SOLUTION: Please help me with this question.2. Suppose a presidential candidate wants to compare the preferences of registered voters in the northeastern United States with those in the sou

Question 1168061: Please help me with this question.2. Suppose a presidential candidate wants to compare the preferences of registered voters
in the northeastern United States with those in the southeastern United States. Such
a comparison would help determine where to concentrate campaign efforts. The candidate hires a professional pollster to randomly choose 1,500 registered voters in the northeast and 2,000 in the southeast and interview each to learn of her or his voting preference.
Construct a 95% confidence interval for the difference between the population proportion of voters in Northeast and the population proportion of voters in Southeast.
Results of Poll
Northeast Southeast
n1 = 1,500 n2 = 2,000
x1 = 546 x2 = 475

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to construct the 95% confidence interval for the difference between the population proportions of voters in the Northeast and the Southeast.
**1. Calculate the Sample Proportions:**
* **Northeast:**
* $n_1 = 1500$ (sample size)
* $x_1 = 546$ (number of voters with a specific preference)
* $\hat{p}_1 = \frac{x_1}{n_1} = \frac{546}{1500} = 0.364$
* **Southeast:**
* $n_2 = 2000$ (sample size)
* $x_2 = 475$ (number of voters with the same specific preference)
* $\hat{p}_2 = \frac{x_2}{n_2} = \frac{475}{2000} = 0.2375$
**2. Calculate the Difference in Sample Proportions:**
* $\hat{p}_1 - \hat{p}_2 = 0.364 - 0.2375 = 0.1265$
**3. Calculate the Standard Error of the Difference in Proportions:**
* The standard error of the difference between two independent proportions is given by:
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$
* Substitute the values:
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{\frac{0.364(1 - 0.364)}{1500} + \frac{0.2375(1 - 0.2375)}{2000}}$
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{\frac{0.364 \times 0.636}{1500} + \frac{0.2375 \times 0.7625}{2000}}$
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{\frac{0.2314}{1500} + \frac{0.1811}{2000}}$
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{0.00015427 + 0.00009055}$
$SE(\hat{p}_1 - \hat{p}_2) = \sqrt{0.00024482}$
$SE(\hat{p}_1 - \hat{p}_2) \approx 0.015647$
**4. Determine the Critical Value for a 95% Confidence Interval:**
* For a 95% confidence interval, the critical z-value ($z_{\alpha/2}$) is 1.96. This is the z-score that leaves 2.5% in each tail of the standard normal distribution.
**5. Calculate the Margin of Error:**
* The margin of error is calculated as:
$Margin \, of \, Error = z_{\alpha/2} \times SE(\hat{p}_1 - \hat{p}_2)$
$Margin \, of \, Error = 1.96 \times 0.015647$
$Margin \, of \, Error \approx 0.030668$
**6. Construct the Confidence Interval:**
* The 95% confidence interval for the difference between the population proportions is:
$(\hat{p}_1 - \hat{p}_2) \pm Margin \, of \, Error$
$0.1265 \pm 0.030668$
* **Lower bound:** $0.1265 - 0.030668 = 0.095832$
* **Upper bound:** $0.1265 + 0.030668 = 0.157168$
**Conclusion:**
The 95% confidence interval for the difference between the population proportion of voters in the Northeast and the population proportion of voters in the Southeast is approximately **(0.0958, 0.1572)**.
This means we are 95% confident that the true difference in the proportion of registered voters with the specific preference between the Northeast and the Southeast lies between 0.0958 and 0.1572. Since the entire interval is above zero, it suggests that the proportion of voters with this preference is significantly higher in the Northeast than in the Southeast.
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