A new manager,hired at a large warehouse,was told to reduce the employee sick leave.The manager introduced a new incentive program for employees with perfect attendance.The manager later compares the sick leave rate prior to the program starting (65 out of 250 employees were sick)to the sick leave rate one month into the program (60 out of 253 employees were sick).Construct a 98% confidence interval for the difference in the proportions of sick employees prior to the program and one month into the program.
A)(-0.0405,0.0862)
B)(0.0668,0.1125)
C)(-0.0764,0.1221)
D)(-0.0668,0.1125)
E)(-0.0527,0.0984)

Question

Quizplus · Accepted Answer

We can use the two-proportion z-test formula to calculate the confidence interval.

p1 = 65/250 = 0.26 (proportion of sick employees prior to the program)
p2 = 60/253 = 0.24 (proportion of sick employees one month into the program)
n1 = 250
n2 = 253

pooled proportion:
p = (x1 + x2) / (n1 + n2) = (65 + 60) / (250 + 253) = 0.253

standard error:
SE = sqrt(p*(1-p)*(1/n1 + 1/n2)) = sqrt(0.253 * 0.747 * (1/250 + 1/253)) = 0.034

z-score (for 98% confidence interval):
z = 2.33

margin of error:
ME = z * SE = 2.33 * 0.034 = 0.079

confidence interval:
(p1 - p2) ± ME = (0.26 - 0.24) ± 0.079 = 0.02 ± 0.079

Therefore, the 98% confidence interval for the difference in proportions is (-0.0668, 0.1125). The best choice is D.

A New Manager,hired at a Large Warehouse,was Told to Reduce