# Log Regression

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The marketing manager for a large nationally franchised lawn service company would like to study the characteristics that differentiate homeowners who do and do not have a lawn service. A random sample of 30 homeowners located in a suburban area near a large city was selected. Use the lawn.xls dataset in SPSS, Minitab or Excel. Predictor variables include household income ($K), lawn size (square feet K), attitude toward outdoor recreational activities, number of teenagers in the household, and age of the head of household.

1. Generate a scatterplot matrix (matrix plot) of the six variables. Comment on anything unusual or problematic.

2. Generate a correlation matrix of the six variables. Comment on substantial correlations.

3. Fit a logistic regression of whether a household has a lawn service (lawnserv) on the other five variables and state the estimated regression equation.

4. Estimate the probability of purchasing a lawn service for a 38-year-old home owner with a family income of $50K, a lawn size of 3,000 square feet, a negative attitude towards outdoor recreation, and one teenager in the household.

5. Test whether the overall regression model is significant.

6. Which predictor variables are significantly different from zero (use alpha=.05)?

7. Estimate a logistic regression model with income, attitude, and teenage as predictors. State the estimated regression equation and indicate which variables are significant.

8. Estimate a logistic regression model with lawnszie, attitude, and teenage as predictors. State the estimated regression equation and indicate which variables are significant.

9. Estimate a logistic regression model with HOH age, attitude, and teenage as predictors. State the estimated regression equation and indicate which variables are significant.

10. Write a short paragraph with your final conclusions.

11. If your objective were to make accurate predictions of the likelihood someone would want lawn service, which model would you suggest? Why?

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###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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