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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?

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up vote 15 down vote accepted

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.

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