Comparing Multiple Means in R

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Repeated Measures ANOVA in R

25 mins

Comparing Multiple Means in R

The repeated-measures ANOVA is used for analyzing data where same subjects are measured more than once. This test is also referred to as a within-subjects ANOVA or ANOVA with repeated measures. The “within-subjects” term means that the same individuals are measured on the same outcome variable under different time points or conditions.

For example, you might have measured 10 individuals’ self-esteem score (the outcome or dependent variable) on three time points during a specific diet to determine whether their self-esteem improved.

This chapter describes the different types of repeated measures ANOVA, including:

  • One-way repeated measures ANOVA, an extension of the paired-samples t-test for comparing the means of three or more levels of a within-subjects variable.
  • two-way repeated measures ANOVA used to evaluate simultaneously the effect of two within-subject factors on a continuous outcome variable.
  • three-way repeated measures ANOVA used to evaluate simultaneously the effect of three within-subject factors on a continuous outcome variable.

The main goal of two-way and three-way repeated measures ANOVA is, respectively, to evaluate if there is a statistically significant interaction effect between two and three within-subjects factors in explaining a continuous outcome variable.

You will learn how to:

  • Compute and interpret the different repeated measures ANOVA in R.
  • Check repeated measures ANOVA test assumptions
  • Perform post-hoc tests, multiple pairwise comparisons between groups to identify which groups are different
  • Visualize the data using box plots, add ANOVA and pairwise comparisons p-values to the plot


Contents:

Related Book

Practical Statistics in R II - Comparing Groups: Numerical Variables

Assumptions

The repeated measures ANOVA makes the following assumptions about the data:

  • No significant outliers in any cell of the design. This can be checked by visualizing the data using box plot methods and by using the function identify_outliers() [rstatix package].
  • Normality: the outcome (or dependent) variable should be approximately normally distributed in each cell of the design. This can be checked using the Shapiro-Wilk normality test (shapiro_test() [rstatix]) or by visual inspection using QQ plot (ggqqplot() [ggpubr package]).
  • Assumption of sphericity: the variance of the differences between groups should be equal. This can be checked using the Mauchly’s test of sphericity, which is automatically reported when using the R function anova_test() [rstatix package]. Read more in Chapter @ref(mauchly-s-test-of-sphericity-in-r).

Before computing repeated measures ANOVA test, you need to perform some preliminary tests to check if the assumptions are met.

Note that, if the above assumptions are not met there are a non-parametric alternative (Friedman test) to the one-way repeated measures ANOVA.

Unfortunately, there are no non-parametric alternatives to the two-way and the three-way repeated measures ANOVA. Thus, in the situation where the assumptions are not met, you could consider running the two-way/three-way repeated measures ANOVA on the transformed and non-transformed data to see if there are any meaningful differences.

If both tests lead you to the same conclusions, you might not choose to transform the outcome variable and carry on with the two-way/three-way repeated measures ANOVA on the original data.

It’s also possible to perform robust ANOVA test using the WRS2 R package.

No matter your choice, you should report what you did in your results.

Prerequisites

Make sure that you have installed the following R packages:

  • tidyverse for data manipulation and visualization
  • ggpubr for creating easily publication ready plots
  • rstatix provides pipe-friendly R functions for easy statistical analyses
  • datarium: contains required data sets for this chapter

Start by loading the following R packages:

library(tidyverse)
library(ggpubr)
library(rstatix)

Key R functions:

  • anova_test() [rstatix package], a wrapper around car::Anova() for making easy the computation of repeated measures ANOVA. Key arguments for performing repeated measures ANOVA:
    • data: data frame
    • dv: (numeric) the dependent (or outcome) variable name.
    • wid: variable name specifying the case/sample identifier.
    • within: within-subjects factor or grouping variable
  • get_anova_table() [rstatix package]. Extracts the ANOVA table from the output of anova_test(). It returns ANOVA table that is automatically corrected for eventual deviation from the sphericity assumption. The default is to apply automatically the Greenhouse-Geisser sphericity correction to only within-subject factors violating the sphericity assumption (i.e., Mauchly’s test p-value is significant, p <= 0.05). Read more in Chapter @ref(mauchly-s-test-of-sphericity-in-r).

One-way repeated measures ANOVA

Data preparation

We’ll use the self-esteem score dataset measured over three time points. The data is available in the datarium package.

# Data preparation
# Wide format
data("selfesteem", package = "datarium")
head(selfesteem, 3)
## # A tibble: 3 x 4
##      id    t1    t2    t3
##   <int> <dbl> <dbl> <dbl>
## 1     1  4.01  5.18  7.11
## 2     2  2.56  6.91  6.31
## 3     3  3.24  4.44  9.78
# Gather columns t1, t2 and t3 into long format
# Convert id and time into factor variables
selfesteem <- selfesteem %>%
  gather(key = "time", value = "score", t1, t2, t3) %>%
  convert_as_factor(id, time)
head(selfesteem, 3)
## # A tibble: 3 x 3
##   id    time  score
##   <fct> <fct> <dbl>
## 1 1     t1     4.01
## 2 2     t1     2.56
## 3 3     t1     3.24

The one-way repeated measures ANOVA can be used to determine whether the means self-esteem scores are significantly different between the three time points.

Summary statistics

Compute some summary statistics of the self-esteem score by groups (time): mean and sd (standard deviation)

selfesteem %>%
  group_by(time) %>%
  get_summary_stats(score, type = "mean_sd")
## # A tibble: 3 x 5
##   time  variable     n  mean    sd
##   <fct> <chr>    <dbl> <dbl> <dbl>
## 1 t1    score       10  3.14 0.552
## 2 t2    score       10  4.93 0.863
## 3 t3    score       10  7.64 1.14

Visualization

Create a box plot and add points corresponding to individual values:

bxp <- ggboxplot(selfesteem, x = "time", y = "score", add = "point")
bxp

Check assumptions

Outliers

Outliers can be easily identified using box plot methods, implemented in the R function identify_outliers() [rstatix package].

selfesteem %>%
  group_by(time) %>%
  identify_outliers(score)
## # A tibble: 2 x 5
##   time  id    score is.outlier is.extreme
##   <fct> <fct> <dbl> <lgl>      <lgl>     
## 1 t1    6      2.05 TRUE       FALSE     
## 2 t2    2      6.91 TRUE       FALSE

There were no extreme outliers.

Note that, in the situation where you have extreme outliers, this can be due to: 1) data entry errors, measurement errors or unusual values.

You can include the outlier in the analysis anyway if you do not believe the result will be substantially affected. This can be evaluated by comparing the result of the ANOVA with and without the outlier.

It’s also possible to keep the outliers in the data and perform robust ANOVA test using the WRS2 package.

