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Hello everybody, I have a test equation with 6 parameters and I would like to test the hypotheses c(6)>=c(5) against H1: c(6)<c(5).
I have performed a Wald test with c(6)=c(5) and I have a p-value equal to 0. How can I confirm that c(6) is larger or equal than c(5)?
Thank you for your answer.

Basically, you just want to do a one-tailed test. See https://en.wikipedia.org/wiki/One-_and_two-tailed_tests for an explanation.

Basically, you just want to do a one-tailed test. See https://en.wikipedia.org/wiki/One-_and_two-tailed_tests for an explanation.

Yes, exactly, thank you. How could I perform it on EViews?

EViews does not have one-tailed tests built in. Go through the steps for a two-tailed test and then compare the t-statistic to the critical value for a one-tailed test from a t-table.

EViews does not have one-tailed tests built in. Go through the steps for a two-tailed test and then compare the t-statistic to the critical value for a one-tailed test from a t-table.

Okok, thank you a lot!
So, do I have to pick the right tail to test if c(6)>c(5) testing on EViews c(6)-c(5)=0?

The tails are symmetric. Of course, if the estimated c(6) is greater than the estimated c(5) then you can't reject the hypothesis no matter what the t-statistic is.

I didn't know all of this, test wald is great, but I go directly to -> Prob(F-statistic)
H0: c(2)=c(3)=0
H1: c(2) != c(3)

if p>5% ---> accept H0
otherwise rejected

maybe false

I didn't know all of this, test wald is great, but I go directly to -> Prob(F-statistic)
H0: c(2)=c(3)=0
H1: c(2) != c(3)

if p>5% ---> accept H0
otherwise rejected

maybe false

I'm not sure if there is a question here. (Do note that H1 isn't quite right, as c(2)=c(3)=17 would lead to a rejection of the null.)

I'm not sure if there is a question here. (Do note that H1 isn't quite right, as c(2)=c(3)=17 would lead to a rejection of the null.)

noted. then H1: each one is != 0
Thank you

I'm not sure if there is a question here. (Do note that H1 isn't quite right, as c(2)=c(3)=17 would lead to a rejection of the null.)

noted. then H1: each one is != 0
Thank you

Actually, the alternative is that either one is not equal to 0.