Question

out of a sample of 100 adults aged 18 to 30, 34 still lived with their parents. based on this, construct a 95% 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 <

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 34$ (number of successes) and $n=100$ (sample size). So, $\hat{p}=\frac{34}{100}=0.34$.

Step2: Find critical value

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$. Then $\alpha/2=0.025$. The critical value $z_{\alpha/2}$ is such that $P(Z>z_{\alpha/2})=\alpha/2$. From the standard normal table, $z_{\alpha/2}=z_{0.025} = 1.96$.

Step3: Calculate margin of error

The margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.34$, $n = 100$, and $z_{\alpha/2}=1.96$ into the formula:

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Step4: Construct confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E$0.34-0.0929=0.2471$ and $0.34 + 0.0929=0.4329$

Answer:

$0.2471