The effect sizes I assume you are considering --- Cramer's V, (phi), Contingency coefficient C, and Cohen's w --- can all be calculated with the chi-square value. But the chi-square is simply calculated from the difference of observed values from expected values. This is way Cohen defines his w in Cohen (1988).
I assume that because there's no inference with these statistics, that it is fine to report them even if some test using the chi-square statistic would not be appropriate. It's like saying the difference between two means is some value, without addressing whether or not you could use a t-test or not in this case.
Because your sample size is large, the Chi-square test is likely to return a low p-value even for a table with small differences from the expected proportions.
To get a sense of the effect size being reported by Cramer's v, it is helpful to look at the proportions in the table. For example, for column 1, you can see that there is not much difference in the proportions for each grade within the rows. Row 0 is about one-half to 1 percent of the observations in each grade. Row 2 is, say, 93 to 97 percent of observations in each grade. And so on.
Whether these kind of differences in proportions are meaningful in your context is up to you. The p-value and Cramer's v give you certain information. The practical importance of your results is something you will have decide.
The following is code for R.
I am getting a slightly different Cramer's v than you, so that's something that you might want to look into.
Input =("
Col1 Grade1 Grade2 Grade3 Grade4 Grade5
0 290 392 932 1812 2854
1 522 421 574 917 1247
2 56789 81296 117971 147811 204480
3 3719 2975 2811 1704 2244
")
Matrix = as.matrix(read.table(textConnection(Input),
header=TRUE,
row.names=1))
Matrix
chisq.test(Matrix)
### Pearson's Chi-squared test
###
### X-squared = 8113.9, df = 12, p-value < 2.2e-16
library(vcd)
assocstats(Matrix)
### Cramer's V : 0.065
prop.table(Matrix, margin=2)
### Grade1 Grade2 Grade3 Grade4 Grade5
### 0 0.004729289 0.004607212 0.007621353 0.011901947 0.013537294
### 1 0.008512720 0.004948051 0.004693837 0.006023226 0.005914858
### 2 0.926108937 0.955479291 0.964698090 0.970882268 0.969903949
### 3 0.060649054 0.034965446 0.022986720 0.011192559 0.010643899
Best Answer
I think you need to rework this question. It all depends on the problem/data which has generated the cross-tab.