THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
In a recent survey of 240 teachers in Richmond,Virginia,77.2% supported standardized national testing of elementary students.In a survey of 162 teachers in Raleigh,North Carolina,64.2% supported national testing.
-What is the upper confidence limit of the 99% confidence interval for the difference between the two population proportions?
A)0.249
B)0.222
C)0.265
D)0.278

Question

Quizplus · Accepted Answer

The upper confidence limit (UCL) for the difference between two population proportions can be calculated using the formula for the confidence interval of the difference between two proportions, which includes the standard error of the difference and the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. Given:- Proportion in Richmond (P1) = 77.2% = 0.772- Proportion in Raleigh (P2) = 64.2% = 0.642- Number of teachers in Richmond (n1) = 240- Number of teachers in Raleigh (n2) = 162The standard error (SE) of the difference between two proportions is calculated using the formula:$$ SE = \sqrt{\frac{P1(1-P1)}{n1} + \frac{P2(1-P2)}{n2}} $$Plugging in the values:$$ SE = \sqrt{\frac{0.772(1-0.772)}{240} + \frac{0.642(1-0.642)}{162}} $$$$ SE \approx \sqrt{\frac{0.176}{240} + \frac{0.229}{162}} $$$$ SE \approx \sqrt{0.000733 + 0.001414} $$$$ SE \approx \sqrt{0.002147} $$$$ SE \approx 0.04633 $$The difference between the two proportions (d) is:$$ d = P1 - P2 = 0.772 - 0.642 = 0.13 $$The upper confidence limit (UCL) is calculated as:$$ UCL = d + (Z * SE) $$$$ UCL = 0.13 + (2.576 * 0.04633) $$$$ UCL \approx 0.13 + 0.119 $$$$ UCL \approx 0.249 $$Therefore, the upper confidence limit of the 99% confidence interval for the difference between the two population proportions is approximately 0.249.