Problem 1

Pick one of the quantitative independent variables (Xi) from the data set below, and define that variable as X. Also, pick one of the dependent variables (Yi) below, and define that as Y.

X = X3 = {9.5, 3.7, 11.7, 7.4, 5.3, 7.4, 7.4, 8.6, 9.1, 11.4, 8.4, 7.3, 11.3, 4.4, 9.3, 10.9, 10.9, 7.7, 7.7, 11.5}

Y = Y3 = {28.4, 21.5, 20.8, 22.2, 21.6, 21.8, 25.2, 22.5, 21.1, 21.7, 21.4, 20.8, 23.0, 17.4, 21.3, 15.1, 17.8, 26.4, 21.6, 22.5}

Calculate as a minimum the below probabilities a through c. Assume the small letter “x” is estimated as the 3rd quartile of the X variable, and the small letter “y” is estimated as the 1st quartile of the Y variable. Interpret the meaning of all probabilities.

Assign X and Y values to their respective vectors.

X <- c(9.5, 3.7, 11.7, 7.4, 5.3, 7.4, 7.4, 8.6, 9.1, 11.4, 8.4, 7.3, 11.3, 4.4, 9.3, 10.9, 10.9, 7.7, 7.7, 11.5)

Y <- c(28.4, 21.5, 20.8, 22.2, 21.6, 21.8, 25.2, 22.5, 21.1, 21.7, 21.4, 20.8, 23.0, 17.4, 21.3, 15.1, 17.8, 26.4, 21.6, 22.5)

Summary of X.

  • The 3rd quartile (Q3) of X is 10.90.
  • There are 20 values in X.
summary(X)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   3.700   7.400   8.500   8.545  10.900  11.700
length(X)
## [1] 20

Summary of Y.

  • The 1st quartile (Q1) of of Y is 21.02.
  • There are 20 values in Y.
summary(Y)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.10   21.02   21.60   21.70   22.50   28.40
length(Y)
## [1] 20

Plot of X and Y.

Answers to Probabilities

Please see below for details of calculations.

Q3 of X = 10.90

Q1 of Y = 21.02

  1. P(X > 10.90 | Y > 21.02) Answer: 1/5 or 0.20
  2. P(X > 10.90, Y > 21.02) Answer: 3/20 or 0.15
  3. P(X < 10.90 | Y > 21.02) Answer: 12/15 or 0.80

Interpretation and Calculation of the Probabilities

Table A: Shows all 20 (x,y) points in the dataset

XY <- data.frame(X,Y)
kable(XY)
X Y
9.5 28.4
3.7 21.5
11.7 20.8
7.4 22.2
5.3 21.6
7.4 21.8
7.4 25.2
8.6 22.5
9.1 21.1
11.4 21.7
8.4 21.4
7.3 20.8
11.3 23.0
4.4 17.4
9.3 21.3
10.9 15.1
10.9 17.8
7.7 26.4
7.7 21.6
11.5 22.5 
nrow(XY)
## [1] 20


(a) P(X > 10.90 | Y > 21.02)

Below calculates the conditional probability that X > 10.90 given Y > 21.02.

Table B: Shows the 15 (x,y) points where Y > 21.02 

XY_subset1 <- subset(XY, Y > 21.02, select=c("X", "Y"))
rownames(XY_subset1) <- c()
kable(XY_subset1, row.names=NA)
X Y
9.5 28.4
3.7 21.5
7.4 22.2
5.3 21.6
7.4 21.8
7.4 25.2
8.6 22.5
9.1 21.1
11.4 21.7
8.4 21.4
11.3 23.0
9.3 21.3
7.7 26.4
7.7 21.6
11.5 22.5 
nrow(XY_subset1)
## [1] 15

Table C: Shows the 3 (x,y) points where X > 10.90 and Y > 21.02 

XY_subset2 <- subset(XY, X > 10.90 & Y > 21.02, select=c("X", "Y"))
rownames(XY_subset2) <- c()
kable(XY_subset2)
X Y
11.4 21.7
11.3 23.0
11.5 22.5 
nrow(XY_subset2)
## [1] 3

Answer: So, P(X > 10.90 | Y > 21.02) is 3/15 or 1/5 or 0.20. 

nrow(XY_subset2)/nrow(XY_subset1)
## [1] 0.2


(b) P(X > 10.90, Y > 21.02)

Below calculates the probability that X > 10.90 and Y > 21.02.

