# Combine 2 groups based on the regression coefficient

I have two groups of people, N=211 in group 1, N= 310 in group 2.

I would like to combine these groups because they share a certain characteristic that I am interested in. The groups are used in a multiple regression, so I only have the regression coefficients and CI's. They are:

group 1: -.05 (-.21 to .11) group 2: -.14 (-.27 to .00)

Is it possible to combine the groups? I did a similar calculation fot two other groups with the Cochrane combine groups formula. However, those groups had a mean, N and SE. Does anyone know if I can combine these 'regression' groups as well?

Thank you for the response, In following your advice I came to the following outcome:

SE= (upper limit-lower limit)/3,93 (Cochrane handbook)
y= width CI/2*1,96
S2= SE*SE


This resulted in:

 SE y   S2 group 1 0,08163  0,3136  0,006663457
group 2 0,06888 0,2646  0,004744454


Applying the inverse method resulted in: S2 combinede = 102,833/360,845= 0,285. Is that correct or am I missing something? And I only have the weighted shared variance now, right? How do I calculate the average weighted regression coeffiecient?

Thanks to mdewey for the correct solution:

w1 1 / v1 150,0722545 b1 -0,05 w2 1 / v2 210,7723914 b2 -0,14

Overall estimate of coefficient y and variance w

sum (wiyi)/sum(wi)
wiyi -37,01174753 y -0,102569757
som(wi) 360,844646 w 0,002771276

You have the estimated coefficient and from its confidence interval you can work out its standard error. Assuming they used a normal approximation you divide the width of the confidence interval by 2 * 1.96. If they used $t$ then replace 1.96 by the corresponding value of $t$ for your degrees of freedom. Square the standard error to get the sampling variance. You now have enough to combine the estimates using the inverse variance weighting method.

Edited in response to additional calculations posted by OP

Your calculation of the sampling variances is correct but you do not need to estimate a shared variance, you need to weight each coefficient by the inverse of its sampling variance then average them. It would be much easier at this point to use your favourite software to do this. There are several packages in R for meta-analysis and you can also use Stata.

So you have two estimates $y_i$ where $i=1,2$ and their sampling variances $v_i$. You then form the weights $w+i = \frac{1}{v_i}$ then your overall estimate $y$ and its sampling variance $w$ are

$$y = \sum \frac{(w_i y_i)}{\sum w_i}$$

$$w = \frac{1}{\sum w_i}$$

• Thank you for the response, In following your advice I came to the following outcome: SE= (upper limit-lower limit)/3,93 (Cochrane handbook) y= width CI/2*1,96 S2= SE*SE This resulted in: SE y S2 group 1 0,08163 0,3136 0,006663457 group 2 0,06888 0,2646 0,004744454 Applying the inverse method resulted in: S2 combinede = 102,833/360,845= 0,285. Is that correct or am I missing something? And I only have the weighted shared variance now, right? How do I calculate the average weighted regression coeffiecient? – PCdL Jan 12 '17 at 8:43
• Hi mdewey thanks for your time, I use comprehensive meta- analysis, which works great for most data. The point is that I need to combine these 2 groups and compare them with group 0 which is the reference group. If I way the coefficient of group 1 and group 2, can I use that as the correlation and variance to enter into CMA? Or do I misrepresent the coefficient as correlation for this particular case? – PCdL Jan 12 '17 at 17:18
• I have edited in how to do the hand calculations, I hope correctly – mdewey Jan 12 '17 at 18:18
• I hope my calculations are correct, I edited the first post. Thank you for your time and effort mdewey. – PCdL Jan 16 '17 at 10:17
• Hi Mdewey, one last question on the topic. Can I regard this averaged coefficient and variance as a standardized correlation? After all group 0 was the reference category in this multinomial regression, which is now turned into a logistic regression with y1=0 w1=0 and y2= -0,102569757 w2=0,002771276. Or am I missing something? – PCdL Jan 16 '17 at 10:50