Question

out of a sample of 500 adults aged 18 to 30, 125 still lived with their parents. based on this, construct a 90% 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.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 125$ (number of successes) and $n=500$ (sample size). So, $\hat{p}=\frac{125}{500}=0.25$.

Step2: Find z - value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. The $z$-value $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.25$, $n = 500$ and $z_{\alpha/2}=1.645$ into the formula.
$E=1.645\sqrt{\frac{0.25\times(1 - 0.25)}{500}}=1.645\sqrt{\frac{0.25\times0.75}{500}}=1.645\sqrt{\frac{0.1875}{500}}=1.645\sqrt{0.000375}\approx1.645\times0.0194\approx0.0320$.

Step4: Construct confidence interval

The confidence interval is $\hat{p}-E$0.25- 0.0320

Answer:

Lower limit: $0.2180$, Upper limit: $0.2820$