Because of the round numbers in the first column (group), I wonder if all
eight numbers are counts of subjects randomly chosen from two populations.
If you randomly assigned 250 subjects to one Method (col 1) and 277 subjects to the other Method (col 2),
and then later each subject was found to be in one of four categories (T1, ..., T4), then it would be appropriate to perform a chi-squared test of
independence on these data, as follows:
DATA = matrix(c(60,60,70,60, 68, 53, 72, 84), nrow=4)
DATA
[,1] [,2]
[1,] 60 68
[2,] 60 53
[3,] 70 72
[4,] 60 84
The chi-squared statistic is
$$Q = \sum_{i=1}^4 \sum_{j=1}^2 \frac{(X_{ij}-E_{ij})^2}{E_{ij}},$$
where $E_{ij}$ are expected counts and $X_{ij}$ are the observed counts
in the matrix above.
Expected counts are obtained by taking "(row total)(col total)/(grand total)" for each cell in the matrix; for example
$E_{11} = \frac{128\,\cdot\, 250}{527} = 60.72106.$ Here is a matrix of all six
expected values:
[,1] [,2]
[1,] 60.72106 67.27894
[2,] 53.60531 59.39469
[3,] 67.36243 74.63757
[4,] 68.31120 75.68880
Then under $H_0,$ we have
$Q \stackrel{aprx}{\sim}\mathsf{Chisq}(\nu = 3),$ where the
degrees of freedom are found as $\nu = (r-1)(c-1) = 3(1) = 3.$
The approximation is useful if all six $E_{ij}$'s are at least 5.
In R the test can be performed as follows:
chisq.test(DATA)
Pearson's Chi-squared test
data: DATA
X-squared = 3.5879, df = 3, p-value = 0.3095
So there is no evidence at the 5% level that the proportions of T1, T2, T3 and T4 differ
across Methods 1 and 2. (You would need $Q > 7.815$ instead of $Q = 3.588$ for
a result significant at the 5% level.) More succinctly, Method does not appear to have made
a significant difference.
qchisq(.95, 3)
[1] 7.814728
Strictly as a 'thought-expteriment', consider this: If you had exactly the same relative proportions among expected and observed counts, but
with more data, you might have had a significant result. For example, let's imagine you had
three times as many subjects with similar proportional resuls:
DTA3 = 3*DATA; DTA3
[,1] [,2]
[1,] 180 204
[2,] 180 159
[3,] 210 216
[4,] 180 252
chisq.test(DATA*3)
Pearson's Chi-squared test
data: DATA * 3
X-squared = 10.764, df = 3, p-value = 0.01307
Notice that if you made bar charts for DATA and for DTA3,
they would look almost exactly the same (except possibly for labels on axes).
Yet one would show data with no significant difference and one would show
highly significant data.
This illustrates that bar charts alone are
almost useless in judging the significance of categorical data. Every bar chart for
categorical data should have a footnote or caption showing the P-value
of the corresponding chi-squared test.
In a chi-squared test the degrees
of freedom are determined by the numbers of levels of the categories, yet
sample size is reflected in the value of $Q$ and in the P-value in ways that may not be
intuitively obvious.
As a simpler example of a similar comparison:
If you throw a coin 10 times and get 7 heads, that is not convincing
evidence the coin is biased; but if you get 697 heads in 1000 tosses,
that would be strong evidence of bias.
