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

Abimael Reyes · Accepted Answer

Final Answer : B, Explanation :  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 = 0.04Squaring both sides, simplifying, and solving for n, we get:n = (1.645 / (2 * 0.04))^2n = 846.41We 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 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

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Learning Objectives

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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) 423B) 846C) 1984D) 845E) 422

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Learning Objectives

Learning Objectives

Related questions

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