X2 Test for Independence - r

Question

So we have
13-17 18-24 25-34 35-44 45-54 55-64 65+
Female 1 45 15 6 2 3 2
Male 2 121 31 7 4 2 3
and the raw data has headers like F.13-20, F.21-35 M.13-20 etc.
How would you do this? It is hard to explain, but we can't find it anywhere.
tab <- matrix(as.numeric(WeekReach[158,3:16]), nrow=2, byrow=TRUE)
colnames(tab) <- c("13-17", "18-24", "25-34", "35-44", "45-54", "55-64", "65+")
rownames(tab) <- c("Female","Male")
Then after this is:
exInd = function() {
n = sum(tab)
p = rowSums(tab)/sum(tab)
q = colSums(tab)/sum(tab)
return(p %o% q * n)}
chiSquaredStatistic = function(E) {
return(sum((tab - E)^2/E))}
E = exInd()
x2 = replicate(1000, {
ageShuffle = sample(age)
genderShuffle = sample(gender)
Xindep = table(ageShuffle, genderShuffle)
chiSquaredStatistic(Xindep, E)})
But we need something to make male and female their own thing - It is hard to explain, the teacher wont even explain it to us #univeristyproblems
Right so this was the solution given by the teacher - note, they do not explain anything.
WeekReach = read.csv("http://staff.scm.uws.edu.au/~lapark/300958/labs/WeeklyReachDemog.csv", as.is=TRUE)
tab = matrix(as.numeric(WeekReach[158,3:16]), nrow=2, byrow=TRUE)
colnames(tab) <- c("13-17", "18-24", "25-34", "35-44",
+ "45-54", "55-64", "65+")
rownames(tab) <- c("Female","Male")
stretchTable = function(tab, variableNames) {
+ tabx = rep(rownames(tab), rowSums(tab))
+ l = ncol(tab)
+ m = nrow(tab)
+ cn = colnames(tab)
+ taby = c()
+ for (a in 1:m) {
+ for (b in 1:l) {
+ taby = c(taby, rep(cn[b], tab[a,b]))
+ }
+ }
+
+ d = data.frame(x = tabx, y = taby)
+ colnames(d) = variableNames
+ return(d)
+ }
tab2 = stretchTable(tab, c("Gender","Age"))
Verify that we the correct values
table(tab2)
This was the 'question'
We showed in the lectures that we can perform a test for independence for a given two way table (two way meaning, has more than one row and column). To perform the test, we need to:
compute the expected values of the table if the rows and columns are independent (code shown in the lecture slides).
shuffle the rows and columns of the table.
To peform the shuffling, we must first untabulate the table. For example, if we start with the table:
A B C
X 2 1 1
Y 1 3 1
We must convert it to the form:
Column Row
A X
A X
A Y
B X
B Y
B Y
B Y
C X
C Y
Write the code to do this table conversion.
Hint: To compute the two column table, the two columns can be computed seperately as gender and age, then combined using tab2 = data.frame(Gender = gender, Age = age). Also, the functions rowSums, colSums, rownames, colnames and rep may be useful (if you are unfamiliar with these functions, read the R documentation on them, e.g help(rowSums)).
Once we have the data in two columns, we shuffle the columns and recompute the table and compute the χ2 value (as shown in the lecture).
Using the above table tab and the hypotheses H0: Gender and Age are independent, HA: Gender and Age are not independent:
Compute the χ2 randomisation distribution.
Compute the χ2 statistic for tab.
Compute the p value of the test.
Finally state the conclusion of the test.
HAVE FUN.

Answer 1

I assume you want to perform a chi-square test of independence, to establish if there are significant differences between the expected and observed frequencies in the male/female groups across the different age brackets.
The following should get you started
df <- read.table(text =
"13-17 18-24 25-34 35-44 45-54 55-64 65+
Female 1 45 15 6 2 3 2
Male 2 121 31 7 4 2 3", header = T)
chisq.test(df)
#
# Pearson's Chi-squared test
#
#data: df
#X-squared = 4.8117, df = 6, p-value = 0.5682
Based on the sample data and the chi-square test results, we fail to reject the null hypothesis, and conclude that there is not enough evidence to infer that there is a statistically significant difference between the male and female frequencies across the different age brackets.