Answer :
To determine whether there is a significant difference in proportions between the first-year and fourth-year students at a public university regarding their support for modifying the student Honor Code, we follow these steps for the hypothesis test:
### Summary of Given Data:
- First-Years (Pop. 1)
- Sample size ([tex]\( n_1 \)[/tex]): 95
- Number answering "yes" ([tex]\( x_1 \)[/tex]): 58
- Fourth-Years (Pop. 2)
- Sample size ([tex]\( n_2 \)[/tex]): 85
- Number answering "yes" ([tex]\( x_2 \)[/tex]): 56
### Step-by-Step Solution:
#### 1. Calculate Sample Proportions:
- Proportion of first-years supporting the change ([tex]\( p_1 \)[/tex]):
[tex]\[ p_1 = \frac{x_1}{n_1} = \frac{58}{95} = 0.6105 \][/tex]
- Proportion of fourth-years supporting the change ([tex]\( p_2 \)[/tex]):
[tex]\[ p_2 = \frac{x_2}{n_2} = \frac{56}{85} = 0.6588 \][/tex]
#### 2. Combined Proportion:
- Combined sample proportion ([tex]\( p_{\text{combined}} \)[/tex]):
[tex]\[ p_{\text{combined}} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{58 + 56}{95 + 85} = \frac{114}{180} = 0.6333 \][/tex]
#### 3. Standard Error of the Difference in Proportions:
- Standard error (SE):
[tex]\[ SE = \sqrt{p_{\text{combined}} \cdot (1 - p_{\text{combined}}) \cdot \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.6333 \cdot (1 - 0.6333) \cdot \left( \frac{1}{95} + \frac{1}{85} \right)} = 0.0719 \][/tex]
#### 4. Test Statistic (Z):
[tex]\[ Z = \frac{p_1 - p_2}{SE} = \frac{0.6105 - 0.6588}{0.0719} = -0.6713 \][/tex]
#### 5. Determine Rejection Region for Z-test:
- For a significance level ([tex]\( \alpha \)[/tex]) of 0.065 and a two-tailed test, we find the critical values using the Z-distribution table or Z-scores corresponding to [tex]\( \alpha/2 \)[/tex].
[tex]\[ -z_{\alpha/2} \text{ and } z_{\alpha/2} \][/tex]
- The critical Z-values:
[tex]\[ z_{\alpha/2} = z_{0.0325} = 1.8453 \][/tex]
- Rejection region in interval notation:
[tex]\[ (-\infty, -1.8453) \cup (1.8453, \infty) \][/tex]
#### 6. Calculate the p-value:
- The p-value is found using the Z-statistic value and the normal distribution.
[tex]\[ p\text{-value} = 2 \left[ 1 - \Phi(|Z|) \right] = 2 \left[ 1 - \Phi(0.6713) \right] = 0.5020 \][/tex]
(Note: [tex]\(\Phi\)[/tex] is the cumulative distribution function of the standard normal distribution.)
#### 7. Hypothesis Test Decision:
- Compare the p-value with the significance level.
[tex]\[ p\text{-value} = 0.5020 \][/tex]
[tex]\[ \alpha = 0.065 \][/tex]
Since the p-value (0.5020) is greater than [tex]\(\alpha\)[/tex] (0.065), we do not reject the null hypothesis ([tex]\( H_0 \)[/tex]).
### Conclusion and Answers:
A. The value of the standardized test statistic: [tex]\[ -0.6713 \][/tex]
B. The rejection region for the standardized test statistic: [tex]\[ (-\infty, -1.8453) \cup (1.8453, \infty) \][/tex]
C. The p-value is: [tex]\[ 0.5020 \][/tex]
D. Your decision for the hypothesis test: Do Not Reject [tex]\( H_0 \)[/tex] (option B).
### Summary of Given Data:
- First-Years (Pop. 1)
- Sample size ([tex]\( n_1 \)[/tex]): 95
- Number answering "yes" ([tex]\( x_1 \)[/tex]): 58
- Fourth-Years (Pop. 2)
- Sample size ([tex]\( n_2 \)[/tex]): 85
- Number answering "yes" ([tex]\( x_2 \)[/tex]): 56
### Step-by-Step Solution:
#### 1. Calculate Sample Proportions:
- Proportion of first-years supporting the change ([tex]\( p_1 \)[/tex]):
[tex]\[ p_1 = \frac{x_1}{n_1} = \frac{58}{95} = 0.6105 \][/tex]
- Proportion of fourth-years supporting the change ([tex]\( p_2 \)[/tex]):
[tex]\[ p_2 = \frac{x_2}{n_2} = \frac{56}{85} = 0.6588 \][/tex]
#### 2. Combined Proportion:
- Combined sample proportion ([tex]\( p_{\text{combined}} \)[/tex]):
[tex]\[ p_{\text{combined}} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{58 + 56}{95 + 85} = \frac{114}{180} = 0.6333 \][/tex]
#### 3. Standard Error of the Difference in Proportions:
- Standard error (SE):
[tex]\[ SE = \sqrt{p_{\text{combined}} \cdot (1 - p_{\text{combined}}) \cdot \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.6333 \cdot (1 - 0.6333) \cdot \left( \frac{1}{95} + \frac{1}{85} \right)} = 0.0719 \][/tex]
#### 4. Test Statistic (Z):
[tex]\[ Z = \frac{p_1 - p_2}{SE} = \frac{0.6105 - 0.6588}{0.0719} = -0.6713 \][/tex]
#### 5. Determine Rejection Region for Z-test:
- For a significance level ([tex]\( \alpha \)[/tex]) of 0.065 and a two-tailed test, we find the critical values using the Z-distribution table or Z-scores corresponding to [tex]\( \alpha/2 \)[/tex].
[tex]\[ -z_{\alpha/2} \text{ and } z_{\alpha/2} \][/tex]
- The critical Z-values:
[tex]\[ z_{\alpha/2} = z_{0.0325} = 1.8453 \][/tex]
- Rejection region in interval notation:
[tex]\[ (-\infty, -1.8453) \cup (1.8453, \infty) \][/tex]
#### 6. Calculate the p-value:
- The p-value is found using the Z-statistic value and the normal distribution.
[tex]\[ p\text{-value} = 2 \left[ 1 - \Phi(|Z|) \right] = 2 \left[ 1 - \Phi(0.6713) \right] = 0.5020 \][/tex]
(Note: [tex]\(\Phi\)[/tex] is the cumulative distribution function of the standard normal distribution.)
#### 7. Hypothesis Test Decision:
- Compare the p-value with the significance level.
[tex]\[ p\text{-value} = 0.5020 \][/tex]
[tex]\[ \alpha = 0.065 \][/tex]
Since the p-value (0.5020) is greater than [tex]\(\alpha\)[/tex] (0.065), we do not reject the null hypothesis ([tex]\( H_0 \)[/tex]).
### Conclusion and Answers:
A. The value of the standardized test statistic: [tex]\[ -0.6713 \][/tex]
B. The rejection region for the standardized test statistic: [tex]\[ (-\infty, -1.8453) \cup (1.8453, \infty) \][/tex]
C. The p-value is: [tex]\[ 0.5020 \][/tex]
D. Your decision for the hypothesis test: Do Not Reject [tex]\( H_0 \)[/tex] (option B).
