MBA700 Homework 2: Health Care Negotiations Case Study and Analysis

This document provides a comprehensive solution to the MBA700 Homework 2 assignment, which focuses on health care negotiations within a business context. The solution addresses various aspects of health care cost analysis, including probability calculations, statistical tests, and confidence intervals. It explores scenarios such as the probability of employees exceeding certain health care claim amounts, the use of Z-tests to evaluate average annual health care claims, and the construction of confidence intervals to determine the range of true average claims. Furthermore, the solution incorporates binomial probability distribution to analyze the probability of employees signing up for disability insurance and employs Z-tests to assess the proportion of employees visiting the emergency room. The assignment covers several statistical concepts and their application in a practical business scenario, demonstrating how to interpret and apply statistical findings to make informed decisions regarding health care costs and employee benefits.

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.

9600 ± 1861.466= [ 7738.534 , 11461.466 ] The upper confidence limit was 11461.466 while the
lower confidence limit was 7738.534. The limits had the interpretation that with 98% certainty
the true average of the annual health claims would be within the range [7738.534, 11461.466]2.
g)
A z-test was performed to investigate the following hypothesis;
H0: μ = 8000
Versus
H1: μ ≠ 8000
The test statistic was given by
9600−8000
4801
√36 =2, the tabulated value from the normal table was
1.96.
Since 2 was greater than 1.96 the null hypothesis has rejected this lead to the conclusion that the
annual health care claims were either less than or greater than $80003.
h)
2 Chen, X., Doerge, R. W., & Heyse, J. F. (2018). Multiple testing with discrete data: Proportion
of true null hypotheses and two adaptive FDR procedures. Biometrical Journal, 60(4), 761-779.
3 Roy, A., & Klein, D. (2018, October). Hypothesis Testing of Equality of Two Mean Intervals.
In Session V (statistical modelling) of the workshop Symbolic Data Analysis, SDA (pp. 59-60).

Binomial probability distribution was used to determine the probability of at least employees
sign up for disability insurance. Since its discrete probability, the total probability was obtained
as follows ( 20
x ) ×0.4x ×0.620− x
where x=10, 11,…, 20. The total probability was given by
summing the following (20
10 )×0.410 × 0.610+ (20
11 )×0.411 ×0.69+ …+(20
12)× 0.412 × 0.68=0.245. The
probability obtained had the interpretation that at least ten employees had 24. 5% chance of
signing for the disability insurance.
i)
A Z test was performed to test the following hypothesis:
H0: π ≤ 0.25
Versus
H1: π > 0.25.
The test was carried out at 5% level of significance. Below is the test statistic
z= 0.25
0.0675 =3.7028. The tabulated value from the normal table was 1.29. Since the test statistic
was greater than the tabulated value the null hypothesis (H0) was rejected in favor of the
alternative hypothesis (H1). This lead to the conclusion that the proportion of employees who
visited the emergency room in the past year exceed 0.25.
j)
The following was used to calculate 95% confidence interval for the employees who visit the
emergency room,

0.4167 ± 0.161=[0.2557 ,0.5777 ] The confidence limits had the interpretation that with 95%
certainty the true proportion of the employees who visit the emergency room lie between the
intervals4. Further, 95 in 100 employees their proportion of going to the emergency room lies
between the limits of the 95% confidence interval constructed.
4 Yang, S., & Black, K. (2019). Using the standard Wald confidence interval for a population
proportion hypothesis test is a common mistake. Teaching Statistics, 41(2), 65-68.