Question

we wish to estimate what percent of adult residents in a certain county are parents. out of 600 adult residents sampled, 486 had kids. based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county. give your answers as decimals rounded to three places. < p <

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 486$ (number of successes, i.e., number of parents) and $n=600$ (sample size). So, $\hat{p}=\frac{486}{600}=0.81$.

Step2: Determine z - value for 95% confidence interval

For a 95% confidence interval, the critical z - value $z_{\alpha/2}$ is 1.96.

Step3: Calculate margin of error

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.81$, $n = 600$, and $z_{\alpha/2}=1.96$ into the formula. First, calculate $1-\hat{p}=1 - 0.81=0.19$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.81\times0.19}{600}=\frac{0.1539}{600}\approx0.0002565$. $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{0.0002565}\approx0.016$. And $E=1.96\times0.016 = 0.03136$.

Step4: Calculate confidence interval

The lower limit of the confidence interval is $\hat{p}-E=0.81 - 0.03136=0.779$. The upper limit is $\hat{p}+E=0.81 + 0.03136=0.841$.

Answer:

$0.779 < p < 0.841$