Final answer:
To estimate the minimum sample size for a 99% confidence level and 4% margin of error without prior information, we would need 1,041 participants. Using a prior estimate of 28%, we would need a sample size of 760 participants. The sample size is larger when no prior estimate is available due to maximum variance assumption at a proportion of 0.5.
Step-by-step explanation:
To estimate the minimum sample size needed for a population proportion with 99% confidence and a margin of error of 4%, the standard formula for sample size estimation is used:
n = (Z^2 * p * (1 - p)) / E^2
Where:
- n is the sample size
- Z is the Z-score corresponding to the desired level of confidence
- p is the estimated proportion of the population
- E is the margin of error
For part a, with no preliminary estimate available, we use p = 0.5 because this gives the maximum sample size. The Z-score for a 99% confidence interval is approximately 2.576.
n = (2.576^2 * 0.5 * 0.5) / 0.04^2
This calculation results in a sample size of approximately 1,040.96, which we would round up to 1,041 participants as we can't have a fraction of a participant.
For part b, using a prior study estimate of 28% (p = 0.28):
n = (2.576^2 * 0.28 * (1 - 0.28)) / 0.04^2
This results in a sample size of approximately 759.69, rounded up to 760 participants.
In part c, when comparing the results from parts a and b, the initial estimate without prior information required a larger sample size than when a preliminary estimate was available. This is because the highest variance of a proportion occurs at 0.5, leading to the largest sample size requirement when no prior estimate is used.