Solution :
Given :
[tex]$n_1=388$[/tex] , [tex]$x_1 = 294, \ n_2 = 566, \ x_2 = 38$[/tex]
Sample proportion, [tex]$(\hat p_1)=\frac{x_1}{n_1}$[/tex]
[tex]$=\frac{294}{388}$[/tex]
= 0.7577
Sample proportion, [tex]$(\hat p_2)=\frac{x_2}{n_2}$[/tex]
[tex]$=\frac{38}{566}$[/tex]
= 0.0671
For 95 % CI, z = 1.96
The confidence interval for the population proportion,
[tex]$p_1-p_2=(\hat p_1-\hat p_2)\pm z \left\{\sqrt{\frac{\hat p_1 \times (1- \hat p_1)}{n_1}+\frac{\hat p_2 \times (1- \hat p_2)}{n_2}}\right\}$[/tex]
[tex]$=(0.7577-0.0671)\pm 1.96 \left\{\sqrt{\frac{0.7577 \times (1- 0.7577)}{388}+\frac{0.0671 \times (1- 0.0671)}{566}}\right\}$[/tex]
[tex]$=0.6906 \pm 1.96 \times \left 0.0241$[/tex]
= [tex]$0.6906 \pm 0.0472 $[/tex]
Lower limit : 0.6906 - 0.0472 = 0.6434
Upper limit : 0.6906 + 0.0472 = 0.7378
