Test the claim that the mean GPA of night students is smaller than 2.8 at the .05 significance level. The null and alternative hypothesis would be: Ho: p= 0.7 Ho:μ = 2.8 Ho: p = 0.7 Ho: = 2.8 Ho:p = 0.7 Ho:μ = 2.8 H₁ p < 0.7 H₁: 2.8 H₁:p 0.7 H₁: <2.8 H₁:p> 0.7 H₁: 2.8 O The test is: right-tailed two-tailed left-tailed Based on a sample of 50 people, the sample mean GPA was 2.75 with a standard deviation of 0.02 The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis

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### Hypothesis Testing Example: Mean GPA of Night Students

#### Problem:
Test the claim that the mean GPA of night students is smaller than 2.8 at the 0.05 significance level.

#### Null and Alternative Hypothesis:
The null and alternative hypothesis would be:
- Null Hypothesis (\( H_0 \)): \( \mu = 2.8 \)
- Alternative Hypothesis (\( H_1 \)): \( \mu < 2.8 \)

This setup corresponds to a **left-tailed test**.

#### Sample Data:
Based on a sample of 50 people, the sample mean GPA was 2.75 with a standard deviation of 0.02.

#### Steps to Solve:

1. **Calculate the Test Statistic**:
   - Formula: 
     \[
     Z = \frac{{\bar{X} - \mu}}{{\frac{\sigma}{\sqrt{n}}}}
     \]
   - Where:
     \(\bar{X}\) = Sample mean = 2.75
     \(\mu\) = Population mean = 2.8
     \(\sigma\) = Sample standard deviation = 0.02
     \(n\) = Sample size = 50

2. **Determine the Critical Value** at the 0.05 significance level for a left-tailed test.

3. **Decision Rule**:
   - Compare the test statistic to the critical value to determine whether to reject the null hypothesis.

#### Calculation Inputs:
- **Test Statistic Calculation**: \( \_\_\_\_\_\@\)
- **Critical Value Lookup**: \( \_\_\_\_\_\@\)

#### Conclusion:
Based on the calculated test statistic and the critical value, we will either:
- **Reject the null hypothesis**
- **Fail to reject the null hypothesis**

Please refer to statistical tables or software to find the exact critical value and to perform the precise calculations needed to find the test statistic. This will help in making the final decision regarding the hypothesis.
Transcribed Image Text:### Hypothesis Testing Example: Mean GPA of Night Students #### Problem: Test the claim that the mean GPA of night students is smaller than 2.8 at the 0.05 significance level. #### Null and Alternative Hypothesis: The null and alternative hypothesis would be: - Null Hypothesis (\( H_0 \)): \( \mu = 2.8 \) - Alternative Hypothesis (\( H_1 \)): \( \mu < 2.8 \) This setup corresponds to a **left-tailed test**. #### Sample Data: Based on a sample of 50 people, the sample mean GPA was 2.75 with a standard deviation of 0.02. #### Steps to Solve: 1. **Calculate the Test Statistic**: - Formula: \[ Z = \frac{{\bar{X} - \mu}}{{\frac{\sigma}{\sqrt{n}}}} \] - Where: \(\bar{X}\) = Sample mean = 2.75 \(\mu\) = Population mean = 2.8 \(\sigma\) = Sample standard deviation = 0.02 \(n\) = Sample size = 50 2. **Determine the Critical Value** at the 0.05 significance level for a left-tailed test. 3. **Decision Rule**: - Compare the test statistic to the critical value to determine whether to reject the null hypothesis. #### Calculation Inputs: - **Test Statistic Calculation**: \( \_\_\_\_\_\@\) - **Critical Value Lookup**: \( \_\_\_\_\_\@\) #### Conclusion: Based on the calculated test statistic and the critical value, we will either: - **Reject the null hypothesis** - **Fail to reject the null hypothesis** Please refer to statistical tables or software to find the exact critical value and to perform the precise calculations needed to find the test statistic. This will help in making the final decision regarding the hypothesis.
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