Question

Human Resource Management researchers examined the impact of environment on employee development. Employees were randomly assigned one of the following four workplace types/conditions: Impoverished (isolated cubicles each with bare minimum equipment - chair, desk & computer), standard (cubicles placed hear each other, equipped with a 'normal level' of office equipment - printer, shelves, manuals, stationery, etc.), enriched (standard cubicles plus regular work-related meetings), super enriched (enriched environment plus regular non-work-related social events).

Employees were also categorized by function they performed for the company (engineering, sales, manufacturing, etc.), because it's possible that functional background may have been a bigger association with test score than working condition.  

After two months, the employees were tested on a variety of work-relevant learning measures. Use the Microsoft Excel "Anova: Two-Factor Without Replication" Data Analysis tool to conduct a 2-way ANOVA test for the data in the following table:

Employee Function	Working Condition
	Impoverished	Standard	Enriched	Super Enriched
Engineering	8	17	22	22
Sales	7	21	24	19
Marketing	15	10	15	21
Finance & Accounting	14	12	19	29
Purchasing	18	19	15	16
Manufacturing	12	11	14	4

1. What level of significance did you choose and why?
2. What are your F_crit values?
3. If there is a significant difference between working conditions, where specifically are the differences?
4. If there is a significant difference among employee functions, where specifically are the differences?

Answer 1

The ANOVA table is given below:

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
Engineering	4	69	17.25	43.58333
Sales	4	71	17.75	55.58333
Marketing	4	61	15.25	20.25
Finance_Accounting	4	74	18.5	57.66667
Purchasing	4	68	17	3.333333
Manufacturing	4	41	10.25	18.91667
Impoverished	6	74	12.33333	17.86667
Standard	6	90	15	21.2
Enriched	6	109	18.16667	17.36667
Super Enriched	6	111	18.5	69.1
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	182	5	36.4	1.225131	0.345243	2.901295
Columns	152.3333	3	50.77778	1.70905	0.207894	3.287382
Error	445.6667	15	29.71111
Total	780	23

The null hypothesis is framed as:

Null Hypothesis

H0 1: There is no significant difference among employee functions.

H0 2: There is no significant difference between working conditions.

Alternative Hypothesis

H1 1: There is significant difference among employee functions.

H1 2: There is significant difference between working conditions.

a) The level of significance which I chose is alpha=0.05, that there is only 5% of chance that the null hypothesis will be rejected if the hypothesis is true. And also .05 is the standard alpha value if no other vaue mentioned in the question.

b) For Employee Function (Rows) the F critical value= 2.901295

And for the working conditions( Coloumn) the F critical value=3.287382

c) For working conditions,the p value =0.207894.

If P-value < alpha(.05) we reject H0 2 or If F > F critic we reject the null hypothesis (H0 2)

Here P- value =0.207894> .05 .Therefore, we accept Null Hypothesis. That is there is no significant difference between working conditions.

d) For employee functions, the P-value=0.345243

Here P- value =0.345243> .05 .Therefore, we accept Null Hypothesis (H0 1). That is there is no significant difference among employee functions.