Both ideas of the null and alternative hypothesis are true. The null hypothesis must always include an equals sign, whether it be $\geq\text{, } \leq\text{, or just}=$. Usually, however, it's just $=$. The alternative hypothesis is what we wish to show.

The null hypothesis in this case is that the proportion of children in economically disadvantaged areas raised in single-parent homes is $30$%.

The alternative hypothesis is that the proportion of children in economically disadvantaged areas raised in single-parent homes is greater than $30$%.

More formally

$$H_0 : p=0.3$$

$$H_a : p \gt 0.3$$

There are two ways you can test this hypothesis if you so wish. Letting $X$ be the number of children raised in single-parent homes, you can use normal approximation to the binomial:

$$P(X\geq317)=1-P(X\lt317)=1-\Phi\left(\frac{316.5-300}{\sqrt{1000\cdot0.3\cdot0.7}}\right)$$

where I used a continuity correction

In R statistical software

> 1-pnorm((316.5-300)/sqrt(1000*.3*.7))
[1] 0.1274333

You could also, using software, find the exact probability using the standard binomial distribution:

$$P(X\geq317)=\sum_{k=317}^{1000} {1000 \choose k}\cdot0.3^k\cdot0.7^{1000-k}$$

> sum(dbinom(317:1000,1000,.3))
[1] 0.1277011

Since $n$ is large, the normal approximation does very well.

At $\alpha=0.05$ we fail to reject the null hypothesis.