Probability and Hypothesis Testing in Employee Health Care Claims

a)
The probability that an employee selected at random has health care claims exceeding $11000
was obtained as follows:
p ¿ This implied that that there was a probability of 0.5 that an employee selected at random
exceeded health care claims of $11000.
b)
Since the sample followed a normal distribution this implied that even the sample of five
employees selected at random also followed a normal distribution with mean $9500 and a
standard deviation of 600 which is obtained as follows 3000
5 =600
Thus the probability of the five employees exceeding $11000 is obtained as below
p (z >
11000−9500
3000
√5 )= p ( z >1.11 )=0.1841
c)
The probability that all of the five employees each exceed $11000 is given below
p (z >11000−9500
600 )= p ( z> 2.5 )=0.0062 .
d)

Since the population follows a normal distribution and it involves random sampling the sample
size can be estimated as follows n=1.962 × 30002
( ± 500 ) 2 =138.23 this implied that an estimated
sample size of approximately 138 employees was required1.
e)
In order to test the claim that the average annual health care claims per employee are less than
$10000 a Z-test is carried out at a 5% level of significance based on the following hypothesis:
H0: the average annual health care claim is $10000
Versus
H1: the average annual health care claim is less than $10000
Where H0 and H1 represent the null hypothesis and alternative hypothesis respectively.
f)
Based on a random sample of 36 employees, a 98% confidence interval was constructed to
determine the range within which the true average of the annual health care claims would lie. It
was obtained as follows,
1 Gregory, K. B., Carroll, R. J., Baladandayuthapani, V., & Lahiri, S. N. (2015). A two-sample
test for equality of means in high dimension. Journal of the American Statistical
Association, 110(510), 837-849.