We want to make an hypothesis test for the mean value &#x03BC;<!-- μ --> of a normal populatio

We want to make an hypothesis test for the mean value $\mu$ of a normal population with known variance ${\sigma }^{2}=13456$, using a sample of size $n=100$ that has sample mean value equal to $562$.
Calculate the $p$-value.
Make the test with significance level $1\mathrm{%}$ about if the population mean value from which the sample comes from is greater than $530$ using the $p$-value.
For the first one, about the $p$-value, do we have to calculate $P\left(\frac{530-562}{\frac{\sigma }{\sqrt{n}}}\right)$?
And for the second we have to check the $p$-value with the significance level, right?
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Lea Johnson
Yes. correct.
If you calculate $\mathrm{\Phi }\left(\frac{530-562}{\sqrt{134.56}}\right)=\mathrm{\Phi }\left(-2.76\right)\approx 0.29\mathrm{%}$
this $p$-value is highly significant so the test is significant at $1\mathrm{%}$ and the hypothesis that the mean is $530$ is rejected.
Lower is the $p$-value and higher is the significance of the test. This because the p-value is the area of the queue. A very low value of $p$ indicates that the quantile (your observed mean) is very far from the centered mean (null hypothesis)