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I have dataframe df.

df <- structure(list(t = c("T1", "T1", "T1", "T1", "T1", "T1", "T1", 
"T1", "T1", "T1", "T2", "T2", "T2", "T2", "T2", "T2", "T2", "T2", 
"T2", "T2", "T3", "T3", "T3", "T3", "T3", "T3", "T3", "T3", "T3", 
"T3", "T4", "T4", "T4", "T4", "T4", "T4", "T4", "T4", "T4", "T4"
), n = c(10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L), m = c(4L, 5L, 4L, 5L, 6L, 4L, 5L, 3L, 5L, 4L, 
2L, 1L, 2L, 3L, 4L, 2L, 1L, 2L, 3L, 1L, 9L, 9L, 9L, 10L, 8L, 
7L, 8L, 9L, 10L, 8L, 5L, 6L, 5L, 6L, 5L, 6L, 4L, 5L, 6L, 5L)), class = "data.frame", row.names = c(NA, 
-40L))

t is the treatment group; the independent variable.

n is the sample size of each subsample.

m is the response; the number of successes out of the possible n; the dependent variable.

m/n is bounded between 0 and 1, therefore I use binomial glm.

mdf <- aov(glm(m/n~t, weights=n, data=df, family=binomial)) #with weights

mdf2 <- aov(glm(m/n~t, data=df, family=binomial)) #without weights

Then I use agricolae::LSD.test to run Fisher's LSD to test for significant differences in mean m/n among t.

However, I get different results with and without using the weights argument, even though all subsamples have 10 individuals

ldf <- LSD.test(mdf, 't')
ldf2 <- LSD.test(mdf2, 't')

output: 
> ldf$groups
    m/n groups
T3 0.87      a
T4 0.53      b
T1 0.45     bc
T2 0.21      c

> ldf2$groups
    m/n groups
T3 0.87      a
T4 0.53      b
T1 0.45      c
T2 0.21      d

In some cases, the differences are even more extreme. Which, is very unexpected to me. Any idea what is happening?

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  • $\begingroup$ A larger sample size impacts significance. Otherwise we would do studies with $n=1$ and save money. $\endgroup$
    – Michael M
    Commented Nov 30, 2023 at 19:43
  • $\begingroup$ So it takes the sample size within each replicate into account to calculate significant rather than how many replicates? What does the default NULL use? $\endgroup$
    – scott.pilgrim.vs.r
    Commented Nov 30, 2023 at 19:49
  • $\begingroup$ Basically. You can test by replicating each row of the data m times (via df[rep(1:nrow(df), each = m), ]) and fit a "simple" logistic regression. This should give you the same inference as the aggregated approach. $\endgroup$
    – Michael M
    Commented Nov 30, 2023 at 19:57
  • $\begingroup$ I guess my question is: If the difference is because of low sub sample size, what does the model do when I don’t pass the optional argument weights ? Is it wrong if I don’t? $\endgroup$
    – scott.pilgrim.vs.r
    Commented Nov 30, 2023 at 20:09
  • 1
    $\begingroup$ I think @MichaelM has already clarified this but - the two GLMs are different. From ?glm, "For a binomial GLM prior weights are used to give the number of trials when the response is the proportion of successes". So basically, mdf is correct, mdf2 is not. $\endgroup$
    – Alex J
    Commented Nov 30, 2023 at 21:57

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