Question

in a survey of 2521 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 $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 1459$ (number of successes) and $n=2521$ (sample size).
$\hat{p}=\frac{1459}{2521}\approx0.579$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.579=0.421$

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 formula is $E=z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
Substitute the values: $E = 2.576\sqrt{\frac{0.579\times0.421}{2521}}$
First calculate $\frac{0.579\times0.421}{2521}=\frac{0.243759}{2521}\approx0.0000967$
Then $\sqrt{0.0000967}\approx0.00983$
$E=2.576\times0.00983\approx0.025$

Step5: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.579$ and $E = 0.025$
$0.579- 0.025=0.554$ and $0.579+0.025=0.604$

Answer:

$(0.554,0.604)$

Interpretation: We are 99% confident that the true proportion of adults who have started paying bills online in the last year lies between 0.554 and 0.604.