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Question: Forecasting a job applicant’s merit rating. A large research and development firm rates the performance of each member of its technical staff on a scale of 0 to 100, and this merit rating is used to determine the size of the person’s pay raise for the coming year. The firm’s personnel department is interested in developing a regression model to help forecast the merit rating that an applicant for a technical position will receive after being employed for 3 years. The firm proposes to use the following second-order model to forecast the merit ratings of applicants who have just completed their graduate studies and have no prior related job experience:

E(y)=β0+β1x1+β2x2+β3x1x2+β4x12+β5x22, where

y = Applicant’s merit rating after 3 years

x1= Applicant’s GPA in graduate school

x2= Applicant’s total score (verbal plus quantitative) on the Graduate Record Examination (GRE)

The model, fit to data collected for a random sample of n=40employees, resulted inSSE=1830.44andSSmodel=4911.5. The reduced modelE(y)=β0+β1x1+β2x2 is also fit to the same data, resulting inSSE=3197.16.

  1. Identify the appropriate null and alternative hypotheses to test whether the complete (second-order) model contributes information for predicting y.
  2. Conduct the test of the hypothesis given in part a. Test using. Interpret the results in the context of this problem.
  3. Identify the appropriate null and alternative hypotheses to test whether the complete model contributes more information than the reduced (first-order) model for predicting y.
  4. Conduct the test of the hypothesis given in part c. Test using. Interpret the results in the context of this problem.
  5. Which model, if either, would you use to predict y? Explain.

Short Answer

Expert verified

Answer

  1. The null and alternate hypothesis for testing whether the second order model contributes information for the prediction of y can be written as
    H0:β4=β5=0while At least one of the parameters β3,β4orβ5or is non-zero.
  2. At a 95% significance levelβ4β50 , the second-order model doesn’t contribute much to the model.
  3. The null and alternate hypotheses for testing whether the complete model contributes more information than the reduced model for the prediction of y can be written asH0:β3=β4=β5=0 while

At least one of the parametersβ3,β4orβ5.

4. At the 95% significance level, at least one of the parameters β3,β4orβ5is non-zero.

5. Since the hypothesis testing concludes the complete model is a better fit for the data, the complete model would be more useful for predicting y.

Step by step solution

01

Given information 

The second order model used by the firm is given as Ey=β0+β1x1+β2x2+β3x1x2+β4x12+β5x22,there are 40 observations and the,SSE=1830.44,SSmodel=4911.5 .

The reduced model is given where and the test needs to be conducted .

02

(a) Hypothesis statement

The Research and development firm is doing a regression analysis to find if certain variables affect their employees’ merit scores or not. The firm first fits a second-order equation and then fits a reduced equation with only one variable.

Here, the null and alternate hypotheses for testing whether the second order model contributes information for the prediction of y can be written asH0:β4=β5=0while

At least one of the parametersβ4orβ5 is non-zero.

03

(b) Hypothesis testing

H0:β4=β5=0

Ha :At least one of the parametersβ4orβ5 is non-zero.

Here,

F-teststatistic=SSEn-k+1=1830.4440-6=53.836

Value of F0.05,39,39is 1.509.

H0 :is rejected if F-statistic>F0.05,39,39.

There is sufficient evidence to reject H0at 95% confidence interval.

Therefore, β4β50meaning that the second order model contributes much to the model.

04

(c) Hypothesis statement

The null and alternate hypothesis for testing whether the complete model contributes more information than the reduced model for the prediction of y can be written asH0:β3=β4=β5=0
while At least one of the parameters β3,β4orβ5is non-zero.

05

(d) Hypothesis testing

H0:β3=β4=β5=0

Ha:At least one of the parameters β3,β4orβ5is non-zero.

Teststatistic=SSER-SSECk-gSSECn-k+1=3197.16-1830.443-11830.4440-3+1=13.4401

Forα=.05,F-Test statistic = 13.4401.

The numerator degree of freedom is k – g =2, and the denominator degree of freedom is n – (k+1) = 36.

We know that it is rejected if

The critical value of F0.05 = 3.37, since F-statistic>F0.05, therefore, at95%confidence interval we reject H0.

Therefore, at least one of the parametersβ3,β4orβ5 is non-zero.

06

(e) Interpretation

The results of the hypothesis testing show that the complete model fits the data better. It would be more useful to predict y.

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