Answer:
a) We need a sample size of at least 705.
b) We need a sample size of at least 1692.
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:

90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey.
We need a sample of size at least n
n is found when
.
So






Rounding up
We need a sample size of at least 705.
b) She does not use any estimate.
Same thing as above, we just use
when do not use any estimate.






Rounding up
We need a sample size of at least 1692.