Question

in a survey of 2350 adults, 723 say they believe in ufos. construct a 99% confidence interval for the population proportion of adults who believe in ufos. a 99% confidence interval for the population proportion is (, ). (round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 723$ (number of successes) and $n=2350$ (sample size).
$\hat{p}=\frac{723}{2350}\approx0.308$

Step2: Find the critical value $z_{\alpha/2}$

For a $99\%$ confidence interval, $\alpha = 1 - 0.99=0.01$, so $\alpha/2=0.005$.
From the standard normal distribution table, $z_{\alpha/2}=z_{0.005} = 2.576$

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 $\hat{p}=0.308$, $n = 2350$, and $z_{\alpha/2}=2.576$
$\hat{p}(1-\hat{p})=0.308\times(1 - 0.308)=0.308\times0.692 = 0.213136$
$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.213136}{2350}}\approx\sqrt{9.07\times10^{-5}}\approx0.00952$
$E=2.576\times0.00952\approx0.0245$

Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E$\hat{p}-E=0.308 - 0.0245=0.2835$
$\hat{p}+E=0.308+0.0245 = 0.3325$

Answer:

$(0.284,0.333)$