The
owner of the organization is trying to improve employee relations. The goal of
the owner is to increase employee satisfaction up to 75%. The chi-square test is
utilized to predict the owner’s expectations. The following are the hypothesis:
H0: Employees
of the organization are satisfied
H1: Employees
of the organization are not satisfied
Level of Significance
Statistic Test of Employee
The
Formula for evaluating the statistic test value is mentioned as follows:
|
Sales
|
Finance
|
Human
Resources
|
Technology
|
Satisfied
|
38
|
12
|
5
|
8
|
Dissatisfied
|
19
|
7
|
3
|
1
|
Total
|
57
|
19
|
8
|
9
|
In
the first step, the predicted values are going to be calculated so that they
can be compared with actual values of survey. According to the available data
the predicted number of employees that are satisfied is 75%. It means that the
remaining 25% number of employees in each department are dissatisfied. Now the
predicted number of satisfied employees in the finance department is evaluated
as 0.75(19) = 14.25. Similarly the predicted number of dissatisfied employees
in the finance department is 0.25(19) = 4.75 (Wegner, 2010). In the below table the details
regarding the expected and observed values are presented:
|
Sales
|
Finance
|
Human
Resources
|
Technology
|
Satisfied
|
38
(42.75)
|
12
(14.25)
|
5
(6)
|
8
(6.75)
|
Dissatisfied
|
19 (14.25)
|
7 (4.75)
|
3 (2)
|
1 (2.25)
|
Total
|
57
|
19
|
8
|
9
|
The
expected & observed values are utilized for evaluating the Chi-Square:
Part B of Employee
Satisfaction:
After
calculating the Chi-square the next step is to find the p-value. The Chi-square
table can be utilized for finding the p-value. As there are 4 departments in
the organization the degree of freedom (df)
would be 3. In the chi-square table, the row of degree of freedom 3, the
nearest value to 5.124 will be selected. The nearest value to 5.124 is 6.251
which indicates the p-value of 0.10. The level of significance of 0.05 is used
to determine whether to accept or reject the hypothesis. As the p-value of 0.10
> 0.05 the H0 will not
be rejected which means that employees are satisfied with the organization (Anderson, Sweeney, & Williams, 2010).
Part C of Employee
Satisfaction:
The
study conducted by Mary L. McHugh (2013) has provided brief overview regarding Chi-square
and how Chi-square can be used for evaluating the group differences. The researcher
has stated that the chi-square has many benefits such as easy calculations and
its robustness in data distribution. However, Chi-square has some drawbacks as
well which include difficulty in interpretation and sample size requirements. The
chi-square can be utilized in different business scenarios such as knowing the
motivational level of the employees and how efficiently each employee’s performance
in the organization (McHugh, 2013).
References of Employee Satisfaction
Anderson, Sweeney, D. J., & Williams, T. A.
(2010). Fundamentals of Business Statistics (illustrated ed.). Cengage
Learning.
McHugh, M.
L. (2013). The Chi-square test of independence. Biochemia Medica, 23(2),
143-149.
Wegner, T.
(2010). Applied Business Statistics: Methods and Excel-based Applications
(illustrated ed.). Juta and Company Ltd.