Question

A teacher wants to develop a model to predict a student’s grade on the final exam from the number of hours spent studying for the final exam and the student’s GPA at the university. The data (for 22 students) follows below.

      PREDICTOR               COEF           STDEV               P-VALUE

      Constant                      -1.30             1.429                0.405

      Hours                           .0793            .0759                0.344

      GPA                             1.11             .7543                0.202

      ANOVA

      SOURCE                                SS                     DF                 MS                F                                 

      Regression                             5.0040         

                                          

      Error                                       1.1548         

      TOTAL  

(a) What is the student’s expected grade if she has a 2.7 GPA and she studies 12 hours for this test?

(b) Interpret the slope coefficient for the variable Hours.

(c) Use the p-value approach to see if GPA is linearly related to the dependent variable? Use alpha = .05. Please include all test parts.

(d) Fill in all missing parts of the ANOVA table.

(e) Does the model explain a significant portion of the variation in final grades? Use alpha = .05. Conduct a formal (all five parts) test.

(f) Calculate and interpret the coefficient of determination.

(g) Based upon the p-values for the individual independent variables, does either variable appear linearly related?

(h) Is the finding in part (g) consistent with would answer to part (e).

Answer 1

(a) Regression Equation:

Grade= -1.30+0.0793Hours+1.11GPA

When GPA=2.7, Hours=12, then expected grade=-1.30+0.0793*12+1.11*2.7=2.6486

(b) If study time is increased by 1 hours and if GPA is kept fixed then expected grade is increased by 0.0793.

(c) Since p-value corresponding GPA=0.202>0.05 so GPA is not linearly related to the dependent variable.

(d)

Source	SS	DF	MS=SS/DF	F
Regression	5.0040	1	5.0040/1=5.0040	5.0040/0.05774=86.6644
Error	1.1548	21-1=20	1.1548/20=0.05774
Total	5.0040+1.1548=6.1588	22-1=21

(e)

Null hypothesis, H₀: Overall model is not significant vs. Alternative hypothesis, H_A: H₀ is not true.

Test statistic=F=86.6644

Critical value=F_0.05,1,20=4.3512

Since value of test statistic>critical value so we reject null hypothesis at 5% level of significance and conclude that the model explains a significant portion of the variation in final grades.

(f) R²=SS_regession/SS_total=5.0040/6.1588=0.8125 i.e. 81.25% of total variation in final grade is explained by this model.

(g) No, since all p-value>0.05.

(h) No, because from (e), we see that the model explains a significant portion of the variation in final grades. Whereas from (g) we see that the individual independent variables are not linearly related with dependent variable.