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 6 10 6 5 10 7 4 9 6 5 9 5 6 10 6 4 9 6 Compute the values identified below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error 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 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 x1-x2 ?x1-x3 x2-x3

Answer 1

Applying one way ANOVA on above data:

X3 X1 X2 10 Total 19 19 10 4 6 19 18 Total=Xi 36 57 = Average-D-EX/n- grand mean Xgrand 5.0000 9.5000 543 0.3000 6.0000 6.8333 0.4000 0.8000 Total observation 18 number of groups between groups degree of freedom df -k-1 - within groups degree of freedom dfwg-N-k Total degree of freedom 0.947368421 15 17 df Total 4.00 5.0000 9.5000 6.0000 0.800 0.300 0.400 20.17 42.67 4.17 67.0000 1.50 6 total 7.5000 sum of square between groups 67.0000 7.5000 74.5000 Σ(ni-1)*S- sum of square within groups SSE total sum of square SSG+SSE- Summary table Source of variation between within total MS 33.5000 0.5000 df 67.0000 7.5000 74.5000 67.000 15.0000 17.0000

sum of square; treatment=	67.00
sum of square; error=	7.50
mean square; treatment=	33.50
mean square; error=	0.50

test statistic =

67.00

p-value is less than .01
conclude there are differences among the mean perception scores for the three groups

Fisher's (LSD) for group i and j =                  (tN-k)*(sp*√(1/ni+1/nj)           =

0.87

Difference	Absolute Value	Conclusion
x1-x2	4.50	significant difference
x1-x3	1.00	significant difference
x2-x3	3.50	significant difference