John Davis, a manager of a supermarket, wants to estimate the proportion of customers that will be using food stamps at his store. How large a sample is required to estimate the true proportion to within 3% with 98% confidence?
1509
Well, John's in luck because I've got the answer! To estimate the proportion of customers using food stamps with a 98% confidence level and a margin of error of 3%, he'll need to calculate the required sample size.
Now, to find that out, we need to know the approximate proportion of customers using food stamps in the population. Do you have any idea about that, or should we just assume it's the same percentage as the number of people who laugh at my jokes?
To determine the sample size required to estimate the proportion within a given margin of error and confidence level, we can use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n is the required sample size
Z is the Z-value corresponding to the desired confidence level
p is the estimated proportion (which we do not have, so we can take p = 0.5)
E is the desired margin of error
In this case, we want a confidence level of 98% (which corresponds to a Z-value of approximately 2.33, using a standard normal distribution table), and a margin of error of 3% (which is equivalent to 0.03). Since we do not have an estimate of the proportion, we can assume that p = 0.5, which provides a conservative estimate.
Plugging these values into the formula, we get:
n = (2.33^2 * 0.5 * (1-0.5)) / 0.03^2
Simplifying the calculation:
n = (5.4289 * 0.25) / 0.0009
n = 1.357225 / 0.0009
n ≈ 1508.03
Therefore, John Davis would need a sample size of approximately 1508 (rounding up for a whole number) in order to estimate the proportion of customers using food stamps within a 3% margin of error and with 98% confidence.
Formula:
n = [(z-value)(p)(q)]/E^2
z-value = 2.33 for 98% confidence
p = .5 if no value is stated
q = 1 - p
E = Maximum error
With your data:
n = [(2.33)(.5)(.5)]/.03^2
I'll let you take it from here. Round to the next whole number.