data set includes data from student evaluations of courses. The summary statistics are nequals=9494​, x overbarxequals=4.194.19​, sequals=0.510.51. Use a 0.010.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.254.25. 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.
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Part 1
What are the null and alternative​ hypotheses?
A.
Upper H 0H0​: muμequals=4.254.25
Upper H 1H1​: muμnot equals≠4.254.25
Your answer is correct.B.
Upper H 0H0​: muμequals=4.254.25
Upper H 1H1​: muμless than<4.254.25
C.
Upper H 0H0​: muμequals=4.254.25
Upper H 1H1​: muμgreater than>4.254.25
D.
Upper H 0H0​: muμnot equals≠4.254.25
Upper H 1H1​: muμequals=4.254.25
Part 2
Determine the test statistic.
enter your response here ​(Round to two decimal places as​ needed.)

### Part 1: Null and Alternative Hypotheses

The null and alternative hypotheses for this test are:

- Null Hypothesis (\( H_0 \)): \( \mu = 4.25 \)
- Alternative Hypothesis (\( H_1 \)): \( \mu \ne 4.25 \)

This corresponds to option A:

\[ \text{A. } H_0: \mu = 4.25 \quad H_1: \mu \ne 4.25 \]

### Part 2: Determine the Test Statistic

To determine the test statistic, we use the formula for the \( t \)-statistic for a sample mean:

\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]

Where:
- \( \bar{x} \) is the sample mean
- \( \mu \) is the population mean under the null hypothesis
- \( s \) is the sample standard deviation
- \( n \) is the sample size

Given:
- \( \bar{x} = 4.19 \)
- \( \mu = 4.25 \)
- \( s = 0.51 \)
- \( n = 9494 \)

Substitute these values into the formula:

\[ t = \frac{4.19 - 4.25}{\frac{0.51}{\sqrt{9494}}} \]

First, compute the standard error of the mean:

\[ \text{Standard Error} = \frac{0.51}{\sqrt{9494}} \]

Let's calculate this step-by-step:

\[ \sqrt{9494} \approx 97.46 \]
\[ \text{Standard Error} = \frac{0.51}{97.46} \approx 0.00523 \]

Now, compute the \( t \)-statistic:

\[ t = \frac{4.19 - 4.25}{0.00523} = \frac{-0.06}{0.00523} \approx -11.47 \]

### Part 3: Determine the P-value and Conclusion

The degrees of freedom (\( df \)) is \( n - 1 \):

\[ df = 9494 - 1 = 9493 \]

For a two-tailed test at a significance level of \( \alpha = 0.01 \), we compare the computed \( t \)-value to the critical values from the \( t \)-distribution table.

Given the large sample size, we can approximate the critical values using the standard normal distribution. For \( \alpha = 0.01 \), the critical values are approximately \( \pm 2.576 \).

Since \( |t| = 11.47 \) is much greater than 2.576, the test statistic falls in the rejection region.

Thus, we reject the null hypothesis.

### Final Conclusion

There is sufficient evidence to reject the claim that the population mean of student course evaluations is 4.25. Therefore, we conclude that the population mean is significantly different from 4.25.

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Do you want details or have any questions?

Here are 5 questions you might want to ask next:
1. How do you interpret the P-value in hypothesis testing?
2. What are Type I and Type II errors in hypothesis testing?
3. Can you explain how to find critical values in a \( t \)-distribution table?
4. How would the conclusion change if the significance level was 0.05?
5. What are the assumptions of a \( t \)-test for a single sample mean?

Tip: Always check the assumptions of the statistical test you are using to ensure the validity of your results.

Explore the hypothesis testing process for student course evaluations with 9494 data points. Learn how to calculate the test statistic and P-value to test the claim that the population mean equals 4.25, with a significance level of 0.01.

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