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I'm conducting a hypothesis test for the difference between two groups. The file dat1 contains all observations of measure for the first group; dat2 contains all observations of measure for the second group. Measures of both groups are normally distributed, so a parametric test is appropriate. The code for a student's t-test would look like this:

    ttest1 <- t.test(x = dat1$measure, y = dat2$measure, alternative 
                     = "two.sided", var.equal = TRUE)

But a Levene test indicates that the groups have differing variances, so the t-statistic needs to be adjusted. So the code looks like this:

    ttest2 <- t.test(x = dat1$measure, y = dat2$measure, alternative 
                     = "two.sided", var.equal = FALSE)

But I need to weight each observation. I can produce a weighted student's t-test using the wtd.t.test() function in the weights package:

    ttest3 <- wtd.t.test(x = dat1$measure, y = dat2$measure, 
                         weight = dat1$weight, weighty = 
         dat2$weight, alternative = "two.tailed")

But there is no var.equal argument that would allow me to use the Welch adjustment to correct for concerns about different variances. I have not been able to find another package that facilitates such a test.

Is there a package that allows for this? If not, I need to figure out how to write a function that has this feature.

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    $\begingroup$ Please explain what you mean by 'weight each observation'. In the Welch version of the tests (default setting var.eq=F), it is not relevant whether $\sigma_1^2$ and $\sigma_2^2$ are the same. If you want to test for equality of variances anyway you can still use an F-test (normal data) or Levene's test. $\endgroup$
    – BruceET
    Jan 28, 2021 at 2:45
  • $\begingroup$ @BruceET - I mention in the post that I used a Levene test and determined the two groups have different variances. This violates the assumption of homogeneity in variance necessary for a student's t-test, and is why I plan to use the Welch test. Explaining what sample weights are may be out of scope, but it's a very standard procedure in survey research (thus the wtd.t.test() function in the popular weights package, which I use in my post). See documentation for ?weights::wtd.t.test() or this link: aapor.org/Education-Resources/For-Researchers/Poll-Survey-FAQ/… $\endgroup$
    – JmQ
    Jan 28, 2021 at 2:52
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    $\begingroup$ Where you able to find a solution for this? I am currently running through the exact same problem, but I have 5 groups to compare. I searched everywhere on the internet and nothing. $\endgroup$
    – mimi
    Oct 8, 2021 at 16:50
  • $\begingroup$ Was there enough data to match the two groups so that the distributions would be similar? $\endgroup$
    – user405514
    Jan 23 at 21:16

1 Answer 1

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The weights::wtd.t.test function does (try to) do the Welch t-test. It has

        vx <- wtd.var(x, weight, na.rm=TRUE)
        vy <- wtd.var(y, weighty, na.rm=TRUE)
        sxy <- sqrt((vx/n)+(vy/n2))

and

        df <- (((vx/n)+(vy/n2))^2)/((((vx/n)^2)/(n-1))+((vy/n2)^2/(n2-1)))

in the code

I don't think that's the correct thing to do, since the degrees of freedom calculation doesn't take weights (and any other features of sampling) into account. More generally, the denominator of the t-test in the code is an estimate of the denominator in a population version rather than an estimate of the standard error of the numerator. It may well be ok for something like Poisson sampling though.

In other options, the survey::svyttest function does a proper survey-weighted comparison of means.

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