I have obtained a Pearson's Chi-Square test statistic value of $0.065$, for $1$ df, and a $p$-value of $0.799$.

Comparing my calculated $\chi^2$ test statistic with the critical value in the $\chi^2$ distribution table value for $1$ df ($0.00393$ for $95\%$ confidence), I would assume my Chi-Square test statistic is significant (because it is greater than the critical value)?

However, the $p$-value being greater than $0.05$, makes me think otherwise. I also doubt this result because the proportions are equal in the cross-tabulated table (no difference between answers in my survey).

Therefore I am confused and do not know whether my $\chi^2$ statistic is significant or not. The only thing I think I am doing wrong is reading the wrong critical value of the $\chi^2$ distribution table. I conducted these Pearson $\chi^2$ tests using SPSS software - I assumed it would be at the $95\%$ confidence, but perhaps I am wrong and it is a standard to set the $99\%$ confidence?


1 Answer 1


If your test statistic follows $\chi^2_1$ distribution, then you have to look at the probability of observing value $0.065$ or greater value under this distribution. Using R software to get the values you can find that probability of observing such or greater value is $0.79$ under null distribution:

> 1-pchisq(0.065, df = 1)
[1] 0.798761

so obviously $0.79 \not \le 0.05$. On the plot below you can see your value marked by blue lines and the critical value that is marked by red lines.

enter image description here

The critical value is

> qchisq(.95, df = 1, lower.tail = TRUE)
[1] 3.841459

under $\chi^2_1$ distribution $5\%$ cases are greater and $95\%$ are lower than this value. You made mistake of looking at value

> qchisq(.05, df = 1)
[1] 0.00393214

i.e. a point where $5\%$ are less than this value and $95\%$ are more than this value, so you were comparing to $p=0.95$ rather than $p=0.05$.

Check also What is the meaning of p values and t values in statistical tests? to learn more about how to understand $p$-values.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.