Question 1 (1 point)

Consider the following data. y is the outcome, x1 and x2 are two predictors of y. y=c('A','A','A','A','A','B','B','B') x1 = c(0,0,0,0,1,1,1,0) x2 = c(33,54,56,42,50,55,77,49) What is the Information Gain value associated with the partition (x2 < 45.5, x2 >= 45.5)?

Question 1 options:

	0.82
	0.75
	0.65

Question 2 (1 point)

Consider the following data. y is the outcome, x1 and x2 are two predictors of y y=c('A','A','A','A','A','B','B','B') x1 = c(0,0,0,0,1,1,1,0) x2 = c(33,54,56,42,50,55,77,49) What is the Information Gain value associated with the partition (x2 < 35.5, x2 >= 35.5)?

Question 2 options:

	0.8620746
	0.76
	0.82

Question 3 (1 point)

Consider the following data. class is the outcome, x and y are two predictors of class

class=c('A','A','A','A','A','B','B','B','B','B') x = c(0,0,0,0,1,1,1,0,1,0) y = c(33,54,56,42,50,55,31,-4,77,49) What is the Gini value associated with the partition (x1 = 0, x1 = 1)?

Question 3 options:

	0.5166667
	0.166667
	0.4166667

Question 4 (1 point)

Consider the following data. y is the outcome, x1 and x2 are two predictors of y.

y <- c(rep('A',10),rep('B',10)) x1 <- c(rep(0,6),rep(1,4),rep(0,8),rep(1,2)) x2 <- c(rep(0,4),rep(1,2),rep(0,2),rep(1,2),rep(0,2),rep(1,6),0,1) What is the Gini index associated with the split x1=0,x1=1?

Question 4 options:

	0.5761905
	0.3761905
	0.4761905

Question 5 (1 point)

Consider the following data: y <- c(rep(0,10),rep(1,10)) x1 <- c(rep(0,4),rep(1,8),rep(0,3),rep(1,5)) x2 <- c(rep(0,4),rep(1,2),rep(0,2),rep(1,2),rep(0,2),rep(1,6),0,1) Build a contingency table of the counts at each combination of factor levels for the predictor x1 (columns) and outcome y (rows). Which of the following tables is the correct table?

Question 5 options:

	x1 y 0 1 0 4 6 1 3 7
	x1 y 0 1 0 3 6 1 4 7
	x1 y 0 1 0 4 7 1 3 6

Question 6 (1 point)

Consider the following data. y is the outcome, x1 and x2 are two predictors of y y <- c(rep('A',10),rep('B',10)) x1 <- c(rep(0,4),rep(1,8),rep(0,3),rep(1,5)) x2 <- c(rep(0,4),rep(1,2),rep(0,2),rep(1,2),rep(0,2),rep(1,6),0,1) what is the Gini index of the split x1=0,x1=1

Question 6 options:

	0.5945055
	0.4945055
	0.3945055

Question 7 (1 point)

Consider the following data. y is the outcome, x1 and x2 are two predictors of y

y <- c(rep('A',10),rep('B',10)) x1 <- c(rep(0,4),rep(1,8),rep(0,3),rep(1,5)) x2 <- c(rep(0,4),rep(1,2),rep(0,2),rep(1,2),rep(0,2),rep(1,6),0,1) what is the Gini index of the split x2=0,x2=1

Question 7 options:

	0.4945455
	0.4545455
	0.5545455

Question 8 (1 point)

What is the main reason why tree-based models are useful?

Question 8 options:

	They can model non-linear relationships between an outcome and predictors
	They produce a smaller fitting error compared to linear regression models
	They are recursive algorithms

Question 9 (1 point)

The value that is predicted in each leaf of a CART regression tree is constant in the region defined by the leaf

Question 9 options:

	True
	Sometimes true, sometimes false
	False

Question 10 (1 point)

The values predicted in the leaves of a classical CART regression tree

Question 10 options:

	minimize the total sum of absolute values of errors
	maximize the total entropy
	minimize the total sum of squared errors

Have finished, do not need anymore! Thank you!