To find the sample size necessary for a given confidence level and margin of error, we can use the formula:
n = (Z^2 * p * (1 - p)) / E^2
Where:
- n is the sample size needed
- Z is the z-score corresponding to the desired confidence level
- p is the estimated proportion
- E is the margin of error
In this case, the researcher wishes to be 95% confident (which corresponds to a z-score of approximately 1.96) and have a margin of error of 0.04.
So, we can plug these values into the formula to find the sample size needed:
n = (1.96^2 * p * (1 - p)) / 0.04^2
Since there is no prior estimate available, we can use the worst-case scenario where p = 0.5, which will maximize the sample size needed for any given margin of error.
n = (1.96^2 * 0.5 * (1 - 0.5)) / 0.04^2
Simplifying the equation:
n = (3.8416 * 0.5 * 0.5) / 0.0016
n = 0.9604 / 0.0016
n = 600.25
Rounding up to the nearest whole number because the sample size cannot be a fraction:
n = 601
Therefore, the researcher would need a sample size of at least 601 to be 95% confident that her estimate of the true proportion of individuals who travel overseas is within 4% of the true proportion.