# Why is my p-value correlated to difference between means in two sample tests?

A colleague has recently made the claim that a large p-value was not more support for the null hypothesis than a low one. Of course, this is also what I learned (uniform distribution under the null hypothesis, we can only reject the null hypothesis...). But when I simulate two random normal distributions (100 samples in each group) in R, my p-value is correlated to the difference (averaged over 30 repetitions) between the two means (with for example a T test or a Mann & Whitney test).

Why is my p-value, above the threshold of 0.05, correlated to the difference between the means of my two groups? With 1000 repetitions for each x (difference between means/2) value. My R code in case this is just a silly mistake.

pvaluetot<-NULL
xtot<-NULL
seqx<-seq(0,5,0.01)
for (x in seqx){
ptemp<-NULL
pmean<-NULL
a<-0

repeat{
a<-a+1
pop1<-rnorm(100,0+x,2)
pop2<-rnorm(100,0-x,2)
pvalue<-t.test(pop1,pop2)\$p.value

ptemp<-c(ptemp,pvalue)
#print(ptemp)
if (a==30)
break
}

pmean<-mean(ptemp)
pvaluetot<-c(pvaluetot,pmean)
xtot<-c(xtot,x)
print(x)
}

pvaluetot
xtot
plot(pvaluetot,xtot)


## 3 Answers

As you said, the p-value is uniformly distributed under the null hypothesis. That is, if the null hypothesis is really true, then upon repeated experiments we expect to find a fully random, flat distribution of p-values between [0, 1]. Consequently, a frequentist p-value says nothing about how likely the null hypothesis is to be true, since any p-value is equally probable under the null.

What you're looking at is the distribution of p-values under an alternative hypothesis. Depending on the formulation of this hypothesis, the resulting p-values can have any non-Uniform, positively skewed distribution between [0, 1]. But this doesn't tell you anything about the probability of the null. The reason is that the p-value expresses the probability of the evidence under the null hypothesis, i.e. $$p(D|H_0)$$, whereas you want to know $$p(H_0|D)$$. These two are related by Bayes' rule: $$p(H_0|D) = \frac{p(D|H_0)p(H_0)}{p(D|H_0)p(H_0)+p(D|\neg H_0)p(\neg H_0)}$$ This means that in order to calculate the probability you're interested in, you need to know and take into account the prior probability of the null being true ($$p(H_0)$$), the prior probability of the null being false ($$p(\neg H_0)$$) and the probability of the data given that the null is false ($$p(D|\neg H_0)$$). This is the purview of Bayesian, rather than frequentist statistics.

As for the correlation you observed: as I said above the p-values will be positively skewed under the alternative hypothesis. How skewed depends what that alternative hypothesis is. In the case of a two-sample t-test, the more you increase the difference between your population means, the more skewed the p-values will become. This reflects the fact that you're making your samples increasingly more different from what is plausible under the null, and so by definition the resulting p-values (reflecting the probability of the data under the null) must decrease.

Why would you expect anything else? You don't need a simulation to know this is going to happen. Look at the formula for the t-statistic: $$t = \frac{\bar{x_1} - \bar{x_2} }{\sqrt{ \frac{s^2_1}{n_1} + \frac{s^2_2}{n_2} }}$$

Obviously if you increase the true difference of means you expect $$\bar{x_1} - \bar{x_2}$$ will be larger. You are holding the variance and sample size constant, so the t-statistic must be larger and thus the p-value smaller.

I think you are confusing a philosophical rule about hypothesis testing with a mathematical fact. If the null hypothesis is true, you would expect a higher p-value. This has to be true in order for hypothesis testing to make any sense.

You should indeed not interpret the p-value as a probability that the null hypothesis is true.

However, a higher p-value does relate to stronger support for the null hypothesis.

### Considering p-values as a random variable

You could consider p-values as a transformation of your statistic. See for instance the secondary x-axis in the graph below in which the t-distribution is plotted with $$\nu=99$$. Here you see that a larger p-value corresponds to a smaller t-statistic (and also, for a two-sided test, there are two t-statistic associated with one p-value).

### Distribution of p-values $$P(\text{p-value}|\mu_1-\mu_2)$$

When we plot the distribution density of the p-values, parameterized by $$\mu_1-\mu_2$$, you see that higher p-values are less likely for $$\mu_1-\mu_2 \neq 0$$. # compute CDF for a given observed p-value and parameter ncp=mu_1-mu_2
qp <- function(p,ncp) {
from_p_to_t <- qt(1-p/2,99)   # transform from p-value to t-statistic
1-pt(from_p_to_t,99,ncp=ncp) + pt(-from_p_to_t,99,ncp=ncp)  # compute CDF for t-statistic (two-sided)
}
qp <- Vectorize(qp)

# plotting density function
p <- seq(0,1,0.001)
plot(-1,-1,
xlim=c(0,1), ylim=c(0,9),
xlab = "p-value", ylab = "probability density")

# use difference between CDF to plot PDF
lines(p[-1]-0.001/2,(qp(p,0)[-1]-qp(p,0)[-1001])/0.001,type="l")
lines(p[-1]-0.001/2,(qp(p,1)[-1]-qp(p,1)[-1001])/0.001,type="l", lty=2)
lines(p[-1]-0.001/2,(qp(p,2)[-1]-qp(p,2)[-1001])/0.001,type="l", lty=3)


The bayes factor, the ratio of the likelihood for different hypotheses is larger for larger p-values. And you could consider higher p-values as stronger support. Depending on the alternative hypothesis this strong support is reached at different p-values. The more extreme the alternative hypothesis, or the larger the sample of the test, the smaller the p-value needs to be in order to be strong support. ### Illustration

See below an example with simulations for two different situations. You sample $$X \sim N(\mu_1,2)$$ and $$X \sim N(\mu_2,2)$$ Let in one case

• $$\mu_i \sim N(i,1)$$ such that $$\mu_2-\mu_1 \sim N(1,\sqrt{2})$$

the other case

• $$\mu_i \sim N(0,1)$$ such that $$\mu_2-\mu_1 \sim N(0,\sqrt{2})$$. In the first case you can see that the probability for $$\mu_1-\mu_2$$ is most likely to be around 1, also for higher p-values. This is because the marginal probability $$\mu_1-\mu_2 \sim N(1,\sqrt{2})$$ is already close to 1 to start with. So a high p-value will be support for the hypothesis $$\mu_1-\mu_2$$ but is is not strong enough.

In the second case you can see that $$\mu_1-\mu_2$$ is indeed most likely to be around zero when the p-value is large. So, you could consider it as some sort of support for the null hypothesis.

So in any of the cases a high p-value is support for the null hypothesis. But, it should not be considered as the probability that the hypothesis is true. This probability needs to be considered case by case. You can evaluate it when you know the joint distribution of the mean and the p-value (that is, you know something like a prior probability for the distribution of the mean).

Sidenote: When you use the p-value in this way, to indicate support for the null hypothesis, then you are actually not using this value in the way that is was intended for. Then you may better just report the t-statistic and present something like a plot of a likelihood function (or bayes factor).