= Solution
An <unbiased estimator> $T$ of a parameter $\theta$ satisfies $\mathbb E_\theta[T]=\theta$ for every permissible value of $\theta$. Since the <sample mean> is linear,
$$
\mathbb E[\widehat\mu]=\frac1n\sum_{i=1}^n\mathbb E[X_i]=\mu,
$$
so $\widehat\mu$ is unbiased.
Use the identity
$$
\sum_{i=1}^n(X_i-\overline X)^2
=\sum_{i=1}^n(X_i-\mu)^2-n(\overline X-\mu)^2.
$$
The two expectations on the right are $n\sigma^2$ and $n\operatorname{Var}(\overline X)=\sigma^2$, respectively. Hence
$$
\mathbb E[\widehat{\sigma}^2]=\frac{n-1}{n}\sigma^2.
$$
Thus
$$
\boxed{\widehat\mu\text{ is unbiased},
\qquad
\widehat{\sigma}^2\text{ has bias }-\frac{\sigma^2}{n}}.
$$
The unbiased <sample variance> instead divides the same sum of squares by $n-1$.
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