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 8 7 6 7 8 6

Marketing Research 10 10 9 9 10 9

Advertising 5 6 5 4 5 5

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 Alpha= .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?

What is your conclusion?

Using Alpha= .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 for (1)-(2)?

Difference Absolute Value Conclusion for (1)-(3)?

Difference Absolute Value Conclusion for (2)-(3)?

Answer 1

Date: 21/04/2019 Answei a) The sum of square treatment is By using Excel software, find sum of square treatment with the help of following steps 1) Import the data. 2) Choose ANOVA: Single Factor from Data Analysis in Data menu. 3) Choose Input Range and Column in Groups by 4) Choose Output Range. 5)Click Ok

Anova: Single Factor SUMMARY Groups Count Sum Average Variance 0.80 0.30 0.40 Manager Research Advertising 42 57 30 7.00 9.50 5.00 ANOVA Source of Variation Between Groups Within Groups Total SS 61.00 MS P-value F crit 30.50 0.50 2 15 17 61.00 6.24E-08 3.68232 7.50 68.50 From the Excel output, the sum of square treatment is 61.00 Te sum of square error is, From Excel output, the sum of square error is 7.50

The mean square treatment is From Excel output, the mean square treatment is 30.50 The mean square error is, FromExcel output, the mean square erroris0.50 b) To test the hypothesis is that there is a significant difference in perception among the three groups at 5% level of significance. The null and alternative hypothesis is, H, : at least one μ is unequal. The F-test statistics is,

From the Excel output in part (a), the F-test statistics is 61.00 The p-value for this test is, From the Excel output in part (a), the p value for this test is 0.0000 less than 0.01 Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0000, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a significant difference in perception among the three groups. The result is statistically significant c) The Fisher's LSD value is, First, compute t critical value then find Fisher's LSD value The t critical value is,

The sample size is small and two-tailed test. Look in the column headed α 0.05 and the row headed in the t distrībution table by using degree of freedom within is 15 and the it critical value is 2.131 The critical value of Fisher's LSD is A. = 2.131x-10.50×1二 = 0.87 The critical value of Fisher's LSD is 0.87 The mean difference table is, Mean 1 Mean 2 Difference Treatments Manager-Research Manager-Advertisin Research-Advertisin 7.00 7.00 9.50 2.50 2.00 4.50 9.50 5.00 5.00

Decision The statistical decision is that the mean difference is greater than critical value of Fisher's LSD which is 0.87, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the mean difference of Manager-Research, Manager Advertising and Research-Advertising is statistically significant. The result is statistically significant