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I have a vector of values and a vector of the reciprocal of their standard error. I'd like to estimate their mean and the standard error on the mean.

At first I used linear regression using the lm() function. Later I found out that the weights library provided a wtd.t.test() function for doing t-tests on weighted data. One side-effect of the test is an estimate of the mean of the data, with a standard error.

I expected both approaches to yield the same result, but they don't.

Here is a short but complete example:

library(weights)
test <- c(1,1,1,1,1,1,2,2,2,3,3,3,4,4)
weight <- c(.5,.5,.5,.5,.5,1,1,1,1,2,2,2,2,2)
wtd.t.test(test, weight=weight)
summary(lm(test~1, weights=weight))

The weighted t-test yields 2.64 +/- 0.27, whereas the linear regression yields 2.64 +/- 0.30. Which is correct (if any)?

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    $\begingroup$ I wonder if the regression is using sampling weights (weights that denote the inverse of the probability that the observation is included because of the sampling design) and the t-test is using importance weights. That is what I get with Stata using your data. $\endgroup$
    – dimitriy
    Commented Jun 18, 2014 at 21:59

2 Answers 2

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Probably because the weight in wtd.t.test is treated as a frequency weight but in lm it is treated as an analytical/proportion weight.

If you try multiply the variable weight by 2 or 3 and refit the t-test and regression, you'll find that the SE of the regression model does not change, while the degrees of freedom in wtd.t.test keeps increasing (and SE shrinking.)

So, I think it's natural that they don't agree.

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It is not clear what wtd.t.test really does when you give it only one set of data. The help page and the output indicates that it is doing a 2 sample test (with a default value of 0 for y). So it is really not doing the same test as lm.

I would expect more similarity if you do a 2 sample test for both. But there is still possibility of difference if wtd.t.test does not use a pooled variance (I could not find anything in the documentation that says whether it assumes equal variance or not, the lm approach does fit a pooled variance).

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