Question

in a survey of 3362 adults, 1443 say they have started paying bills online in the last year. construct a 99% confidence interval for the population proportion. interpret the results. 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 $x = 1443$ (number of successes) and $n=3362$ (sample size).
$\hat{p}=\frac{1443}{3362}\approx0.429$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.429=0.571$

Step3: Find $z$-value for 99% confidence interval

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

Step4: Calculate margin of error $E$

The formula for margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
Substitute the values: $E = 2.576\sqrt{\frac{0.429\times0.571}{3362}}$
First calculate $\frac{0.429\times0.571}{3362}=\frac{0.2459}{3362}\approx0.0000732$
Then $\sqrt{0.0000732}\approx0.00856$
$E=2.576\times0.00856\approx0.022$

Step5: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute the values: $0.429-0.022 < p<0.429 + 0.022$
$0.407

Answer:

$(0.407,0.451)$