$P$-value for testing an inequality?

Question

I may just be searching up the wrong thing but I can't find anything for this online. Let's say I want to get a $p$ value for the hypothesis that $$\mu > \mu_0$$ for a population mean parameter. How do I go about getting a $p$ value? The way I thought this was done was by getting a test statistic (which in this case does not have an absolute value sign afaik):$$d = \frac{\hat{\mu} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$$

And then getting the $p$ value by calculating ($\sigma$ is known in this case): $$p = \Pr(D < d), D \sim \mathcal{N}(0,1)$$

What's wrong here, and how do I calculate the $p$ value?

Answer 1

you are almost there. The p-value is the probability of the queue

$$p_{value}=P(D<d)$$

and you can get it using Z-tables

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Example:

Suppose we have the following system to be verified:

$$ \begin{cases} H_0:\mu \geq2 \\ H_1: \mu<2 \end{cases}$$

let's suppose normality, $\sigma=1$,$n=4$.

The sample mean should be less then 2, otherwise there is nothing to prove and the p-value will be very high.

So let's suppose $\overline{X}_4=1$

Your statistic is

$(1-2)\times \sqrt{4}=-2$

and the p-value is $P(Z<-2)\approx2.28\%$