Many colleges and universities hold a “blood drive” for organizations such as the Red Cross. To help with

the planning of an upcoming blood drive, we want to estimate the proportion, p, of students and faculty who
would be willing to donate a pint of blood. The Red Cross would like us to estimate p with 99% confidence
such that the margin of error is no greater than 0.04. How large should the random sample be if
(A) the proportion has always been at most 0.24?
(B) no other information is given?

To estimate the proportion, p, of students and faculty who are willing to donate blood with a desired level of confidence and margin of error, we can use the formula:

n = (Z^2 * p * (1-p)) / (E^2)

where:
n is the required sample size,
Z is the Z-score corresponding to the desired confidence level,
p is the estimated proportion,
(1-p) is the complement of the estimated proportion,
and E is the desired margin of error.

(A) If we assume that the proportion has always been at most 0.24, we can use this value for p.

First, we need to find the Z-score corresponding to a 99% confidence level. The Z-score for a 99% confidence level is approximately 2.575. This value can be obtained using a Z-table or by using statistical software.

Using the given values:
p = 0.24,
Z = 2.575, and
E = 0.04.

Plugging these values into the formula, we can calculate the required sample size:

n = (2.575^2 * 0.24 * (1-0.24)) / (0.04^2)
n ≈ 245.77

Therefore, to estimate the proportion with a 99% confidence level and a margin of error no greater than 0.04, the required sample size should be at least 246.

(B) If no other information is given about the proportion, we can assume the worst-case scenario where the proportion is 0.5. This is because using 0.5 as the estimated proportion results in the largest possible required sample size, providing the most conservative estimate.

Using the worst-case scenario value:
p = 0.5,
Z = 2.575, and
E = 0.04.

Plugging these values into the formula, we can calculate the required sample size:

n = (2.575^2 * 0.5 * (1-0.5)) / (0.04^2)
n = 642.93

Therefore, if no other information is given about the proportion, the required sample size should be at least 643 to estimate the proportion with a 99% confidence level and a margin of error no greater than 0.04.