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Thetloving dats sre fom an experiment desined to inetate the perceptin df capora

ID: 3361514 • Letter: T

Question

Thetloving dats sre fom an experiment desined to inetate the perceptin df caporate tcl valbues among Individuals who sae n markting Thee ous are consldered management, research and advertising (higher scores indicate higher ethical values). Marketing Managers Marketing Research Advertising 10 10 10 10 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 -.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 answe What is your conclusion? -Select your answer

Explanation / Answer

Step 1

Null Hypothesis Ho : µ1 =µ2 =µ3

Alternative Hypothesis : µ1 µ2 µ3

Step 2

Degrees of freedom between = k - 1 = 3 - 1 = 2

Degrees of freedom Within = n - k = 18 - 3 = 15

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.682

Step 3

Grand Mean = G / N = 3+4.5+10 / 3 = 5.833

SST = ( Xi - GrandMean)^2 = (4-5.833)^2 + (3-5.833)^2 + (2-5.833)^2 + ……..& so on = 170.5

SS Within = (Xi - Mean of Xi ) ^2 =,(4-3)^2 + (3-3)^2 + (2-3)^2 + ……..& so on = 7.5

SS Between = SST - SS Within = 170.5 - 7.5 = 163

Step 4

Mean Square Between = SS Between / df Between = 163/2 = 81.5

Mean Square Within = SS Within / df Within = 7.5/15 = 0.5

Step 5

F Cal = MS Between / Ms Within = 81.5/0.5 = 163

We got |F cal| = 163 & |F Crit| =3.682

MAKE DECISION

Hence Value of |F cal| > |F Crit|and Here We Reject Ho


Treatments ONE WAY ANOVA Mean = X /n Treatment1 4 3 2 3 4 2 3 Treatment2 5 5 4 4 5 4 4.5 Treatment3 10 11 10 9 10 10 10

Step 1

Null Hypothesis Ho : µ1 =µ2 =µ3

Alternative Hypothesis : µ1 µ2 µ3

Step 2

Degrees of freedom between = k - 1 = 3 - 1 = 2

Degrees of freedom Within = n - k = 18 - 3 = 15

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.682

Step 3

Grand Mean = G / N = 3+4.5+10 / 3 = 5.833

SST = ( Xi - GrandMean)^2 = (4-5.833)^2 + (3-5.833)^2 + (2-5.833)^2 + ……..& so on = 170.5

SS Within = (Xi - Mean of Xi ) ^2 =,(4-3)^2 + (3-3)^2 + (2-3)^2 + ……..& so on = 7.5

SS Between = SST - SS Within = 170.5 - 7.5 = 163

Step 4

Mean Square Between = SS Between / df Between = 163/2 = 81.5

Mean Square Within = SS Within / df Within = 7.5/15 = 0.5

Step 5

F Cal = MS Between / Ms Within = 81.5/0.5 = 163

We got |F cal| = 163 & |F Crit| =3.682

MAKE DECISION

Hence Value of |F cal| > |F Crit|and Here We Reject Ho

Mean n Std. Dev 3.0 6 0.89 Treat1 4.5 6 0.55 Treat2 10.0 6 0.63 Treat3 5.8 18 3.17 Total ANOVA table Source SS    df MS F    p-value Treatment 163.00 2 81.500 163.00 6.68E-11 Error 7.50 15 0.500 Total 170.50 17


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