Submit your assignment to Gradescope on the subject Canvas LMS.

In a study of the predictors of domestic violence, a survey was conducted among women, aged 18 and over, who attended a general practitioner in Metropolitan Melbourne between November 1993 and February 1994, and who had been in a relationship during the previous 12 months. The “response” variable was whether or not the woman responded positively to questions about physical domestic violence or mental abuse experienced during the previous  12 months.   The predictor variables considered included the following:

The responses from 1316 women, for whom there were no missing values, have been stored in file (domviolence .csv) which can be read into R using the command

domviolence  <-  read .csv(file="C:/subjects/MAST90139/data/domviolence .csv")

provided that domviolence .csv is on the path C:/subjects/MAST90139/data/. The file can be found in the subject Canvas Module section.

The variable dv is the response variable: 0 = no; 1 = yes.

1. Carry out a series of data analysis using glm(),  anova(),  summary() and step() etc. to argue that the“best” logistic model is of the form

logit(θ)   =   β0 + β1 ×ms.2 + β2 ×ms.3 + β3 ×ms.4 + β4 ×ms.5 + β5 ×ms.6

+β6 ×smok + β7 ×falc + β8 ×reg.2 + β9 ×reg.3 + β10 ×reg.4 +β11 ×age + β12 ×educ + β13 ×ms.2 : falc + β14 ×ms.3 : falc +β15 ×ms.4 : falc + β16 ×ms.5 : falc + β17 ×ms.6 : falc

where estimates of all βj   in the above model can be obtained from the arguing process. Present your work and results in no more than 300 words (excluding R outputs; the R outputs

should be presented as an appendix).                                                                           [15]

Some suggestions:

(a) You can start the process by fitting a model0 that includes all predictor variables as having main effects only.  Then use anova(model0,  test="Chi") alike commands re- peatedly to remove any non-significant predictors.   Note categorical predictors, e.g. reg, should be treated as factors here and be included in the model in the form of e.g. factor(reg). Denote as model1 after this process. Note model1 may also be obtained by using commands model0 .5  =  step(model0) and anova(model0 .5,  test="Chi") to remove any non-significant predictors.

(b) Even though age and education are observed as categorical variables, they are actually numerical variables. Thus, replace factor(age) and factor(educ) with age and educ in model1 to get model2. Then fit model2 to see whether it is significantly different from model1 or not in terms of goodness of fit. Accept model2 if it is not significantly different from model1, because model2 is simpler than model1 in terms of model complexity.

(c) Expand model2 to model3 that contains all predictors in model2 plus all their first-order interaction terms.  Perform model comparisons by anova( . . .)   to simplify model3. Alternatively,  use  step(model3)  to select the stepwise  “best” model.   Denote the resultant model as model4  and fit it.   Then use anova(model4,  test="Chi")  and summary(model4) to see whether it can be further simplified. You can use this eventu- ally obtained model as the“best” model.

2. Suppose the “best” model is indeed the one shown above.  Interpret this “best” model in terms of odds ratios for each predictor in this mode.  There will be many odds ratios to be presented here.  So it should be helpful to use tables or diagrams in your description. Specifically, you can answer the question in following form:                                           [15]