Question

Based on telephone interviews in October 2017 of a random sample of 1028 adults, Gallup News Service reported that 617 favored stricter gun control laws. Interviews of 1041 adults in March 2018 showed that 694 favored stricter gun control laws. Let p1 and p2 be the respective proportions in 2017 and 2018 who favored stricter gun control laws.

(a) Give point estimates of p1 and p2.

(b) Find an approximate 95% confidence interval for p1-p2.

Answer 1

a) The points estimates for p1 and p2 are computed here as:

p1 = 617 / 1028 = 0.6002
p2 = 694/ 1041 = 0.6667

Therefore 0.6002 and 0.6667 are the required point estimates here.

b) From standard normal tables, we have:
P(-1.96 < Z < 1.96) = 0.95

The pooled proportion first is computed here as:

Pa 617 + 694 F = 0.6336 1028 + 1041

Therefore now the standard error is computed here as:

w11) SE = P(1-P) = 0.6336(1 – 0.6336)(1028 + 1041) = 0.0212

The 95% confidence interval now is computed here as:

P1 - p2 SE
$0.6002 - 0.6667 \pm 1.96*0.0212$

$-0.0665 \pm 0.0415$

This is the required 95% confidence interval for the difference in proportions.