Answer:
a) The minimum sample size is 601.
b) The minimum sample size is 2401.
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 zscore that has a pvalue of
.
The margin of error is:

For this problem, we have that:
We dont know the true proportion, so we use
, which is when we are are going to need the largest sample size.
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)
This is n for which M = 0.04. So






Rounding up
The minimum sample size is 601.
b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
Now we want n for which M = 0.02. So






The minimum sample size is 2401.