In part (a) above we already know that there are 3 points where X > 10.90 and Y > 21.02. See Table C.

We also know that there is a total of 20 points in the XY universe. See Table A.

Answer: So, P(X > 10.90 and Y > 21.02) is 3/20 or 0.15.

nrow(XY_subset2)/nrow(XY)
## [1] 0.15


(c) P(X < 10.90 | Y > 21.02)

Below calculates the conditional probability that X < 10.90 given that Y > 21.02.

Table D: Shows the 12 (x,y) points where X < 10.90 and Y > 21.02 

XY_subset3 <- subset(XY, X < 10.90 & Y > 21.02, select=c("X", "Y"))
rownames(XY_subset3) <- c()
kable(XY_subset3)
X Y
9.5 28.4
3.7 21.5
7.4 22.2
5.3 21.6
7.4 21.8
7.4 25.2
8.6 22.5
9.1 21.1
8.4 21.4
9.3 21.3
7.7 26.4
7.7 21.6 
nrow(XY_subset3)
## [1] 12

In part (a) above, we already know that there are 15 (x,y) points where Y > 21.02. See Table B.

Answer: So, P(X < 10.90 | Y > 21.02) is 12/15 or 0.80. 

nrow(XY_subset3)/nrow(XY_subset1)
## [1] 0.8


Table E: Shows count of each segment 

Y_less_or_equal_Q1 <- subset(XY, Y <= 21.02, select=c("X", "Y"))
Y_greater_Q1 <- subset(XY, Y > 21.02, select=c("X", "Y"))
row1 <- c(nrow(subset(Y_less_or_equal_Q1, X <= 10.90)), nrow(subset(Y_less_or_equal_Q1, X > 10.90)))
row2 <- c(nrow(subset(Y_greater_Q1, X <= 10.90)), nrow(subset(Y_greater_Q1, X > 10.90)))
rows <- rbind(row1, row2)
rows <- cbind(rows, rowSums(rows))
rows <- rbind(rows, colSums(rows))
rownames(rows) <- c('X <= 1st quartile', 'X > 1st quartile', 'Total')
colnames(rows) <- c('Y <= 3rd quartile', 'Y > 3rd quartile', 'Total')
kable(rows)
Y <= 3rd quartile Y > 3rd quartile Total
X <= 1st quartile 4 1 5
X > 1st quartile 12 3 15
Total 16 4 20

Does splitting the training data in this fashion make them independent?

Let A be the new variable counting those observations above the 1st quartile for X, and let B be the new variable counting those observations above the 1st quartile for Y.

Does P(AB) = P(A)P(B)?

Check mathematically, and then evaluate by running a Chi Square test for independence.

  • The 1st quartile of X is 7.40 (information under “Summary of x” section).
  • The 1st quartile of Y is 21.02 (information under “Summary of Y” section)


Plot of (x, y)

  • The blue vertical line is X = 7.40, which is the 1st quartile of X.
  • The red horizontal line is Y = 21.02, which is the 1st quartile of Y.

