校正后的成对比较如何产生 p.value < 然后是单个 t.test？

Question

Hi suppose I have these results

df <- structure(list(len = c(4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 
5.2, 7, 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7, 
16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5, 19.7, 
23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3, 23.6, 18.5, 
33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5, 25.5, 26.4, 22.4, 
24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23), supp = structure(c(2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("OJ", 
"VC"), class = "factor"), dose = structure(c(1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("D0.5", "D1", "D2"
), class = "factor")), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 
7L, 8L, 9L, 10L, 31L, 32L, 33L, 34L, 35L, 36L, 37L, 38L, 39L, 
40L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 41L, 42L, 
43L, 44L, 45L, 46L, 47L, 48L, 49L, 50L, 21L, 22L, 23L, 24L, 25L, 
26L, 27L, 28L, 29L, 30L, 51L, 52L, 53L, 54L, 55L, 56L, 57L, 58L, 
59L, 60L), class = "data.frame") 

df$int <- interaction(df$supp, df$dose)
e <- pairwise.t.test(df$len, df$int, p.adjust.method="BH")

so from the output

        OJ.D0.5          VC.D0.5            OJ.D1     VC.D1            OJ.D2  
VC.D0.5 0.00285          -                  -         -                -      
OJ.D1   0.00000079391014 0.00000000000984   -         -                -      
VC.D1   0.04207          0.00000243821908 **0.00088** -                -      
OJ.D2   0.00000000042891 0.00000000000001   0.04645   0.00000089414918 -      
VC.D2   0.00000000042891 0.00000000000001   0.04474   0.00000085310153 0.96089

the comparison of, VC.D1 vs OJ.D1 = 0.00088

however a single t.test

t.test(df[df$supp == "VC" & df$dose == "D1", ]$len, 
       df[df$supp == "OJ" & df$dose == "D1", ]$len)

yields a p.value = p-value = 0.001038

so I most have messed up somewhere because shouldn't an adjusted p value be greater than a single uncorrected p value?

Answer 1

Solution

You'll get the same results when you set p.adjust.method = "none" and pool.sd = FALSE :

pairwise.t.test(df$len, df$int, p.adjust.method = "none", pool.sd = FALSE)$p.value[3,3]
# 0.001038376

t.test(df[df$supp == "VC" & df$dose == "D1", ]$len, 
       df[df$supp == "OJ" & df$dose == "D1", ]$len)$p.value
# 0.001038376

Notes

1. Just a reminder to always carefully read documentation and perform some sanity checks, to make sure the function does what you think it does.
2. This only illustrates where the difference comes from. How to run it in your case will have to depend on your familiarity with the data.

Explanation

The comparison becomes much easier when we simply don't apply multiple testing correction. In that case, they should have the same p-value, right? So let's compare using p.adjust.method = "none" . When running pairwise.t.test we now get 0.00059 ... closer, but still not right.

The problem stems from the pool.sd argument. This forces the use of a common standard deviation across all comparisons. This is useful in general (if the assumption is met), but does lead to different p-values.

When we look at the underlying code, this becomes clear:

if (pool.sd) {
        METHOD <- "t tests with pooled SD"
        xbar <- tapply(x, g, mean, na.rm = TRUE)
        s <- tapply(x, g, sd, na.rm = TRUE)
        n <- tapply(!is.na(x), g, sum)
        degf <- n - 1
        total.degf <- sum(degf)
        pooled.sd <- sqrt(sum(s^2 * degf)/total.degf)
        compare.levels <- function(i, j) {
            dif <- xbar[i] - xbar[j]
            se.dif <- pooled.sd * sqrt(1/n[i] + 1/n[j])
            t.val <- dif/se.dif
            if (alternative == "two.sided") 
                2 * pt(-abs(t.val), total.degf)
            else pt(t.val, total.degf, lower.tail = (alternative == 
                "less"))
        }
    }

Amongst others, a total degrees of freedom is calculated across the tests ( total.degf ) which is then used to calculate a pooled standard deviation ( pooled.sd ).

when we set pool.sd = FALSE , the code simply uses the t.test function:

    else {
        METHOD <- if (paired) 
            "paired t tests"
        else "t tests with non-pooled SD"
        compare.levels <- function(i, j) {
            xi <- x[as.integer(g) == i]
            xj <- x[as.integer(g) == j]
            t.test(xi, xj, paired = paired, alternative = alternative, 
                ...)$p.value
        }
    }