Normality assumption

The normality assumption can be checked by computing Shapiro-Wilk test for each time point. If the data is normally distributed, the p-value should be greater than 0.05.

selfesteem %>%
  group_by(time) %>%
  shapiro_test(score)
## # A tibble: 3 x 4
##   time  variable statistic     p
##   <fct> <chr>        <dbl> <dbl>
## 1 t1    score        0.967 0.859
## 2 t2    score        0.876 0.117
## 3 t3    score        0.923 0.380

The self-esteem score was normally distributed at each time point, as assessed by Shapiro-Wilk’s test (p > 0.05).

Note that, if your sample size is greater than 50, the normal QQ plot is preferred because at larger sample sizes the Shapiro-Wilk test becomes very sensitive even to a minor deviation from normality.

QQ plot draws the correlation between a given data and the normal distribution. Create QQ plots for each time point:

ggqqplot(selfesteem, "score", facet.by = "time")

From the plot above, as all the points fall approximately along the reference line, we can assume normality.

Assumption of sphericity

As mentioned in previous sections, the assumption of sphericity will be automatically checked during the computation of the ANOVA test using the R function anova_test() [rstatix package]. The Mauchly’s test is internally used to assess the sphericity assumption.

By using the function get_anova_table() [rstatix] to extract the ANOVA table, the Greenhouse-Geisser sphericity correction is automatically applied to factors violating the sphericity assumption.

Computation

res.aov <- anova_test(data = selfesteem, dv = score, wid = id, within = time)
get_anova_table(res.aov)
## ANOVA Table (type III tests)
## 
##   Effect DFn DFd    F        p p<.05   ges
## 1   time   2  18 55.5 2.01e-08     * 0.829

The self-esteem score was statistically significantly different at the different time points during the diet, F(2, 18) = 55.5, p < 0.0001, eta2[g] = 0.83.

where,

  • F Indicates that we are comparing to an F-distribution (F-test); (2, 18) indicates the degrees of freedom in the numerator (DFn) and the denominator (DFd), respectively; 55.5 indicates the obtained F-statistic value
  • p specifies the p-value
  • ges is the generalized effect size (amount of variability due to the within-subjects factor)

Post-hoc tests

You can perform multiple pairwise paired t-tests between the levels of the within-subjects factor (here time). P-values are adjusted using the Bonferroni multiple testing correction method.

# pairwise comparisons
pwc <- selfesteem %>%
  pairwise_t_test(
    score ~ time, paired = TRUE,
    p.adjust.method = "bonferroni"
    )
pwc
## # A tibble: 3 x 10
##   .y.   group1 group2    n1    n2 statistic    df           p    p.adj p.adj.signif
## * <chr> <chr>  <chr>  <int> <int>     <dbl> <dbl>       <dbl>    <dbl> <chr>       
## 1 score t1     t2        10    10     -4.97     9 0.000772    0.002    **          
## 2 score t1     t3        10    10    -13.2      9 0.000000334 0.000001 ****        
## 3 score t2     t3        10    10     -4.87     9 0.000886    0.003    **

All the pairwise differences are statistically significant.

Report

We could report the result as follow:

The self-esteem score was statistically significantly different at the different time points, F(2, 18) = 55.5, p < 0.0001, generalized eta squared = 0.82.

Post-hoc analyses with a Bonferroni adjustment revealed that all the pairwise differences, between time points, were statistically significantly different (p <= 0.05).

# Visualization: box plots with p-values
pwc <- pwc %>% add_xy_position(x = "time")
bxp + 
  stat_pvalue_manual(pwc) +
  labs(
    subtitle = get_test_label(res.aov, detailed = TRUE),
    caption = get_pwc_label(pwc)
  )

Two-way repeated measures ANOVA

Data preparation

We’ll use the selfesteem2 dataset [in datarium package] containing the self-esteem score measures of 12 individuals enrolled in 2 successive short-term trials (4 weeks): control (placebo) and special diet trials.

Each participant performed all two trials. The order of the trials was counterbalanced and sufficient time was allowed between trials to allow any effects of previous trials to have dissipated.

The self-esteem score was recorded at three time points: at the beginning (t1), midway (t2) and at the end (t3) of the trials.

The question is to investigate if this short-term diet treatment can induce a significant increase of self-esteem score over time. In other terms, we wish to know if there is significant interaction between diet and time on the self-esteem score.

The two-way repeated measures ANOVA can be performed in order to determine whether there is a significant interaction between diet and time on the self-esteem score.

Load and show one random row by treatment group:

# Wide format
set.seed(123)
data("selfesteem2", package = "datarium")
selfesteem2 %>% sample_n_by(treatment, size = 1)
## # A tibble: 2 x 5
##   id    treatment    t1    t2    t3
##   <fct> <fct>     <dbl> <dbl> <dbl>
## 1 4     ctr          92    92    89
## 2 10    Diet         90    93    95
# Gather the columns t1, t2 and t3 into long format.
# Convert id and time into factor variables
selfesteem2 <- selfesteem2 %>%
  gather(key = "time", value = "score", t1, t2, t3) %>%
  convert_as_factor(id, time)
# Inspect some random rows of the data by groups
set.seed(123)
selfesteem2 %>% sample_n_by(treatment, time, size = 1)
## # A tibble: 6 x 4
##   id    treatment time  score
##   <fct> <fct>     <fct> <dbl>
## 1 4     ctr       t1       92
## 2 10    ctr       t2       84
## 3 5     ctr       t3       68
## 4 11    Diet      t1       93
## 5 12    Diet      t2       80
## 6 1     Diet      t3       88

In this example, the effect of “time” on self-esteem score is our focal variable, that is our primary concern.

However, it is thought that the effect “time” will be different if treatment is performed or not. In this setting, the “treatment” variable is considered as moderator variable.

Summary statistics

Group the data by treatment and time, and then compute some summary statistics of the score variable: mean and sd (standard deviation).

selfesteem2 %>%
  group_by(treatment, time) %>%
  get_summary_stats(score, type = "mean_sd")
## # A tibble: 6 x 6
##   treatment time  variable     n  mean    sd
##   <fct>     <fct> <chr>    <dbl> <dbl> <dbl>
## 1 ctr       t1    score       12  88    8.08
## 2 ctr       t2    score       12  83.8 10.2 
## 3 ctr       t3    score       12  78.7 10.5 
## 4 Diet      t1    score       12  87.6  7.62
## 5 Diet      t2    score       12  87.8  7.42
## 6 Diet      t3    score       12  87.7  8.14

Visualization

Create box plots of the score colored by treatment groups:

bxp <- ggboxplot(
  selfesteem2, x = "time", y = "score",
  color = "treatment", palette = "jco"
  )
bxp

Chek assumptions

Outliers

selfesteem2 %>%
  group_by(treatment, time) %>%
  identify_outliers(score)
## [1] treatment  time       id         score      is.outlier is.extreme
## <0 rows> (or 0-length row.names)

There were no extreme outliers.

