In an exercise I have the output of R for a simple linear regression that displays
Residual standard error: 0.1011 on 23 degrees of freedom
And I have to do the following hypothesis test: $$H_0: \sigma^2=0.01\text{ vs }H_1: \sigma^2\ne 0.01$$ with a significativity of $1-\gamma=0.1$
So I did this:
Assuming that $SS_E=\text{Sum of Squared Errors}=\sum (y-\hat{y})^2$
Assuming that $\hat{\sigma}^2=\frac{SS_E}{n-2}$ so $SS_E=(n-2)\hat{\sigma}^2$
We can write $$\frac{SS_E}{\sigma^2}=\frac{(n-2)\hat{\sigma}^2}{\sigma^2}\sim\chi^2_{n-2}$$
So, for the test, I get an acceptation region of $$\chi^2_{n-2}\left(\frac{1-\gamma}{2}\right)\le\frac{(n-2)\hat{\sigma}^2}{\sigma^2}\le\chi^2_{n-2}\left(\frac{1+\gamma}{2}\right)$$
And, due to the fact that $\frac{(n-2)\hat{\sigma}^2}{\sigma^2}=\frac{23\cdot(0.1011)^2}{0.01}\approx23.509$ this converts to $$\chi^2_{23}\left(0.05\right)=13.091\le 23.509 \le 35.172=\chi^2_{23}\left(0.95\right)$$
So we should reject the alternative hypothesis.
What's throwing me off is a note on this exercise that suggest
It may be useful to know that 1.714 is the 0.95-quantile of $t_{23}$
And I really can't understand where the t-Student should fit in here. Am I missing something or doing something wrong?