Problem 1
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=theta=(N+1)/2.
set.seed(999)
a = 1
n = 12
X = runif(10000, min = 1, max = n)
Y = rnorm(10000, mean = (n+1)/2 ,sd = (n+1)/2)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. 5 points a. P(X>x | X>y) b. P(X>x, Y>y) c. P(X<x | X>y)
#Standard Deviation for X
X_sd = (n-a)/sqrt(n)
#Standard Deviation for Y
Y_sd = (n+1)/2Y_mu = (n+1)/2
#To calculate the 1st quartile of Y, we subtract the product of the z score and sd, from the mean of Y.
y = Y_mu + qnorm(.25)*Y_sd
x = ((n-a)/2)+a\[Conditional \space Probability \\P(F|E)= \frac{P(F \cap E)}{P(E)} \] \[ f(x) = \frac{{n_1 \choose x} {n_3 - n_1\choose n_2 - x}}{{n_3 \choose n_2}}. \]
\[Standard \space Deviation \space of \space Uniform \space Distibution \\ \sigma = \frac{b-a}{\sqrt{12}}\]
We note that y<x, therefore X>x is a subset of X>y. This tells us that X>x is the intersection of the two sets. Thus, P(X>x | X>y) = P(X>x)/P(X>y)
#We can apply the conditional probability formula
P_a = round(( ((n-x)/n) / ((n-y)/n) ),2)
print("This expression describes the probability that X is larger than x, given that it is larger than y")## [1] "This expression describes the probability that X is larger than x, given that it is larger than y"print(paste0("P(X>x | X>y)= ", P_a))## [1] "P(X>x | X>y)= 0.56"This expression describes the circumstance that X is larger than x, and Y is larger that y. Assuming these are independent yields the below results. P(X>x) = .5 and P(Y>y) = .75 P(X>x, Y>y) = P(X>x) * P(Y>y) = .375
This expression describes the chance of X being less than x, given that it is greater that y.
P_c = round((x-y)/(n-y),2)
P_c## [1] 0.445 points. Investigate whether P(X>x and Y>y)=P(X>x)P(Y>y) by building a table and evaluating the marginal and joint probabilities.
df = data_frame(X,Y) %>%
mutate( Xtophalf = X>x, YtopHalf= Y>y)%>%
group_by(Xtophalf, YtopHalf)%>%
summarise(countx = n())%>%
pivot_wider(names_from = YtopHalf, values_from = countx)%>%
rename(X_Y_Top_Half = 1)## Warning: `data_frame()` was deprecated in tibble 1.1.0.
## Please use `tibble()` instead.print("As expected the joint probability table suggests that the variables are independent")## [1] "As expected the joint probability table suggests that the variables are independent"print(df)## # A tibble: 2 x 3
## # Groups: X_Y_Top_Half [2]
## X_Y_Top_Half `FALSE` `TRUE`
## <lgl> <int> <int>
## 1 FALSE 1307 3770
## 2 TRUE 1197 37265 points. Check 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?
#1st step is to construct the expected value matrix
exp_m = matrix(c(1250,1250,3750,3750), ncol = 2)\[f(x) = \frac{{2514 \choose x}{7496 \choose 5077 - x}}{{10000 \choose 5077}}.\]
sum = 0
for (i in 1037:5077){
sum = sum + dhyper(i,2514,7496,5077-i)
}
print("The Null Hypothesis is that there is no correlation between the variable, the p-value shown below is to high to reject that hypothesis. In other words, we have no reason to believe they are correlated.")## [1] "The Null Hypothesis is that there is no correlation between the variable, the p-value shown below is to high to reject that hypothesis. In other words, we have no reason to believe they are correlated."print(sum)## [1] 0.1226736