Question

I would prefer a handwritten answer to this question with detailed steps

1. 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 them forecast the merit rating that an applicant for a technical position will receive after being employed three 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 = 40 employees resulted in SSE = 1,830.44 and SS(model) = 4,911.5. The reduced model E(y) = β₀ + β₁x₁ + β₂x₂ is also fit to the same data, resulting in SSE = 3197.16.

1. Identify the appropriate null and alternative hypotheses to test whether the complete model contributes more information than the reduced (first-order) model for the prediction of y. (6 pts)

2. Conduct the test of hypothesis given in part (a). Test using α = .05. Interpret the results in the context of the problem.   (10 points)

3. Which model, if either, would you use to predict y. Explain.   (5 points)     

Answer 1

(a) The hypothesis being tested is:

H0: β1 = β2 = β3 = β4 = β5 = 0

H1: At least one βi ≠ 0

(b)

Source	SS	df	MS	F	p-value
Regression	4911.5	5	982.3	18.246	8.89837E-09
Residual	1830.4	34	53.836471
Total	6741.9	39

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the model is significant.

(c) r2 = 4911.5/6741.9 = 0.73

The model will be preferred because it explained more variation in the data than the previous model.