Answer :
To test the hypothesis [tex]\( H_0: (p_1 - p_2) = 0 \)[/tex] against the alternative hypothesis [tex]\( H_a: (p_1 - p_2) > 0 \)[/tex] at a significance level of [tex]\( \alpha = 0.02 \)[/tex], let's break down the solution step by step:
### Step-by-Step Solution
1. Sample Sizes & Successes:
- Sample size for population 1, [tex]\( n_1 = 70 \)[/tex]
- Sample size for population 2, [tex]\( n_2 = 70 \)[/tex]
- Number of successes in sample 1, [tex]\( x_1 = 53 \)[/tex]
- Number of successes in sample 2, [tex]\( x_2 = 43 \)[/tex]
2. Sample Proportions:
- Proportion of successes in sample 1, [tex]\( \hat{p}_1 = \frac{x_1}{n_1} = \frac{53}{70} \approx 0.7571 \)[/tex]
- Proportion of successes in sample 2, [tex]\( \hat{p}_2 = \frac{x_2}{n_2} \approx \frac{43}{70} \approx 0.6143 \)[/tex]
3. Pooled Proportion:
- The pooled sample proportion is calculated as:
[tex]\[ \hat{p}_{\text{pool}} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{53 + 43}{70 + 70} = \frac{96}{140} \approx 0.6857 \][/tex]
4. Standard Error:
- The standard error of the difference in proportions is calculated using the pooled proportion:
[tex]\[ SE = \sqrt{\hat{p}_{\text{pool}} \cdot (1 - \hat{p}_{\text{pool}}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \][/tex]
[tex]\[ SE \approx \sqrt{0.6857 \cdot (1 - 0.6857) \left(\frac{1}{70} + \frac{1}{70}\right)} \approx 0.0785 \][/tex]
5. Test Statistic:
- The test statistic (Z) for the difference in proportions is:
[tex]\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.7571 - 0.6143}{0.0785} \approx 1.8205 \][/tex]
6. P-Value:
- Since the alternative hypothesis is testing if [tex]\( p_1 \)[/tex] is greater than [tex]\( p_2 \)[/tex], the P-value is determined from the Z-table (standard normal distribution) as:
[tex]\[ P\text{-value} = 1 - \Phi(Z) \approx 1 - \Phi(1.8205) \approx 0.0343 \][/tex]
### Conclusion
(a) The test statistic is [tex]\( \boxed{1.8205} \)[/tex].
(b) The P-value is [tex]\( \boxed{0.0343} \)[/tex].
With an alpha level [tex]\( \alpha = 0.02 \)[/tex], the P-value [tex]\( 0.0343 \)[/tex] is greater than [tex]\( 0.02 \)[/tex], so we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means there is not enough evidence to conclude that the proportion of successes in population 1 is greater than the proportion of successes in population 2 at the 0.02 significance level.
### Step-by-Step Solution
1. Sample Sizes & Successes:
- Sample size for population 1, [tex]\( n_1 = 70 \)[/tex]
- Sample size for population 2, [tex]\( n_2 = 70 \)[/tex]
- Number of successes in sample 1, [tex]\( x_1 = 53 \)[/tex]
- Number of successes in sample 2, [tex]\( x_2 = 43 \)[/tex]
2. Sample Proportions:
- Proportion of successes in sample 1, [tex]\( \hat{p}_1 = \frac{x_1}{n_1} = \frac{53}{70} \approx 0.7571 \)[/tex]
- Proportion of successes in sample 2, [tex]\( \hat{p}_2 = \frac{x_2}{n_2} \approx \frac{43}{70} \approx 0.6143 \)[/tex]
3. Pooled Proportion:
- The pooled sample proportion is calculated as:
[tex]\[ \hat{p}_{\text{pool}} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{53 + 43}{70 + 70} = \frac{96}{140} \approx 0.6857 \][/tex]
4. Standard Error:
- The standard error of the difference in proportions is calculated using the pooled proportion:
[tex]\[ SE = \sqrt{\hat{p}_{\text{pool}} \cdot (1 - \hat{p}_{\text{pool}}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \][/tex]
[tex]\[ SE \approx \sqrt{0.6857 \cdot (1 - 0.6857) \left(\frac{1}{70} + \frac{1}{70}\right)} \approx 0.0785 \][/tex]
5. Test Statistic:
- The test statistic (Z) for the difference in proportions is:
[tex]\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.7571 - 0.6143}{0.0785} \approx 1.8205 \][/tex]
6. P-Value:
- Since the alternative hypothesis is testing if [tex]\( p_1 \)[/tex] is greater than [tex]\( p_2 \)[/tex], the P-value is determined from the Z-table (standard normal distribution) as:
[tex]\[ P\text{-value} = 1 - \Phi(Z) \approx 1 - \Phi(1.8205) \approx 0.0343 \][/tex]
### Conclusion
(a) The test statistic is [tex]\( \boxed{1.8205} \)[/tex].
(b) The P-value is [tex]\( \boxed{0.0343} \)[/tex].
With an alpha level [tex]\( \alpha = 0.02 \)[/tex], the P-value [tex]\( 0.0343 \)[/tex] is greater than [tex]\( 0.02 \)[/tex], so we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means there is not enough evidence to conclude that the proportion of successes in population 1 is greater than the proportion of successes in population 2 at the 0.02 significance level.
