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
7	5	9
6	5	10
5	4	9
6	4	8
7	5	9
5	4	9

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

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
X1-x2		- Select your answer -No significant differenceSignificant differenceItem 10
X1-x3		- Select your answer -No significant differenceSignificant differenceItem 12
X2-x3		- 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	63.00	2	31.50	63.00	0.0000
Within	7.50	15	0.50
Total	70.50	17

a)

sum of sq;treatment=	63.00
sum of sq; error=	7.50
mean sq;treatment=	31.50
mean square; error=	0.50

b)

test statistic =	63.00
p value is less than 0.01

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

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	1.50	significant difference
x3-x1	3.00	significant difference
x3-x2	4.50	significant difference