${H}_{0}:\mu =0$ against${H}_{A}:\mu >0.$

Had the following question on my exam today and I'm just wondering if I did this correctly.

From a normally distributed population with mean $\mu $ and variance ${\sigma}^{2}$ a sample has been drawn:

$\mathbf{X}=(0.05,4.35,-0.48,-0.63,1.17,2.01).$.

a) Test ${H}_{0}:\mu =0$ against ${H}_{A}:\mu >0$ on a 5% significance level under the assumption that σ2 is unknown.

c) Do the same test, given that $\sigma =\mathrm{1.5.}$.

Solution: I know that ${H}_{0}$ is to be rejected if $T\ge c,$, where T is the teststatistic and ${F}_{{t}_{n-1}}=1-\alpha ,$, where $\alpha $ is the significance level.

So lets first compute them for a), where we need to use a t-distribution since σ is unknown, so we have that $c={F}_{{t}_{5}}^{-1}(0.95)\approx \mathrm{2.015.}$. Also, the estimator for ${\sigma}^{2}$ is

${s}^{2}=\frac{1}{n-1}(\sum _{k=1}^{n}{X}_{k}^{2}-\frac{1}{n}{\left(\sum _{k=1}^{n}{X}_{k}\right)}^{2})=\frac{1}{5}(24.961-\frac{1}{6}(41.861))=3.405,$

so, $s=\sqrt{3.405}=\mathrm{1.845.}$. We also have that $\overline{X}=6.47/6=\mathrm{1.078.}$.

The teststatistic is

$T=\frac{\overline{X}-{\mu}_{0}}{s/\sqrt{n}}=\frac{1.078-0}{1.845/\sqrt{6}}=\frac{1.078}{0.753}=1.43\le c,$

which means that we can not reject H0.

For b) we do an identical approach but wtith $s=\sigma =1.5$ and we use a z-test instead. Here we have that $c={\mathrm{\Phi}}^{-1}(0.95)=\mathrm{1.644.}$ The teststatistic in this case is

$Z=\frac{\overline{X}-{\mu}_{0}}{\sigma /\sqrt{n}}=\frac{1.078}{1.5/\sqrt{6}}=1.76\ge c,$

thus we reject ${H}_{0}.$.

Is this correct?

What my professor does in the solutions is he just makes a 95% confidenceinterval and just arrives to the same conclusions as I've done. Is his method correct without using the teststatistics?