Suppose that .06 of each of two populations possess a given characteristic.Samples of size 400 are randomly drawn from each population.What is the probability that the differences in sample proportions will be greater than 0.02?
A)0.4535
B)0.9535
C)0.1170
D)0.5465
E)0.4650

Question

Quizplus · Accepted Answer

To solve this, we use the formula for the standard error (SE) of the difference between two proportions: SE = sqrt[(p1(1-p1)/n1) + (p2(1-p2)/n2)], where p1 and p2 are the proportions in the two populations, and n1 and n2 are the sample sizes. Given p1 = p2 = 0.06 and n1 = n2 = 400, SE = sqrt[(0.06*0.94/400) + (0.06*0.94/400)] = sqrt[2*(0.06*0.94/400)] = sqrt[0.1128/400] = sqrt[0.000282] ≈ 0.0168. The Z-score for a difference of 0.02 is Z = (0.02 - 0)/0.0168 ≈ 1.19. Using the standard normal distribution, the probability of a Z-score being more than 1.19 is approximately 0.1170 (one-tailed test), indicating the probability that the difference in sample proportions will be greater than 0.02.

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