Monotonicity issues on p value

The methods of statistical testing are widely used in medical research. However, there is a well-known problem, which is that a large sample size gives a small p-value. In this section, we will provide an explicit explanation of this phenomenon with respect to simple hypothesis tests.

Consider the following null hypothesis $H_0$ and its alternative hypothesis $H_1$; $$\begin{eqnarray*} H_0: \mathbb E[ X_i] &=&m_0, \\ H_1: \mathbb E[ X_i] &>&m_0, \\ \end{eqnarray*}$$ where $\mathbb E[ X_i]$ means the expectation of random samples $X_i$ from a normal distribution whose variance $\sigma _0 ^2$ is known. In this situation, the test statistic is given by $$Z^{\text{test}} := \frac{\overline{X_{n}} -m_0 }{\sqrt{\sigma _0 ^2/n}},$$ where $\overline{X_{n}} := \sum_{i=1,\cdots,n} X_i /n$ is normally distributed with mean $m_0$ and standard deviation $\sigma_0/\sqrt{n}$. Under the null hypothesis, $Z^{\text{test}}$ is normally distributed with mean $0$ and a standard deviation $1$ (standard normal distribution). The null hypothesis is rejected if $Z^{\text{test}} >z_{2\alpha}$ , where $z_{2\alpha}$ is a percentile point of the normal distribution, e.g., $z_{0.025}=1.96 .$

Suppose that the true distribution of $X_1, \cdots, X_n$ is a normal distribution with mean $m_0 + \epsilon$ and variance $\sigma _0 ^2$, where $\epsilon$ is an arbitrary fixed positive number. Then $$\begin{eqnarray*} Z^{\text{test}} &=&\frac{\overline{X_{n}} -(m_0+\epsilon -\epsilon) }{\sqrt{\sigma _0 ^2/n}}\\ &=& Z^{\text{Truth}} + \frac{\epsilon}{\sqrt{\sigma _0 ^2/n}}\\ \end{eqnarray*}$$ where $Z^{\text{Truth} }:=(\overline{X_{n}} -(m_0+\epsilon ))/\sqrt{\sigma _0 ^2/n}$.

In the following, we calculate the probability with which we reject the null hypothesis $H_0$ with confidence level $\alpha$. $$\begin{eqnarray*} \text{Prob}(Z^{\text{test}} >z_{2\alpha} ) &=&\text{Prob} (Z^{\text{Truth} } + \frac{\epsilon}{\sqrt{\sigma _0 ^2/n}} >z_{2\alpha})\\ &=&\text{Prob} (Z^{\text{Truth} } >z_{2\alpha} - \frac{\epsilon}{\sqrt{\sigma _0 ^2/n}})\\ &=&\text{Prob} (Z^{\text{Truth} } >z_{2\alpha} - \frac{\epsilon}{\sigma _0 }\sqrt{n} )\\ \end{eqnarray*}$$ Note that $\epsilon /\sigma _0$ is called the effect size.

Thus, if $z_{2\alpha} - \epsilon \sqrt{n} /\sigma _0 < z_{2(1-\beta)}$, i.e., if $n > ( z_{2\alpha}- z_{2(1-\beta)})^2 \sigma _0 ^2 \epsilon ^{-2}$, then the probability that the null hypothesis is rejected is greater than $1 - \beta$.

For example, consider the case $\sigma _0 =1$, $\alpha =0.05$, and $(1-\beta) =\alpha$, then $z_{2\alpha}=1.28$ and in this case, for all $\epsilon>0$, if $n > 7 \epsilon ^{-2}$ then the probability in which above hypothesis test concludes that the difference of the observed mean from the hypothesized mean is significant is greater than $0.95$. This means that almost always the p-value is less than 0.05. Thus a large sample size induces a small p-value.

For example,

• if $\epsilon =1$ then by taking a sample size such that $n > 7$, then almost always the conclusion of the test will be that the observed difference is statistically significant. Similarly,

• if $\epsilon =0.1$ then by taking a sample size such that $n > 700$, then almost all tests will reach the conclusion that the difference is significant; and

• if $\epsilon =0.01$ then by taking sample size so that $n > 70000$, then the same problem will arise.

This phenomenon also means that in large samples statistical tests will detect very small differences between populations.

By above consideration we can get the result ``significance difference’’ with respect to any tiny difference $\epsilon$ by collecting a large enough sample $n$, and thus we must not use the statistical test.