Question

in a survey of 3333 adults, 1474 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 $x = 1474$ (number of successes) and $n=3333$ (sample size).
$\hat{p}=\frac{1474}{3333}\approx0.442$

Step2: Calculate $q$

$q = 1-\hat{p}=1 - 0.442=0.558$

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

For 99% confidence interval, $z_{\alpha/2}=2.576$ (from standard normal distribution table)

Step4: Calculate margin of error ($E$)

Margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
$E = 2.576\sqrt{\frac{0.442\times0.558}{3333}}$
First calculate $\frac{0.442\times0.558}{3333}=\frac{0.246636}{3333}\approx0.000074$
$\sqrt{0.000074}\approx0.0086$
$E=2.576\times0.0086\approx0.022$

Step5: Calculate confidence interval

Confidence interval is $\hat{p}-E$0.442 - 0.022=0.420$
$0.442+0.022 = 0.464$

Answer:

$(0.420,0.464)$