CODEWITHSUNDEEP
rpairwise.wilcox.test
Effigy
Effigy

Reputation: 155

R wilcoxon test on groups

I would like to perform a wilcoxon test on a paired sample and I am wondering, if my code is correct for what I would like to test. I want to know if there is a significant difference between my dependent variable mean moisture (=Feuchte) and my independend variable distance (=Transtyp) grouped by kettlehole (Soll). The hypothesis is, that with increasing distance there is a significant decrease in moisture for each kettlehole.

This is my dataframe

df <- structure(list(Datum = structure(c(18703, 18703, 18703, 18703, 
18724, 18724, 18724, 18724, 18730, 18730, 18730, 18730, 18744, 
18744, 18744, 18744, 18758, 18758, 18758, 18758, 18774, 18774, 
18774, 18774), class = "Date"), Soll = c("1192", "1192", "149", 
"149", "1192", "1192", "149", "149", "1192", "1192", "149", "149", 
"1192", "1192", "149", "149", "1192", "1192", "149", "149", "1192", 
"1192", "149", "149"), Transtyp = structure(c(1L, 2L, 1L, 2L, 
1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
1L, 2L, 1L, 2L), .Label = c("2", "5"), class = "factor"), Feuchte = c(36.15, 
36.6518518518519, 37.66, 37.8310344827586, 28.7625, 30.128125, 
27.271875, 23.0645161290323, 31.903125, 32.15625, 31.740625, 
29.9875, 14.6290322580645, 14.6516129032258, 15.058064516129, 
13.159375, 13.675, 13.7896551724138, 12.390625, 9.690625, 16.2586206896552, 
17.441935483871, 24.24375, 20.24375)), row.names = c(NA, -24L
), class = c("tbl_df", "tbl", "data.frame"))

This is my code so far:

df %>% ungroup() %>%
                  split(.$Soll)%>%
                  map_df( ~broom::tidy(wilcox.test(Feuchte ~ Transtyp, data = .x, paired = T, )), .id = "Soll")

Am I really testing what I want to test as described above? The results are confusing to me. Also, I know you can also use a "," instead of "~". What is the difference between those two and which one do I need and why? I am really stuck and I cant find a good explanation. Thanks a lot in advance!

Cheers

Upvotes: 2

Views: 1471

Answers (1)

Dave2e
Dave2e

Reputation: 24069

Yes, it appears you are performing the calculation correctly. When to use the ~ versus the , is dependent on what form your data is in.
In your example above, your data frame has 1 column of dependent values (Feuchte) and a column of independent variables (Transtyp) so the formula style is correct "y ~ x" (y as a function of x).
On the other hand if you have 2 separate vectors of data then you need to use the , format (y1 compared to y2).

To demonstrate using your example:

df %>% ungroup() %>%
    split(.$Soll)%>%
    map_df( ~broom::tidy(wilcox.test(Feuchte ~ Transtyp, data = .x, paired = T, )), .id = "Soll")

# A tibble: 2 × 5
#  Soll  statistic p.value method                          alternative
#  <chr>     <dbl>   <dbl> <chr>                           <chr>      
#1 1192          0  0.0313 Wilcoxon signed rank exact test two.sided  
#2 149          20  0.0625 Wilcoxon signed rank exact test two.sided

Now extracting Transtyp==2 and Transtyp==5 from when Sol=1192:

sg<-df %>% ungroup() %>% split(.$Soll)
wilcox.test(sg$`1192`$Feuchte[sg$`1192`$Transtyp==2], sg$`1192`$Feuchte[sg$`1192`$Transtyp==5], paired = TRUE)

#   Wilcoxon signed rank exact test
#
#data:  sg$`1192`$Feuchte[sg$`1192`$Transtyp == 2] and sg$`1192`$Feuchte[sg$`1192`$Transtyp == 5]
#V = 0, p-value = 0.03125
#alternative hypothesis: true location shift is not equal to 0

As you can see the V=0 and value =0.0313 in both cases for Soll==1192.

Upvotes: 3

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