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Use the normal distribution to find a confidence interval for a difference in proportions $p_{1}-p_{2}$ given the relevant sample results. Assume the results come from random samples.
A $90 \%$ confidence interval for $p_{1}-p_{2}$ given that $\hat{p}_{1}=0.75$ with $n_{1}=540$ and $\hat{p}_{2}=0.69$ with $n_{2}=230$
Give the best estimate for $p_{1}-p_{2}$, the margin of error, and the confidence interval.
Round your answer for the best estimate to two dechlal places and round your answers for the margin of error and the confidence interval to three decimal places.
Best estimate:
i
Margin of error:
i
Confidence interval : to
i
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Step 1 ：The best estimate for \(p_{1}-p_{2}\) is the difference in the sample proportions, \(\hat{p}_{1}-\hat{p}_{2}\). So, the best estimate = \(0.75 - 0.69 = 0.06\).

Step 2 ：Next, calculate the standard error (SE) for the difference in proportions. The formula for the standard error of the difference in proportions is: \(SE = \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\).

Step 3 ：Substitute the given values into the formula: \(SE = \sqrt{\frac{0.75(1-0.75)}{540} + \frac{0.69(1-0.69)}{230}} = \sqrt{\frac{0.1875}{540} + \frac{0.2139}{230}} = \sqrt{0.000347 + 0.000930} = 0.035\).

Step 4 ：For a 90% confidence interval, the z-score is 1.645. The margin of error (ME) is calculated as the product of the z-score and the standard error. So, \(ME = 1.645 * 0.035 = 0.057575\).

Step 5 ：Finally, calculate the confidence interval as the best estimate ± the margin of error. So, the confidence interval is \(0.06 ± 0.058\).

Step 6 ：\(\boxed{\text{Therefore, the best estimate for } p_{1}-p_{2} \text{ is } 0.06, \text{ the margin of error is } 0.058, \text{ and the confidence interval is from } 0.002 \text{ to } 0.118.}\)