In a large school district, 16 of 85 randomly selected high school seniors play a varsity sport. in the same district, 19 of 67 randomly selected high school juniors play a varsity sport. a 95 percent confidence interval for the difference between the proportion of high school seniors who play a varsity sport in the school district and high school juniors who play a varsity sport in the school district is to be calculated. what is the standard error of the difference?
Accepted Solution
A:
First of all, you need to calculate the sample proportions: p₁ = 16 / 85 = 0.188 p₂ = 19 / 67 = 0.284
The (approximated) standard error of the difference is given by the formula: SE = [tex] \sqrt{ \frac{ p_{1} (1 - p_{1}) }{ n_{1} } + \frac{p_{2} (1 - p_{2})}{n_{2}} } [/tex] = [tex]\sqrt{ \frac{ 0.188 (1 - 0.188) }{ 85 } + \frac{0.284 (1 - 0.284)}{67}} }[/tex]