Assuming that sample proportion is 0.35, what sample size is needed to have a 5% margin of error and 99% confidence?(1 point)
To calculate the sample size needed with a 5% margin of error and 99% confidence level when the sample proportion is 0.35, you can use the following formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-value for the desired confidence level (2.576 for 99% confidence)
p = sample proportion (0.35)
E = margin of error (0.05)
Plugging in the values:
n = (2.576^2 * 0.35 * (1-0.35)) / 0.05^2
n = (6.635776 * 0.35 * 0.65) / 0.0025
n = (1.8367416 * 0.65) / 0.0025
n = 1.1936614 / 0.0025
n = 477.46456
Therefore, a sample size of 478 is needed to have a 5% margin of error and 99% confidence level when the sample proportion is 0.35.