A study reported in the Journal of Small Business Management concluded that self-employed individuals do not experience higher job satisfaction than individuals who are not self-employed. In this study, job satisfaction is measured using 18 items, each of which is rated using a Likert-type scale with 1–5 response options ranging from strong agreement to strong disagreement. A higher score on this scale indicates a higher degree of job satisfaction. The sum of the ratings for the 18 items, ranging from 18–90, is used as the measure of job satisfaction. Suppose that this approach was used to measure the job satisfaction for lawyers, physical therapists, cabinetmakers, and systems analysts. The results obtained for a sample of 10 individuals from each profession follow. Lawyer Physical Therapist Cabinetmaker Systems Analyst 44 55 54 44 42 78 65 73 74 80 79 71 42 86 69 60 53 60 79 64 50 59 64 66 45 62 59 41 48 52 78 55 64 55 84 76 38 50 60 62 At the ɑ = .05 level of significance, test for any difference in the job satisfaction among the four professions. check_circle
Sorry I am bad at this so can you clarify what you put as the input
i asked the question pasted below and am getting different results when I do it in Excel. For the anova table in excel do you use one factor, two factor with replication or two factor without replication? If you could be a bit more thorough im sorry I just want to make sure I am doing it right. For example instead of what you got for mean square I got 969.433 using one factor anova in excel and when I tried the two factor it didnt work or get what you did. Just please be more detailed in the response so I can grasp it. Thank you so much and im sorry for any inconveniance.
A study reported in the Journal of Small Business Management concluded that self-employed individuals do not experience higher job satisfaction than individuals who are not self-employed. In this study, job satisfaction is measured using 18 items, each of which is rated using a Likert-type scale with 1–5 response options ranging from strong agreement to strong disagreement. A higher score on this scale indicates a higher degree of job satisfaction. The sum of the ratings for the 18 items, ranging from 18–90, is used as the measure of job satisfaction. Suppose that this approach was used to measure the job satisfaction for lawyers, physical therapists, cabinetmakers, and systems analysts. The results obtained for a sample of 10 individuals from each profession follow.
|
Lawyer |
Physical Therapist |
Cabinetmaker |
Systems Analyst |
|
44 |
55 |
54 |
44 |
|
42 |
78 |
65 |
73 |
|
74 |
80 |
79 |
71 |
|
42 |
86 |
69 |
60 |
|
53 |
60 |
79 |
64 |
|
50 |
59 |
64 |
66 |
|
45 |
62 |
59 |
41 |
|
48 |
52 |
78 |
55 |
|
64 |
55 |
84 |
76 |
|
38 |
50 |
60 |
62 |
At the ɑ = .05 level of significance, test for any difference in the job satisfaction among the four professions.
Expert Answer
Use the excel to form the Anova table:
| Source | Sum of squares | Degree of freedom | Mean square | F-value |
| Between treatments | 1939.4 | 3 | 646.4667 | 4.86614 |
| Within treatments | 4782.6 | 36 | 132.85 | |
| Total | 6722 | 39 |
Null hypothesis: H0 : μ1=μ2=μ3=μ4H0 : μ1=μ2=μ3=μ4
Alternative hypothesis: HA : At-least one of the μμ is not equal.
For this analysis, the significance level is 0.05.
Expert Answer
ANOVA: The analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the population means (μ) of more than two independents (unrelated) groups.
The null and alternative hypothesis are,
H0: There is no significant difference between the treatment means (μ).
H1: At least one treatment mean differs significantly.
OR
Given,
Level of significance = αα = 0.05
Here we use one-way ANOVA. for a single factor.
By using excel software, calculate the ANOVA from the following steps,
1) Enter the data into excel software.
2) Go to Data < data analysis tool pack < Anova: Single-factor.
3) Insert input range.
4) Insert output range.
5) Put the value of the level of significance.
6) Select the label in the first row.
7) OK.
Excel output:
Criteria: if p_value < 0.05 then we may reject H0 at 5% level of significance.
Decision: Here we may reject the null hypothesis at 5% level of significance. because p-value < 0.05
conclusion: There is enough evidence to reject the null hypothesis.
Interpretation: Yes, there is a difference in job satisfaction among the four professions.
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