Question

we wish to estimate what percent of adult residents in a certain county are parents. out of 300 adult residents sampled, 234 had kids. based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. give your answers as decimals rounded to three places.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 234$ (number of successes, i.e., adults with kids) and $n=300$ (sample size). So, $\hat{p}=\frac{234}{300}=0.78$.

Step2: Find z - value for 99% confidence interval

For a 99% confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$. Then $\alpha/2=0.005$. The $z$-value $z_{\alpha/2}=z_{0.005}$. Looking up in the standard - normal table, $z_{0.005} = 2.576$.

Step3: Calculate the 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 $\hat{p}=0.78$, $n = 300$, and $z_{\alpha/2}=2.576$ into the formula. First, $1-\hat{p}=1 - 0.78=0.22$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.78\times0.22}{300}=\frac{0.1716}{300}\approx0.000572$. And $E = 2.576\times\sqrt{0.000572}\approx2.576\times0.0239\approx0.062$.

Step4: Construct the confidence interval

The confidence interval for the proportion $p$ is $\hat{p}-E

Answer:

$0.718 < p<0.842$