In this analysis assignment, we will use the Chi-Square to analyze whether a management
training program is related to the promotion of managers and use the ANOVA to analyze
whether the satisfaction rate of employees in four different offices is the same.
In the Chi-square testing, the total number of employees who didn’t promote is 40
employees. And among the 40 employees, it is expected that 17.6 employees didn’t participate
in the training program and 22.4 employees participated in the program. However, the real
outcome is that among 40 employees, 27 employees, which is 67.5% of the total employees
who didn’t promote, didn’t participate in the training program, while 13 employees, which is
32.5% of the total employees who didn’t promote, participated in the program.
In addition, the total number of employees who promoted is 60 employees. And among
the 60 employees, it is expected that 26.4 employees didn’t participate in the program and 33.6
Pearson Chi-square < 0.05, chance is not the
only factor that causes differences.
employees participated in the program. However, the real outcome is that among 60 employees,
17 employees, which is 28.3% of the total employees who promoted, didn’t participate in the
program and 43 employees, which is 71.7% of the employees who promoted, participate in the
Therefore, the management training program is more efficient than expected in helping
employees to promote. Clearly, the management training program is related to the promotion
2. ANOVA In the ANOVA test, we analyze the satisfaction rate data of different location of
offices to see if the satisfaction rate in four offices are the same. According to the charts, the p-
value is 0.034 which is less than 0.05. Therefore, we reject the null hypothesis that data is from
a sample population with the same mean. So, the satisfaction rate in four offices are not the
same. Moreover, with the scatter gram below, we can figure out that office 3 has the smallest
variance (which means the data within the group has smallest difference) while it contains the
smallest data value; the office 2 has the largest variance (which means the data within group
has largest difference) while it contains the biggest data value.
The P value in first subset >0.05, don’t have distinct
differences; The P value in second subset <0.05,
have distinct differences
0 1 2 3 4 5
mothly satisfaction rating of different offices
XXXXXXXXXXX ARE 112
Analysis Assignment A
For this analysis assignment, we were introduced to the program SPSS. This is a statistical tool to help us find certain statistics about a given data set. For Data Set 1, we used the Chi-Square test to see if there was a relation between a management training program and promotion. The most important statistical values for this data are the “count” and “expected count.” The “count” showed the relationship between who was promoted and who was not for each variable. The “expected count” showed the difference between who was actually promoted and who was expected to be promoted. The difference between “expected” and “actual” can also be referred to as “residual.” All the variables have the same residual value of 9.4, however, half are negative, and the other half is positive. The employees that had a positive residual value were the ones that were not promoted and did not have management training and the ones that were promoted and did have management training. The negative residual value were the employees that were not promoted and had management training and the ones that were promoted and did not have management training.
For Data Set 2, we used a different test, the ANOVA test, to see monthly employee
satisfaction rates at offices with different locations. The test results tell us that the p-value equals 0.034, thus all office employee satisfactions are different. There are four different office locations that are placed into two different “subsets.” Each of these subsets are classified as homogeneous, meaning that each office in a certain subset is alike. From the ANOVA test, we were able to identify which offices had homogenous employee satisfaction rates. The test also told us which offices had differing employee satisfaction rates. Subset 1 contained offices 1,3, and 4, while subset 2 contained offices 2,3, and 4. This tells us that the only offices that varied in employee satisfaction were office 1 and 2.
XXXXXXXXXX ARE 112 Analysis 6/7/17
Promoted * Management_Training_Program Crosstabulation
Total No Yes Promoted No Count 27 13 40
Expected Count 17.6 22.4 40.0 % within Promoted 67.5% 32.5% 100.0% % within Management_Training_Program 61.4% 23.2% 40.0% % of Total 27.0% 13.0% 40.0% Residual 9.4 -9.4
Yes Count 17 43 60 Expected Count 26.4 33.6 60.0 % within Promoted 28.3% 71.7% 100.0% % within Management_Training_Program 38.6% 76.8% 60.0% % of Total 17.0% 43.0% 60.0% Residual -9.4 9.4
Total Count 44 56 100 Expected Count 44.0 56.0 100.0 % within Promoted 44.0% 56.0% 100.0% % within Management_Training_Program 100.0% 100.0% 100.0% % of Total 44.0% 56.0% 100.0%
From the Chi squared analysis and the management training program crosstabulation there is a statistical importance of the expected and observed promotions. The crosstabulation showed that an expected 17.6 employees who did not complete the training program would bot be promoted. The actual amount of employees who were not trained and did not get promoted was 27. This value is proven to be significantly different because the Chi squared analysis resulted in a less than one in a thousand chance of this data occurring without there being a correlation. Because of the Chi squared test result the management training program and promotion increased the amount of those promoted, expectedly 33.6 and actually 43, and had the reverse effect on those who did not get the management training and were not promoted. These differences in the data are statistically relevant; going to the management training would increase the chances of getting a promotion.
Value df Asymptotic
Significance (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square 14.942a 1 .000 Continuity Correctionb 13.395 1 .000 Likelihood Ratio 15.211 1 .000 Fisher’s Exact Test .000 .000 N of Valid Cases 100 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 17.60. b. Computed only for a 2×2 table
XXXXXXXXXX ARE 112 Analysis 6/7/17
From the ANOVA analysis there is used to determine if the level of employ satisfaction is the same in the four offices. About the same level of satisfaction is seen in offices 1, 3, and 4. Offices 2, 3, and 4 also have very similar levels of satisfaction across the offices. There is a noticeable difference in the satisfaction of office 1 and 2. Office 1 had 39.25 and office 2 had a
satisfaction of 45.88. The ANOVA analysis gives the manager the insight of the noticeable difference in satisfaction of offices 1 and 2. These differences call to the manager’s attention that they should address the offices and determine why the satisfaction is varying.
Monthly_Satisfaction_Rating Tukey HSDa
Office Code N Subset for alpha = 0.05 1 2
1 8 39.25 3 8 39.38 39.38 4 8 41.38 41.38 2 8 45.88 Sig. .813 .053 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 8.000.
- Analysis Assignment (1)
- ARE 112 SPSS A
- ARE112 AA