# Severity and S-value

This post is under constant updating…

The severity principle proposed by Deborah Mayo is used for accepting a null hypothesis H when:

1.  there is no evidence to reject H and
2.  H passes the test with high severity.

It seems to me that it is quite similar to my purpose (the s-value), but the measures involved in steps 1. and 2. are different.

An example from the normal distribution with known variance $\sigma^2 = 1$ follows. Consider $n=100$ and the hypotheses

$H_0: \mu \leq 10$ vs $H_1: \mu > 10$,

a) If the sample average is 5, then 5+50*1/10 = 10. That is, we have no evidence against $H_0$ (since the sample average is corroborating $H_0$) and strong evidence against $H_1$ (since the sample average is very far away from $H_1$). The s-value for $H0$ is one and the s-value for $H_1$ is almost zero (i.e., $H_0$ passes with high severity).

b) If  the sample average is 15, then 15-50*1/10 = 10. That is, we have strong evidence against $H_0$ (since the sample average is very far away from $H_0$) and no evidence against $H_1$ (since the sample average is corroborating $H_1$). The s-value for $H_0$ is almost zero and the s-value for $H_1$ is one.

c) If the sample average is 9.9 , then 9.9+1*1/10 = 10. That is, we have no evidence against $H_0$ (since the sample average  is corroborating $H_0$) and also a not strong evidence against $H_1$ (since the sample average is near $H_1$). The s-value for $H_0$ is one and the s-value for $H_1$ is approx 0.3 (i.e., $H_0$ does not pass with high severity)

d) If the sample average is 10.1, then 10.1-1*1/10 = 10. That is, we do not have strong evidence against $H_0$ (since the sample average is near $H_0$) and also we do not have evidence against $H_1$ (since the sample average is corroborating $H_1$). The s-value for $H_0$ is approx 0.3 and the s-value for $H_1$ is one.

My impression is that the Deborah’s approach tries to accept H even when the data do not corroborate with H.

Deborah’s severity:

For my example $H_0: \mu \leq 10$ vs $H_1: \mu > 10$:

$SEV(\mu < m) = P(\bar{X} > \bar{x}; \mu \geq m) = 1 - P(\bar{X} \leq \bar{x}; \mu \geq m)$

Notice that

$P(\bar{X} \leq \bar{x}; \mu \geq m)$

is the p-value for the following “inverted”  hypotheses

$H_0^*: \mu \geq m$  vs  $H_1^*: \mu < m$

Then, SEV is high whenever the p-value for the “inverted” hypotheses is low. This means that the event “$\mu \geq m$” is quite improbable for the observed data. The p-value for the inverted hypotheses $H_0^*$ vs $H_1^*$ is:

a) very small for m=10
b) not applied (since we reject the null)
c) not small for m=10
d) not small for m=10.

That is, we have the same conclusions as I showed above with the s-value.  If there some think wrong please let me know.

All the best,
Alexandre.