Question

The director of Human Resources at a large company wishes to determine if newly instituted training has been effective in reducing work-related injuries. In a random sample of 100 employees taken from the six months before this training began (group 1), she found 15 had suffered a work-related injury. Using a random sample of 150 employees from the six months since the training began (group 2), she found 12 had suffered a work-related injury. The 99% large sample confidence interval estimate of the difference in population proportions, p1 – p2, is:

(–0.038, 0.178). (–0.006, 0.030). (–0.012, 0.152). (–0.028, 0.168).

Answer 1

Here, , n1 = 100 , n2 = 150
p1cap = 0.15 , p2cap = 0.08

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.15 * (1-0.15)/100 + 0.08*(1-0.08)/150)
SE = 0.042

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.15 - 0.08 - 2.58*0.042, 0.15 - 0.08 + 2.58*0.042)
CI = (-0.038 , 0.178)

Option A