So there isn't any contradiction here, everywhere there are $df$ independent normally distributed random variables in the background somehow, and $df$ sometimes equals $n - 1$, sometimes $k - 1$ (here $k$ is the $k$ from the Pearson's $\chi^2$ test above, not the $k$ in your question), sometimes some other formula, depends on the test. $\chi^2$ test of independence: $df = \left( r - 1 \right) \cdot \left( c - 1 \right)$ (where $r$ is the number of rows, $c$ is the number of columns in the contingency table)įor more examples, please check the $\chi^2$-test wikipedia page.Suppose we have a random sample of size n. $\chi^2$ test that the variance of a normally distributed population has a given value based on a sample variance: $df = n - 1$ (where $n$ is the sample size) The chi-squared distribution (usually written ) is a sampling distribution derived from the normal distribution.Example: Reporting a chi-square test There was no significant relationship between handedness and nationality, 2 (1, N 428) 0.44, p. phenotype resistant susceptible P value expected. Pearson's $\chi^2$ test (of goodness of fit): $df = k - 1$ (or $df = k - p - 1$), where $k$ is the number of classes (and $p$ is the number of estimated parameters of the investigated distribution, if any) Report the chi-square alongside its degrees of freedom, sample size, and p value, following this format: 2 (degrees of freedom, N sample size) chi-square value, p p value). In a chi-square analysis this is the number of classes in the data set minus one.There are quite a lot of $\chi^2$ tests, and all have different formula to calculate the value of $df$ (degrees of freedom). Many test statistics follow the $\chi^2$ distribution.
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