Question

The following data are from an experiment designed to investigate the perception of corporate ethical values among individuals who are in marketing. Three groups are considered: management, research and advertising (higher scores indicate higher ethical values).

Marketing Managers	Marketing Research	Advertising
10	10	7
9	10	8
8	9	7
9	9	6
10	10	7
8	9	7

a. Compute the values identified below (to 2 decimal, if necessary).

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

b. Use  to test for a significant difference in perception among the three groups.

Calculate the value of the test statistic (to 2 decimals).

The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6

What is your conclusion?

- Select your answer -Cannot conclude there are differences among the mean perception scores for the three groupsConclude the mean perception scores for the three groups are not all the sameItem 7

c. Using , determine where differences between the mean perception scores occur.

Calculate Fisher's LSD value (to 2 decimals).

Test whether there is a significant difference between the means for marketing managers (1), marketing research specialists (2), and advertising specialists (3).

Difference	Absolute Value	Conclusion
		- Select your answer -No significant differenceSignificant differenceItem 10
		- Select your answer -No significant differenceSignificant differenceItem 12
		- Select your answer -No significant differenceSignificant differenceItem 14

Answer 1

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Source	SS	df	MS	F	P value
Between	21.00	2	10.500	21.000	0.000
Within	7.50	15	0.5000
Total	28.50	17

a)

sum of sq;treatment=	21.00
sum of sq; error=	7.50
mean sq;treatment=	10.50
mean square; error=	0.50

b)

test statistic =	21.00
p value is less than 0.01

Conclude the mean perception scores for the three groups are not all the same

c)

critical value of t with 0.05 level and N-k=15 degree of freedom=				tN-k=	2.131
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =					0.87

Difference	Absolute Value	Conclusion
x1-x2	0.50	not significant difference
x3-x1	2.00	significant difference
x3-x2	2.50	significant difference