A data set includes data from student evaluations of courses. The summary statistics are n = 84, x= 4.17, s= 0.55. Use a 0.05 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. What are the null and alternative hypotheses? Ο Α. Ηρ: μ = 4.25 H₁: μ<4.25 OC. Ho: 4.25 H₁: μ> 4.25 Determine the test statistic. (Round to two decimal places as needed.) C ( B. Ho: H =4.25 H₁: μ#4.25 O D. Ho: μ#4.25 H₁: p=4.25

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Determine the test statistic _____________ (Round to two decimal places as needed.)

Determine the P-value ______________ (Round to three decimal places as needeed.

 

State the final conclusion that addresses the original claim:


Reject or Not reject   H0. There is sufficent or insufficent evidence to conclude that the mean of the population of ratings is less than, greater than, or equal to   4.25.

**Educational Content on Hypothesis Testing**

A data set includes data from student evaluations of courses. The summary statistics are as follows: the sample size \( n = 84 \), the sample mean \( \bar{x} = 4.17 \), and the sample standard deviation \( s = 0.55 \). The task is to use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. It is assumed that a simple random sample has been selected.

### Steps to Perform Hypothesis Testing

1. **State the Null and Alternative Hypotheses**

   * The null hypothesis (\( H_0 \)): \( \mu = 4.25 \)
   * The alternative hypothesis (\( H_1 \)): \( \mu \neq 4.25 \)

2. **Determine the Test Statistic**

   The test statistic is calculated to compare the sample data with the null hypothesis. The box provided in the image is intended for the user to calculate and fill in the test statistic value, rounded to two decimal places.

3. **Evaluate the P-Value**

   Using the calculated test statistic, determine the p-value to assess the strength of the evidence against the null hypothesis.

4. **Make a Conclusion**

   Based on the p-value and the predetermined significance level (0.05), decide whether to reject or fail to reject the null hypothesis.
   
   - If the p-value is less than 0.05, reject the null hypothesis.
   - If the p-value is greater than 0.05, do not reject the null hypothesis.

This analysis is crucial for understanding whether the population mean significantly differs from the claimed value of 4.25, based on the sample data.
Transcribed Image Text:**Educational Content on Hypothesis Testing** A data set includes data from student evaluations of courses. The summary statistics are as follows: the sample size \( n = 84 \), the sample mean \( \bar{x} = 4.17 \), and the sample standard deviation \( s = 0.55 \). The task is to use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. It is assumed that a simple random sample has been selected. ### Steps to Perform Hypothesis Testing 1. **State the Null and Alternative Hypotheses** * The null hypothesis (\( H_0 \)): \( \mu = 4.25 \) * The alternative hypothesis (\( H_1 \)): \( \mu \neq 4.25 \) 2. **Determine the Test Statistic** The test statistic is calculated to compare the sample data with the null hypothesis. The box provided in the image is intended for the user to calculate and fill in the test statistic value, rounded to two decimal places. 3. **Evaluate the P-Value** Using the calculated test statistic, determine the p-value to assess the strength of the evidence against the null hypothesis. 4. **Make a Conclusion** Based on the p-value and the predetermined significance level (0.05), decide whether to reject or fail to reject the null hypothesis. - If the p-value is less than 0.05, reject the null hypothesis. - If the p-value is greater than 0.05, do not reject the null hypothesis. This analysis is crucial for understanding whether the population mean significantly differs from the claimed value of 4.25, based on the sample data.
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