In: Statistics and Probability
A company applies an aptitude test to all those who apply for
work as programmers. The results of these tests are analyzed
according to their precedence (general advertising, advertisements
for specialized magazines, employment agencies, personal
recommendations or people who come spontaneously). With a
significance level of 0.025, determine if there is a difference
between the results of these tests according to their
precedence.
Your work should have detailed procedure, in this exercise you can use the tool offered by Excel for the required test, but you have to have the appropriate analysis.
General Pub | 36 | 47 | 38 | 51 | 62 | 78 | 60 | 47 | 49 | 53 | 26 | 38 | 61 | 39 | 43 |
Rev Esp. | 58 | 64 | 62 | 47 | 71 | 90 | 65 | 82 | 61 | 59 | |||||
Ag Emp. | 47 | 59 | 48 | 81 | 66 | 50 | 42 | 53 | |||||||
Rec Per. | 67 | 61 | 82 | 97 | 65 | 72 | 54 | 59 | 58 | ||||||
Spontaneous | 38 | 47 | 80 | 41 | 38 | 66 | 50 |
The appropriate test is one way ANOVA as five groups are involved in the study.
The statistical hypothesis is,
Ho: There is no difference between the test results according to the distinct five precedence. Ha: There is a difference between the test results according to the distinct five precedence.
Perform the test in Excel:
Enter the data in the Excel spreadsheet.
Go to data analysis add-in in the DATA tab. Select ANOVA:single factor and click OK.
Enter range of the entire data set. Select the labels and enter the output range. Click Ok.
The obtained output is shown below:
Anova: Single Factor |
||||||
SUMMARY |
||||||
Groups |
Count |
Sum |
Average |
Variance |
||
General Pub |
15 |
728 |
48.53333 |
169.6952 |
||
Rev Esp. |
10 |
659 |
65.9 |
152.9889 |
||
Ag Emp. |
8 |
446 |
55.75 |
159.9286 |
||
Rec per. |
9 |
615 |
68.33333 |
186 |
||
Spontaneous |
7 |
360 |
51.42857 |
253.2857 |
||
ANOVA |
||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
Between Groups |
3248.561 |
4 |
812.1401 |
4.53488 |
0.003736 |
2.583667 |
Within Groups |
7879.848 |
44 |
179.0874 |
|||
Total |
11128.41 |
48 |
Decision: Since the p-value (0.003736) is less than the significance level 0.025, so the researcher rejects the null hypothesis.
Conclusion: There is a sufficient evidence to support that there is a significant difference between the test results according to the distinct five precedence.