A random sample of $n=12$ observations from a normal population produced the following estimates: estimate of population mean $= 47.1$ and estimate of population variance $=4.7$. Test the null hypothesis $H_0: m = 48$ against the alternative $H_a: m \neq 48$ at the $5\%$ level.
So this looked pretty simple, I used the normal distribution with the values $\frac{(48-47.1)12^{1/2}}{{4.7}^{1/2}}=1.44$ and read from table that that goes to $0.663%$, So as $0.6628-0.5=0.1628$ is $>0.05$ It is valid. But the question was worth loads of marks so I must be missing something or have done it wrong?
It should be: $$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{47.1-48}{\sqrt{4.7}/\sqrt{12}}=-1.44.$$ $$t_{0.025;11}=2.2.$$ $$|t|<t_{0.025;11} \Rightarrow \text{Do not reject } H_0.$$ You must use $t$-distribution, because $n=12$ is a small sample.