Use a large sample when individual times need not be normal

The sample mean is $\bar x=4626/150=30.84$ hours. The unbiased variance estimate is $s^2=[147691-4626^2/150]/149=33.7259060403\ldots\ \mathrm{hours}^2$, or 33.7 to 3 s.f. The source first prints 33.7259; use the unrounded value in later calculations.

Let $\mu$ be the population mean completion time in hours. Test $H_0:\mu=30$ against $H_1:\mu>30$. For a large independent random sample with finite variance, the CLT and estimated variance give $\bar X\approx N(30,s^2/150)$ under $H_0$. Thus $(\bar X-30)/\sqrt{s^2/150}$ is approximately standard normal. This is an approximation, not an exactly normal population model.

$z=(30.84-30)/\sqrt{s^2/150}=1.771507938\ldots$ and $p=P(Z\ge z)=0.03823813692\ldots\approx0.0382$. Since $p\le0.05$, reject $H_0$: sufficient evidence at 5% that the mean completion time exceeds 30 hours. The teacher strategy chooses the p-value method because all summary inputs are available. The source Z-Test Stats screen uses $\mu_0=30$, estimated SD 5.8074004511485 from rounded variance 33.7259, $\bar x=30.84$, $n=150$ and the greater-than alternative; its outputs are $z=1.771508097$, $p=0.0382380757$. Retain this source decimal separately from the independent calculation; both give the printed 0.0382 and same decision.

No assumption of a normally distributed individual completion time is needed for the stated large-sample approximation. The teacher says “No assumption is needed”; interpret that in this population-shape sense. Random, suitably independent observations, finite variance and a sufficiently good CLT approximation are still required.