Answer
Confidence interval: $0.019\lt p_1-p_2\lt0.097$
Notice that the confidence interval does not contain $p_1-p_2=0$ -> $p_1=p_2$. So, we can conclude that there is a difference in the proportion of individuals who believe it is morally wrong for unwed women to have children.
Work Step by Step
- Proportion;
- Independent sampling.
$N_1,n_1~and~p_1$ refer to "earned in excess" group and $N_2,n_2~and~p_2$ refer to "earned less than" group.
$p̂ _1=\frac{x_1}{n_1}=\frac{710}{1205}=0.5892$ and $p̂ _2=\frac{x_2}{n_2}=\frac{695}{1310}=0.5305$
Requirements:
$n_1p̂ _1(1-p̂ _1)=1205\times0.5892(1-0.5892)=291.7\geq10$
$n_2p̂ _2(1-p̂ _2)=1310\times0.5305(1-0.5305)=326.3\geq10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.5892-0.5305)-1.96\sqrt {\frac{0.5892(1-0.5892)}{1205}+\frac{0.5305(1-0.5305)}{1310}}=0.020$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.5892-0.5305)+1.96\sqrt {\frac{0.5892(1-0.5892)}{1205}+\frac{0.5305(1-0.5305)}{1310}}=0.097$
