# Likelihood ratio test vs. p-value under the null hypothesis

I am learning about hypothesis testing and I have a conceptual confusion about the differences between a likelihood ratio test and the p-values derived from testing the likelihood of the data under the null. My confusion comes from the fact that the LRT employs the null and the alternative, whereas the typical approach employs just the null. Therefore, I have difficulty seeing how these approaches are the same.

Let's take a toy example where the null hypothesis is that the data is normally distributed with mean of 0 and variance of 1. Now if we have a data set of $${x_i}$$, we can look at the mean of $$x_i$$ and look at how likely it is for the mean of these data points to deviate a certain amount from $$0$$.

In the likelihood ratio test, we would look at how likely the data is given that the actual mean is 0, and then we would look at how likely the data is given that the actual mean is equal to the empirical mean of the data set, and we would compare them.

How is it that two such different approaches are both appropriate for testing the null hypothesis? It seems that by not even considering the alternative hypothesis, the "likelihood under null hypothesis alone" approach seems to lack some key information.

Can you clarify this confusion? When is the LRT appropriate and more useful than simply testing the null hypothesis?

Comment: I think you are mixing up (at least) two foundational approaches to testing a null hypothesis. If you can answer a few of my questions below, maybe that will help you answer your own question.

To begin, as you suggest, let's test the null hypothesis $$H_0: \mu = \mu_0 = 0,$$ given data $$n = 9: X_1, X_2. \dots, X_9$$ chosen randomly from a normal population with mean $$\sigma = 1.$$ You will use the sample mean $$\bar X = \frac 1 n \sum_{i=1}^n,$$ and the test statistic $$Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{9}} = \frac{\bar X - 0}{1/3} = 3\bar(X).$$ You will reject $$H_0$$ if $$|Z| \ge 1.96.$$

This is an example of the test procedure you mentioned first. If $$\bar X = 3.38,$$ what is the P-value? Do I reject $$H_0$$ at the 5% level. There is one null value for $$H_0;$$ what is it? How do I get $$\sigma = 1?$$ Is there one specific alternative value of $$\mu?$$

This test might also be viewed as a likelihood ratio test. If you do that, what are the two likelihood functions; give their means and their standard deviations, and say how do you get them? How would you find the Type I and Type II errors for your specific LR test? What is your test statistic? What is the rejection region?

• Hi Bruce, would you be able to name the two "foundational approaches" so I can read more about them? To answer your questions: we do reject H0 at the 5% level, the P-value is equivalent to the 2*(1-CDF(3.38)), there is not one alternative value of mu.
– bGe
Commented Jun 13, 2022 at 15:09
• The two likelihood functions are: PDF(3.38)/PDF(3.38 - MLE(mean)), where the PDF is for the Normal(0, 1) since we assume that variance=1 for both scenarios and are simply testing the mean shift in the alternative hypothesis, and MLE(mean) = 3.38. The rejection region is the most extreme 5% of the normal distribution which is equivalent to the rejection region for the first test.
– bGe
Commented Jun 13, 2022 at 15:11