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Question: Study of supervisor-targeted aggression. 鈥淢oonlighters鈥 are workers who hold two jobs at the same time. What are the factors that impact the likelihood of a moonlighting worker becoming aggressive toward his/her supervisor? This was the research question of interest in the Journal of Applied Psychology (July 2005). Completed questionnaires were obtained from n = 105 moonlighters, and the data were used to fit several multiple regression models for supervisor-directed aggression score 1y2. Two of the models (with R2-values in parentheses) are given below:

a. Interpret the R2-values for the models.

b. Give the null and alternative hypotheses for comparing the fits of models 1 and 2.

c. Are the two models nested? Explain.

d. The nested F-test for comparing the two models resulted in F = 42.13 and p-value < .001. What can you conclude from these results?

e. A third model was fit, one that hypothesizes all possible pairs of interactions between self-esteem, history of aggression, interactional injustice at primary job, and abusive supervisor at primary job. Give the equation of this model (model 3).

f. A nested F-test to compare models 2 and 3 resulted in a p-value > .10. What can you conclude from this result?

Short Answer

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Answer

a. The R2 values for the two models are 0.101 and 0.555 respectively. Higher value of R2denotes that the model is a good fit for the data and that approximately what percentage of the variation in the variables is explained by the model. This number is very less for model 1 (only 10%) indicating that the model is not a good fit while for model 2 the value of R2 is 55.5% which is still able to explain around 55% variation in the model.

b. The null and alternate hypothesis to test whether the complete model contributes more information for the prediction of y than the reduced model can be written as

H0: 尾5 = 尾6 = 尾7 = 尾8 = 0 whileHa: At least one of 尾 parameters are nonzero.

c. Two models are nested if one model contains all the terms of the second model and at least one additional term. IN the question model 1 contains 4 terms while model 2 contains 9 terms where 4 are additional terms (尾5, 尾6, 尾7, and 尾8). Hence the two models are nested models where model 1 is reduced model and model 2 is complete model.

d. The nested F-test for comparing the two models resulted in F = 42.13 and p-value < .001. H0 is rejected if p-value < 伪. For 伪 = 0.10 or 0.05, p-value <0.001 is very less. Hence, H0 is rejected.

e. The equation for model 3 with all possible pairs of interaction can be written as 0+1x1+2x2+3x3+4x4+5x5+6x6+7x7+8x8+9x5x6+10x7x8+11x5x7+12x5x8+13x6x7+14x6x8..

f. The nested F-test for comparing model 2 and 3 resulted in p-value > .10. H0 is rejected if p-value < 伪. For 伪 = 0.10 or 0.05, p-value > 0.10. Hence, H0 is not rejected.

Step by step solution

01

Interpretation of R2

The R2 values for the two models are 0.101 and 0.555 respectively. Higher value of R2denotes that the model is a good fit for the data and that approximately what percentage of the variation in the variables is explained by the model. This number is very less for model 1 (only 10%) indicating that the model is not a good fit while for model 2 the value of R2 is 55.5% which is still able to explain around 55% variation in the model.

02

Hypotheses

The null and alternate hypothesis to test whether the complete model contributes more information for the prediction of y than the reduced model can be written as

H0: 尾5 = 尾6 = 尾7 = 尾8 = 0 whileHa: At least one of 尾 parameters are nonzero.

03

Nested models

Two models are nested if one model contains all the terms of the second model and at least one additional term. IN the question model 1 contains 4 terms while model 2 contains 9 terms where 4 are additional terms (尾5, 尾6, 尾7, and 尾8). Hence the two models are nested models where model 1 is reduced model and model 2 is complete model.

04

Hypothesis testing

The nested F-test for comparing the two models resulted in F = 42.13 andp-value < .001.

H0 is rejected if p-value < 伪. For 伪 = 0.10 or 0.05, p-value <0.001 is very less

Hence, H0 is rejected.

05

Model 3 equation

The equation for model 3 with all possible pairs of interaction can be written as0+1x1+2x2+3x3+4x4+5x5+6x6+7x7+8x8+9x5x6+10x7x8+11x5x7+12x5x8+13x6x7+14x6x8..

06

Hypothesis testing

The nested F-test for comparing model 2 and 3 resulted in p-value > .10.

H0 is rejected if p-value < 伪. For 伪 = 0.10 or 0.05, p-value > 0.10.

Thus, H0 is not rejected.

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Most popular questions from this chapter

Suppose you fit the model y =0+1x1+1x22+3x2+4x1x2+to n = 25 data points with the following results:

^0=1.26,^1= -2.43,^2=0.05,^3=0.62,^4=1.81s^1=1.21,s^2=0.16,s^3=0.26, s^4=1.49SSE=0.41 and R2=0.83

  1. Is there sufficient evidence to conclude that at least one of the parameters b1, b2, b3, or b4 is nonzero? Test using a = .05.

  2. Test H0: 尾1 = 0 against Ha: 尾1 < 0. Use 伪 = .05.

  3. Test H0: 尾2 = 0 against Ha: 尾2 > 0. Use 伪 = .05.

  4. Test H0: 尾3 = 0 against Ha: 尾3 鈮 0. Use 伪 = .05.

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d. The p-value of the subset F-test for comparing the two models for successful cabin crews was reported in the article as p 6 .05. Formally test the null hypothesis using 伪 = .05. What do you conclude?

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