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Hi,

I'm running a multiple linear regression on the dependent variable KMLIT and independent variables CYL, ENGCM3, HP and WTKG
I need to do a t-test to see if one unit increase in HP has greater detrimental effect than one unit increase in WTKG.

With reparameterization:
Test β3>β4 = β3-β4=0
Let Φ=β3-β4

H0: Φ = 0 β3-β4=0 β3=β4
H1: Φ > 0 β3-β4>0 β3>β4

I ran the estimation: kmlit c cyl engcm3 hp (hp+wtkg)
However the result is that I can reject the null with the P-value but fail to reject using the t-stat.

All of the four independent variables have a negative coefficient, which I think is the problem. Will I need to change the signs somehow?

Dependent Variable: KMLIT
Method: Least Squares
Date: 05/27/10 Time: 02:05
Sample: 1 300
Included observations: 300

Coefficient Std. Error t-Statistic Prob.

C 14.13693 0.422440 33.46494 0.0000
CYL -0.058797 0.115682 -0.508266 0.6116
ENGCM3 -6.07E-05 0.000150 -0.404790 0.6859
HP -0.008236 0.003475 -2.369906 0.0184
WTKG -0.003715 0.000426 -8.718605 0.0000

R-squared 0.780580 Mean dependent var 7.377235
Adjusted R-squared 0.777605 S.D. dependent var 2.241106
S.E. of regression 1.056879 Akaike info criterion 2.965043
Sum squared resid 329.5129 Schwarz criterion 3.026773
Log likelihood -439.7565 Hannan-Quinn criter. 2.989748
F-statistic 262.3631 Durbin-Watson stat 1.015431
Prob(F-statistic) 0.000000

Im using version 6

Why not simply run the unrestricted regression, then do a wald test?

What Gareth says is true, but your method also works. There is no difference between the t-test results and the p-value, although remember that the p-value is given for a two-tailed test.

OK, so what would I put as the restriction in the Wald-test?

Oh I probably misinterpreted the p-value.
HP+WTKG -0.003715 0.000426 -8.718605 0.0000

For H0: Φ=0, H1: Φ>0
The p-value is 0.0000 from the standard normal table with -8.71
But for the right tail its 1-0.000= 1
So at 0.05 significance i have

1>0.05 Fail to Reject null
-8.71<1.645 fail Reject null

Correct?

c(3)=c(4)

or whatever coefficient numbers are appropriate in the original parameterization.

Startz,

I know I'm being a pain in the ass now, but just to make sure: so it doesn't make any difference that the coefficient are negative, it will just be absolute value anyway?

Actually, you've done a nice job of carefuly explaining your question. It's a good model for all of us.

The sign doesn't matter, except that the "t" can come out positive or negative and one rejects on a one-tailed test while the other doesn't.

The sign doesn't matter, except that the "t" can come out positive or negative and one rejects on a one-tailed test while the other doesn't.

Wouldn't you say it is very important though, whether to reject or fail to reject null? Atleast for my purposes it is.

The alternative hypothesis states that b3>b4 or numerically -0.008236>-0.003715.

The sign doesn't matter, except that the "t" can come out positive or negative and one rejects on a one-tailed test while the other doesn't.

Wouldn't you say it is very important though, whether to reject or fail to reject null? Atleast for my purposes it is.

The alternative hypothesis states that b3>b4 or numerically -0.008236>-0.003715.

Well yes, so if the estimated value of b3-b4 is significantly greater than zero you reject.

I also have a question regarding the wald test...

I'm running a regression with two dummies. One which is 1 when returns of the S&P 500 are positive (usa_p) and the other is pretty logical: 1 when returns of the S&P 500 are negative (usa_n). And then I want to find out whether a negative return of the S&P 500 has a bigger or smaller effect on other indices, for instance of the FTSE100.

My regression looks as follows: uk=c(1)+c(2)*usa*usa_p+c(3)*usa*usa_n

When i want to compare these two coefficients is it possible that i just do a Wald test c(2)=c(3) and after that i could tell whether the coefficients are significantly different or not?

Thanks in advance!