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

The standard error of the difference is 0.06943

• Before calculating the standard error of the difference, we need to geet the proportion first as shown

• If in a large school district, 16 of 85 randomly selected high school seniors play a varsity sport, then the proportion of selected seniors is

• $p_1=\frac{16}{85}= 0.19$

• Similarly, if at the same district, 19 of 67 randomly selected high school juniors to play a varsity sport, then the proportion of selected students will be expressed as:

• $p_1=\frac{19}{67}= 0.28$

Get the standard error of the difference. Using the formula below;

$e=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}}$

Substitute the resulting parameters into the formula to have:

$e=\sqrt{\frac{0.19(1-0.19)}{85} +\frac{0.28(1-0.28)}{67}}\\e=\sqrt{\frac{0.19(0.81)}{85} +\frac{0.28(0.72)}{67}}\\e=\sqrt{0.001811 +0.003009}\\e=\sqrt{0.00482}\\e= 0.06943$

Hence the standard error of the difference is 0.06943

Learn more about standard error here: https://brainly.com/question/13329039