Question

in a survey of 2345 adults, 748 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 sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 748$ (number of successes) and $n=2345$ (sample size).
$\hat{p}=\frac{748}{2345}\approx0.319$

Step2: Calculate $z$-value for 99% confidence level

For 99% confidence level, the $z$-value $z_{\alpha/2}$ is 2.576 (from standard normal distribution table).

Step3: Calculate the margin of error $E$

Margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.319$, $n = 2345$ and $z_{\alpha/2}=2.576$
$E=2.576\sqrt{\frac{0.319\times(1 - 0.319)}{2345}}$
First calculate inside the square - root: $0.319\times(1 - 0.319)=0.319\times0.681 = 0.217$
Then $\sqrt{\frac{0.217}{2345}}\approx\sqrt{0.0000925}\approx0.0096$
$E=2.576\times0.0096\approx0.025$

Step4: Calculate the confidence interval

The confidence interval is $(\hat{p}-E,\hat{p} + E)$
Substitute $\hat{p}=0.319$ and $E = 0.025$
Lower limit: $0.319-0.025=0.294$
Upper limit: $0.319 + 0.025=0.344$

Answer:

$(0.294,0.344)$