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I understand that a Type II error that arises from a hypothesis test indicates a failure to reject the null hypothesis $H_0$ when $H_a$ in reality is true. But when I try to interpret a Type II error in the context of an actual problem, I have a question on the wording I can use.

Suppose I conduct a study to see if mean systolic blood pressure of patients receiving a treatment ($\mu_T$) is lower than that of patients taking a placebo, ($\mu_P$). I would have:

$$ H_0: \mu_T = \mu_P \\ H_a: \mu_T < \mu_P $$

In this context, I would say that a Type II error involves "believing the treatment does not lower means systolic blood pressure even though in reality it does". But is this, in effect, saying I accept $H_0$ instead of failing to reject $H_0$? Is it better to say a Type II error is "failing to find any evidence the treatment lowers mean systolic blood pressure even though in reality it does"?

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A Type II error is when one fails to reject $\sf{H_0}$ when actually $\sf{H_1}$ is true. The following table summarises hypothesis truths and errors. \begin{array}{c|c}&\sf{accept\,H_0}&\sf{reject\,H_0}\\\hline\sf{H_0\,true}&\checkmark&\sf{type \,I\,error}\\\hline\sf{H_1\,true}&\sf{type\,II\,error}&\checkmark\end{array} In this case, the Type II error would be we do not reject the claim that mean systolic blood pressures are equal when actually the one for placebo is greater. To accept and to fail to reject convey the same meaning, so you are correct.

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  • $\begingroup$ So it's fine to say "a type II error is believing $H_0$ while $H_a$ is actually true" even though that that implies accepting $H_0$? I though accepting $H_0$ was something you never do, since you can only fail to reject $H_0$. $\endgroup$ – Andrew Li Apr 20 '19 at 19:33
  • $\begingroup$ They are indeed different, but for the purposes of interpretation they give out the same meaning. Accepting $\sf{H_0}$ is not statistically sound, so don't use the word 'accept' when talking about type II errors (or the power) $\endgroup$ – TheSimpliFire Apr 20 '19 at 19:35
  • $\begingroup$ Okay thanks, just wanted to verify. $\endgroup$ – Andrew Li Apr 20 '19 at 19:36

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