Determine the sample size n needed to construct a 99% confidence interval to estimate the population proportion when p= .42 and the margin of error equals 5%
To determine the sample size (n) needed to construct a 99% confidence interval to estimate the population proportion, you can use the following formula:
n = [(Z^2 * p * q) / E^2]
Where:
- n represents the required sample size
- Z represents the Z statistic corresponding to the desired level of confidence (in this case, 99%). Z can be obtained from a standard normal distribution table or using a calculator.
- p represents the estimated population proportion (in this case, p = 0.42)
- q represents the complement of the estimated population proportion (q = 1 - p)
- E represents the desired margin of error (in this case, 5% or 0.05)
First, calculate q:
q = 1 - p
q = 1 - 0.42
q = 0.58
Next, substitute the values into the formula:
n = [(Z^2 * p * q) / E^2]
n = [(Z^2 * 0.42 * 0.58) / (0.05)^2]
Since you didn't provide the value of the Z statistic for a 99% confidence level, let's assume it is approximately 2.576 (which is the Z statistic value for a 99% confidence level).
n = [(2.576^2 * 0.42 * 0.58) / (0.05)^2]
n = 1461.78
Therefore, the sample size (n) needed to construct a 99% confidence interval to estimate the population proportion, with a margin of error of 5%, is approximately 1462.