Question

ANOVA

An experiment was conducted to determine if motivational techniques in management significantly impacted employee stress levels compared to no intervention. An employee survey was given to three groups: one group under a manager who started using a transformational motivation technique, one group under a manager who began using a transformational motivation technique, and then a control group with no new technique used. Stress levels were self-reported by the employees as shown below. Use alpha = 0.05

Transformational	Transactional	Control
0	2	6
7	5	5
3	3	8
5	0	9
2	1	5

1. What is your null hypothesis?
2. What is your alternative hypothesis?
3. After running the single factor analysis of variance on the data, what is the p-value for between groups?
4. Using the 95% significance level, is there a significance difference in stress levels between the three groups? Yes or No
5. Do you reject or accept the null hypothesis?

Answer 1

The Hypotheses are:

Null: There is no different between average stress level of all 3 groups

Ho: 1 = 2 = 3

Alternate: There is atleast one group whose stress level is different from other 2.

We will use excel to solve this. Go to DATA tab in excel, under that select DATA ANALYSIS.

Open it and select ANOVA: SINGLE FACTOR

Select this data in excel, and just run it you will get below results:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Transformational	5	17	3.4	7.3
Transactional	5	11	2.2	3.7
Control	5	33	6.6	3.3
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	51.73333333	2	25.86666667	5.426573427	0.020960972	3.885293835
Within Groups	57.2	12	4.766666667
Total	108.9333333	14

Now here p-value = 0.02 which is less than 0.05.

Hence, we can reject the Null Hypotheses and conclude that there is atleast one group whose stress level is significantly different from other 2.