Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've come across two different forms of a skewness-adjusted t-statistic, which was developed originally by Johnson (1978): $$ J = t + \frac{gt^2}{3n} + \frac{g}{6n} $$

and $$ J = t + \frac{gt^2}{3\sqrt{n}} + \frac{g}{6\sqrt{n}}, $$

where $t$ is the conventional t-statistic, $n$ is the number of observations, and $g$ is the skewness estimate. The null hypothesis is zero mean.

Could you advise me what's difference between the two forms?

Many thanks, Dave

share|improve this question
    
Hey Dave, and welcome to the site. I have fixed the formatting of your question, and just want to let you know that equation can be typed within \$ \$ or \$\$ \$\$ to enable TeX. –  Stefan Hansen Oct 30 '12 at 16:31
    
wow, excellent functionality! thanks Stefan. –  Dave Oct 30 '12 at 16:32
    
Can you point to the sources where you found this, particularly the definition of $g$? I found Johnson's original paper, but his form doesn't match either of those given, as far as I can tell. –  Jonathan Christensen Oct 30 '12 at 18:21
    
Hi Jonathan, $g$ is defined as follows: $$g = \frac{\sum_{i=1}^n (x_i-\overline{x})^3}{n\sigma(x)^3}$$ As far as I can understand Johnson's statistic given in his equation (2.5) is equivalent to the second form I have written. So $g$ corresponds to $\mu_3$ in his equation, and $\mu=0$. –  Dave Oct 30 '12 at 18:36
    
The issue I'm having is that when $\overline{x}$ is very negative, and the data is highly skewed ($g$), the second form yields an adjusted t-statistic that is positive, while the conventional t-statistic is negative. This doesn't happen with the first form (the adjustment is smaller). –  Dave Oct 30 '12 at 18:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.