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I am reading "An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements by J. R.Taylor", and I read the following formula in an exercise:

$$\chi^2_\text{adjusted}=\sum_{k=1}^2\frac{\left (|O_k-E_k|- \frac{1}{2}\right )^2}{E_k}$$

but I don't know when using it! In particular:

  • Is $\chi^2_\text{adjusted}$ used when $d=\text{degrees of freedom}=1$ and $k=2$?
  • Is $\chi^2_\text{adjusted}$ used when $d=\text{degrees of freedom}=1$?

The exercise is:

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    $\begingroup$ This question appears to be off-topic because it belongs on stats.stackexchange.com $\endgroup$
    – BMS
    Commented Jul 3, 2014 at 15:37
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    $\begingroup$ So the text says it's an adjustment applicable when you have one degree of freedom, and gives a reference which presumably motivates the choice and perhaps extends it to other cases. (Perhaps, wild-guessing, the adjustment term is $\frac 1{d+1}$, which would vanish with many degrees of freedom.) To the library with you! $\endgroup$
    – rob ♦
    Commented Jul 3, 2014 at 16:20

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The $\chi^2$ statistic is independent of the number of degrees of freedom. But converting that statistic to some type of $p$-value does depend on the degrees of freedom. That is, you calculate $\chi^2$, then with that number and the degrees of freedom you look it up the $p$-value in a $\chi^2$ table.

As for the "corrected" version of this test, you may find the WP page for Yates continuity correction helpful. The extra 0.5 term in the numerator is there to compensate from modeling a discreet distribution using a continuous distribution. Some claim it overcompensates in certain circumstances. True, but I've found it gives closer to exact results in nearly all case.

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  • $\begingroup$ of course, I know it.. I use $\mathcal{P}_d(\tilde\chi^2 \geq \tilde\chi_O^2)$ (with $\tilde{\chi}_O^2=\displaystyle \frac{\chi^2}{d}$ and $\chi^2=\sum_{k=1}^n \displaystyle\frac{(O_k-E_k)^2}{E_k}$); therefore Why Do Taylor uses $\chi^2_\text{corrected}$ in exercise? $\endgroup$
    – mle
    Commented Jul 3, 2014 at 15:57
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    $\begingroup$ en.wikipedia.org/wiki/Yates's_correction_for_continuity $\endgroup$
    – BMS
    Commented Jul 3, 2014 at 19:52

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