help("dhyper")
dhyper(3,8,12,10)[1] 0.2400572phyper(2,8,12,10)[1] 0.08490117phyper(5,8,12,10,lower.tail = F)[1] 0.08490117Lab 6 - with solutions
Please note that there is a file on Canvas called Getting started with R which may be of some use. This provides details of setting up R and Rstudio on your own computer as well as providing an overview of inputting and importing various data files into R. This should mainly serve as a reminder.
Recall that we can clear the environment using rm(list=ls()) It is advisable to do this before attempting new questions if confusion may arise with variable names etc.
Example 1
In this example we will see how to calculate probabilities from the hypergeometric distribution.
- We will calculate the probability as seen in Example 3.8 in the lecture notes. Recall Example 3.8:
“Suppose we have a box of 20 marbles in which 8 are blue and 12 are red. Ten marbles are taken from the box without replacement. Find the probability of obtaining exactly 3 blue marbles.”
- In this case, let \(X\) denote the number of blue marbles, i.e. \(X\sim\text{hypergeometric}(20, 8, x)\) and therefore we want to find \(P(X=3)\). We further will calculate \(P(X\leq 2)\) and \(P(X\geq 6)\). Use the help function to understand the parameters.
Exercise 1
If we add 5 blue marbles to the box in the example above, but keep the sample the same (10), what is \(P(X>4)\)?
Solution
phyper(4,13,12,10,lower.tail = F)[1] 0.7158837Exercise 2
Suppose we now have a situation where we have 10 green and 20 white marbles and 15 are selected. Let \(Y\) denote the number of green marbles selected, i.e. \[Y\sim\text{hypergeometric}(30,10,y).\] Calculate \(P(Y=5)\) and \(P(Y<3)\).
Solution
- \(P(Y=5):\)
dhyper(5,10,20,15)[1] 0.3001499- \(P(Y<3):\)
phyper(2,10,20,15)[1] 0.02508746Example 2
In this example we will see how to use Fisher’s Exact Test in R. Firstly, remind yourself why we would use this test over the \(\chi^2\) test - see the lecture notes or full details. This is important when interpreting the R output.
We will consider Example 3.9 in the lecture notes where we investigate whether there is a relationship between attendance and exam results. Clearly we have independence, i.e. each member can only contribute to one entry in the table:
| Pass | Fail | Totals | |
|---|---|---|---|
| Attendance \(\mathbf{>50\%}\) | 10 | 2 | 12 |
| Attendance \(\mathbf{\leq50\%}\) | 5 | 7 | 12 |
| Totals | 15 | 9 | 24 |
Recall the hypotheses:
- \(H_0\): there is no association between attendance and passing the exam;
- \(H_1\): if attendance \(>50\%\) students are more likely to pass (one-tailed).
Firstly input the data into R and create a matrix:
Row1<-c(10,2)
Row2<-c(5,7)
AttendanceMatrix<-matrix(c(Row1,Row2), byrow=T,nrow=2,
dimnames=list(c(">50","50 or less"),c("Pass","Fail")))
AttendanceMatrix Pass Fail
>50 10 2
50 or less 5 7- Next we visualise the data to get a feeling for potential relationships.
barplot(AttendanceMatrix,
beside=TRUE,
ylim=c(0, 30),legend=T, col=c("yellow","blue"),
xlab="Pass/Fail",
ylab="Frequency")
- Next we perform the Chi-square test of association, but we will see that the expected frequency assumption fails:
result<-chisq.test(AttendanceMatrix, correct=T)
result
Pearson's Chi-squared test with Yates' continuity correction
data: AttendanceMatrix
X-squared = 2.8444, df = 1, p-value = 0.09169result$expected Pass Fail
>50 7.5 4.5
50 or less 7.5 4.5- As the expected frequency assumption fails (we have independence by inspection) we now proceed with Fisher’s exact test. The help function should be used to make sure we perform an appropriate one-tailed test.
help("fisher.test")
fisher.test(AttendanceMatrix, alternative = "greater")
Fisher's Exact Test for Count Data
data: AttendanceMatrix
p-value = 0.04469
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
1.044487 Inf
sample estimates:
odds ratio
6.394505 - Looking at the statistic for this test in the output tables, we see that we have a p-value of \(0.045<0.05\), therefore we reject \(H_0\) and conclude that evidence suggests that if students attend more than 50% of lectures then they are more likely to pass the exam. This result confirms our manual calculations in the lecture.
Exercise 3
The table below contains information on the number of males and females earning low, medium and high salaries in a particular institution.
| Low | Medium | High | Totals | |
|---|---|---|---|---|
| Male | 6 | 58 | 3 | 67 |
| Female | 3 | 44 | 10 | 57 |
| Totals | 9 | 102 | 13 |
Investigate whether there is an association between salary level and male and female earnings, justifying the method you use.
Solution
- We first input the data and create a matrix
Row3<-c(6,58,3)
Row4<-c(3,44,10)
SalariesMatrix<-matrix(c(Row3,Row4), byrow=T,nrow=2, dimnames=list(c("Male","Female"),c("Low","Medium", "High")))
SalariesMatrix Low Medium High
Male 6 58 3
Female 3 44 10- It is good practice to visualise the data.
barplot(SalariesMatrix,
beside=TRUE,
ylim=c(0, 70),legend=T, col=c("yellow","blue"),
xlab="Salaries",
ylab="Frequency")
- Next we perform the Chi-square test of association.
result2<-chisq.test(SalariesMatrix)
result2
Pearson's Chi-squared test
data: SalariesMatrix
X-squared = 5.9229, df = 2, p-value = 0.05174- However, when we check the expected frequencies assumption we see it fails.
result2$expected Low Medium High
Male 4.862903 55.1129 7.024194
Female 4.137097 46.8871 5.975806Hence we use Fisher’s exact test as more than 20% of expected counts are <5 (independence is still required).
The help function should be used to make sure we perform an appropriate two-tailed test.
help("fisher.test")fisher.test(SalariesMatrix)
Fisher's Exact Test for Count Data
data: SalariesMatrix
p-value = 0.04497
alternative hypothesis: two.sidedThis is a significant result, hence there seems to be an association between salary and male and female employees.
Visual inspection of the barplots indicates that there are more female higher earners, but more male middle earners.