EViews.com

Hi all,

I want to show that a coefficient is zero (so the nullyhypothesis c unequals zero). Using the t-stats calculated by eviews and comparing them with the quantils (eg 95 %) from the student-t-distribution leads to results not very reasonable. You have to accept for very low t-stats which means you have to accept for very high standard deviations.

Does anybody knows an alternative procedure on this?

Best regards

Georg

I am not sure I follow, but the null hypothesis should be "c is equal to zero". If you have a very low t-value, it already means that you fail to reject the null hypothesis and therefore the coefficent is not significantly different from zero. In other words, the confidence interval around this coefficent includes zero and you have not enough statistical evidence to support the alternate hypothesis.

Many thanks for the fast response.

My post is related on the question if real estate is an inflation hedge. In my model I regressed inflation on real estate returns. The coefficient I get from eviews is for example -5.0, the standard error is 4.1. My test statistic is (-5-1)/4.1=-1.46. My cutting point from the t-distribution is 1.708.

For the null hypothesis that the coefficient is equal to one I would accept this market to be a hedge.

Alternatively eviews could give me a coefficient of 1.1 and a standard error of 0.05. Then my t-statistic is 2 and in this market inflation is no hedge.

This sounds not reasonable to me.

Alterntatively I could use the null hypothesis that the coefficient is smaller then one. I can reject this hypotheses only for quite strong numbers. Even for a coefficient of one with a very low standard error I have to accept this hypotheses. Additionally I am not abel to identify partial hedges (with a coefficient between 0 and 1.)

Regards

Georg

This is a very good example that shows the difference between "accepting" and "failing to reject" the null hypothesis.

Although the estimation yields -5.0, the results tell you with 95% confidence that the true value of this parameter (approx.) lies somewhere between -13.2 and 3.2. Since zero falls into this interval, the coefficent is not statistically significant at all. It is also indistinguishable from any other value within this range (including 1). In the second case, however, parameter uncertainty is much lower and the coefficent is estimated to be somewhere between 1.0 and 1.2 (i.e. statistically significant). Therefore, you have enough statistical evidence to reject the null hypothesis in favor of the alternative and conclude that its value is even significantly different from 1.

Many thanks!

This sounds reasonable. Do you have an idea how to go on with this?

In a paper on this item they say it is a hedge when the coffiecient is more then two standard deviations away from zero.

Best regards

Georg

The critical value for a 95% confidence interval is 1.96, which is probably why the paper mentions the "two" standard deviations as a threshold...

Many thanks!