Question

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?

Answer 1

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 = $\sqrt{ \frac{ p_{1} (1 - p_{1}) }{ n_{1} } + \frac{p_{2} (1 - p_{2})}{n_{2}} }$
     =  $\sqrt{ \frac{ 0.188 (1 - 0.188) }{ 85 } + \frac{0.284 (1 - 0.284)}{67}} }$

    = 0.0695

Answer 2

Answer:

Standard Error of the difference = 0.0695

Step-by-step explanation:

Its given that : In a large school district, 16 of 85 randomly selected high school seniors play a varsity sport

$\implies P_1=\frac{16}{85}\\\\n_1=85$

Also, in the same district, 19 of 67 randomly selected high school juniors play a varsity sport

$\implies P_1=\frac{19}{67}\\\\n_2=67$

Now, finding the standard error of the difference :

$\text{Standard Error = }\sqrt{\frac{P_1(1-P_1)}{n_1}+\frac{P_2(1-P_2)}{n_2}}\\\\\implies \text{Standard Error = }\sqrt{\frac{\frac{16}{85}(1-\frac{16}{85})}{85}+\frac{\frac{19}{67}(1-\frac{19}{67})}{67}}\\\\\implies\text{Standard Error = }0.0695$

Hence, Standard Error of the difference = 0.0695