Determine the minimum sample size required in order to be 95% confident that our estimate is within 4% of the true percent of all US households using e-mail. Assume no prior estimate available.
Formula to find sample size:
n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is found using a z-table for 95% confidence, p = .5 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = .04 (4% in the problem).
Plug values into the formula and calculate n.
I hope this will help get you started.
To determine the minimum sample size required to estimate the true percentage of all US households using email within a certain margin of error with a 95% confidence level, you can use the following formula:
n = (Z^2 * p * (1 - p)) / E^2
Where:
n: Sample size
Z: Z-value corresponding to the desired confidence level (for a 95% confidence level, Z ≈ 1.96)
p: Estimated percentage of households using email (since no prior estimate is available, you can assume p = 0.5 for a conservative estimate)
E: Margin of error (in decimal form, for example, 4% is 0.04)
Plugging in the values, we can calculate the minimum sample size as follows:
n = (1.96^2 * 0.5 * (1 - 0.5)) / 0.04^2
n = (3.8416 * 0.5 * 0.5) / 0.0016
n = 0.9604 / 0.0016
n ≈ 600
Therefore, the minimum sample size required to be 95% confident that the estimate is within 4% of the true percentage of all US households using email is approximately 600.