[tex]\[10.9, 9.8, 11.7, 8.9, 6.8, 9.3, 13.4, 8.4\][/tex]
#### Step-by-Step Calculation:
1. Sample Mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[\bar{x} = 9.900000000000002\][/tex]
2. Sample Standard Deviation (s):
[tex]\[ s = 2.0605131122403186 \][/tex]
3. Sample Size (n):
[tex]\[ n = 8 \][/tex]
4. Population Mean ([tex]\(\mu\)[/tex]) assumed under null hypothesis:
[tex]\[ \mu = 11.0 \][/tex]
5. Test Statistic (t-score):
[tex]\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \\
t = \frac{9.900000000000002 - 11.0}{2.0605131122403186 / \sqrt{8}} \\
t = -1.509949060133879
\][/tex]
6. Degrees of Freedom (df):
[tex]\[ df = n - 1 = 8 - 1 = 7 \][/tex]
7. P-value: Using the calculated t-score and degrees of freedom, the two-tailed p-value is:
[tex]\[
p = 2 \cdot P(T \geq |t|) \\
p = 2 \cdot P(T \geq 1.509949060133879) \\
p = 0.175 \\
\][/tex]
8. P-value Rounded to Three Decimal Places:
[tex]\[ P = 0.175 \][/tex]
Thus, the P-value is [tex]\(0.175\)[/tex] (rounded to three decimal places).
### Final Answer:
The P-value calculated from the sample data is [tex]\( \boxed{0.175} \)[/tex].
