Answer :
Let's address the outlined question step-by-step.
1. Identify the null and alternative hypotheses:
We are testing the claim that the population mean is equal to 3.50. The null hypothesis ([tex]\(H_0\)[/tex]) always states that there is no effect or no difference, and in this case, it posits that the mean is 3.50. The alternative hypothesis ([tex]\(H_1\)[/tex]) is what we suspect might be true instead of [tex]\(H_0\)[/tex].
Therefore, we have:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3.50\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu \neq 3.50\)[/tex]
These hypotheses correspond to option D from the question.
2. Determine the test statistic:
The test statistic for evaluating the mean when the population standard deviation is unknown is the t-statistic, calculated as follows:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Given:
- [tex]\( n = 85 \)[/tex]
- [tex]\( \bar{x} = 3.45 \)[/tex]
- [tex]\( s = 0.65 \)[/tex]
- [tex]\( \mu = 3.50 \)[/tex]
Plugging in these values, we get:
[tex]\[ t = \frac{3.45 - 3.50}{\frac{0.65}{\sqrt{85}}} \approx -0.71 \][/tex]
Therefore, the test statistic is approximately [tex]\(-0.71\)[/tex].
3. Determine the degrees of freedom:
The degrees of freedom ([tex]\(df\)[/tex]) for this test is given by:
[tex]\[ df = n - 1 = 85 - 1 = 84 \][/tex]
4. Determine the critical t-value for a two-tailed test at [tex]\( \alpha = 0.01 \)[/tex]:
For a two-tailed test with a significance level [tex]\( \alpha = 0.01 \)[/tex] and [tex]\( df = 84 \)[/tex], the critical t value can be found from t-distribution tables or statistical software. The critical t value is approximately [tex]\( \pm 2.64 \)[/tex].
5. Calculate the p-value:
The p-value is the area under the t-distribution curve beyond the absolute value of the test statistic. For our test statistic [tex]\( t = -0.71 \)[/tex] and [tex]\( df = 84 \)[/tex], the p-value can be looked up or calculated using statistical software.
The p-value for our t-statistic is approximately 0.480.
6. State the final conclusion:
To determine whether to reject the null hypothesis, we compare the p-value to the significance level [tex]\( \alpha = 0.01 \)[/tex]:
- If [tex]\( \text{p-value} < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( \text{p-value} \geq \alpha \)[/tex], fail to reject the null hypothesis.
Since [tex]\( 0.480 \geq 0.01 \)[/tex], we fail to reject the null hypothesis.
Conclusion:
Based on the sample data and a significance level of 0.01, there is not sufficient evidence to reject the null hypothesis. Therefore, we fail to reject the claim that the population mean of student course evaluations is 3.50.
The final step-by-step results are:
1. Null and alternative hypotheses: [tex]\( H_0: \mu = 3.50 \)[/tex] and [tex]\( H_1: \mu \neq 3.50 \)[/tex].
2. Test statistic: [tex]\( t \approx -0.71 \)[/tex].
3. Degrees of freedom: [tex]\( df = 84 \)[/tex].
4. Critical t-value: [tex]\( \pm 2.64 \)[/tex].
5. P-value: [tex]\( \approx 0.480 \)[/tex].
6. Conclusion: Fail to reject the null hypothesis.
1. Identify the null and alternative hypotheses:
We are testing the claim that the population mean is equal to 3.50. The null hypothesis ([tex]\(H_0\)[/tex]) always states that there is no effect or no difference, and in this case, it posits that the mean is 3.50. The alternative hypothesis ([tex]\(H_1\)[/tex]) is what we suspect might be true instead of [tex]\(H_0\)[/tex].
Therefore, we have:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3.50\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu \neq 3.50\)[/tex]
These hypotheses correspond to option D from the question.
2. Determine the test statistic:
The test statistic for evaluating the mean when the population standard deviation is unknown is the t-statistic, calculated as follows:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Given:
- [tex]\( n = 85 \)[/tex]
- [tex]\( \bar{x} = 3.45 \)[/tex]
- [tex]\( s = 0.65 \)[/tex]
- [tex]\( \mu = 3.50 \)[/tex]
Plugging in these values, we get:
[tex]\[ t = \frac{3.45 - 3.50}{\frac{0.65}{\sqrt{85}}} \approx -0.71 \][/tex]
Therefore, the test statistic is approximately [tex]\(-0.71\)[/tex].
3. Determine the degrees of freedom:
The degrees of freedom ([tex]\(df\)[/tex]) for this test is given by:
[tex]\[ df = n - 1 = 85 - 1 = 84 \][/tex]
4. Determine the critical t-value for a two-tailed test at [tex]\( \alpha = 0.01 \)[/tex]:
For a two-tailed test with a significance level [tex]\( \alpha = 0.01 \)[/tex] and [tex]\( df = 84 \)[/tex], the critical t value can be found from t-distribution tables or statistical software. The critical t value is approximately [tex]\( \pm 2.64 \)[/tex].
5. Calculate the p-value:
The p-value is the area under the t-distribution curve beyond the absolute value of the test statistic. For our test statistic [tex]\( t = -0.71 \)[/tex] and [tex]\( df = 84 \)[/tex], the p-value can be looked up or calculated using statistical software.
The p-value for our t-statistic is approximately 0.480.
6. State the final conclusion:
To determine whether to reject the null hypothesis, we compare the p-value to the significance level [tex]\( \alpha = 0.01 \)[/tex]:
- If [tex]\( \text{p-value} < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( \text{p-value} \geq \alpha \)[/tex], fail to reject the null hypothesis.
Since [tex]\( 0.480 \geq 0.01 \)[/tex], we fail to reject the null hypothesis.
Conclusion:
Based on the sample data and a significance level of 0.01, there is not sufficient evidence to reject the null hypothesis. Therefore, we fail to reject the claim that the population mean of student course evaluations is 3.50.
The final step-by-step results are:
1. Null and alternative hypotheses: [tex]\( H_0: \mu = 3.50 \)[/tex] and [tex]\( H_1: \mu \neq 3.50 \)[/tex].
2. Test statistic: [tex]\( t \approx -0.71 \)[/tex].
3. Degrees of freedom: [tex]\( df = 84 \)[/tex].
4. Critical t-value: [tex]\( \pm 2.64 \)[/tex].
5. P-value: [tex]\( \approx 0.480 \)[/tex].
6. Conclusion: Fail to reject the null hypothesis.