Normality assumption

Compute Shapiro-Wilk test for each combinations of factor levels:

selfesteem2 %>%
  group_by(treatment, time) %>%
  shapiro_test(score)
## # A tibble: 6 x 5
##   treatment time  variable statistic      p
##   <fct>     <fct> <chr>        <dbl>  <dbl>
## 1 ctr       t1    score        0.828 0.0200
## 2 ctr       t2    score        0.868 0.0618
## 3 ctr       t3    score        0.887 0.107 
## 4 Diet      t1    score        0.919 0.279 
## 5 Diet      t2    score        0.923 0.316 
## 6 Diet      t3    score        0.886 0.104

The self-esteem score was normally distributed at each time point (p > 0.05), except for ctr treatment at t1, as assessed by Shapiro-Wilk’s test.

Create QQ plot for each cell of design:

ggqqplot(selfesteem2, "score", ggtheme = theme_bw()) +
  facet_grid(time ~ treatment, labeller = "label_both")

From the plot above, as all the points fall approximately along the reference line, we can assume normality.



Computation

res.aov <- anova_test(
  data = selfesteem2, dv = score, wid = id,
  within = c(treatment, time)
  )
get_anova_table(res.aov)
## ANOVA Table (type III tests)
## 
##           Effect  DFn  DFd    F        p p<.05   ges
## 1      treatment 1.00 11.0 15.5 2.00e-03     * 0.059
## 2           time 1.31 14.4 27.4 5.03e-05     * 0.049
## 3 treatment:time 2.00 22.0 30.4 4.63e-07     * 0.050

There is a statistically significant two-way interactions between treatment and time, F(2, 22) = 30.4, p < 0.0001.

Post-hoc tests

A significant two-way interaction indicates that the impact that one factor (e.g., treatment) has on the outcome variable (e.g., self-esteem score) depends on the level of the other factor (e.g., time) (and vice versa). So, you can decompose a significant two-way interaction into:

  • Simple main effect: run one-way model of the first variable (factor A) at each level of the second variable (factor B),
  • Simple pairwise comparisons: if the simple main effect is significant, run multiple pairwise comparisons to determine which groups are different.

For a non-significant two-way interaction, you need to determine whether you have any statistically significant main effects from the ANOVA output.

Procedure for a significant two-way interaction

Effect of treatment. In our example, we’ll analyze the effect of treatment on self-esteem score at every time point.

Note that, the treatment factor variable has only two levels (“ctr” and “Diet”); thus, ANOVA test and paired t-test will give the same p-values.

# Effect of treatment at each time point
one.way <- selfesteem2 %>%
  group_by(time) %>%
  anova_test(dv = score, wid = id, within = treatment) %>%
  get_anova_table() %>%
  adjust_pvalue(method = "bonferroni")
one.way
## # A tibble: 3 x 9
##   time  Effect      DFn   DFd      F       p `p<.05`      ges   p.adj
##   <fct> <chr>     <dbl> <dbl>  <dbl>   <dbl> <chr>      <dbl>   <dbl>
## 1 t1    treatment     1    11  0.376 0.552   ""      0.000767 1      
## 2 t2    treatment     1    11  9.03  0.012   *       0.052    0.036  
## 3 t3    treatment     1    11 30.9   0.00017 *       0.199    0.00051
# Pairwise comparisons between treatment groups
pwc <- selfesteem2 %>%
  group_by(time) %>%
  pairwise_t_test(
    score ~ treatment, paired = TRUE,
    p.adjust.method = "bonferroni"
    )
pwc
## # A tibble: 3 x 11
##   time  .y.   group1 group2    n1    n2 statistic    df       p   p.adj p.adj.signif
## * <fct> <chr> <chr>  <chr>  <int> <int>     <dbl> <dbl>   <dbl>   <dbl> <chr>       
## 1 t1    score ctr    Diet      12    12     0.613    11 0.552   0.552   ns          
## 2 t2    score ctr    Diet      12    12    -3.00     11 0.012   0.012   *           
## 3 t3    score ctr    Diet      12    12    -5.56     11 0.00017 0.00017 ***

Considering the Bonferroni adjusted p-value (p.adj), it can be seen that the simple main effect of treatment was not significant at the time point t1 (p = 1). It becomes significant at t2 (p = 0.036) and t3 (p = 0.00051).

Pairwise comparisons show that the mean self-esteem score was significantly different between ctr and Diet group at t2 (p = 0.12) and t3 (p = 0.00017) but not at t1 (p = 0.55).

Effect of time. Note that, it’s also possible to perform the same analysis for the time variable at each level of treatment. You don’t necessarily need to do this analysis.

The R code:

# Effect of time at each level of treatment
one.way2 <- selfesteem2 %>%
  group_by(treatment) %>%
  anova_test(dv = score, wid = id, within = time) %>%
  get_anova_table() %>%
  adjust_pvalue(method = "bonferroni")
# Pairwise comparisons between time points
pwc2 <- selfesteem2 %>%
  group_by(treatment) %>%
  pairwise_t_test(
    score ~ time, paired = TRUE,
    p.adjust.method = "bonferroni"
    )
pwc2

After executing the R code above, you can see that the effect of time is significant only for the control trial, F(2, 22) = 39.7, p < 0.0001. Pairwise comparisons show that all comparisons between time points were statistically significant for control trial.

Procedure for non-significant two-way interaction

If the interaction is not significant, you need to interpret the main effects for each of the two variables: treatment and time. A significant main effect can be followed up with pairwise comparisons.

In our example (see ANOVA table in res.aov), there was a statistically significant main effects of treatment (F(1, 11) = 15.5, p = 0.002) and time (F(2, 22) = 27.4, p < 0.0001) on the self-esteem score.

Pairwise paired t-test comparisons:

# comparisons for treatment variable
selfesteem2 %>%
  pairwise_t_test(
    score ~ treatment, paired = TRUE, 
    p.adjust.method = "bonferroni"
    )
# comparisons for time variable
selfesteem2 %>%
  pairwise_t_test(
    score ~ time, paired = TRUE, 
    p.adjust.method = "bonferroni"
    )

All pairwise comparisons are significant.

Report

We could report the result as follow:

A two-way repeated measures ANOVA was performed to evaluate the effect of different diet treatments over time on self-esteem score.

There was a statistically significant interaction between treatment and time on self-esteem score, F(2, 22) = 30.4, p < 0.0001. Therefore, the effect of treatment variable was analyzed at each time point. P-values were adjusted using the Bonferroni multiple testing correction method. The effect of treatment was significant at t2 (p = 0.036) and t3 (p = 0.00051) but not at the time point t1 (p = 1).

Pairwise comparisons, using paired t-test, show that the mean self-esteem score was significantly different between ctr and Diet trial at time points t2 (p = 0.012) and t3 (p = 0.00017) but not at t1 (p = 0.55).

# Visualization: box plots with p-values
pwc <- pwc %>% add_xy_position(x = "time")
bxp + 
  stat_pvalue_manual(pwc, tip.length = 0, hide.ns = TRUE) +
  labs(
    subtitle = get_test_label(res.aov, detailed = TRUE),
    caption = get_pwc_label(pwc)
  )

Three-way repeated measures ANOVA

Data preparation

We’ll use the weightloss dataset [datarium package]. In this study, a researcher wanted to assess the effects of Diet and Exercises on weight loss in 10 sedentary individuals.

The participants were enrolled in four trials: (1) no diet and no exercises; (2) diet only; (3) exercises only; and (4) diet and exercises combined.

Each participant performed all four trials. The order of the trials was counterbalanced and sufficient time was allowed between trials to allow any effects of previous trials to have dissipated.

Each trial lasted nine weeks and the weight loss score was measured at the beginning (t1), at the midpoint (t2) and at the end (t3) of each trial.

Three-way repeated measures ANOVA can be performed in order to determine whether there is a significant interaction between diet, exercises and time on the weight loss score.

Load the data and show some random rows by groups:

# Wide format
set.seed(123)
data("weightloss", package = "datarium")
weightloss %>% sample_n_by(diet, exercises, size = 1)
## # A tibble: 4 x 6
##   id    diet  exercises    t1    t2    t3
##   <fct> <fct> <fct>     <dbl> <dbl> <dbl>
## 1 4     no    no         11.1   9.5  11.1
## 2 10    no    yes        10.2  11.8  17.4
## 3 5     yes   no         11.6  13.4  13.9
## 4 11    yes   yes        12.7  12.7  15.1
# Gather the columns t1, t2 and t3 into long format.
# Convert id and time into factor variables
weightloss <- weightloss %>%
  gather(key = "time", value = "score", t1, t2, t3) %>%
  convert_as_factor(id, time)
# Inspect some random rows of the data by groups
set.seed(123)
weightloss %>% sample_n_by(diet, exercises, time, size = 1)
## # A tibble: 12 x 5
##   id    diet  exercises time  score
##   <fct> <fct> <fct>     <fct> <dbl>
## 1 4     no    no        t1     11.1
## 2 10    no    no        t2     10.7
## 3 5     no    no        t3     12.3
## 4 11    no    yes       t1     10.2
## 5 12    no    yes       t2     13.2
## 6 1     no    yes       t3     15.8
## # … with 6 more rows

In this example, the effect of the “time” is our focal variable, that is our primary concern.

It is thought that the effect of “time” on the weight loss score will depend on two other factors, “diet” and “exercises”, which are called moderator variables.

Summary statistics

Group the data by diet, exercises and time, and then compute some summary statistics of the score variable: mean and sd (standard deviation)

weightloss %>%
  group_by(diet, exercises, time) %>%
  get_summary_stats(score, type = "mean_sd")
## # A tibble: 12 x 7
##   diet  exercises time  variable     n  mean    sd
##   <fct> <fct>     <fct> <chr>    <dbl> <dbl> <dbl>
## 1 no    no        t1    score       12  10.9 0.868
## 2 no    no        t2    score       12  11.6 1.30 
## 3 no    no        t3    score       12  11.4 0.935
## 4 no    yes       t1    score       12  10.8 1.27 
## 5 no    yes       t2    score       12  13.4 1.01 
## 6 no    yes       t3    score       12  16.8 1.53 
## # … with 6 more rows

Visualization

Create box plots:

bxp <- ggboxplot(
  weightloss, x = "exercises", y = "score",
  color = "time", palette = "jco",
  facet.by = "diet", short.panel.labs = FALSE
  )
bxp

Check assumptions

Outliers

weightloss %>%
  group_by(diet, exercises, time) %>%
  identify_outliers(score)
## # A tibble: 5 x 7
##   diet  exercises time  id    score is.outlier is.extreme
##   <fct> <fct>     <fct> <fct> <dbl> <lgl>      <lgl>     
## 1 no    no        t3    2      13.2 TRUE       FALSE     
## 2 yes   no        t1    1      10.2 TRUE       FALSE     
## 3 yes   no        t1    3      13.2 TRUE       FALSE     
## 4 yes   no        t1    4      10.2 TRUE       FALSE     
## 5 yes   no        t2    10     15.3 TRUE       FALSE

There were no extreme outliers.

Normality assumption

Compute Shapiro-Wilk test for each combinations of factor levels:

weightloss %>%
  group_by(diet, exercises, time) %>%
  shapiro_test(score)
## # A tibble: 12 x 6
##   diet  exercises time  variable statistic     p
##   <fct> <fct>     <fct> <chr>        <dbl> <dbl>
## 1 no    no        t1    score        0.917 0.264
## 2 no    no        t2    score        0.957 0.743
## 3 no    no        t3    score        0.965 0.851
## 4 no    yes       t1    score        0.922 0.306
## 5 no    yes       t2    score        0.912 0.229
## 6 no    yes       t3    score        0.953 0.674
## # … with 6 more rows

The weight loss score was normally distributed, as assessed by Shapiro-Wilk’s test of normality (p > .05).

Create QQ plot for each cell of design:

ggqqplot(weightloss, "score", ggtheme = theme_bw()) +
  facet_grid(diet + exercises ~ time, labeller = "label_both")

From the plot above, as all the points fall approximately along the reference line, we can assume normality.

Computation

res.aov <- anova_test(
  data = weightloss, dv = score, wid = id,
  within = c(diet, exercises, time)
  )
get_anova_table(res.aov)
## ANOVA Table (type III tests)
## 
##                Effect  DFn  DFd       F        p p<.05   ges
## 1                diet 1.00 11.0   6.021 3.20e-02     * 0.028
## 2           exercises 1.00 11.0  58.928 9.65e-06     * 0.284
## 3                time 2.00 22.0 110.942 3.22e-12     * 0.541
## 4      diet:exercises 1.00 11.0  75.356 2.98e-06     * 0.157
## 5           diet:time 1.38 15.2   0.603 5.01e-01       0.013
## 6      exercises:time 2.00 22.0  20.826 8.41e-06     * 0.274
## 7 diet:exercises:time 2.00 22.0  14.246 1.07e-04     * 0.147

From the output above, it can be seen that there is a statistically significant three-way interactions between diet, exercises and time, F(2, 22) = 14.24, p = 0.00011.

Note that, if the three-way interaction is not statistically significant, you need to consult the two-way interactions in the output.

In our example, there was a statistically significant two-way diet:exercises interaction (p < 0.0001), and two-way exercises:time (p < 0.0001). The two-way diet:time interaction was not statistically significant (p = 0.5).

Post-hoc tests

If there is a significant three-way interaction effect, you can decompose it into:

  • Simple two-way interaction: run two-way interaction at each level of third variable,
  • Simple simple main effect: run one-way model at each level of second variable, and
  • simple simple pairwise comparisons: run pairwise or other post-hoc comparisons if necessary.

Compute simple two-way interaction

You are free to decide which two variables will form the simple two-way interactions and which variable will act as the third (moderator) variable. In the following R code, we have considered the simple two-way interaction of exercises*time at each level of diet.

Group the data by diet and analyze the simple two-way interaction between exercises and time:

# Two-way ANOVA at each diet level
two.way <- weightloss %>%
  group_by(diet) %>%
  anova_test(dv = score, wid = id, within = c(exercises, time))
two.way
## # A tibble: 2 x 2
##   diet  anova     
##   <fct> <list>    
## 1 no    <anov_tst>
## 2 yes   <anov_tst>
# Extract anova table
get_anova_table(two.way)
## # A tibble: 6 x 8
##   diet  Effect           DFn   DFd     F        p `p<.05`   ges
##   <fct> <chr>          <dbl> <dbl> <dbl>    <dbl> <chr>   <dbl>
## 1 no    exercises          1    11 72.8  3.53e- 6 *       0.526
## 2 no    time               2    22 71.7  2.32e-10 *       0.587
## 3 no    exercises:time     2    22 28.9  6.92e- 7 *       0.504
## 4 yes   exercises          1    11 13.4  4.00e- 3 *       0.038
## 5 yes   time               2    22 20.5  9.30e- 6 *       0.49 
## 6 yes   exercises:time     2    22  2.57 9.90e- 2 ""      0.06

There was a statistically significant simple two-way interaction between exercises and time for “diet no” trial, F(2, 22) = 28.9, p < 0.0001, but not for “diet yes” trial, F(2, 22) = 2.6, p = 0.099.

Note that, it’s recommended to adjust the p-value for multiple testing. One common approach is to apply a Bonferroni adjustment to downgrade the level at which you declare statistical significance.

This can be done by dividing the current level you declare statistical significance at (i.e., p < 0.05) by the number of simple two-way interaction you are computing (i.e., 2).

Thus, you only declare a two-way interaction as statistically significant when p < 0.025 (i.e., p < 0.05/2). Applying this to our current example, we would still make the same conclusions.

Compute simple simple main effect

A statistically significant simple two-way interaction can be followed up with simple simple main effects.

In our example, you could therefore investigate the effect of time on the weight loss score at every level of exercises or investigate the effect of exercises at every level of time.

You will only need to consider the result of the simple simple main effect analyses for the “diet no” trial as this was the only simple two-way interaction that was statistically significant (see previous section).

Group the data by diet and exercises, and analyze the simple main effect of time. The Bonferroni adjustment will be considered leading to statistical significance being accepted at the p < 0.025 level (that is 0.05 divided by the number of tests (here 2) considered for “diet:no” trial.

# Effect of time at each diet X exercises cells
time.effect <- weightloss %>%
  group_by(diet, exercises) %>%
  anova_test(dv = score, wid = id, within = time)
time.effect
## # A tibble: 4 x 3
##   diet  exercises anova     
##   <fct> <fct>     <list>    
## 1 no    no        <anov_tst>
## 2 no    yes       <anov_tst>
## 3 yes   no        <anov_tst>
## 4 yes   yes       <anov_tst>
# Extract anova table
get_anova_table(time.effect) %>%
  filter(diet == "no")
## # A tibble: 2 x 9
##   diet  exercises Effect   DFn   DFd     F        p `p<.05`   ges
##   <fct> <fct>     <chr>  <dbl> <dbl> <dbl>    <dbl> <chr>   <dbl>
## 1 no    no        time       2    22  1.32 2.86e- 1 ""      0.075
## 2 no    yes       time       2    22 78.8  9.30e-11 *       0.801

There was a statistically significant simple simple main effect of time on weight loss score for “diet:no,exercises:yes” group (p < 0.0001), but not for when neither diet nor exercises was performed (p = 0.286).

Compute simple simple comparisons

A statistically significant simple simple main effect can be followed up by multiple pairwise comparisons to determine which group means are different.

Group the data by diet and exercises, and perform pairwise comparisons between time points with Bonferroni adjustment:

# Pairwise comparisons
pwc <- weightloss %>%
  group_by(diet, exercises) %>%
  pairwise_t_test(score ~ time, paired = TRUE, p.adjust.method = "bonferroni") %>%
  select(-df, -statistic) # Remove details
# Show comparison results for "diet:no,exercises:yes" groups
pwc %>% filter(diet == "no", exercises == "yes") %>%
  select(-p)     # remove p columns
## # A tibble: 3 x 9
##   diet  exercises .y.   group1 group2    n1    n2        p.adj p.adj.signif
##   <fct> <fct>     <chr> <chr>  <chr>  <int> <int>        <dbl> <chr>       
## 1 no    yes       score t1     t2        12    12 0.000741     ***         
## 2 no    yes       score t1     t3        12    12 0.0000000121 ****        
## 3 no    yes       score t2     t3        12    12 0.000257     ***

In the pairwise comparisons table above, we are interested only in the simple simple comparisons for “diet:no,exercises:yes” groups. In our example, there are three possible combinations of group differences. We could report the pairwise comparison results as follow.

All simple simple pairwise comparisons were run between the different time points for “diet:no,exercises:yes” trial. The Bonferroni adjustment was applied. The mean weight loss score was significantly different in all time point comparisons when exercises are performed (p < 0.05).

Report

A three-way repeated measures ANOVA was performed to evaluate the effects of diet, exercises and time on weight loss. There was a statistically significant three-way interaction between diet, exercises and time, F(2, 22) = 14.2, p = 0.00011.

For the simple two-way interactions and simple simple main effects analyses, a Bonferroni adjustment was applied leading to statistical significance being accepted at the p < 0.025 level.

There was a statistically significant simple two-way interaction between exercises and time for “diet no” trial, F(2, 22) = 28.9, p < 0.0001, but not for “diet yes”" trial, F(2, 22) = 2.6, p = 0.099.

There was a statistically significant simple simple main effect of time on weight loss score for “diet:no,exercises:yes” trial (p < 0.0001), but not for when neither diet nor exercises was performed (p = 0.286).

All simple simple pairwise comparisons were run between the different time points for “diet:no,exercises:yes” trial with a Bonferroni adjustment applied. The mean weight loss score was significantly different in all time point comparisons when exercises are performed (p < 0.05).

# Visualization: box plots with p-values
pwc <- pwc %>% add_xy_position(x = "exercises")
pwc.filtered <- pwc %>% 
  filter(diet == "no", exercises == "yes")
bxp + 
  stat_pvalue_manual(pwc.filtered, tip.length = 0, hide.ns = TRUE) +
  labs(
    subtitle = get_test_label(res.aov, detailed = TRUE),
    caption = get_pwc_label(pwc)
  )

Summary

This chapter describes how to compute, interpret and report repeated measures ANOVA in R. We also explain the assumptions made by repeated measures ANOVA tests and provide practical examples of R codes to check whether the test assumptions are met.



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Comments ( 60 )

  • Hanna
    10 Feb 2020

    What an excellent resource! Thank you for providing this website to the public. I will recommend it to my students, too.

    Reply
    • Kassambara
      10 Feb 2020

      Thank you for your positive feedback, highly appreciated!

      Reply
  • Tom
    17 Feb 2020

    Thank you so much, helps a lot! One thing that might be relevant for some people is when to use between subjects instead of within subjects. But, the main thing is that this is the best source for doing these types of analyses I could find on the web, so great job!

    Reply
    • Kassambara
      17 Feb 2020

      Thank you for your positive feedback! I’ll take your suggestion into account in the next update

      Reply
  • TomO
    27 Feb 2020

    Hi! Thanks for such a wonderfully written intro to the topic with easy to read R code – a breath of fresh air!

    I had a question about the following section:

    “It’s also possible to keep the outliers in the data and perform robust ANOVA test using the WRS2 package.”

    I’ve been reading the Mair et al (2018) paper that introduces WRS2 titled “Robust Statistical Methods Using WRS2” and specifically section 6.1 on Repeated Measurement and Mixed ANOVA Designs that introduces the function I believe you are referring to (rmanova). However, the authors state that the function provides a “robust heteroscedastic repeated measurement ANOVA based on the trimmed means” which defaults to 20%. Would this trimmed mean not remove the outliers, rather than keep them?

    I’d be grateful for any help you could offer, maybe I’ve got the wrong function or the wrong end of the stick?

    Cheers!

    Reply
  • Richard
    01 Mar 2020

    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases

    Reply
    • Kassambara
      01 Mar 2020

      Hi, could you please provide a reproducible example?

      Reply
      • Anton
        09 Mar 2020

        I am also having this problem, though the dataset you use in the example works fine :(.

        Data:

        id treatment t1 t2 t3
        1 Diet 3 7 7
        2 Diet 5 7 7
        3 Diet 2 7 8
        4 Diet 4 3 4
        5 Diet 5 4 4
        6 Diet 3 4 4
        7 Diet 5 9 9
        8 Diet 3 4 4
        9 Diet 3 8 8
        10 Diet 5 4 4
        147 ctr 5 3 3
        148 ctr 4 6 6
        149 ctr 7 8 8
        150 ctr 3 2 2
        151 ctr 7 4 4
        152 ctr 3 6 6
        153 ctr 5 5 5
        154 ctr 4 7 7
        155 ctr 5 1 1
        157 ctr 6 5 5

        zna <- read.csv("zero_na.csv")
        zna % gather(key = “time”, value = “score”, t1, t2, t3) %>%
        convert_as_factor(id, time)

        res.aov <- anova_test(
        data = zna, dv = score, wid = id,
        within = c(treatment, time)
        )
        get_anova_table(res.aov)

        Reply
    • Jake
      09 Mar 2020

      I got this error message when I tried to do two-way anova. My case was different from the example because I had different participants in all (three) groups; I think this is why I got an error.

      Eventually, I analysed my data using lme function and setting a unique random effect to each participant.

      Reply
  • May
    02 Mar 2020

    Hello! Thank’s a lot for this article, it helps a lot. I’m using three-way anova for my dataset and when I run function ‘anova_test’ (opy from this text here) i get an error: ‘Each row of output must be identified by a unique combination of keys.
    Keys are shared for 715 rows:’. Maybe you could help and tell me what’s going on?

    My function looks like:
    aov <- anova_test(data=rtdata, dv=rt, wid=part, within=c(mod_sm, con, resp))
    where:
    rtdata – data in long format
    part – factor, participant code from 1 to 12; each participant got 60 trials, so his code repeats 60 times in dataset
    mod_sm – text, levels 'pos' and 'neg'
    con – text, 'n' and 'y'
    resp – factor, 1 and 2

    Thank you!

    Reply
    • Marie
      19 Jun 2020

      Hey May, did you manage to figure out how to fix this issue? I have exactly the same issue and my data structure is similar.

      Reply
  • Anton
    09 Mar 2020

    I have been trying to do a two-way repeated samples and spent many hours until I worked out what was wrong – my control and treatment groups aren’t the same, so don’t have the same ids repeated. So it is not “ctr” followed by “Diet” for the same subjects but separate subject

    Can you comment on how this would work? What kind of test would it be? Thanks!

    Reply
  • Odi Makov
    24 Mar 2020

    Very informative!
    I’ll recommend this guide everywhere

    Reply
    • Kassambara
      24 Mar 2020

      Thank you for your positive feedback, highly appreciated!

      Reply
  • Hannah PW
    16 Apr 2020

    Hi, I’ve been trying for days to run the 3-way repeated measures ANOVA for my dataset in a long format with 2x treatment groups (temp & region) and then a time group (T0, T1, T2, T3). Whenever I run the ANOVA, I get the same error message that has been mentioned above and in many other R forums (StackExchange, etc.), although I’ve not as yet seen a solution as my data does not contain any null data and all my groups are in factor format (I’ve also tried them in numeric format without luck):

    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases

    Can you help or offer any suggestions that I might not have tried?

    Reply
      • Hans Bourgeois
        24 Apr 2020

        Hi Kassambara,

        Thank you for your work.

        It seems there are a conflict with the packtage “tidyverse” and “rstatix”.
        This conflict may explain the errors :
        – “Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
        0 (non-NA) cases”
        – “Error: Unknown column `ID`”

        To remedy this i use the package “conflict_prefer”.

        Moreover, the first error also appears when i have unused columns (Factor or Numeric) in my dataframe. When i remove unused columns, the anova_test work fine.

        Reply
  • Jason Weber
    29 Apr 2020

    Hi great article!
    Tried to produce graph report of one-way anova and error at bottom of code came up.
    PS_tried to install pubr but got nothing.
    Thx JW

    # Repeated One-Way ANOVA
    res.aov <- anova_test(data = V3, dv=Data, wid=Name, within = Rep)
    get_anova_table(res.aov)

    # pairwise comparisons
    pwc %
    pairwise_t_test(
    Data ~ Rep, paired = TRUE,
    p.adjust.method = “bonferroni”
    )
    pwc

    # Visualization: box plots with p-values
    pwc % add_xy_position(x = “Rep”)
    bxp +
    stat_pvalue_manual(pwc) +
    labs(
    subtitle = get_test_label(res.aov, detailed = TRUE),
    caption = get_pwc_label(pwc)

    Error in bxp + stat_pvalue_manual(pwc) + labs(subtitle = get_test_label(res.aov, :
    non-numeric argument to binary operator

    Reply
    • Kassambara
      29 Apr 2020

      Hi,

      It should be possible to install pubr using `devtools::install_github(“kassambara/pubr”)`.

      In order efficiently I need a reproducible example with a demo data.

      Thanks

      Reply
  • Elizabeth
    04 May 2020

    Really great examples that are well explained.

    My issue is when trying to visualize the final box plot with the p-values for any of the examples. I receive the following error each time:

    Error in f(…) :
    Can only handle data with groups that are plotted on the x-axis
    In addition: Warning messages:
    1: Ignoring unknown parameters: hide.ns
    2: Ignoring unknown aesthetics: xmin, xmax, annotations, y_position

    Please advise and what the issue may be. Thank you

    Reply
    • Kassambara
      04 May 2020

      Please make sure you have the latest version of the rstatix and the ggpubr packages

      Reply
  • Qian Yang
    05 May 2020

    Hi Kassambara,
    Thanks a lot for your input, it’s really helpful. I have one question that the output of two-way repeated ANOVA analysis in R is not consistent with the results obtained in SPSS, what kind of reason do you think can explain this difference? thanks in advance.

    Reply
    • Kassambara
      06 May 2020

      Hi!

      After checking, I found that the output of SPSS and rstatix are the same when considering the rstatix raw output (res.aov).

      When displaying the ANOVA table using the function get_anova_table(res.aov), the sphericity correction is automatically applied when the sphericity can’t be assumed.

      The sphericity assumption can be checked using the Mauchly’s test of sphericity, which is automatically reported when using the R function rstatix::anova_test()

      library(rstatix)

# Data preparation
data("selfesteem2", package = "datarium")
selfesteem2 <- selfesteem2 %>%
  gather(key = "time", value = "score", t1, t2, t3) %>%
  convert_as_factor(id, time)
head(selfesteem2, 3)
      ## # A tibble: 3 x 4
##   id    treatment time  score
##   <fct> <fct>     <fct> <dbl>
## 1 1     ctr       t1       83
## 2 2     ctr       t1       97
## 3 3     ctr       t1       93
      # Anova result
res.aov <- anova_test(
  data = selfesteem2, dv = score, wid = id,
  within = c(treatment, time)
)

# Display raw results
res.aov
      ## ANOVA Table (type III tests)
## 
## $ANOVA
##           Effect DFn DFd      F        p p<.05   ges
## 1      treatment   1  11 15.541 2.00e-03     * 0.059
## 2           time   2  22 27.369 1.08e-06     * 0.049
## 3 treatment:time   2  22 30.424 4.63e-07     * 0.050
## 
## $`Mauchly's Test for Sphericity`
##           Effect     W     p p<.05
## 1           time 0.469 0.023     *
## 2 treatment:time 0.616 0.089      
## 
## $`Sphericity Corrections`
##           Effect   GGe      DF[GG]    p[GG] p[GG]<.05   HFe      DF[HF]    p[HF] p[HF]<.05
## 1           time 0.653 1.31, 14.37 5.03e-05         * 0.705 1.41, 15.52 2.81e-05         *
## 2 treatment:time 0.723  1.45, 15.9 1.25e-05         * 0.803 1.61, 17.66 4.82e-06         *
      # Display ANOVA table
# Sphericity correction is applied by default if 
# the the Mauchly's test is significant
# Read this: https://www.datanovia.com/en/lessons/mauchlys-test-of-sphericity-in-r/
get_anova_table(res.aov)
      ## ANOVA Table (type III tests)
## 
##           Effect  DFn   DFd      F        p p<.05   ges
## 1      treatment 1.00 11.00 15.541 2.00e-03     * 0.059
## 2           time 1.31 14.37 27.369 5.03e-05     * 0.049
## 3 treatment:time 2.00 22.00 30.424 4.63e-07     * 0.050
      # No correction
get_anova_table(res.aov, correction = "none")
      ## ANOVA Table (type III tests)
## 
##           Effect DFn DFd      F        p p<.05   ges
## 1      treatment   1  11 15.541 2.00e-03     * 0.059
## 2           time   2  22 27.369 1.08e-06     * 0.049
## 3 treatment:time   2  22 30.424 4.63e-07     * 0.050
      Reply
  • Leon
    27 May 2020

    Hey, great tutorial! This is by far the easiest and most understandable approach I have found.
    I am working on my bachelor thesis and using R instead of SPSS. I need to calculate several 2 way repeated Anovas and rstatix seems more than handy atm. However, I cannot find any explanation of the output: What is the value of DFd? I calculated a oneway repeated Anova with 1 factor of 3 levels. DFn seems to be the degree of freedom between the levels (2) and DFd’s value is 30.
    I have 16 participants and I dont know how the degrees of freedom could go up this far… Though, my statistics courses were 2 years ago…

    Thanks in advance!

    Reply
    • Kassambara
      28 May 2020

      Hi, thank you for the positive feedback!

      DFn is the Degrees of Freedom in the numerator (i.e. DF for the effect of the experiment); DFd Degrees of Freedom in the denominator (i.e., DF for the residual, DFd Degrees of Freedom in the denominator (i.e., DF error)). This is explained in the documentation of anova_summary(), but obviously, we should clarify it in the current blog post.

      If only one factor is repeated measures, the number of degrees of freedom (DFd) can be calculated as (n-1)(a-1), where n is the number of participants and a is the number of levels of the repeated measures factor.

      If two factors are repeated measures, the number of degrees of freedom (DFd) for the two-way interaction term can be calculated as (n-1)(a-1)(b-1) where n is the number of participants, a is the number of levels one factor, and b is the number of levels of the other factor. Note that, a sphericity correction is applied, when the sphericity assumtion is not met, resulting in a fractional degres of freedom.

      Reply
      • Leon
        28 May 2020

        Thank you so much for the fast response!
        (16-1) (3-2) = 30 should be fine then.

        Due to the pandemic I was not able to test all participants and I am allowed to set the alpha level to 0.10 instead of 0.05. Is it possible to adjust it within your package, so that get_anova_table() marks p<.10 with * ?

        Even if not, your package worked great so far. Thanks again!

        Reply
        • Kassambara
          28 May 2020

          It’s possible to do something like this: 

          res <- get_anova_table(res.aov)
res[, "p<.05"] <- ifelse(res$p <= 0.1, "*", "")
res
          Reply
  • Gene Lu
    28 May 2020

    Hi,
    Thank you very much for your website! It helps a lot with my analysis works.
    I have one question, after running the rstatix::anova_test() code, it only shows the ANOVA table. I would like to check the sphericity assumption for my data but could not find it in the anova_test result. (I have tried to use your example data (selfesteem2) to do the analysis and can check the sphericity assumption in the ANOVA table. So I think the problem is from my data.) Do you have any idea what would be the possible reasons? (I run the two-wat rmANOVA) Thank you very much again!!

