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 long format data frame
df <- tibble(before = c(1.2, 2.0, 0.8, 0.6, 1.3, 3.2, 0.9), # original values
             after = c(1.6, 2.8, 0.7, 0.6, 1.5, 2.7, 1.1)) %>% 
  pivot_longer(cols = before:after, # making this long format
               names_to = "sample", # creating a column called "sample" that has before and after
               values_to = "value") %>% # creating a column called value that has the values
  mutate(sample = fct_relevel(sample, "before", "after")) %>% # making sure that before appears in the data frame first
  arrange(sample) # arranging the data frame so that before and after are all sitting 

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

wilcox.test(value ~ sample, # formula
            data = df, # data
            paired = TRUE) # argument for a paired test

Warning in wilcox.test.default(x = DATA[[1L]], y = DATA[[2L]], ...): cannot
compute exact p-value with zeroes

    Wilcoxon signed rank test with continuity correction

data:  value by sample
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