A surprising number of young adults (ages $19$to $25$) still live in their parents’ homes. A random sample by the National Institutes of Health included $2253$men and $2629$women in this age group. The survey found that $986$of the men and $923$of the women lived with their parents.

(a) Construct and interpret a $99\%$confidence interval for the difference in population proportions (men minus women).

(b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

Given

$n_1=$Sample size $=2253$

$x_1=$Number of successes $=986$

$n_2=$Sample size $=2629$

$x_2=$Number of successes $=923$

$c=$Confidence level$=99\%=0.99$

Conditions

Random, Independent $10\%$condition , and Normal are the three conditions for computing a hypothesis test for the population proportion $p$. (large counts). Because the samples are separate random samples, I'm satisfied.

Independent: Satisfied because the $2253$U.S. men and the $2629$U.S. women account for less than $10\%$of the total population of the United States. Normal: Happy because the first sample has 983 successes and $2253-983=1270$failures, while the second sample has $923$successes and $2629-923=1706$failures, all of which are at least $10$percent .

Because all of the conditions have been met, a confidence interval for $p_1-p_2$can be calculated.

Confidence interval

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{986}{2253} \\ \approx 0.438 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{923}{2629} \\ \approx 0.351 \end{gathered}$$

For confidence level $1-\alpha =0.99$, determine $z_{\alpha /2}=z_{0.005}$using table II (look up $0.005$in the table, the $z$-score is then the found $z$-score with opposite sign):

$z_{\alpha /2}=2.575$

The endpoints of the confidence interval for $p_1-p_2$are then:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.438-0.351)-2.575\sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}$

$\approx 0.051$

$\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.438-0.351)+2.575\sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}$

$\approx 0.123$

Given

$n_1=$Sample size $=2253$

$x_1=$Number of successes $=986$

$n_2=$Sample size $=2629$

$x_2=$Number of successes $=923$

$c=$Confidence level $=99\%=0.99$

Result exercise part(a)

$(0.051,0.123)$

The confidence interval does not contain $0$, then it is very unlikely that the population proportions are equal and thus there is convincing evidence of a difference between the population proportions.