Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
