Suppose by complete randomization design, I reach into a decision that there is at least difference between two treatment means. That is, my hypothesis is

$$H_o:\mu_1=\mu_2=\mu_3=\mu_4=\mu_5$$ $$H_1: \mu_i\ne \mu_j\quad \text{for at least one pair $(i,j)$},$$

and from ANOVA test, I have rejected the null hypothesis.

Next, my research interest is whether the $i$th treatment effect is equal to zero, that is,

$$H_0:\mu_i=0$$ $$H_1:\mu_i\ne 0, \quad i=1,2,3,4,5.$$

How can I test the above hypothesis that whether an individual treatment mean significant or not?


Simultaneous confidence interval seems to me a possible solution. The formula to compute $r$ simultaneous confidence intervals is:

$$\bar y_{i.}-t_{\alpha/(2r),N-5}\sqrt\frac{MS_E}{n}\le \mu_i\le \bar y_{i.}+t_{\alpha/(2r),N-5}\sqrt\frac{MS_E}{n}.$$

Suppose I got $14.5\le \mu_3\le 40$. Can I reject or fail to reject the $H_0$ from here?

  • $\begingroup$ Can you include some more background information? Why do you want to show that the means are non-zero, as opposed to comparing the means to each other (which is the usual post-hoc comparison for ANOVA)? If you really want to show the means are non-zero, then why use the ANOVA omnibus test in the first place? $\endgroup$ – Frans Rodenburg Oct 5 '18 at 0:38
  • $\begingroup$ @FransRodenburg I have just edited my post. $\endgroup$ – user81411 Oct 5 '18 at 3:53

You can perform 5 separate T-tests for the aforementioned hypotheses, given the fact that you have the sample standard deviations.

  1. Get the T-statistic: $T=\frac{\bar{Z}-\mu}{\frac{\sigma}{n}}$
  2. Test it against the Null Hypothesis with the mean of 0: $\mu=0$
  3. See where you $H_a$ falls
  4. Get the $\alpha$
  • $\begingroup$ But this may inflate type-1 error rate since I have 5 tests in total. $\endgroup$ – user81411 Oct 5 '18 at 3:44
  • $\begingroup$ Use Bonferoni correction or similar $\endgroup$ – Stepan Ulyanin Oct 5 '18 at 13:37

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