A survey is planned to determine the mean
annual family medical expenses of employees of a
large company. The management of the company wishes to
be 95% confident that the sample mean is correct to within $50
of the population mean annual family medical expenses.
A previous study indicates that the standard deviation
is approximately $400.
a. How large a sample is necessary?
b. If management wants to be correct to within $25 how
many employees need to be selected?
To determine the sample size needed for this survey, we can use the formula for sample size calculation with a known population standard deviation. The formula is:
n = (Z * σ / E)^2
Where:
n = required sample size
Z = Z-score corresponding to the desired confidence level
σ = population standard deviation
E = desired margin of error
a. To calculate the required sample size for a margin of error of $50 and a confidence level of 95%, we need to find the corresponding Z-score. The Z-score can be found using a Z-table or a statistical calculator. For a 95% confidence level, the Z-score is approximately 1.96.
Using the formula:
n = (Z * σ / E)^2
n = (1.96 * 400 / 50)^2
n ≈ 15.38^2
n ≈ 236.03
Therefore, a sample size of at least 237 employees is necessary to estimate the mean annual family medical expenses with a margin of error of $50 and a 95% confidence level.
b. If management wants to reduce the margin of error to $25, we need to recalculate the sample size using the updated margin of error (E = $25).
n = (Z * σ / E)^2
n = (1.96 * 400 / 25)^2
n ≈ 31.36^2
n ≈ 984.13
Therefore, in order to estimate the mean annual family medical expenses with a margin of error of $25 and a 95% confidence level, we would need a sample size of at least 985 employees.