An insurance company estimates 40 percent of its claims have errors. The insurance company wants to estimate with 90 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 5 percent of the actual?
Try this formula:
n = [(z-value)^2 * p * q]/E^2
Find the z-value for 90% confidence using a z-table. p = .4; q = .6 (q = 1-p); and E = .05
Round n to the next highest whole number.
To determine the sample size needed, we can use the formula for estimating a proportion with confidence:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level
p = estimated proportion of claims with errors
E = margin of error
In this case, the confidence level is 90% (equivalent to a Z-score of 1.645, which corresponds to the cumulative probability of 0.95 from the standard normal distribution). The estimated proportion of claims with errors is 40%, which can be expressed as 0.4. The desired margin of error is 5%, which can be expressed as 0.05.
Plugging these values into the formula, we get:
n = (1.645^2 * 0.4 * (1-0.4)) / 0.05^2
Simplifying the equation, we have:
n = (2.7056 * 0.24) / 0.0025
n = 0.649344 / 0.0025
n ≈ 259.7376
Since you cannot have a fractional sample size, you will need at least 260 claims to estimate the proportion of claims with errors with a 90% confidence level and a margin of error within 5%.