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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 8 7 6 7 8 6
Marketing Research 10 10 9 9 10 9
Advertising 5 6 5 4 5 5
1. Compute the values identified below (to 1 decimal, if necessary).
a. Sum of Squares, Treatment?
b. Sum of Squares, Error?
c. Mean Squares, Treatment?
d. Mean Squares, Error?
2. Use Alpha= .05 to test for a significant difference in perception among the three groups.
3. Calculate the value of the test statistic (to 2 decimals)?
4. The p-value is?
5. What is your conclusion?
6. Using Alpha= .05, determine where differences between the mean perception scores occur.
7. Calculate Fisher's LSD value (to 2 decimals)?
8. 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 for (1)-(2)?
Difference Absolute Value Conclusion for (1)-(3)?
Difference Absolute Value Conclusion for (2)-(3)?

Respuesta :

akiran007

Answer:

Part 1

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. the F value lies in the rejection region > 3.6823

3. The value of the test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

Since p-value < α, H0 is rejected.

6. Between x`2 and x`3

7. Fisher's Least Significant Difference value almost 0.869

8.There is a significant difference between the means

Step-by-step explanation:

Summary of Data

                          Treatments

                       1             2             3                 Total

n                      6             6             6                   18

∑x                   42          57           30                  129

Mean              7            9.5           5                  7.167

∑x2              298         543        152                    993

Sd.D       0.8944     0.5477     0.6325           2.0073

ANOVA Table

Source                                  SS              df                  MS

Between-treatments           61              2                   30.5       F = 61

Error                                     7.5           15                     0.5

Total                                     6             8.5                     17

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. Using alpha= 0.05 the F value lies in the rejection region i.e F > 3.6823

x1` -x2`= 7-9.5= -2.5 Not significant as difference <3.68

x1`- x3`= 7-5= 2 Not significant as difference <3.68

x2` -x3`= 9.5-5= 4.5 Significant as difference > 3.68

3. The value of the F test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

Since p-value < α, H0 is rejected.

6. Using alpha= .05, differences occurs between x2` and x3` as their difference is greater than 3.68

7. Fisher's Least Significant Difference value almost 0.869

Least Significant Difference= t( 0.025,15) √2s²/r s²= 0.50 r= 6 =n1=n2=n3

Least Significant Difference= 2.13 √ 2*0.50/ 6

=0.869

8.There is a significant difference between the means

x1` -x2`= 7-9.5= -2.5 Significant as difference > Least Significant Difference

x1`- x3`= 7-5= 2 Significant as difference > Least Significant Difference

x2` -x3`= 9.5-5= 4.5 Significant as difference > Least Significant Difference