Matched pairs

Some data:

Matched pairs 1/2

Data are comparison of 2 drugs for effectiveness at reducing pain.
- 12 subjects (cases) were arthritis sufferers
- Response is #hours of pain relief from each drug.
In reading example, each child tried only one reading method.
But here, each subject tried out both drugs, giving us two measurements.
Possible because, if you wait long enough, one drug has no influence over effect of other.

Matched pairs 2/2

Advantage: focused comparison of drugs. Compare one drug with another on same person, removes a lot of variability due to differences between people.
Matched pairs, requires different analysis.
Design: randomly choose 6 of 12 subjects to get drug A first, other 6 get drug B first.

Packages

library(tidyverse)
library(smmr) # for a sign test later

Reading the data

Values aligned in columns:

my_url <- 
  "http://ritsokiguess.site/datafiles/analgesic.txt"
pain <- read_table(my_url)
pain

glimpse(pain)

Rows: 12
Columns: 3
$ subject <dbl> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
$ druga   <dbl> 2.0, 3.6, 2.6, 2.6, 7.3, 3.4, 14.9, 6.6, 2.3, 2.0, 6.8, 8.5
$ drugb   <dbl> 3.5, 5.7, 2.9, 2.4, 9.9, 3.3, 16.7, 6.0, 3.8, 4.0, 9.1, 20.9

Paired t-test

with(pain, t.test(druga, drugb, paired = TRUE))


    Paired t-test

data:  druga and drugb
t = -2.1677, df = 11, p-value = 0.05299
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 -4.29941513  0.03274847
sample estimates:
mean difference 
      -2.133333

P-value is 0.053.
Not quite evidence of difference between drugs.

t-testing the differences

Likewise, you can calculate the differences yourself and then do a 1-sample t-test on them.

pain %>% mutate(diff = druga - drugb) -> pain
pain

t-test on the differences

then throw them into t.test, testing that the mean is zero, with same result as before:

with(pain, t.test(diff, mu = 0))


    One Sample t-test

data:  diff
t = -2.1677, df = 11, p-value = 0.05299
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 -4.29941513  0.03274847
sample estimates:
mean of x 
-2.133333

Same P-value (0.053) and conclusion.

Assessing normality

1-sample and 2-sample t-tests assume (each) group normally distributed.
Matched pairs analyses assume (theoretically) that differences normally distributed.
How to assess normality? A normal quantile plot.

The normal quantile plot (of differences)

ggplot(pain,aes(sample=diff))+stat_qq()+stat_qq_line()

Points should follow the straight line. Bottom left one way off, so normality questionable here: outlier.

What to do instead?

Matched pairs \(t\)-test based on one sample of differences
the differences not normal (enough)
so do sign test on differences, null median 0:

sign_test(pain, diff, 0)

$above_below
below above 
    9     3 

$p_values
  alternative    p_value
1       lower 0.07299805
2       upper 0.98071289
3   two-sided 0.14599609

Did we need to worry about that outlier?

Bootstrap sampling distribution of sample mean differences:

tibble(sim = 1:10000) %>% 
  rowwise() %>% 
  mutate(my_sample = list(sample(pain$diff, replace = TRUE))) %>% 
  mutate(my_mean = mean(my_sample)) %>% 
  ggplot(aes(sample = my_mean)) + stat_qq() + stat_qq_line()

Yes we did; this is clearly skewed left and not normal.

Comments

no evidence of any difference between drugs (P-value 0.1460)
in \(t\)-test, the low outlier difference pulled mean difference downward and made it look more negative than it should have been
therefore, there really isn’t any difference between the drugs.