Answer:
The 99% confidence interval for the population proportion is between (0.5341, 0.5843). The interpretation is that we are 99% sure that the true proportion of adults who have started paying bills online in the last year is between these two values.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
In a survey of 2591 adults, 1449 say they have started paying bills online in the last year. This means that ![n = 2591, p = \frac{1449}{2591} = 0.5592](https://tex.z-dn.net/?f=n%20%3D%202591%2C%20p%20%3D%20%5Cfrac%7B1449%7D%7B2591%7D%20%3D%200.5592)
99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5592 - 2.575\sqrt{\frac{0.5592*0.4408}{2591}} = 0.5341](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.5592%20-%202.575%5Csqrt%7B%5Cfrac%7B0.5592%2A0.4408%7D%7B2591%7D%7D%20%3D%200.5341)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5592 + 2.575\sqrt{\frac{0.5592*0.4408}{2591}} = 0.5843](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.5592%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.5592%2A0.4408%7D%7B2591%7D%7D%20%3D%200.5843)
The 99% confidence interval for the population proportion is between (0.5341, 0.5843). The interpretation is that we are 99% sure that the true proportion of adults who have started paying bills online in the last year is between these two values.