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.

Explanation:

Step1: Calculate sample proportion

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

Step2: Find critical - value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. Then $\alpha/2=0.005$. The critical - value $z_{\alpha/2}$ corresponding to a right - tail area of 0.005 in the standard normal distribution is $z_{0.005} = 2.576$.

Step3: Calculate 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:

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Step4: Construct confidence interval

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

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Answer:

Lower limit: $0.2592$, Upper limit: $0.4008$