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	6
6	5	7
5	4	6
6	4	5
7	5	6
5	4	6

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 a=.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 - Select your answer -

less than .01

between .01 and .025

between .025 and .05

between .05 and .10

greater than .10

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

c. Using a=.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
x1 - x2		- Select your answer -No significant differenceSignificant difference
x1 - x3		- Select your answer -No significant differenceSignificant difference
x2 - x3		- Select your answer -No significant differenceSignificant difference

Answer 1

Applying ANOVA:

Source of variation		SS	df	MS	F	p vlaue
treatments		9.000	2	4.5000	9.000	0.003
error		7.500	15	0.5000
total		16.500	17

a)

sum of square; treatment=	9.00
sum of square; error=	7.50
mean square; treatment=	4.50
mean square; error=	0.50

b)

test statistic =

9.00

p value is less than 0.01

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

c)

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
x1-x3	0.00	not significant difference
x2-x3	1.50	significant difference