Regarding your approach:
Confidence interval is a random interval $I$ constructed from the data that contains the unknown parameter of interest $\theta$ with specified probability $1-\alpha$, $\mathbb{P}(\theta\in I)=1-\alpha$. Your interval $(300−1.96∗30,300+1.96∗30)$ is not random, it is constructed from population parameters, so it is not really a confidence interval.
Let's try to solve the problem "from scratch." Assume that the distribution of running times is normal $\mathcal{N}(\mu, \sigma^2)$, with mean $\mu=300$ and standard deviation $\sigma=30$. Given the context of the problem, the normality assumption is reasonable (although we should keep in mind that we work under this assumption). Given the data $X=230$, our null hypothesis, the lawnmower came from Company A, can be formalized as follows: $H_0: X\sim \mathcal{N}(\mu,\sigma^2)$.
To test this hypothesis, we need to choose a statistic $s$, a function of data, with the following property: the larger $s$, the more tempting to reject the null. It seems intuitive to reject $H_0$ whenever $$s(X)=|X-\mu|$$ is large, i.e. whenever $X$ is far from the mean $\mu_0$. So, our rejection region is
$$
s(X)>c.
$$
How do we chose $c$? We chose $c$ to control the probability of type I error, i.e. an error of rejecting $H_0$ when it is true. To construct a test of size $\alpha$, we chose $c$ such that
$$
\mathbb{P}(\mbox{Reject } H_0|H_0)=\alpha,
$$
or
$$
\alpha=\mathbb{P}(|X-\mu|>c \hspace{1mm}| \hspace{1mm} X\sim \mathcal{N}(\mu,\sigma^2))=\mathbb{P}\left(\left.|Z|>\frac{c}{\sigma} \right| Z\sim\mathcal{N}(0,1)\right)=2\Phi\left(-\frac{c}{\sigma}\right),
$$
where $\Phi$ is the standard normal CDF. Solving this equation for $c$ yields
$$
c=-\sigma\Phi^{-1}\left(\frac{\alpha}{2}\right).
$$
For example, if we want to construct a test of size $\alpha=0.05$, then $c\approx58.8$. Since the value of the test statistic $s(X)=70$, our size $0.05$ test will reject the null.
Reporting the p-value is more informative then simply reporting whether the test accepts or rejects the null. Recall the the p-value is the smallest size $\alpha^*$ at which the test rejects the null. To find the p-value, we need to solve
$$
s(X)=-\sigma\Phi^{-1}\left(\frac{\alpha^*}{2}\right)
$$
for $\alpha^*$. The solution is
$$
\mbox{p-value}\equiv \alpha^*=2\Phi\left(\frac{|X-\mu|}{\sigma}\right)\approx0.0196.
$$
This p-value is small (less than the usual 0.05), and, therefore, the data provides strong evidence against the null hypothesis.
