QUESTION 4 100 points Save Answer Include all of your answers in a SINGLE docume
ID: 3355980 • Letter: Q
Question
QUESTION 4 100 points Save Answer Include all of your answers in a SINGLE document and attach that document at the end of THIS problem. Label each answer clearly. Please begin each problem on a new page. Point values are indicated in the question body below.] Problem 4: The Buyer's Guide section of the web site for Car and Driver magazine provides reviews and road tests for cars, trucks, SUVs, and vans. The average ratings for overall quality, vehicle styling, braking, handling, fuel economy, interior comfort, acceleration, dependability, fit and finish, transmission, and ride are summarized for each vehicle using a scale ranging from 1 (worst) to 10 (best). Using these data, a marketing firm has developed a regression model using ratings of handling, dependability, and fit and finish to predict overall quality of sports cars. The Excel output for the model is shown below. SUMMARY OUTPUT ession Statistics 0.9259 0.8573 0.8145 0.1946 MultipleR uare Adjusted R Square Standard Error Observations ANOVA MS Si nce Regression Residual Total 3 2.2761 0.7587 20.0272 0.0001503 0.3788 2.6550 10 0.0379 13 Coefficients Sta ndard Er t Stat er 95% Lower 95.0% 95.0% 0.6362 1.3202-4.48190.0064 -3.5778 2.3055 -3.5778 2.3055 0.4582 0.1132 4.0466 0.0023 0.2059 0.7105 0.2059 0.7105 0.3385 0.1615 2.0953 0.0626-0.0215 0.6984-0.0215 0.6984 0.4868 p-value Lower 95% Intercept Handling Dependability Fit and Finish 2564 0.1034 2.4802 0.0325 0.0261 0.4868 0.0261 Use this information to answer the questions below. Remember to show your work and/or explain your reasoning a). Use the output above to evaluate how well the model predicts the overall quality rating. (Hint: You should discuss at least three items from the output.) [10 pts] b). If you were trying to find the best regression model using the given variables, would you eliminate any of the explanatory variables used above? Why or why not? Briefly explain. [8 pts] c). True or False: Adding an explanatory variable to a regression model will always result in R equal or greater than the R of the original model. Very briefly justify your response. [4 pts d). Another sports car rated by Car and Driver is the Acura RSX. The ratings for handling, dependability, and fit and finish for the Acura RSX were 9.03, 8.32, and 8.09 respectively. Estimate the overall rating for this car. [6 pts] e). The overall rating reported by Car and Driver for the Acura RSX was 8.25. Compute the residual value for the estimate determined in part d. 4 pts]Explanation / Answer
a) R^2 = 85.73%
So, these independent variable explain 85.73% of the variation in overall rating. THus, it is a good value.
Overall p value (0.00015) < alpha. THus null is accepted telling us overall model is not significant.
Also, some more predictors can be added into the model.
b)
The p-value for each term tests the null hypothesis that the coefficient is equal to zero (no effect). A low p-value (< 0.05) indicates that you can reject the null hypothesis. In other words, a predictor that has a low p-value is likely to be a meaningful addition to your model because changes in the predictor's value are related to changes in the response variable.
Dependability has p > 0.05 and thus it should be eliminated. Others have p < 0.05 and are meanigful.
c) The adjusted R squared increases only if the new term improves the model more than would be expected by chance and it can also decrease with poor quality predictors. If you add more and more useless variables to a model, adjusted r-squared will decrease. If you add more useful variables, adjusted r-squared will increase.
Adjusted R2 will always be less than or equal to R2. You only need R2 when working with samples. In other words, R2 isn’t necessary when you have data from an entire population.
So, false
d) y = -0.6362 + 0.4582 * 9.03 + 0.3385 * 8.32 + 0.2564 * 8.09
Overall rating = 8.39
e) In regression analysis, the difference between the observed value of the dependent variable (y) and the predicted value () is called the residual
Residual = 8.25 - 8.39 = - 0.14
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.