According to the information given, using the z-distribution, it is found that a sample of 1066 is needed.
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, according to the z-distribution.
In which
z is the z-score that has a p-value of .
The margin of error is:
In this problem, the estimate is of 52%, hence .
95% confidence level
So , z is the value of Z that has a p-value of , so .
The minimum sample size is <u>n for which M = 0.03</u>, hence:
Rounding up, a sample of 1066 is needed.
A similar problem is given at brainly.com/question/14515126