Final answer:
To estimate the sample size with a 20% proportion and a 4% margin of error, assuming a 95% confidence level (z-score of 1.96), the formula for the margin of error is rearranged to solve for sample size, resulting in an estimated sample size of 601.
Step-by-step explanation:
The question is asking to estimate the sample size used in a survey given a proportion of respondents owning a cell phone and the margin of error. The estimation of the sample size can be calculated using the formula for the margin of error E for a proportion in a population:
E = z * sqrt((p(1-p))/n)
Where:
- E is the margin of error,
- z is the z-score corresponding to the desired confidence level,
- p is the sample proportion, and
- n is the sample size.
We can rearrange this formula to solve for the sample size n:
n = (z^2 * p(1-p)) / E^2
Given that the margin of error E is 4%, and the sample proportion p is 20%, we would first need to determine the appropriate z-score for the confidence level (often 1.96 for 95% confidence). However, this z-score is not provided. Assuming a 95% confidence level and using the z-score of 1.96, the calculation would be:
n = (1.96^2 * 0.20(1-0.20)) / 0.04^2
n = 600.25
Rounding to the next higher value gives us a sample size of 601.