Question

in a survey of 2686 adults, 1459 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

The sample proportion $\hat{p}$ is calculated as $\hat{p}=\frac{x}{n}$, where $x = 1459$ (number of successes) and $n=2686$ (sample size).
$\hat{p}=\frac{1459}{2686}\approx0.543$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.543=0.457$

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 $E$ is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$
Substitute the values:
$E=2.576\sqrt{\frac{0.543\times0.457}{2686}}$
First calculate $0.543\times0.457 = 0.248$
Then $\frac{0.248}{2686}\approx0.0000923$
$\sqrt{0.0000923}\approx0.0096$
$E=2.576\times0.0096\approx0.025$

Step5: Calculate confidence interval

The confidence interval is $(\hat{p}-E,\hat{p} + E)$
Substitute the values:
$0.543- 0.025=0.518$
$0.543+0.025 = 0.568$

Answer:

$(0.518,0.568)$

Interpretation: We are 99% confident that the true proportion of adults who have started paying bills online in the last year lies between 0.518 and 0.568.