How to determine the expected chi^2 value?

Question

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

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)
The noise distribution for each data point is normally distributed
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.

BruceET · Answer

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")

How to determine the expected chi^2 value?

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