# standard error of group means (help please)

#### sunny3333

##### New Member
Hi All,

I have the following summarized data:
month sample size mean variance weight
1 10 20 3 .1
2 20 30 5 .2
3 25 20 6 .4
...
12 40 50 10 .1
where the 12 weights sum to 1. I don't have the original more detailed data.

What I'm interested in is the standard error of the weighted average of the monthly means, i.e.,
0.1*20 + 0.2*30 + 0.4*20 + ... + 0.1*50.

Do any of you know how to calculate the standard error of the weighted mean from the above statistics? I would greatly appreciate any help.
Thanks so much.

Sunny

#### BGM

##### TS Contributor
Suppose we have the follow $$k$$ group random sample with group size $$n_j$$ is known:

$$X_{i,j}, ~~ i = 1, 2, \ldots, n_j, ~~ j = 1, 2, \ldots, k$$

We also have the assumption that they are independent, with a homogeneous variance inside a group:

$$Var[X_{i,j}] = \sigma_j^2, ~~ i = 1, 2, \ldots, n_j, ~~ j = 1, 2, \ldots, k$$

Now we consider the following proposed grand weighted mean estimator:

$$\sum_{j=1}^k \frac {w_j} {n_j} \sum_{i=1}^{n_j} X_{i,j}$$

where the constants $$w_j$$ are known.

Then the variance will be

$$Var\left[ \sum_{j=1}^k \frac {w_j} {n_j} \sum_{i=1}^{n_j} X_{i,j} \right]$$

$$= \sum_{j=1}^k \frac {w_j^2} {n_j^2} Var\left[ \sum_{i=1}^{n_j} X_{i,j} \right]$$

$$= \sum_{j=1}^k \frac {w_j^2} {n_j^2} (n_j \sigma_j^2)$$

$$= \sum_{j=1}^k \frac {w_j^2 \sigma_j^2} {n_j}$$

Therefore the theorectical standard deviation (standard error of the estimator) will be

$$\sqrt{\sum_{j=1}^k \frac {w_j^2 \sigma_j^2} {n_j}}$$

Suppose the now you have already collect the data. Then you can use your method to estimate $$\sigma_j^2$$ by $$\hat{\sigma}_j^2$$

and we can estimate the standard error by the plug-in version:

$$\sqrt{\sum_{j=1}^k \frac {w_j^2 \hat{\sigma}_j^2} {n_j}}$$

#### sunny3333

##### New Member
Thank you BGM for your response.

That's what I did in the first place. But I realized later that it might be only a part of what I needed. The s.e. I calculated in this way is very small, but in my data the group means are very different, so the calculated s.e. doesn't seem to make sense.

I'm now thinking along the line of computing the total sum of squares (TSS) which is the sum of error sum of squares (ESS) and group sum of squares (GSS). I think the part you gave corresponds to ESS.

Because in my data the group means are very different, I think I should not ignore GSS. Here is a link to the calculation I found from a previous post http://www.talkstats.com/showthread.php/7130-standard-deviation-of-multiple-sample-sets‏ . The same link is in wjt's response near the bottom of the thread. But I don't know how to incorporate the weights into it. Do you or anyone else have any insights into it?

Thanks,
Sunny