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Sagot :
Sure, let's work through the problem step-by-step.
### (a) Write the claim mathematically and identify [tex]\( H_0 \)[/tex] and [tex]\( H_a \)[/tex]
The claim made by the dean is that the mean number of classroom hours per week for full-time faculty is [tex]\(10.0\)[/tex]. This can be written mathematically as:
[tex]\[ \mu = 10.0 \][/tex]
When testing this claim, we set up the null hypothesis [tex]\(H_0\)[/tex] and the alternative hypothesis [tex]\(H_a\)[/tex].
- The null hypothesis [tex]\(H_0\)[/tex] represents the claim made by the dean:
[tex]\[ H_0: \mu = 10.0 \][/tex]
- The alternative hypothesis [tex]\(H_a\)[/tex] represents what we want to test against the null hypothesis. In this case, since we want to test if the mean is different from [tex]\(10.0\)[/tex] (either greater than or less than), we use a two-tailed test:
[tex]\[ H_a: \mu \neq 10.0 \][/tex]
So, the correct statements for [tex]\(H_0\)[/tex] and [tex]\(H_a\)[/tex] are:
[tex]\[ \boxed{H_0: \mu = 10.0, \quad H_a: \mu \neq 10.0} \][/tex]
This corresponds to option E.
### (b) Use technology to find the P-value
Given the sample data for the number of classroom hours:
[tex]\[ 10.7, 8.2, 11.4, 7.5, 4.4, 11.2, 12.2, 9.0 \][/tex]
We can summarize the results of our calculations as follows:
- The sample mean ([tex]\(\bar{x}\)[/tex]) is:
[tex]\[ 9.325 \][/tex]
- The sample standard deviation ([tex]\(s\)[/tex]) is:
[tex]\[ 2.589 \][/tex]
- The t-statistic (t) is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ t = \frac{9.325 - 10.0}{2.589 / \sqrt{8}} \approx -0.737 \][/tex]
- Degrees of freedom ([tex]\(df\)[/tex]):
[tex]\[ df = n - 1 = 8 - 1 = 7 \][/tex]
- Using a two-tailed test, the P-value is:
[tex]\[ P = 0.485 \][/tex]
### Conclusion:
Given the significance level [tex]\(\alpha = 0.05\)[/tex]:
Since the P-value [tex]\((0.485)\)[/tex] is greater than [tex]\(\alpha (0.05)\)[/tex], we fail to reject the null hypothesis [tex]\(H_0\)[/tex]. This means there is not enough evidence to reject the dean's claim that the mean number of classroom hours per week for full-time faculty is [tex]\(10.0\)[/tex].
Thus, the final solution is:
[tex]\[ H_0: \mu = 10.0 \][/tex]
[tex]\[ H_a: \mu \neq 10.0 \][/tex]
[tex]\[ \text{P-value} = 0.485 \][/tex]
Since [tex]\( \text{P-value} > 0.05 \)[/tex], we do not reject [tex]\( H_0 \)[/tex].
### (a) Write the claim mathematically and identify [tex]\( H_0 \)[/tex] and [tex]\( H_a \)[/tex]
The claim made by the dean is that the mean number of classroom hours per week for full-time faculty is [tex]\(10.0\)[/tex]. This can be written mathematically as:
[tex]\[ \mu = 10.0 \][/tex]
When testing this claim, we set up the null hypothesis [tex]\(H_0\)[/tex] and the alternative hypothesis [tex]\(H_a\)[/tex].
- The null hypothesis [tex]\(H_0\)[/tex] represents the claim made by the dean:
[tex]\[ H_0: \mu = 10.0 \][/tex]
- The alternative hypothesis [tex]\(H_a\)[/tex] represents what we want to test against the null hypothesis. In this case, since we want to test if the mean is different from [tex]\(10.0\)[/tex] (either greater than or less than), we use a two-tailed test:
[tex]\[ H_a: \mu \neq 10.0 \][/tex]
So, the correct statements for [tex]\(H_0\)[/tex] and [tex]\(H_a\)[/tex] are:
[tex]\[ \boxed{H_0: \mu = 10.0, \quad H_a: \mu \neq 10.0} \][/tex]
This corresponds to option E.
### (b) Use technology to find the P-value
Given the sample data for the number of classroom hours:
[tex]\[ 10.7, 8.2, 11.4, 7.5, 4.4, 11.2, 12.2, 9.0 \][/tex]
We can summarize the results of our calculations as follows:
- The sample mean ([tex]\(\bar{x}\)[/tex]) is:
[tex]\[ 9.325 \][/tex]
- The sample standard deviation ([tex]\(s\)[/tex]) is:
[tex]\[ 2.589 \][/tex]
- The t-statistic (t) is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ t = \frac{9.325 - 10.0}{2.589 / \sqrt{8}} \approx -0.737 \][/tex]
- Degrees of freedom ([tex]\(df\)[/tex]):
[tex]\[ df = n - 1 = 8 - 1 = 7 \][/tex]
- Using a two-tailed test, the P-value is:
[tex]\[ P = 0.485 \][/tex]
### Conclusion:
Given the significance level [tex]\(\alpha = 0.05\)[/tex]:
Since the P-value [tex]\((0.485)\)[/tex] is greater than [tex]\(\alpha (0.05)\)[/tex], we fail to reject the null hypothesis [tex]\(H_0\)[/tex]. This means there is not enough evidence to reject the dean's claim that the mean number of classroom hours per week for full-time faculty is [tex]\(10.0\)[/tex].
Thus, the final solution is:
[tex]\[ H_0: \mu = 10.0 \][/tex]
[tex]\[ H_a: \mu \neq 10.0 \][/tex]
[tex]\[ \text{P-value} = 0.485 \][/tex]
Since [tex]\( \text{P-value} > 0.05 \)[/tex], we do not reject [tex]\( H_0 \)[/tex].
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