Using R, generate a random variable X that has 10,000 random uniform numbers from 1 to N, where N can be any number of your choosing greater than or equal to 6. Then generate a random variable Y that has 10,000 random normal numbers with a mean of \(\mu =\sigma (N+1)/2\).
creating the Training set:
set.seed(4242)
N <- 7
n <- 10000
mu <- sigma <- (N+1)/2
tset <- data.frame(X = runif(n,min = 1, max=N), Y=rnorm(n,mean=mu, sd=sigma))Calculate as a minimum the below probabilities a through c. Assume the small letter “x” is estimated as the median 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.
x = median(tset$X)
y = quantile(tset$Y, 0.25)
y## 25%
## 1.244764Pa1 <- tset %>%
filter(X>x, X>y)%>%
nrow()/n
Pa2 <- tset %>%
filter(X>y, X>y)%>%
nrow()/n
Pa1/Pa2 ## [1] 0.5213764b.P(X>x,X>y)
Probability of P(A>B) so this states all possible X values are greater than x and all possible Y values are greater than y
Pb2 <- tset %>%
filter(X>x, Y>y)%>%
nrow()/n
Pb2 ## [1] 0.377c.P(X<x|X>y)
Probability A given B -> P(A|B) stating X is greater than it’s own median AND greater than Q1 of Y
Pb3 <- tset %>%
filter(X<x, X>y)%>%
nrow()/n
Pb3 ## [1] 0.459Investigate whether P(X>x and Y>y)=P(X>x)P(Y>y) by building a table and evaluating the marginal and joint probabilities.
# Joint probability
matrix<- tset%>%
mutate(P1 = ifelse(X>x, "X>X", "X<X"), P2 = ifelse(Y>y, "Y>y", "Y<y"))%>%
group_by(P1, P2)%>%
summarise(count=n(), probability = count/n)
matrix## # A tibble: 4 x 4
## # Groups: P1 [2]
## P1 P2 count probability
## <chr> <chr> <int> <dbl>
## 1 X<X Y<y 1270 0.127
## 2 X<X Y>y 3730 0.373
## 3 X>X Y<y 1230 0.123
## 4 X>X Y>y 3770 0.377# Building marginal probability
matrix2<- matrix %>%
ungroup()%>%
group_by(P1)%>%
summarise(count=sum(count), probability = sum(probability))%>%
mutate(P2 = "Total")%>%
bind_rows(matrix)
matrix2## # A tibble: 6 x 4
## P1 count probability P2
## <chr> <int> <dbl> <chr>
## 1 X<X 5000 0.5 Total
## 2 X>X 5000 0.5 Total
## 3 X<X 1270 0.127 Y<y
## 4 X<X 3730 0.373 Y>y
## 5 X>X 1230 0.123 Y<y
## 6 X>X 3770 0.377 Y>ymatrix3 <- matrix2 %>%
ungroup()%>%
group_by(P2)%>%
summarise(count=sum(count), probability = sum(probability))%>%
mutate(P1 = "Total")%>%
bind_rows(matrix2)
matrix3## # A tibble: 9 x 4
## P2 count probability P1
## <chr> <int> <dbl> <chr>
## 1 Total 10000 1 Total
## 2 Y<y 2500 0.25 Total
## 3 Y>y 7500 0.75 Total
## 4 Total 5000 0.5 X<X
## 5 Total 5000 0.5 X>X
## 6 Y<y 1270 0.127 X<X
## 7 Y>y 3730 0.373 X<X
## 8 Y<y 1230 0.123 X>X
## 9 Y>y 3770 0.377 X>X# Table
matrix3 %>% select(-count) %>% spread(P1, probability) %>% rename(`Compare` = P2)## # A tibble: 3 x 4
## Compare Total `X<X` `X>X`
## <chr> <dbl> <dbl> <dbl>
## 1 Total 1 0.5 0.5
## 2 Y<y 0.25 0.127 0.123
## 3 Y>y 0.75 0.373 0.377Check to see if independence holds by using Fisher’s Exact Test and the Chi Square Test. What is the difference between the two? Which is most appropriate?
# Have to set up the dataframe to work with the fisher test
pre.q4 <- matrix3%>%
filter(P1 != "Total", P2 != "Total")%>%
select(-probability)%>%
spread(P1, count)%>%
as.data.frame()
# move values from P2 column to row names
row.names(pre.q4) = pre.q4$P2
#create the matrix
q4 <- pre.q4 %>%
select(-P2)%>%
as.matrix()Running the fisher test
fisher.test(q4,simulate.p.value = TRUE)##
## Fisher's Exact Test for Count Data
##
## data: q4
## p-value = 0.3678
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.9522537 1.1436981
## sample estimates:
## odds ratio
## 1.043608Running the Chi Square test
chisq.test(q4)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: q4
## X-squared = 0.8112, df = 1, p-value = 0.3678