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
8	10	9
7	10	10
6	9	9
7	9	8
8	10	9
6	9	9

1. Compute the values identified below (to 1 decimal, if necessary).
   

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

   
   
2. Use  = .05 to test for a significant difference in perception among the three groups.
   
   Calculate the value of the test statistic (to 2 decimals).
   
   
   The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6
   
   What is your conclusion?
   SelectConclude the mean perception scores for the three groups are not all the sameCannot conclude there are differences among the mean perception scores for the three groupsItem 7
   
3. Using  = .05, 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
   1 -  2		SelectSignificant differenceNo significant differenceItem 10
   1 -  3		SelectSignificant differenceNo significant differenceItem 12
   2 -  3		SelectSignificant differenceNo significant 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	10.50000	21.000	0.0000
Within	7.50	15	0.50000
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 treatment mean 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	2.50	significant difference
x3-x1	2.00	significant difference
x3-x2	0.50	not significant difference