Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A sample was collected to find out the probability of a sports car being owned b

ID: 3233120 • Letter: A

Question

A sample was collected to find out the probability of a sports car being owned by a family based on age, gender, income, and marital status. Use the following table and excel output to answer the questions below it: How many dummy variables are in the model? What is the proportion of variation in y explained by the x variables? What's the sample sire? Does the overall model have utility at the alpha = 0.05 level? State the multiple regression model with intercept and coefficients as per the output above? Which predictor variables are significant at the alpha = 0.05 level? Estimate the probability of owning a sport's car for a 25 year old single male that earns $50,000 With an alpha = 0.05 and repeat with an alpha = 0.10 Estimate the probability of owning a sport's car for a 32 year old married female that earns $80,000 With an alpha = 0.05 and repeat with an alpha = 0.10

Explanation / Answer

Answer to the questions below:

10. There are 2 dummy variables x2 and x4 ( gender and Marital status)

11. x explains 75.9% of y ( look at the rsquare)

12. the sample size is 1826

13. Yes, the overall model is significant , as pvalue of ANOVA regression
is less than .05

14. The regression model is:
log(p/1-p) = .758 -.0063*x1+.05*x2+.0000008266*x3-.266*x4

15.
Variable 1,3, and 4 have pvalue<.05 So only they are significant at alpha = .05
significance level

16.

At alpha = .10 every variable is significant

log(p/1-p) = .758 -.0063*25+.05*1+.0000008266*50000-.266*0 = .69
p = 1/(1+e^-.69) = .666

At alpha = .05,4, 1 and 3 variables are significant as they have pvalue<.05
log(p/1-p) = .758 -.0063*25+.0000008266*50000-.266*0 = .641
p = 1/(1+e^-.641) = .655


17)

At alpha = .10 every variable is significant

log(p/1-p) = .758 -.0063*32+.05*0+.0000008266*80000-.266*1 = .36
p = 1/(1+e^-.36) = .59

At alpha = .05,4, 1 and 3 variables are significant as they have pvalue<.05
log(p/1-p) = .758 -.0063*32+.05*0+.0000008266*80000-.266*1 = .36
p = 1/(1+e^-.36) = .59

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote