Having trouble running friedman test on ordinal data in R - r

Question

I am attempting to run a friedman test on ordinal data in R and am getting errors. The data can be found here on dropbox https://www.dropbox.com/s/gh8crh18y1ueriy/seltoutput.xlsx?dl=0.
As a description of the data:
group1: group assignments, 2 levels
time1: time points, 2 levels
loameasure: ordinal data, 5 levels
distmeasure: continuous data
vectemp: participant IDs
After importing the data I run the following to correctly format:
selt$loameasure<-factor(selt$loameasure)
selt$distmeasure<-as.numeric(selt$distmeasure)
selt$time1<-factor(selt$time1)
Then I run:
friedman_test(formula = loameasure ~ time1 | vectemp, data = selt)
Then I get the error:
Error in friedman.test.default(c(3L, 2L, 3L, 2L, 2L, 5L, 2L, 1L, 3L, 4L, :
not an unreplicated complete block design
I thought that loameasure and time1 had to be factors but I did try them as numeric and I get a similar error:
Error in friedman.test.default(c(3, 2, 3, 2, 2, 5, 2, 1, 3, 4, 2, 2, 4, :
not an unreplicated complete block design
I've been playing around with this for days and haven't been able to figure out what my problem is. I would love some assistance! Thank you in advance!

Answer 1

As far as I can anticipate a Friedman test is not appropriate in your situation. I would suggest to perform a two-way ANOVA test for unbalanced designs with Type-III sums of square method.
The assumptions of Normality of residuals and homogenity are given.
I have tried to guide you how to perform the test and the meaning of some steps. It is not complete (lacking of interpretation etc..) But this should be a begin and direction for you.
We want to know if loameasure depends on group1 and time1
We will perform a two-way anova with two factors
Dependent variable: loameasure
Independent variable: group1 and time1
library(readxl)
# load your data
df <- read_excel("C:/Users/coding/Downloads/seltoutput.xlsx",
col_types = c("numeric", "numeric", "numeric"))
# Prepare data
# group1 to factor
df$group1 <- factor(df$group1,
levels = c(0, 1),
labels = c("Group_0", "Group_1"))
# time1 to factor
df$time1 <- factor(df$time1,
levels = c(1, 2),
labels = c("Time_1", "Time_2"))
----------------------------------------------------------------------------
# Visualize
library("ggpubr")
ggboxplot(df, x = "time1", y = "loameasure", color = "group1",
palette = c("#00AFBB", "#E7B800"))
ggline(df, x = "time1", y = "loameasure", color = "group1",
add = c("mean_se", "dotplot"),
palette = c("#00AFBB", "#E7B800"))
-----------------------------------------------------------------------------
# first decide if balanced or unbalnced design
table(df$group1, df$time1)
# Output
# Time_1 Time_2
# Group_0 20 20
# Group_1 29 29
# Here it is a unbalance design
# An unbalanced design has unequal numbers of subjects in each group!
## We will perform two-way ANOVA test in R for unbalanced designs !!!!!!!!!!!
# The recommended method are the Type-III sums of squares.
# you need `car` package
library(car)
# Our 2 way anova of unbalanced design (SS Type-III)
df_anova <- aov(loameasure ~ group1 * time1, data = df)
Anova(df_anova, type = "III")
## Output
# Anova Table (Type III tests)
# Response: loameasure
# Sum Sq Df F value Pr(>F)
# (Intercept) 120.050 1 83.9312 1.116e-14 ***
# group1 0.700 1 0.4891 0.48607
# time1 62.500 1 43.6960 2.301e-09 ***
# group1:time1 5.716 1 3.9963 0.04849 *
# Residuals 134.452 94
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# Nomrality check ---------------------------------------------------------
# call residuals (difference between each indivdual and their group1/time1 combination mean)
res <- df_anova$residuals
# Histogram of residuals: Residuals should be normally distributed
hist(res,main="Histogram of residuals", xlab = "Residuals")
# # Extract the residuals
# Run Shapiro-Wilk test
shapiro.test(x = res )
# Output
# data: res
# W = 0.97708, p-value = 0.08434
# P is > 0.05 therefore normality can be assumed.
# Homogenity test ---------------------------------------------------------
# Levene's test for equality of variances (in `car` package)
library(car)
leveneTest(loameasure~ time1 * group1,data=df)
# Output:
# Levene's Test for Homogeneity of Variance (center = median)
# Df F value Pr(>F)
# group 3 0.3196 0.8112
# 94
# P is > 0.05 therefore equal variances can be assumed.