OPTIONAL practice problem - Do a Wilcoxon signed rank test by hand

1. Description

In this practice problem, you’ll do a Wilcoxon signed rank test comparing paired samples to each other by hand, then double check your work with R. This is the non-parametric version of the paired t-test.

2. Steps

a. Values

Start with these values:

Before	After
1.2	1.6
2.0	2.8
0.8	0.7
0.6	0.6
1.3	1.5
3.2	2.7
0.9	1.1

b. Calculate the differences between samples

Then, calculate the differences between those values. You should get something that looks like this:

Differences
-0.4
-0.8
0.1
0.0
-0.2
0.5
-0.2

c. Take out 0, arrange by magnitude

Then, omit all observations of 0 and arrange your numbers by magnitude. You should get something that looks like this:

Ordered magnitude
0.1
-0.2
-0.2
-0.4
0.5
-0.8

d. Give each value a “sign” and a “rank”

This is the “signed rank” part of the test: assign each value a sign (+ or -), then rank (1, 2, 3, 4, etc.) You should get something that looks like this:

Ordered magnitude	Signed rank
0.1	+ 1
-0.2	- 2
-0.2	- 2
-0.4	- 4
0.5	+ 5
-0.8	- 6

e. Sum the magnitudes of the + and - values

Sum the magnitudes of the + and - values. This will be the -W and +W statistic. You then choose the lower statistic.

In this case, you should get:

+W = 1 + 5 = 6
-W = 2 + 2 + 4 + 6 = 14

Pick the lowest one: +W = 6.

e. Run the test in R to check your work.

Start with this data frame:

# creating a data frame with "before" and "after" values
# each row is an individual with a measurement taken "before" and "after"
df <- tibble(before = c(1.2, 2.0, 0.8, 0.6, 1.3, 3.2, 0.9), 
             after = c(1.6, 2.8, 0.7, 0.6, 1.5, 2.7, 1.1)) 

Then run a Wilcoxon signed rank test (note the paired = TRUE argument).

wilcox.test(x = df$before,
            y = df$after,
            paired = TRUE) # argument for a paired test

Warning in wilcox.test.default(x = df$before, y = df$after, paired = TRUE):
cannot compute exact p-value with zeroes

    Wilcoxon signed rank test with continuity correction

data:  df$before and df$after
V = 6, p-value = 0.4017
alternative hypothesis: true location shift is not equal to 0

f. Try a one-sample test.

Compare the following values to \(\mu = 7\):

Sample
2.8
8.3
4.5
6.7
8.1
7.2
7.8

First, calculate the difference between each observation and the \(\mu\), which is 7:

Sample	Difference
2.8	-4.2
8.3	1.3
4.5	-2.5
6.7	-0.3
8.1	1.1
7.2	0.2
7.8	0.8

Then order each difference and assign a sign and rank:

Ordered magnitude	Signed rank
0.2	+ 1
-0.3	- 2
0.8	+ 3
1.1	+ 4
1.3	+ 5
-2.5	- 6
-4.2	- 7

Then sum the magnitudes:

+W = 1 + 3 + 4 + 5 = 13
-W = 2 + 6 + 7 = 15

For this example, W = 13.

Then, double check your work in R:

wilcox.test(c(2.8, 8.3, 4.5, 6.7, 8.1, 7.2, 7.8),
            mu = 7)

    Wilcoxon signed rank exact test

data:  c(2.8, 8.3, 4.5, 6.7, 8.1, 7.2, 7.8)
V = 13, p-value = 0.9375
alternative hypothesis: true location is not equal to 7