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.) interpret your results. choose the correct answer below. a. with 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. b. the endpoints of the given confidence interval show that adults pay bills online 99% of the time. c. with 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The 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 $q = 1-\hat{p}$

$q=1 - 0.429=0.571$

Step3: Find the critical value $z_{\alpha/2}$

For a 99% confidence interval, $\alpha=1 - 0.99 = 0.01$, and $\alpha/2=0.005$.
From the standard normal distribution table, $z_{\alpha/2}=z_{0.005} = 2.576$

Step4: Calculate the margin of error $E$

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

Step5: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.429$ and $E = 0.022$
$0.429-0.022=0.407$
$0.429 + 0.022=0.451$

A confidence interval for a population proportion gives a range of values within which the true population proportion is likely to lie. A 99% confidence level means that if we were to construct many such intervals from different samples, 99% of them would contain the true population proportion.
Option A is incorrect because it refers to the sample proportion (the confidence interval is for the population proportion). Option B is incorrect because it misinterprets the confidence level as a proportion of time adults pay bills online. Option C is correct as it correctly states that with 99% confidence, the population proportion lies within the interval endpoints.

Answer:

A 99% confidence interval for the population proportion is $(0.407,0.451)$

For the interpretation: