Question

Gautam is interested in learning what types of ethical messages are most effective.

He has three conditions, integrity, confidentiality, professional behavio, and

he randomly assigns 20 people to each of these conditions. After each person reads the message, they rate its effectiveness

on a 1-10 scale.He runs a one-way ANOVA and finds the following results:

Condition Mean

Integrity               6

Confidentiality           5

Professional Behavior   9

Source SS df MS F

Between   3000 2

Within

Total 11,500 59

a)Complete the one-way ANOVA table.

b) Determine the critical F at a=.01

What conclusion should Gautam make?

c) Is it appropriate to calculate a Tukey HSD test for this problem? Why? If it is appropriate, calculate a Tukey HSD test for each of the three possible comparisons between two means. What groups are significantly different from each other at a=.01

(integrity, confidentiality, professional behavior) according to this test?

Answer 1

a)

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	3000	2	1500.0000	10.0588	0.0002	4.9981
Within Groups	8500	57	149.1228
Total	11500	59
			Level of significance			0.01

b)

f-critical = 4.9981 < f-stat=10.0588,so null hypothesi rejected that all means are equal

c)

yes,it is appropriate as null hypothesi is rejcted

MSwithin=149.1228

std error of difference = sqrt [MSwithin / 2 (1/n_i+1/n_j)]

q(2,57,0.01)=3.7687

CR=q*std error

if abs difference>CR,means are different ,otherwise not

	Sample	Sample			Absolute	Std. Error	Critical
Group	Mean	Size		Comparison	Difference	of Difference	Range	Results
integrity	6	20		Group 1 to Group 2	1	2.730593406	10.29078737	Means are not different
Confidentiality	5	20		Group 1 to Group 3	3	2.730593406	10.29078737	Means are not different
professional	9	20		Group 2 to Group 3	4	2.730593406	10.29078737	Means are not different
Other Data
Level of significance	0.01
Numerator d.f.	2
Denominator d.f.	57
MSW	149.122807
Q Statistic	3.7687