Find the sample size that produces the margin of error ± 3.0%
well i guess i'll answer my own question............
To find the sample size that produces a margin of error of ±3.0%, you need to know three things: the desired level of confidence, the population size (if applicable), and the estimated proportion (if applicable).
First, let's assume you don't have any information about the population size or the estimated proportion. In this case, you can use the conservative estimate where the proportion is assumed to be 0.5 (maximum variability).
The formula to find the sample size without knowing the population size and the estimated proportion is:
n = (Z^2 * p * q) / E^2.
Where:
- n is the sample size
- Z is the Z-score corresponding to the desired level of confidence (e.g., for a 95% confidence level, Z = 1.96)
- p is the estimated proportion (use 0.5 if unknown)
- q is (1 - p)
- E is the margin of error (expressed as a proportion)
In our case, the margin of error ± 3.0% can be expressed as a proportion of 0.03 (i.e., 3.0% / 100).
So, substituting these values into the formula:
n = (1.96^2 * 0.5 * 0.5) / 0.03^2
n ≈ 1068.4444
Since the sample size must be a whole number, rounding up to the nearest whole number, the sample size that produces a margin of error of ±3.0% is approximately 1069.
It's worth noting that if you have information about the population size, you can use a modified formula called the finite population correction factor:
n' = n / (1 + (n/N))
Where:
- n' is the adjusted sample size
- n is the sample size obtained from the formula above
- N is the population size
But if the population size is very large (e.g., N ≥ 10,000), you can use the formula without the finite population correction factor, as it won't significantly impact the results.