Consider a data set containing the IQs of 36 men and the IQs of 36 women. Can use this data to test the following?
\[ H_0: \mu_{M} - \mu_{F} = 0 \\ H_A: \mu_{M} - \mu_{F} \ne 0 \]
## IQ sex ## 24 112 male ## 4 113 male ## 44 120 female ## 43 119 female ## 52 118 female ## 30 111 male
(ds <- d %>%
group_by(sex) %>%
summarize(mean = mean(IQ),
s = sd(IQ),
n = n()))
## # A tibble: 2 × 4 ## sex mean s n ## <fctr> <dbl> <dbl> <int> ## 1 female 118 6.50 36 ## 2 male 115 3.48 36
diff(ds$mean)
## [1] -3.39
sqrt(ds$s[1]^2/sqrt(ds$n[1]) + ds$s[2]^2/sqrt(ds$n[2]))
## [1] 3.01
min(ds$n[1] - 1, ds$n[2] - 1)
## [1] 35
Data were collected from schools in a large city on a set of thirty-six children who were identified as gifted children soon after they reached the age of four.
head(gifted)
## score fatheriq motheriq speak count read edutv cartoons ## 1 159 115 117 18 26 1.9 3.00 2.00 ## 2 164 117 113 20 37 2.5 1.75 3.25 ## 3 154 115 118 20 32 2.2 2.75 2.50 ## 4 157 113 131 12 24 1.7 2.75 2.25 ## 5 156 110 109 17 34 2.2 2.25 2.50 ## 6 150 113 109 13 28 1.9 1.25 3.75
If there is a natural pairing between observations in two groups of size n, it can make more sense to analyze them as a single sample of n differences.
gifted %>% mutate(diff = fatheriq - motheriq) %>% select(fatheriq, motheriq, diff)
## fatheriq motheriq diff ## 1 115 117 -2 ## 2 117 113 4 ## 3 115 118 -3 ## 4 113 131 -18 ## 5 110 109 1 ## 6 113 109 4 ## 7 118 119 -1 ## 8 117 120 -3 ## 9 111 128 -17 ## 10 122 120 2 ## 11 111 117 -6 ## 12 112 120 -8 ## 13 119 126 -7 ## 14 120 114 6 ## 15 114 129 -15 ## 16 111 118 -7 ## 17 111 115 -4 ## 18 115 111 4 ## 19 126 111 15 ## 20 115 109 6 ## 21 114 124 -10 ## 22 115 122 -7 ## 23 115 118 -3 ## 24 112 121 -9 ## 25 115 124 -9 ## 26 117 118 -1 ## 27 116 128 -12 ## 28 114 119 -5 ## 29 116 123 -7 ## 30 111 117 -6 ## 31 112 117 -5 ## 32 115 111 4 ## 33 111 101 10 ## 34 119 113 6 ## 35 111 121 -10 ## 36 114 123 -9
\[ H_0: \mu_{diff} = 0 \\ H_A: \mu_{diff} \ne 0 \]
(gs <- gifted %>% mutate(diff = fatheriq - motheriq) %>% summarize(mean = mean(diff), s = sd(diff), n = n()))
## mean s n ## 1 -3.39 7.45 36
(t_obs <- (gs$mean - 0)/(gs$s/sqrt(gs$n)))
## [1] -2.73
gs$s/sqrt(gs$n)
## [1] 1.24
sqrt(ds$s[1]^2/sqrt(ds$n[1]) + ds$s[2]^2/sqrt(ds$n[2]))
## [1] 3.01
While the point estimate is the same in the paired and independent tests, if the data is paired, the dependency leads to a smaller SE.
This principle is widely used in experiment design, e.g. pre- and post-test.