Question

at 95% confidence, how large should be taken to obtain a margin of error of 0.031 for the estimation of a popular proportion? Assume that past data are not available for developing a planning value for p*. Round up to the next whole number

Answer 1

To achieve a 95% confidence interval with a margin of error of 0.031, the sample size required is 999 when no prior data is available for the population proportion.

Step-by-step explanation:

To determine the sample size needed for a 95% confidence interval with a margin of error of 0.031, when no prior information about the population proportion p* is available, we can use the formula for the sample size of a proportion:

n = (Z^2 * p*(1 - p*)) / E^2

Where Z is the Z-value corresponding to the confidence level, p* is the estimated proportion, and E is the desired margin of error. Since no prior information is available, we use p* = 0.5 for the most conservative estimate, which maximizes the sample size. The Z-value for a 95% confidence interval is approximately 1.96. Substituting the values into the formula:

n = (1.96^2 x 0.5 x (1 - 0.5)) / 0.031^2

n = (3.8416 * 0.25) / 0.000961

n = 998.65

Since we need a whole number of individuals in our sample and we cannot survey a fraction of a person, we round up to the next whole number, which gives us 999 as our sample size.