Question

out of a sample of 300 adults aged 18 to 30, 99 still lived with their parents. based on this, construct a 99% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents. give your answers rounded to 4 decimal places. < select an answer < next question

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 99$ (number of successes) and $n=300$ (sample - size). So, $\hat{p}=\frac{99}{300}=0.33$.

Step2: Find z - value for 99% confidence interval

The confidence level is 99%, so the significance level $\alpha = 1 - 0.99=0.01$. Then $\alpha/2=0.005$. The z - value $z_{\alpha/2}=z_{0.005}=2.576$.

Step3: Calculate the margin of error

The formula for the margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.33$, $n = 300$, and $z_{\alpha/2}=2.576$ into the formula:

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Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ELower limit: $0.33-0.0698 = 0.2602$.
Upper limit: $0.33 + 0.0698=0.3998$.

Answer:

$0.2602< p<0.3998$