Question

Which of the following is the correct margin of error for a $99\%$confidence interval for the difference in the proportion of male and female college students who worked for pay last summer?

a. $2.5760.851(0.149)550+0.851(0.149)500$

b. $2.5760.851(0.149)1050$

c. $2.5760.880(0.120)550+0.820(0.180)500$

d. $1.9600.851(0.149)550+0.851(0.149)500$

e. $1.9600.880(0.120)550+0.820(0.180)500$

Short answer

Option (c) is correct.

Step-by-step solution

Given Information

It is given that $x_1=410$

$x_2=484$

$n_1=500$

$n_2=550$

$c=0.99$

Calculation

Sample proportion is ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{410}{500}=0.82$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{484}{550}=0.88$

When $1-\alpha =0.99$, using table $z_{a/2}=z0.005$

$z$score is $z_{a/2}=2.576$

Margin of Error is $E=z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=2.576\times \sqrt{\frac{0.82(1-0.82)}{500}+\frac{0.88(1-0.88)}{550}}$

$=2.576\times \sqrt{\frac{0.88(1-0.88)}{550}+\frac{0.82(1-0.82)}{500}}$

$=2.576\times \sqrt{\frac{0.88(0.120)}{550}+\frac{0.82(0.180)}{500}}$

Option (c) is correct.