You wish to construct a 90% confidence interval to compare two proportions.If we wish to obtain a margin of error of at most 0.04,what sample size should we take from each group? Assume equal sample sizes.
A)423
B)846
C)1984
D)845
E)422

Question

Quizplus · Accepted Answer

The margin of error for a 90% confidence interval can be calculated using the formula:

ME = 1.645 * sqrt [ (p1 * (1 - p1) / n) + (p2 * (1 - p2) / n) ]

where ME is the margin of error, p1 and p2 are the sample proportions being compared, and n is the sample size for each group.

Since we don't have any information about the sample proportions, we can assume that the worst-case scenario is when both proportions are equal to 0.5. Thus, the margin of error formula simplifies to:

ME = 1.645 * sqrt [ (0.5 * (1 - 0.5) / n) + (0.5 * (1 - 0.5) / n) ]
ME = 1.645 * sqrt ( 0.25 / n )
ME = 0.04

Squaring both sides, simplifying, and solving for n, we get:

n = (1.645 / (2 * 0.04))^2
n = 846.41

We need to round up to the nearest whole number, so the answer is 846. We need to take a sample size of 846 from each group. Thus, the best choice is B) 846.

You Wish to Construct a 90% Confidence Interval to Compare