Let $$Y=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}$$
and suppose our population is modeled by a normal random variable with mean 0 and variance 3. I wan to find the p.d.f. of $Y$ in the case of $n=4$ samples.
I believe the result is the chi-squared distribution with 4 degrees of freedom. Is this result correct, or would it be 3 degrees of freedom? I believe $n$ gives the number of d.o.f., but would it be $n-1$?
You say $X_1,X_2,X_3,X_4\sim\operatorname N(0,3).$ You do not say how they are jointly distributed but I will assume you meant they are independent. In that case you have $$ \frac 1 3 \sum_{i=1}^4 X_i^2 \sim\chi_4^2. $$ Important: The denominator $3$ is there because it's the variance, NOT because it's $n-1.$
Suppose $X_1,\ldots,X_n\sim\text{i.i.d.} \operatorname N(\mu,\sigma^2).$ Let $\overline X = \dfrac {X_1+\cdots+X_n} n.$
\begin{align} \text{Then } & \frac 1 {\sigma^2}\sum_{i=1}^n (X_i-\mu)^2 \sim \chi^2_n \\[10pt] \text{and } & \frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \overline X)^2 \sim \chi^2_{n-1}. \end{align}