Question

180 out of 438 CA residents with a college degree oppose off-shore drilling. 126 out of 389 CA residents without a college degree oppose off-shore drilling. Create a 99% confidence interval for the difference in proportions of those with a college degree and those without a college degree who oppose off-shore drilling. O (-.00502, .16908) O 0.00102, 17309) O 1.02047, .15277) O (.03211, 14199)

Answer 1

Solution :

We construct 99% confidence interval for difference between population proportions $p_1-p_2$ ​​​​​​

From given data,

$X_1=180$​​​​​​,

M1 = 438

$X_2=126$​​​​​​,

N2 = 389

C=0.99, $\alpha=0.01$

Sample proportion 1=$p_1$= $\frac{X_1}{N_1}$

= $\frac{180}{438}$

= 0.411

Sample proportion 2 =$p_2$= $\frac{X_2 }{N_2}$

= $\frac{126}{389}$

= 0.324

The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}$ =2.576 .... (From z-table)

The 99% confidence interval calculated using formula,

$(p_1-p_2)\pm{z_c} *\sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1}+\frac{\hat{p_2}(1-\hat{p_2})}{n_2}}$

$=(0.411-0.324)\pm2.576*\sqrt{\frac{0.411*(1-0.411)}{438}+\frac{0.324*(1-0.324)}{389}}$

$=0.087\pm2.576*\sqrt{\frac{0.411*0.589}{438}+\frac{0.324*0.676}{389}}$$=0.087\pm2.576*\sqrt{\frac{0.242079}{438}+\frac{0.219024}{389}}$

$=0.087\pm2.576*\sqrt{0.0005526917808+0.0005630437017}$$=0.087\pm2.576*\sqrt{0.0011157354825}$

$=0.087\pm0.08604$

99% C. I. =(0. 087-0.08604, 0.87+0.08604)

= (0.001, 0.173)

99% confidence interval for difference in proportion of those with a college degree and those without a college degree who oppose off-shore drilling is, (0.001, 0.173)

i.e. Option B) ( 0.00102, 0.17309)