Answer
Confidence interval: $-0.0577\lt p̂_1-p̂ _2\lt-0.0093$
We are 95% confident that the difference between the two population proportions is between -0.0577 and -0.0093.
Work Step by Step
$n_1~and~p_1$ refer to experimental group and $n_2~and~p_2$ refer to control group.
For the requirements see item (a).
$p̂ _1=0.0388$ and $p̂ _2=0.0723$ (item (a))
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.0388-0.0723)-1.96\sqrt {\frac{0.0388(1-0.0388)}{696}+\frac{0.0723(1-0.0723)}{678}}=-0.0577$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.0388-0.0723)+1.96\sqrt {\frac{0.0388(1-0.0388)}{696}+\frac{0.0723(1-0.0723)}{678}}=-0.0093$
