 # H 0 </msub> : &#x03BC;<!-- μ --> = 0 against H A </ Abram Boyd 2022-06-26 Answered
${H}_{0}:\mu =0$ 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:
$\mathbf{X}=\left(0.05,4.35,-0.48,-0.63,1.17,2.01\right).$.
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 $\sigma =1.5.$.
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 $c={F}_{{t}_{5}}^{-1}\left(0.95\right)\approx 2.015.$. Also, the estimator for ${\sigma }^{2}$ is
${s}^{2}=\frac{1}{n-1}\left(\sum _{k=1}^{n}{X}_{k}^{2}-\frac{1}{n}{\left(\sum _{k=1}^{n}{X}_{k}\right)}^{2}\right)=\frac{1}{5}\left(24.961-\frac{1}{6}\left(41.861\right)\right)=3.405,$
so, $s=\sqrt{3.405}=1.845.$. We also have that $\overline{X}=6.47/6=1.078.$.
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 $c={\mathrm{\Phi }}^{-1}\left(0.95\right)=1.644.$ 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?
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Yes, your solution is indeed correct. You are doing exactly what is required to, while your professor is solving a bit more general problem, namely constructing the 95% confidence interval and then checks whether given values are inside this interval or no.
This, I think, is just a manner of habit. I, personally, am used to the one presented in your post.

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