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 μ s² σ σ² x̄ p s p̂

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 99$ (number of successes, i.e., number of adults living with parents) and $n = 300$ (sample size). So, $\hat{p}=\frac{99}{300}=0.33$.

Step2: Determine z - value for 99% confidence interval

For a 99% confidence interval, the significance level $\alpha=1 - 0.99 = 0.01$. Then $\frac{\alpha}{2}=0.005$. 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 $\hat{p}=0.33$, $n = 300$, and $z_{\alpha/2}=2.576$ into the formula. First, $1-\hat{p}=1 - 0.33 = 0.67$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.33\times0.67}{300}=\frac{0.2211}{300}\approx0.000737$. And $E = 2.576\sqrt{0.000737}\approx2.576\times0.02715\approx0.0699$.

Step4: Calculate the confidence interval

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

Answer:

The 99% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents is $(0.2601, 0.3999)$