Question

In one school district, there are 135 elementary school (K-5) teachers, of which 19 are male (or male-identifying). In a neighboring school district, there are 117 elementary teachers, of which 20 are male. A policy researcher would like to calculate the 90% confidence interval for the difference in proportions of male teachers.

To keep the signs consistent for this problem, we will calculate all differences as p1−p2. That is, start with the percentage from the first school district and then subtract the percentage from the second district. Failing to do so may end up with “correct” answers being marked as wrong.

Point estimate for the percentage males in the first district:
      ˆp1=
Point estimate for the percentage males in the second district:
      ˆp2=
Point estimate for the difference in percentages between the two districts:
      ˆp1−ˆp2=

Estimate of the standard error for this sampling distribution (distribution of differences):
      √ˆp1(1−ˆp1)n1+ˆp2(1−ˆp2)n2=

Critical value for the 90% confidence level:
      zc.v.=

90% margin of error:
      M.E.=

90% confidence interval:

≤p1−p2≤

Answer 1

p1cap = X1/N1 = 19/135 = 0.1407

p2cap = X2/N2 = 20/117 = 0.1709

p1cap - p2cap = 0.1407 - 0.1709 = -0.0302

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.1407 * (1-0.1407)/135 + 0.1709*(1-0.1709)/117)
SE = 0.0459

For 0.9 CI, z-value = 1.64

ME = z * SE

= 1.64 * 0.0459
= 0.0753

Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.1407 - 0.1709 - 1.64*0.0459, 0.1407 - 0.1709 + 1.64*0.0459)
CI = (-0.1055 , 0.0451)

-0.1055 ≤p1−p2≤ 0.0451