Using the z-distribution, it is found that the needed sample sizes are:
a) 242
b) 1842
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 .
The margin of error is:
99% confidence level, hence, z is the value of Z that has a p-value of
, so
.
Item a:
The estimate is:
The sample size is <u>n for which M = 0.03</u>, then:
Rounding up, a sample of 242 is needed.
Item b:
No prior estimates, hence is used.
Rounding up, a sample of 1842 is needed.
For more on the z-distribution, you can check brainly.com/question/25404151