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 90% confident that the sample mean is correct to within plus or minus$40 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately $341. a. How large a sample is necessary?b. If management wants to be correct to within (plus or minus $25), how many employees need to be selected?
Accepted Solution
A:
Answer:Part A:n=196.66≅197 employeesPart B:n=503.45≅503 employeesStep-by-step explanation:Part A:The formula we are going to use to find a sample size is gien below::[tex]n=\frac{Z^2*S^2}{E^2}[/tex]Where:n is the sample sizeZ is the distributionS is the standard deviationE is the marginS=341, E=40Z is calculated as:Alpha=1-0.90Alpha=0.1Alpha/2=0.1/2Alpha/2=0.05From Standard distribution table Z at Alpha/2 is 1.645[tex]n=\frac{1.645^2*341^2}{40^2}[/tex]n=196.66≅197 employeesPart B:S=341, E=25, Z=1.645[tex]n=\frac{1.645^2*341^2}{25^2}[/tex]n=503.45≅503 employees