In a survey, the planning value for the population proportion is p* = 0.26. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05? (Round your answer up to nearest whole number.)
In a survey, the planning value for the population proportion is p* = 0.26. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05?
Answer: To find the required sample size, we can use the following formula:
n = (Z^2 * p*(1-p*)) / (E^2)
where:
n is the sample size
Z is the z-score associated with the desired confidence level (95% confidence level has a z-score of 1.96)
p* is the planning value for the population proportion
E is the desired margin of error
Plugging in the given values, we get:
n = (1.96^2 * 0.26 * (1 - 0.26)) / (0.05^2)
n = 377.34
Rounding up to the nearest whole number, we get:
n = 378
Therefore, we need to take a sample size of 378 to provide a 95% confidence interval with a margin of error of 0.05 when the planning value for the population proportion is p* = 0.26.