Question

in a survey of 3460 adults, 1446 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 $x = 1446$ (number of successes) and $n=3460$ (sample size).
$\hat{p}=\frac{1446}{3460}\approx0.418$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.418=0.582$

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 $E$

Margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
Substitute the values:
$E = 2.576\sqrt{\frac{0.418\times0.582}{3460}}$
First calculate $\frac{0.418\times0.582}{3460}=\frac{0.243276}{3460}\approx0.0000703$
Then $\sqrt{0.0000703}\approx0.0084$
$E=2.576\times0.0084\approx0.022$

Step5: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.418$ and $E = 0.022$
$0.418-0.022=0.396$
$0.418 + 0.022=0.440$

Answer:

$(0.396,0.440)$