Answer :
Certainly! Let's go through the steps to determine if the normal sampling distribution can be used and then test the claim about the difference between the two population proportions at the given level of significance ([tex]\(\alpha = 0.01\)[/tex]) using the provided sample statistics.
### Step 1: Check if the normal approximation can be used
To use the normal approximation for the sampling distribution, we need to verify the following conditions:
- [tex]\( n_1 \cdot \hat{p}_1 \geq 5 \)[/tex]
- [tex]\( n_1 \cdot (1 - \hat{p}_1) \geq 5 \)[/tex]
- [tex]\( n_2 \cdot \hat{p}_2 \geq 5 \)[/tex]
- [tex]\( n_2 \cdot (1 - \hat{p}_2) \geq 5 \)[/tex]
Where:
- [tex]\(\hat{p}_1\)[/tex] is the sample proportion of successes for the first sample.
- [tex]\(\hat{p}_2\)[/tex] is the sample proportion of successes for the second sample.
Given:
- [tex]\(x_1 = 105\)[/tex]
- [tex]\(n_1 = 168\)[/tex]
- [tex]\(x_2 = 47\)[/tex]
- [tex]\(n_2 = 199\)[/tex]
We first calculate the sample proportions:
[tex]\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{105}{168} \approx 0.625 \][/tex]
[tex]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{47}{199} \approx 0.236 \][/tex]
Next, we check the conditions:
[tex]\[ n_1 \cdot \hat{p}_1 = 168 \cdot 0.625 = 105 \geq 5 \][/tex]
[tex]\[ n_1 \cdot (1 - \hat{p}_1) = 168 \cdot (1 - 0.625) = 168 \cdot 0.375 = 63 \geq 5 \][/tex]
[tex]\[ n_2 \cdot \hat{p}_2 = 199 \cdot 0.236 \approx 47 \geq 5 \][/tex]
[tex]\[ n_2 \cdot (1 - \hat{p}_2) = 199 \cdot (1 - 0.236) = 199 \cdot 0.764 \approx 152 \geq 5 \][/tex]
Since all the conditions are satisfied, we can use the normal approximation.
### Step 2: Formulate the hypotheses
The null and alternative hypotheses are:
- [tex]\(H_0: p_1 = p_2\)[/tex] (The population proportions are equal)
- [tex]\(H_a: p_1 \neq p_2\)[/tex] (The population proportions are not equal)
### Step 3: Calculate the test statistic
We need to calculate the pooled sample proportion [tex]\(\hat{p}\)[/tex], the standard error (SE), and the z-test statistic:
[tex]\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{105 + 47}{168 + 199} \approx 0.414 \][/tex]
[tex]\[ SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.414 \cdot (1 - 0.414) \left(\frac{1}{168} + \frac{1}{199}\right)} \approx 0.0516 \][/tex]
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.625 - 0.236}{0.0516} \approx 7.534 \][/tex]
### Step 4: Determine the critical value
For a two-tailed test at [tex]\(\alpha = 0.01\)[/tex], the critical value is found using the standard normal distribution.
[tex]\[ z_{\text{critical}} = \pm 2.576 \][/tex]
### Step 5: Make the decision
Compare the absolute value of the test statistic with the critical value:
[tex]\[ |z| = 7.534 > 2.576 \][/tex]
Since [tex]\(|z|\)[/tex] is greater than the critical value, we reject the null hypothesis.
### Conclusion
There is sufficient evidence at the [tex]\( \alpha = 0.01 \)[/tex] significance level to reject the claim that the population proportions [tex]\( p_1 \)[/tex] and [tex]\( p_2 \)[/tex] are equal. Therefore, we conclude that there is a significant difference between the two population proportions.
### Step 1: Check if the normal approximation can be used
To use the normal approximation for the sampling distribution, we need to verify the following conditions:
- [tex]\( n_1 \cdot \hat{p}_1 \geq 5 \)[/tex]
- [tex]\( n_1 \cdot (1 - \hat{p}_1) \geq 5 \)[/tex]
- [tex]\( n_2 \cdot \hat{p}_2 \geq 5 \)[/tex]
- [tex]\( n_2 \cdot (1 - \hat{p}_2) \geq 5 \)[/tex]
Where:
- [tex]\(\hat{p}_1\)[/tex] is the sample proportion of successes for the first sample.
- [tex]\(\hat{p}_2\)[/tex] is the sample proportion of successes for the second sample.
Given:
- [tex]\(x_1 = 105\)[/tex]
- [tex]\(n_1 = 168\)[/tex]
- [tex]\(x_2 = 47\)[/tex]
- [tex]\(n_2 = 199\)[/tex]
We first calculate the sample proportions:
[tex]\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{105}{168} \approx 0.625 \][/tex]
[tex]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{47}{199} \approx 0.236 \][/tex]
Next, we check the conditions:
[tex]\[ n_1 \cdot \hat{p}_1 = 168 \cdot 0.625 = 105 \geq 5 \][/tex]
[tex]\[ n_1 \cdot (1 - \hat{p}_1) = 168 \cdot (1 - 0.625) = 168 \cdot 0.375 = 63 \geq 5 \][/tex]
[tex]\[ n_2 \cdot \hat{p}_2 = 199 \cdot 0.236 \approx 47 \geq 5 \][/tex]
[tex]\[ n_2 \cdot (1 - \hat{p}_2) = 199 \cdot (1 - 0.236) = 199 \cdot 0.764 \approx 152 \geq 5 \][/tex]
Since all the conditions are satisfied, we can use the normal approximation.
### Step 2: Formulate the hypotheses
The null and alternative hypotheses are:
- [tex]\(H_0: p_1 = p_2\)[/tex] (The population proportions are equal)
- [tex]\(H_a: p_1 \neq p_2\)[/tex] (The population proportions are not equal)
### Step 3: Calculate the test statistic
We need to calculate the pooled sample proportion [tex]\(\hat{p}\)[/tex], the standard error (SE), and the z-test statistic:
[tex]\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{105 + 47}{168 + 199} \approx 0.414 \][/tex]
[tex]\[ SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.414 \cdot (1 - 0.414) \left(\frac{1}{168} + \frac{1}{199}\right)} \approx 0.0516 \][/tex]
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.625 - 0.236}{0.0516} \approx 7.534 \][/tex]
### Step 4: Determine the critical value
For a two-tailed test at [tex]\(\alpha = 0.01\)[/tex], the critical value is found using the standard normal distribution.
[tex]\[ z_{\text{critical}} = \pm 2.576 \][/tex]
### Step 5: Make the decision
Compare the absolute value of the test statistic with the critical value:
[tex]\[ |z| = 7.534 > 2.576 \][/tex]
Since [tex]\(|z|\)[/tex] is greater than the critical value, we reject the null hypothesis.
### Conclusion
There is sufficient evidence at the [tex]\( \alpha = 0.01 \)[/tex] significance level to reject the claim that the population proportions [tex]\( p_1 \)[/tex] and [tex]\( p_2 \)[/tex] are equal. Therefore, we conclude that there is a significant difference between the two population proportions.