Table F: Shows the 13 (x,y) points where X > 7.40 

A <- subset(XY, X > 7.40, select=c("X", "Y"))
rownames(A) <- c()
kable(A)
X Y
9.5 28.4
11.7 20.8
8.6 22.5
9.1 21.1
11.4 21.7
8.4 21.4
11.3 23.0
9.3 21.3
10.9 15.1
10.9 17.8
7.7 26.4
7.7 21.6
11.5 22.5 
nrow(A)
## [1] 13


Table G: Shows the 15 (x,y) points where X > 7.40 

B <- subset(XY, Y > 21.02, select=c("X", "Y"))
rownames(B) <- c()
kable(B)
X Y
9.5 28.4
3.7 21.5
7.4 22.2
5.3 21.6
7.4 21.8
7.4 25.2
8.6 22.5
9.1 21.1
11.4 21.7
8.4 21.4
11.3 23.0
9.3 21.3
7.7 26.4
7.7 21.6
11.5 22.5 
nrow(B)
## [1] 15


Table H: Shows the 10 (x,y) points where X > 7.40 and Y > 21.02 

AB <- subset(XY, X > 7.40 & Y > 21.02, select=c("X", "Y"))
rownames(AB) <- c()
kable(AB)
X Y
9.5 28.4
8.6 22.5
9.1 21.1
11.4 21.7
8.4 21.4
11.3 23.0
9.3 21.3
7.7 26.4
7.7 21.6
11.5 22.5 
nrow(AB)
## [1] 10

P(A) = 13/20 or 0.65

nrow(A)/nrow(XY)
## [1] 0.65

P(B) = 15/20 or 0.75

nrow(B)/nrow(XY)
## [1] 0.75

P(AB) = 10/20 = 0.50

nrow(AB)/nrow(XY)
## [1] 0.5

Does P(AB)=P(A)P(B)?

P(AB) = 0.50 P(A)P(B) = 0.4875

As you can see P(AB) != P(A)P(B), which means that A and B are not independent. So, splitting the data in this fashion does not make them independent, 

prob_A <- 0.65
prob_B <- 0.75
prob_A * prob_B
## [1] 0.4875


Evaluate by running a Chi-Square test for independence


For this test, I broke down the data points into 2 categories. Category 1 has 2 levels: A and Not-A. Category 2 has 2 levels: B and Not-B. A data point (x,y) is in A if X > 7.40 otherwise it’s in Not-A. A data point (x,y) is in B if Y > 21.02 otherwise it’s in Not-B. So each (x,y) point is assigned to a specific level for each category.

Table H: Contingency table by 2 Categories

XY$A[XY$X > 7.40] <- 'A'
XY$B[XY$Y > 21.02] <- 'B'
XY$A[XY$X <= 7.40] <- 'Not-A'
XY$B[XY$Y <= 21.02] <- 'Not-B'
AB_table <- table(XY$A, XY$B)
kable(AB_table)
B Not-B
A 10 3
Not-A 5 2

The null hypothesis is category 1 and category 2 are independent (no relationship exists between the 2 categories)

The alternative is category 1 and category 2 are not independent (a relationship exists between the 2 categories)

The significance level is 0.05.

result <- chisq.test(AB_table)
## Warning in chisq.test(AB_table): Chi-squared approximation may be incorrect
result
## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  AB_table
## X-squared = 0, df = 1, p-value = 1

The results of the Pearson’s chi-squared test shows a chi-square statistic of 0, which suggests that there is no difference between the actual and expected data, and that any difference between the actual and expected data is due to chance. Below you will see the expected and actual data.

Expected data:

result$expected
##        
##            B Not-B
##   A     9.75  3.25
##   Not-A 5.25  1.75

Actual data:

AB_table
##        
##          B Not-B
##   A     10     3
##   Not-A  5     2

The p-value is 1, which is greater than the significance level of 0.05, which means that we fail to reject the null hypothesis that the 2 categories are independent.

The result suggests that there is no significant relationship between the two categories (they are independent); however, chi-square tests are sensitive to sample size and frequency count of each cell. The contingency table should have at least a frequency of 5 when doing a chi-square test . As you can see, the actual data has frequencies that are less than 5. In addition, the chi-square test also presents a warning stating that the chi-squared approximation may be incorrect.

I would have to conclude that the result of this chi-square test is not reliable since we do not have sufficient data to perform this test.

Sources:

https://stattrek.com/chi-square-test/independence.aspx

https://www.statisticssolutions.com/using-chi-square-statistic-in-research/