Solved – Why is the probability of the chi square statistic equal to 0

chi-squared-testlogisticregression

Binary logistic regression in R

I have derived the chi square statistic and degrees of freedom for my model (200.7839, 8, respectively) however, when I attempt to determine the probability associated with the chi square statistic, I get a value of 0 with no decimal places – just "0". I was expecting a probability <0.05 so I can reject the null hypotheses. Please see my code below. Does anyone have any suggestions as to why R would return a value of 0 with no decimal places? or is the value simply 0.00000 etc and therefore significant? Cheers!

modelChi<-Model1.0$null.deviance - Model1.0$deviance
modelChi
chidf<-Model1.0$df.null - Model1.0$df.residual
chidf
chisq.prob<- 1 - c(modelChi, chidf)
chisq.prob

Best Answer

That Chi-squared statistic is gigantic. With 8 degrees of freedom, the a chi-squared statistic of 21.96 is associated with a p-value of 0.005. So a very large statistic like 200, with 8 degrees of freedom has a p-value so small that R returns zero, (i.e. close to zero). It is certainly less than .05, the level you are trying to test at, which is achieved with a chi-squared test statistic of only 2.73.

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Solved – Chi Square Analysis Percentages

Chi-squared calculations in the usual form is for counts. If you scale all the counts down to percentages, you scale the chi-square down as well.

Example

            Agegroup1   Other
Obs count     40         920
Exp %         3.1%      96.9%
Exp count     29.76      930.24

Chi-square = (40-29.76)^2/29.76 + (920-930.24)^2/930.24 = 3.636

            Agegroup1     Other
Obs %         4.17%      95.83%
Exp %         3.1%       96.9%

Not actually chi-square: (.0417-.031)^2/0.031 + (.9583-.969)^2/0.969 = 0.00381

If the calculations are done exactly, the second calculation gives 1/960$^\text{th}$ of the correct values, for this example ($960=40+920$). In this case, rounding error in the second calculation means we got a little over that.

It's possible to adjust to make it work with percentages if you know the totals by which the values were scaled, but it's easy to just scale back to counts and do it with the counts anyway.

There are many elementary tutorials on the various forms of chi-square test (goodness of fit, homogeneity / independence, and a number of others) that might be used on such data.

Solved – Pearson and deviance GOF test for logistic regression in SAS and R

As has been discussed elsewhere on this site, the deviance statistic does not have a $\chi^2$ distribution. Any statistic where the d.f. increases as the sample size has a degenerate distribution.

For goodness of fit set up easy directed hypotheses such as linearity and additivity or use the 1 d.f. test in the R rms package residuals.lrm function.

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