    Reply
    • Kassambara
      28 May 2020

      Note that the the Mauchly’s Test of Sphericity is only reported for variables or effects with >2 levels because sphericity necessarily holds for effects with only 2 levels Read more: Mauchly’s Test of Sphericity in R. Does your within-measures variable contains only two levels?

      Reply
  • Gene Lu
    29 May 2020

    Hi, thank you very much for your response! I found the solution from your answer in the previous question asked by Jack. My data is a mixed two-way ANOVA design (have 3 treatments). After changing the analysis method I get the Mauchly’s Test of Sphericity in my result. Thank you very much for your help!

    Reply
  • X32
    30 May 2020

    There are non-parametric method for 2/3 way repeated ANOVA. It’s WTS (Wald-Type Statistics), permuted WTS, ATS (ANOVA-Type Statistics) and ART (Aligned-Rank Transform) ANOVA. Both methods are available in R, not to mention Conover’s ANOVA (on “classic” ranks). Please check: nparLD, GFD, rankFD, ARTool and MANOVA packages.

    Reply
    • Kassambara
      30 May 2020

      Thank you for your input, very useful and instructive. Obviously, we’ll write on these packages!

      Reply
  • Sara
    08 Jun 2020

    Hi! Thanks for this beautiful and useful explanation.
    I have a question regarding treating outliers:

    You say that if outliers are NOT extreme you keep in the analysis right?

    If I want to remove outliers from my dataset and compute anova, which is the command to remove them?

    Thank you in advance.

    Sara

    Reply
    • Kassambara
      08 Jun 2020

      Thank you Sara, for your positive feedback, highly appreciated!

      Once you identified outliers, you can filter out the corresponding samples (before computing ANOVA), by using the R function dplyr::filter().

      Reply
  • Viper
    08 Jun 2020

    Excellent guide and your hard work is very much appreciated.
    I am getting the same error code as several others:
    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases

    I have tried using the Mixed ANOVA, but that will not run either. I also tried using the “conflict_prefer”as a poster suggested, but that does not help either. There are no NA’s in any of my data columns. I have three categorical IVs (speaker, complexity, format) and not sure where I am going wrong.

    I will try and post a reproducible R script here later.

    Thanks for the help and thanks for the excellent guides.

    Reply
  • Pi
    11 Jun 2020

    Great article! Only one aspect, a non-significant p-value for the Shapiro test doesn’t guarantee that the data is normally distributed. It may happen that the data you have is not enough to reject the null. The q-q plot is always a good double check as you did!

    Reply
    • Kassambara
      12 Jun 2020

      Thank you for your input!

      Reply
  • Daniel
    12 Jun 2020

    Hi,
    Thank for build so amazing tool!
    My question is: My data is not perfectly normal and I have some outliers, but results are consistent. In these cases, I am more calm if I can see model residuals. Any way to obtain anova_test residuals?

    Reply
    • Kassambara
      12 Jun 2020

      The returned object by the anova_test() function has an attribute called args, which is a list holding the arguments used to fit the ANOVA model, including: data, dv, within, between, type, model, etc.

      You can access to the residuals as follow: residuals(attr(res.aov, "args")$model)

      Reply
      • Daniel
        13 Jun 2020

        Are you the best? maybe… jejeje
        Thanks!

        Reply
  • Danny
    13 Jun 2020

    Hi, I’m trying to run Two-way repeated measures ANOVA but I kept getting this message:
    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases
    And I don’t see any missing data for my data. This is what I did:

    res.aov <- anova_test(
    data = Tab14_4, dv = Behavior, wid = Subject,
    within = c(Group, Interval)
    )

    Pls help!

    Reply
  • Peggy
    13 Jun 2020

    Hi, I’m also trying to run Two-way repeated measures ANOVA like Danny but I kept getting this message:
    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases
    And I don’t see any missing data for my data. This is what I did:

    anova_test(data = mydata2,dv = Score,wid = Subjects,within = c(Drinks,Gender))

    May I sent my demo to you?
    I really need your help!

    Reply
    • Kassambara
      14 Jun 2020

      Hi, In order to fix this issue I need a reproducible example with a data. Would you like to send your data with a reproducible R script?

      Reply
  • Lauren
    14 Jun 2020

    Hi there, I am also experiencing the same issue as Peggy and Danny. I have spent hours trying to get this to work (after it had successfully worked many times the other day) and have had no luck. It works perfectly fine with the data presented on this site, but is no longer working with my data. I also have no missing cases present within my dataset.

    res.aov <- anova_test(data = newanova, dv = score, wid = id, within = c(treatment, time))

    Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) :
    0 (non-NA) cases

    Please help!

    Reply
    • Kassambara
      14 Jun 2020

      Hi, In order to fix this issue I need a reproducible example with a data. Would you like to send your data with a reproducible R script?

      Reply
    • Laurianne
      19 Jun 2020

      Have you find a solution? I am in the same situation.

      Reply
  • Dongze chen
    18 Jun 2020

    If the data is an imbalance between groups, Does this still work?
    Cause in my data, the first group has 100 samples, the second group only has 10 samples
    I am looking forward to your reply! Thank you very much!

    Reply
  • Laurianne Schoenberger
    19 Jun 2020

    Hi ! Thank you so much for this explanation. I found this so helpful.
    However, I have some issues when doing this with my data. I removed the NA from my dataset. I tried to do as your “example 1: Reproducible R script using R built-in data” but I am not sure this works… I had error mesage while running “pubr::render_r()”.
    My independent variable is secondhand smoke (SHS yes or no) and cognitive test score at 2 points of time.

    When I compute the anova test. I got this message:
    “Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, …) : 0 (non-NA) cases”

    # Computation
    res.aov2 <- anova_test(data = ds, dv = score, wid = ID,
    within = c(SHSCAT, time))
    get_anova_table(res.aov2)

    Then when I want to see the effect of independent variable at each time point, I got an error message:
    "Error: Problem with `mutate()` input `data`. x 0 (non-NA) cases i Input `data` is `map(.data$data, .f, …)`"

    # Effect of SHS smoke at each time point
    one.way % group_by(time) %>%
    anova_test(dv = score, wid = ID, within = SHSCAT) %>%
    get_anova_table() %>%
    adjust_pvalue(method = “bonferroni”)

    I really need your help. I tried to find other codes, to watch videos, but nothing fixed my problem.
    Thank you for time.

    Reply
    • Kassambara
      19 Jun 2020

      Please, provide your data and the R script you used so that we can help, thanks

      Reply
  • Chiara
    21 Jun 2020

    Hello,
    thank you for the great article! I would like to use this procedure in my master’s thesis. Do you have some references to literature for me ?
    Thank you so much!

    Reply
    • Chiara
      21 Jun 2020

      I need references for the pairwise paired t-tests and the one way ANOVA with repeated measures. Thank you so much!

      Reply

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Course Curriculum

Teacher
Alboukadel Kassambara
Role : Founder of Datanovia
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