in a survey of 1002 people, 701 said that they voted in a recent presidential election. voting records show that 61% of eligible voters actually did vote. find a 99% confidence interval estimate of the proportion of people who say that they voted

The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).

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.

In which

z is the z-score that has a p-value of .

In a survey of 1002 people, 701 said that they voted in a recent presidential election.

This means that

99% confidence level

So , z is the value of Z that has a p-value of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).

y∈r, 8> y≥0

hope i !

p.s the first bit only matters if you know what that means.

p.p.s brainliest would really