Question

We are interested in estimating the difference between means in two groups of individuals, μ1−μ2. We initially guess that the standard deviation for group1 and group2 will be 15 and 5 respectively. Assume that we will be able to recruit twice as many individuals in group1 as in group2, i.e., n1=2n2. Given this, what is the total number of individuals (n1+n2) required to estimate μ1−μ2 to within 0.1 units with 95% confidence?

• A. 160,000
• B. 14,000
• C. 270,000
• D. 190,000

Answer 1

Here given is

n1 = 2n2

$\sigma _1$ = 15

$\sigma _2$ = 5

Here mergin of error = 0.1

confidence interval = 0.95

critical test statistic = 1.96

standard error = sqrt [15/n12 + 5/n22]

0.1= 1.96 * sqrt [152/n1 + 52/n2]

0.1 = 1.96 * sqrt [152/2n2 + 52/n2]

n2 = 52822

n1 = 2 * 52822

total n = 52 822 * 3 = 158466 or option A is correct here