Problem1


Small: 1Y 3N
Medium: 4Y 2N
Large: 1Y 4N
Info(D)=I(6.9)=0.97095
I(1,3)=-(1/4)log2(1/4)-(3/4)log2(3/4)=0.81128
I(4,2)=-(4/6)log2(4/6)-(2/6)log2(2/6)=0.91829
I(1,4) )=-(1/5)log2(1/5)-(4/5)log2(4/5)=0.72193
Info(A1)=(4/15)I(1,3)+(6/15)I(4,2)+(5/15)I(1,4)=0.82430

Gain(A2)=Info(D)-Info(A2)=0.14665
A2 has the larger information gain, so A2 is the test attribute at root level.

Problem2
(1)
A1:
A1=Low->no error rate=4/9
A1=high->yes error rate=1/2
A1 error rate = 7/15
A2
A2=hot->yes error rate=2/5
A2=mild->yes error rate =2/5
A2=cold->no error rate = 1/5
A2 error rate=5/15
A3
A3=medium->yes error rate=2/6
A3=large-> no error rate=1/5
A3=small->yes error rate=2/4
A3 error rate = 5/15

Choose A2
A2=hot->yes
A2=mild->yes
A2=cold->no


(2)
Y N Y N

Problem 3

6Y and 9N
P(Y)=6/15=0.4
P(N)=9/15=0.6
P(A1=high|Y)=2/6
P(A2=hot|Y)=3/6
P(A3=large|Y)=1/6
P(x|Y)=0.027

P(A1=high|N)=4/9
P(A2=hot|N)=1/9
P(A3=large|N)=3/9
P(X|N)=0.016
P(Y)*P(X|Y)=0.0108
P(N)*P(X|N)=0.0096
The instance belong to Y.

Problem 4

Problem 5
Null Hyphothesis: Attrbutes A1 and A2 are independent(no correlation)



BY USING X^2(D,B)
DF=3
P-value also bigger then o.o5
Do not reject the null

Problem6
(1)




Problem 7
(1)


(2)





(3) A1 and A2 has the strongest correlation.

Problem 8
(1)


(2)



Problem 9
(1)


(2)
lowerq = 23
upperq = 50
iqr = 50-23=27
mild.threshold.upper = (iqr * 1.5) + upperq =90.5
mild.threshold.lower = lowerq - (iqr * 1.5) =-17.5
result = which(data > mild.threshold.upper | data < mild.threshold.lower)
which is: 19,27,92,97,99

(3)


R code:
data <- read.table("e1-p9.csv", header=TRUE,sep = ",")
attach(data)

quantile(data$A1)

source("https://raw.githubusercontent.com/talgalili/R-code-
snippets/master/boxplot.with.outlier.label.r") # Load the function
lab <- as.character(1:nrow(data))
boxplot.with.outlier.label(data$A1, label_name = data$A1)
legend(x = "topright", legend = "Outliers", pch = 1, col = "black")


Problem 10.


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