You wish to construct a 95% confidence interval to compare the mean measurement for two groups.A small pilot study yields sample standard deviations of 10 and 15 for Group 1 and Group 2,respectively.If we wish to obtain a margin of error of at most 5,what sample size should we take from each group? Assume equal sample sizes.
A)46
B)50
C)48
D)47
E)49

Question

Quizplus · Accepted Answer

To determine the sample size needed for each group to achieve a margin of error (ME) of at most 5 with a 95% confidence interval, we use the formula for the margin of error in comparing two means: $ME = z \times \sqrt{\frac{{s_1}^2}{n} + \frac{{s_2}^2}{n}}$, where $s_1$ and $s_2$ are the sample standard deviations for groups 1 and 2, respectively, $n$ is the sample size for each group, and $z$ is the z-score corresponding to the desired confidence level (approximately 1.96 for 95%). Rearranging the formula to solve for $n$, and plugging in the given standard deviations (10 and 15) and the desired margin of error (5), we solve for $n$. This calculation typically results in a value that needs to be rounded up to the nearest whole number to ensure the margin of error is not exceeded, leading to the correct answer of 50.

You Wish to Construct a 95% Confidence Interval to Compare