Answer:
The 99% confidence interval has the higher value of z, so it has the largest margin of error
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
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
So the higher the value of Z, the higher the margin of error.
90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The 99% confidence interval has the higher value of z, so it has the largest margin of error