Question 5: Consider the following table (The plot is just a snapshot of part of the data, but the model was created using the whole data set): > m5.1 < 1m(logprice~total. full, data=Housing) > summary(m5.1)" Call: 1m(formula = Coefficients: logprice total. full, data = Housing) Estimate Std. Error t value Pr(>|t|) (Intercept) 1.111e+01 2.115e-02 525.38 var(logprice) = 0.6399318 > mean (Housing $total.full) = 149410.8 > var (Housing $total. full) = 11106410992 0.0025- Fitted values vs Leverage of 50 points in the data Leverage 0.0020- 0.0015- 0.0010- 0.0005- 11.5 12.0 Fitted.values 12.5 A) i) Calculate the cutoff value for leverage points and the cut off of the cook's distance based on the whole data size: B) Consider the 140th data point and based on m5.1: > Housing[140,c(1,9)] total.full 140 0 logprice 12.54083 Calculate i) The leverage of this point ii) The fitted value of this point The residual of this point iv) The standardized residual of this point. v) The cook's distance of this point C) Do you classify the point (X140, Y140) to be: i) A Good leverage iii) A Bad leverage point Why so? ii) An outlier but Not a leverage point iv) Not a leverage point nor an outlier D) Given the following: > LEVERAGE 4/length(logprice), "YES","NO") > OUTLIER 2,"YES","NO") > table(LEVERAGE, OUTLIER) LEVERAGE NO NO 2687 YES 114 OUTLIER YES 193 6 Based on model m5.1, how many outliers (not leverage) points does this data have?