Question

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?

Answer 1

Answer:

Part A:

n=196.66≅197 employees

Part B:

n=503.45≅503 employees

Step-by-step explanation:

Part A:

The formula we are going to use to find a sample size is gien below::

$n=\frac{Z^2*S^2}{E^2}$

Where:

n is the sample size

Z is the distribution

S is the standard deviation

E is the margin

S=341, E=40

Z is calculated as:

Alpha=1-0.90

Alpha=0.1

Alpha/2=0.1/2

Alpha/2=0.05

From Standard distribution table Z at Alpha/2 is 1.645

$n=\frac{1.645^2*341^2}{40^2}$

n=196.66≅197 employees

Part B:

S=341, E=25, Z=1.645

$n=\frac{1.645^2*341^2}{25^2}$

n=503.45≅503 employees