= Solution
For an <affine variety> $X\subseteq\mathbb A_k^n$ with <vanishing ideal> $I(X)$, its <Zariski tangent space> at $p$ is
$$
T_pX=\left\{v\in k^n:
\sum_{i=1}^n\frac{\partial h}{\partial x_i}(p)v_i=0
\text{ for every }h\in I(X)\right\}.
$$
The <dimension from minimum tangent dimension> is
$$
\dim X=\min_{p\in X}\dim_kT_pX.
$$
Now let $X=Z(f)$ with $f$ irreducible. The <Strong Hilbert Nullstellensatz> gives $I(X)=(f)$, so
$$
T_pX=\ker\left(v\mapsto\nabla f(p)\mathbin\cdot v\right).
$$
Its dimension is at least $n-1$, and it is exactly $n-1$ wherever $\nabla f(p)\ne0$.
It remains to find such a point in positive characteristic $q$. If every partial derivative of $f$ were the zero polynomial, every exponent occurring in $f$ would be divisible by $q$. Since the algebraically closed field $k$ is <perfect field>[perfect], all coefficients have $q$th roots and $f$ would be a $q$th power, contradicting irreducibility. Thus some partial derivative $f_{x_i}$ is nonzero. If the gradient nevertheless vanished at every point of $Z(f)$, the Nullstellensatz would imply $f_{x_i}\in(f)$. This is impossible because $0\leq\deg f_{x_i}<\deg f$. The <smooth point on an irreducible hypersurface in positive characteristic> therefore exists, and
$$
\boxed{\dim Z(f)=n-1.}
$$
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