Answer
-.18 to -.1006
Work Step by Step
We will estimate the population proportion with 95% confidence using a two-sample z-interval with the given information:
$n_{1}$ = 1084
$x_{1}$ = 520
$n_{2}$ = 1308
$x_{2}$ = 811
sample proportion=$\frac{Number of Succeses}{Pop. Size}$
$\frac{520}{1084}$ = .479 | $\frac{811}{1308}$ = .62
z score is 1.96 since the confidence interval is 95%
$p_{1}-$$p_{2}$ +1.96 sqrt ($\frac{p_{1}(1- p_{1})}{n_{1}}$+ $\frac{p_{2}(1- p_{2})}{n_{2}}$) =
(.479 -.62 )+1.96 sqrt ($\frac{.479(1- .479)}{1084}$+ $\frac{.62(1- .62)}{1308}$) =-.18
(.479 -.62 )-1.96 sqrt ($\frac{.479(1- .479)}{1084}$+ $\frac{.62(1- .62)}{1308}$) =-.1006
This means that we are 95% confident that the true difference between the proportion of male and female internet users who used the internet to obtain health information about a medical condition in the last year is between -.18 and -.1006.
