Question

out of a sample of 500 adults aged 18 to 30, 135 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 $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 135$ (number of successes) and $n=500$ (sample size). So, $\hat{p}=\frac{135}{500}=0.27$.

Step2: Find $z -$ value

For a $90\%$ confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z -$ value $z_{\alpha/2}=z_{0.05}$. From the standard normal table, $z_{0.05} = 1.645$.

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.27$, $n = 500$, and $z_{\alpha/2}=1.645$ into the formula:

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Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.27$ and $E = 0.0327$:
$0.27-0.0327=0.2373$ and $0.27 + 0.0327=0.3027$

Answer:

$0.2373< p<0.3027$