Please show me the process, thank you. A consumer group rates 54 cars on Cost/Mi

Question

Please show me the process, thank you.

A consumer group rates 54 cars on Cost/Mile, a Road-Test score given by an expert driver, Predicted Reliability based on historical repair records for the cars, and a Value Score given by owners of the cars. Above is multiple regression output 26-31 (with deletions) from Minitab relating these variables. Use it to answer Questions Regression Analysis: Value Score versus Cost/Mile, Road-Test Score, Predicted Reliability Model Summary SR-sq R-sq(adj) 0.0721260 CoefficientS Term Constant Cost/Mile Road-Test Score Predicted Reliability 0.1651 0.0102 Coef SE Coef T-Value P-Value 0.000 13.42 1.2444 0.0927 -2.043 0.105 0.01138 0.00123 9.25 0.000 26. The variance of the model residuals is 0.00491. The variance of Value Score is 0.0726. Based on these, the R-squared for this model is A. 93.2% B. 6.8% C. 87.2% D. 12.8% E. Cannot be computed based on the given information. 10 27. The T-Value for Predicted Reliability is: A. 12.53 B. 16.26 C. 4.170 D. 1.626 E. Cannot be computed based on the given information. 28. The P-Value for the Cost/Mile (S1000) variable is A.0.10 B. Between 0.05 and 0.10 C. Between 0.01 and 0.05 D. 0.01 29. Another variable that was recorded is Price. It has a correlation of 0.945 with Cost/Mile. Adding Price to this model A. Must increase the adjusted R-squared. B. Must decrease the multiple R-squared. C. Should not be done D. Both B and C. E. None of the above

Explanation / Answer

Question 26

R - squared = (Variance of value score - Variance of model variance)/ Variance of value score

R - squared = (0.0726 - 0.00491)/ 0.0726 = 0.932 = 93.2% [option a is correct]

Question 27

T -value for Predicted reliability is

T = coefficient / SE Coeff. = 0.1651/ 0.0102 = 16.2 [ option b is correct]

Question 28

P value for Cost/mile variable is

first we will calculate the T - value

T = -2.043/ 0.105 = -19.46

p- value shall be less than 0.01 [option D is correct]

Question 29

Here it is given that price is highly correlated with cost/mile so we can say that the model has multicollinearity proble.

Multicollinearity doesn’t affect how well the model fits. In fact, if you want to use the model to make predictions, both models produce identical results for fitted values and prediction intervals.

So, adding a new model will not secerly affect the r- square value but may decrease the r-squared value. So option (b) and (c) are both correct.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.