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A data set includes data from student evaluations of courses. The summary statistics are:

[tex]\[ n=83, \overline{x}=4.18, s=0.66 \][/tex]

Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

State the final conclusion that addresses the original claim.

A. Fail to reject. There is not sufficient evidence to conclude that the mean of the population of student course evaluations is equal to 4.25.
B. Reject. There is sufficient evidence to conclude that the mean of the population of student course evaluations is not equal to 4.25.



Answer :

GinnyAnswer
Certainly! Let's work through the steps systematically to address the question and arrive at the final conclusion.

### Step 1: State the Hypotheses

- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 4.25\)[/tex] (The population mean is equal to 4.25)
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu \neq 4.25\)[/tex] (The population mean is not equal to 4.25)

### Step 2: Identify the Test Statistic

To determine how the sample mean compares to the population mean, we use the t-test for the sample:

The formula for the t-statistic [tex]\( t \)[/tex] is:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]

Given:
- [tex]\( n = 83 \)[/tex]
- [tex]\( \bar{x} = 4.18 \)[/tex]
- [tex]\( s = 0.66 \)[/tex]
- [tex]\( \mu = 4.25 \)[/tex]

### Step 3: Calculate the Test Statistic

As per our calculations:
[tex]\[ t \approx -0.966 \][/tex]

### Step 4: Determine Degrees of Freedom

Degrees of freedom ([tex]\( \text{df} \)[/tex]):
[tex]\[ \text{df} = n - 1 = 83 - 1 = 82 \][/tex]

### Step 5: Calculate the P-value

For a two-tailed test, the p-value is twice the area under the t-distribution curve to the right of [tex]\( |t| \)[/tex]:
[tex]\[ p \approx 0.337 \][/tex]

### Step 6: Compare P-value with Significance Level

The provided significance level ([tex]\( \alpha \)[/tex]) is:
[tex]\[ \alpha = 0.01 \][/tex]

We compare the p-value to [tex]\( \alpha \)[/tex]:
- Since [tex]\( p = 0.337 \)[/tex] is greater than [tex]\( 0.01 \)[/tex], we do not reject the null hypothesis.

### Step 7: Draw a Conclusion

Based on the p-value and the comparison with the significance level:
- We fail to reject the null hypothesis.

### Final Conclusion

There is not sufficient evidence to conclude that the mean of the population of student course evaluations is not equal to 4.25.

Thus, we state:

Fail to reject. There is not sufficient evidence to conclude that the mean of the population of student course evaluations is equal to 4.25 is not correct.

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