My advisor is able to determine that a chi^2 value will be near 30 in this particular circumstance. I calculated a chi^2 for a dataset and fitted line to be 31.89 which validates his claim. How is he able to do this? The only features of my data that I can think of that may give someone an expected value

  1. The data set has 29 degrees of freedom; is this why it is near 30? (In a chi^2 table with various levels of significance the values usually run higher than 30, about mid 30s - 40s for 29 degrees of freedom)
  2. The noise distribution for each data point is normally distributed
  3. If neither #1 & #2 what information do I need to provide about the dataset?

I am looking for an intuitive explanation.

Edit/More context: I fit a line through some data in matlab. The data is generated by the function y = (1/2)x +- noise (i.e. plus or minus some noise value). The noise values are randomly sampled from a normal distribution with mean/mu = 0 and a standard deviation/sigma = 6.

The chi^2 value is obtained by calculating the sum of the differences, ( y_i - y(x_i) )^ 2 and dividing by sigma^2. y(x_i) is the fitted line... You can think of this as SSE divided by sigma squared.

  • $\begingroup$ We’ll need a lot more detail about what you’re doing that gives a $\chi^2$ value. $\endgroup$
    – Dave
    Jul 21, 2020 at 2:09
  • $\begingroup$ @Dave Apologies! Just added more context. $\endgroup$
    – Bwoods
    Jul 21, 2020 at 2:21
  • $\begingroup$ How does $\chi^2$ arise from this? $\endgroup$
    – Dave
    Jul 21, 2020 at 2:27
  • $\begingroup$ The sum of the differences, ( y_i - y(x_i) )^ 2 divided by sigma^2 @Dave $\endgroup$
    – Bwoods
    Jul 21, 2020 at 2:37
  • $\begingroup$ $\chi^2(29)$ is the distribution of the sum of squares of $29$ independent standard Normal variables. Your advisor recognized that the statistic you are using will behave approximately like such a sum of squares (because it's the sum of squares of standardized values). At that point, they would have recalled that the expected square of a standard Normal variable must coincide with its variance (which by definition is $1$) and that variances of sums of independent variables add, whence they would immediately have concluded the expectation of your statistic should be near $29.$ $\endgroup$
    – whuber
    Jul 21, 2020 at 18:34

1 Answer 1


Here are some fragmentary answers based on what you have told us about your data and analysis.

If $X \sim \mathsf{Chisq}(\nu = k),$ then $E(X) = k$ and $Var(X) = 2k.$ [See Wikipedia or your text or class notes for some details of chi-squared distributions.]

P-value. If you're doing a chi-squared test for which the null distribution is (approximately) $\mathsf{Chisq}(29),$ and the observed value of the test statistic is $X = 31.89,$ then you can use software to find that $P(X \ge 31.89)= 0.3247,$ which would not lead you to reject the null hypothesis.

This is the P-value of the chi-squared test. (You would reject at the 5% level if the P-value is below $0.05=5\%.)$ [Computation using R statistical software in which pchisq is the CDF of a chi-squared distribution.]

1 - pchisq(31.89, 29)
[1] 0.3247224

Critical value. Using printed tables of chi-squared distributions, you could find the critical value $c = 42.557$ of the chi-squared test, for which $P(X \ge c) = 0.05.$

If the chi-squared test statistic is greater than or equal to $c,$ you will reject the null hypothesis at the 5% level. The critical value can also be found using R, where qchisq is the inverse CDF (or 'quantile function') of a chi-squared distribution:

qchisq(.95, 29)
[1] 42.55697

Graph. Below is a plot of the density function of $\mathsf{Chisq}(29).$ The solid vertical line shows the observed value $X = 31.89.$ The P-value is the area under the density curve to the right of this line. The dotted vertical line shows the the critical value $c = 42.557;$ the area under the density curve to the right of this line is the significance level $5\%.$

curve(dchisq(x, 29), 0, 55, col="blue", lwd=2, ylab="PDF", 
   main="Density of CHISQ(29)")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 abline(v=31.89, lwd=2)
 abline(v=42.557, lwd=2, lty="dotted", col="red")

enter image description here

  • 1
    $\begingroup$ Apologies! Just added more context in an edit of my original question. @BruceET $\endgroup$
    – Bwoods
    Jul 21, 2020 at 2:22
  • 1
    $\begingroup$ Thanks. It seems you are doing a goodness-of-fit test to see if the normal distribution of the xs is $\mathsf{Norm}(\mu = 1/2, \sigma=6).$ Beyond that it seems I guessed correctly about the use of $\mathsf{Chisq}(29).$ If you need further clarification about the procedure, it may be best for you to ask your instructor or a TA. $\endgroup$
    – BruceET
    Jul 21, 2020 at 3:05
  • $\begingroup$ What is E(X)=ν in this context? What are E and v? I understand X∼Chisq(ν=k) is the chisq distribution with k degrees of freedom and Var(X)=2k is variance of chi. $\endgroup$
    – Bwoods
    Jul 21, 2020 at 3:32
  • 1
    $\begingroup$ Sorry for not explaining. The letter $\nu$ (Greek 'nu') is (pretty much) standard notation for degrees of freedom. The mean of a chi-squared distribution happens to be equal to its degrees of freedom (see Wikipedia link) for the last sentence. $\endgroup$
    – BruceET
    Jul 21, 2020 at 3:56
  • 1
    $\begingroup$ Substantially correct. To prevent confusion with a normal distribution, I'd avoid using the word normal here. Perhaps say that an observed value of 31.89 is 'consistent with' observations from $\mathsf{Chisq}(29).$ $\endgroup$
    – BruceET
    Jul 21, 2020 at 21:10

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