Question

Exhibit 4
A 2016 study of 70 elementary school kids found that 30 of them spent significant time outside during the summer. A similar study in 1980 of 70 kids had found that 40 of them spent significant time outside during the summer.

What is the point estimate of difference in proportion of kids who played outside in 1980 and 2016?

Answer 1

p1 = 40/70 = 0.5715

p2 = 30/70 = 0.4285

Confidence Interval = (p1-p2) +/- z*( p1(1-p1)/n1 + p2(1-p2)/n2))^0.5

let's take a 95% CI => z-value = 1.96

(0.5715 -0.4285) +/- 1.96*(0.5715*0.4285)*(1/70 + 1/70)^0.5

= 0.143 +/- 0.0811

=(0.0619, 0.2241)