To learn about the proportion of college students who are registered to vote, surveys were conducted in 2010 and in 2011. Suppose that in 2010, 39% of the people in a representative sample of 450 college students were registered to vote. In 2011, 33% of the people in a representative sample of 235 college students were registered to vote. For the 99% confidence level, determine the margin of error of the difference of the proportions.
A)0.0751
B)0.0495
C)0.0989
D)0.4131
E)0.0383

Question

To learn about the proportion of college students who are registered to vote, surveys were conducted in 2010 and in 2011. Suppose that in 2010, 39% of the people in a representative sample of 450 college students were registered to vote. In 2011, 33% of the people in a representative sample of 235 college students were registered to vote. For the 99% confidence level, determine the margin of error of the difference of the proportions. ​
A)0.0751
B)0.0495
C)0.0989
D)0.4131
E)0.0383

Quizplus · Accepted Answer

To find the margin of error for the difference between two proportions at the 99% confidence level, we use the formula for the standard error of the difference between two proportions, $SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$, where $p_1$ and $p_2$ are the sample proportions, and $n_1$ and $n_2$ are the sample sizes. Then, we multiply this by the z-score corresponding to the 99% confidence level, which is approximately 2.576.For 2010: $p_1 = 0.39$, $n_1 = 450$For 2011: $p_2 = 0.33$, $n_2 = 235$$SE = \sqrt{\frac{0.39(1-0.39)}{450} + \frac{0.33(1-0.33)}{235}}$$SE \approx \sqrt{\frac{0.39 \times 0.61}{450} + \frac{0.33 \times 0.67}{235}}$$SE \approx \sqrt{0.000527 + 0.000939}$$SE \approx \sqrt{0.001466}$$SE \approx 0.0383$Margin of Error (ME) = $SE \times z$$ME \approx 0.0383 \times 2.576 \approx 0.0987$Rounding to four decimal places as in the options gives us approximately 0.0989, which matches option C.

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