Question

A random sample of 388 married couples found that 294 had two or more personality preferences in common. In another random sample of 566 married couples, it was found that only 38 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common.

Answer 1

Solution :

Given :

$n_1=388$ , $x_1 = 294, \ n_2 = 566, \ x_2 = 38$

Sample proportion, $(\hat p_1)=\frac{x_1}{n_1}$

                                        $=\frac{294}{388}$

                                       = 0.7577

Sample proportion, $(\hat p_2)=\frac{x_2}{n_2}$

                                        $=\frac{38}{566}$

                                       = 0.0671

For 95 % CI, z = 1.96

The confidence interval for the population proportion,

$p_1-p_2=(\hat p_1-\hat p_2)\pm z \left\{\sqrt{\frac{\hat p_1 \times (1- \hat p_1)}{n_1}+\frac{\hat p_2 \times (1- \hat p_2)}{n_2}}\right\}$

           $=(0.7577-0.0671)\pm 1.96 \left\{\sqrt{\frac{0.7577 \times (1- 0.7577)}{388}+\frac{0.0671 \times (1- 0.0671)}{566}}\right\}$

$=0.6906 \pm 1.96 \times \left 0.0241$

= $0.6906 \pm 0.0472$

Lower limit : 0.6906 - 0.0472 = 0.6434

Upper limit : 0.6906 + 0.0472 = 0.7378