1
$\begingroup$

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?

EDIT:

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?

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Commented Oct 5, 2018 at 0:38
  • $\begingroup$ @FransRodenburg I have just edited my post. $\endgroup$
    – user81411
    Commented Oct 5, 2018 at 3:53

1 Answer 1

0
$\begingroup$

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 $\alpha$
$\endgroup$
2
  • $\begingroup$ But this may inflate type-1 error rate since I have 5 tests in total. $\endgroup$
    – user81411
    Commented Oct 5, 2018 at 3:44
  • $\begingroup$ Use Bonferoni correction or similar $\endgroup$ Commented Oct 5, 2018 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.