In: Statistics and Probability
With double-digit annual percentage increases in the cost of
health insurance, more and more workers are likely to lack health
insurance coverage (USA Today, January 23, 2004). The following
sample data provide a comparison of workers with and without health
insurance coverage for small, medium, and large companies. For the
purposes of this study, small companies are companies that have
fewer than 100 employees. Medium companies have 100 to 999
employees, and large companies have 1000 or more employees. Sample
data are reported for 50 employees of small companies, 75 employees
of medium companies, and 100 employees of large companies.
Health Insurance
Size of Company yes no total
Small 36 14 50
Medium 66 9 75
Large 87 13 100
a.) Conduct a test of independence to determine whether employee
health insurance coverage is independent of the size of the
company. Use = .05. Compute the value of the test
statistic (to 2 decimals).
b.) the p-value is _____________
c.) What is your conclusion _________________
d.) The USA Today article indicated employees of small companies
are more likely to lack health insurance coverage. Calculate the
percentages of employees without health insurance based on company
size (to the nearest whole number).
Small _____ %
Medium ______ %
Large ______ %
Based on the percentages above what do you conclude
___________________
a)
Observed Frequencies | |||||||
0 | |||||||
0 | yes | no | Total | ||||
small | 36 | 14 | 50 | ||||
medium | 66 | 9 | 75 | ||||
large | 87 | 13 | 100 | ||||
Total | 189 | 36 | 225 | ||||
Expected frequency of a cell = sum of row*sum of column / total sum | |||||||
Expected Frequencies | |||||||
yes | no | Total | |||||
small | 189*50/225=42 | 36*50/225=8 | 50 | ||||
medium | 189*75/225=63 | 36*75/225=12 | 75 | ||||
large | 189*100/225=84 | 36*100/225=16 | 100 | ||||
Total | 189 | 36 | 225 | ||||
(fo-fe)^2/fe | |||||||
small | 0.857 | 4.500 | |||||
medium | 0.143 | 0.750 | |||||
large | 0.1071 | 0.5625 |
Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe = 6.92
.........
b)
Level of Significance = 0.05
Number of Rows = 3
Number of Columns = 2
Degrees of Freedom=(#row - 1)(#column -1) = (3- 1 ) * ( 2- 1 )
= 2
p-Value = 0.0314
c)
Decision: p-value < α , Reject Ho
conclusion : so it is concluded that employee health insurance coverage is not independent of the size of the company.
d)
small = 14 / 50 *100 = 28%
medium = 9/75 *100 = 27%
large =13/100 *100 = 13%
large company has less percentage of non insurance peolple while small is having large % of non insurance