As I assume you know, with such a large value you have to reject the assumption of randomness (if you cared to test it).

Regarding any errors, only you can verify them by carefully reviewing the data matrix.

Adding to what Gioacchino de Candia said, you could show us the contingency table. ;-)

The chi-square value may get larger as the sample size gets relatively large. As Bruce Weaver requests, probably the best way to get good advice is to simply provide the contingency table.

Below is an example I just made up in R. Like your data, the df=4, n=247, and the chi-square value comes out close to your value. But also note that this is a pretty clear (extreme ?) case of association between two variables.

You might want to also explore reporting an effect size statistic (Cramer's V) or report the data as percentages, possibly within the columns or rows. This explains the data in a way that doesn't focus on the relatively large sample size.

Matrix = as.matrix(read.table(header=TRUE, row.names=1, text="

County Pass Fail

Salem 42 5

Cumberland 42 6

CapeMay 7 42

Atlantic 8 43

Ocean 9 43

"))

sum(Matrix)

### 247

chisq.test(Matrix)

### Pearson's Chi-squared test

###

### X-squared = 125.46, df = 4, p-value < 2.2e-16

Here it is Gioacchino de Candia Bruce Weaver

Anik Ghosh what you posted confirms what i said.

Among other things, there was no need, given the type of "survey" you are conducting and the distribution of responses.

A couple of notes:

1) All the "expected counts" in your table are greater than 5. This is a good thing when conducting the usual chi-square test, (despite the fact that you have a 0 count in the table). Because of this 0, you might conduct the the test by Monte Carlo simulation. Conditioning on fixed margins for the table with Monte Carlo, I get a p-value of approximately 0.00001.

2) You may want to treat your variables as ordinal in nature. That is, it is the case that Somewhat < Agree