Posted on April 6, 2021 by finnstats in R bloggers | 0 Comments
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Repeated Measures of ANOVA in R, in this tutorial we are going to discuss one-way and two-way repeated measures of ANOVA.
In this case, the same individuals are measured the same outcome variable under different time points or conditions.
This test is also known as a within-subjects ANOVA or ANOVA with repeated measures.
1. No significant outlier
To identify the outlier, you can use the box plot method or the three sigma limit method.
2. Normality Assumption
The outcome of the dependent variable should be normally distributed in each cell of the design.
Based on the Shapiro Wilks test can use it for the same.
3. Assumption of Sphericity
These can be checked based on the anova_test function. The variance of the differences between groups should be equal.
Sphericity can be checked using Mauchly’s test of sphericity, which is automatically reported when using the R function anova_test()
The above assumptions are not met then you can use an alternative of Repeated Measures ANOVA in the nonparametric method (Friedman Test).
library(xlsx) library(rstatix) library(reshape) library(tidyverse) library(dplyr) library(ggpubr) library(plyr) library(datarium)
data<-read.xlsx("D:/RStudio/data.xlsx",sheetName="Sheet1") data <- data %>% gather(key = "time", value = "score", T0, T1, T2) %>% convert_as_factor(id, time) data.frame(head(data, 3)) id time score 1 S1 T0 7.454333333 2 S2 T0 8.030000000 3 S3 T0 9.911666667
The objective is to identify any significant difference between time points exit or not.
summary% group_by(time) %>% get_summary_stats(score, type = "mean_sd") data.frame(summary) time variable n mean sd 1 T0 score 24 7.853 3.082 2 T1 score 24 9.298 2.090 3 T2 score 24 5.586 0.396
Create a box plot and add points corresponding to individual values:
You can use a box plot also for outlier identification.
Based on boxplot no outlier detected in the dataset.
outlier% group_by(time) %>% identify_outliers(score) data.frame(outlier) [1] time id score is.outlier is.extreme (or 0-length row.names)
These indicates no outlier in the data set.
You can use Shapiro-Wilk’s test for normality checking, Please note if the data set is small then Shapiro-Wilk’s test is ideal otherwise go for the QQ plot.
normality% group_by(time) %>% shapiro_test(score) data.frame(normality) time variable statistic p 1 T0 score 0.9540460218 0.33073045010 2 T1 score 0.9868670126 0.98282491871 3 T2 score 0.9261641361 0.08004275234
The normality assumption can be checked by computing the Shapiro-Wilk test for each time point. If the data is normally distributed, the p-value should be greater than 0.05.
Tested data was normally distributed at each time point, as assessed by Shapiro-Wilk’s test (p > 0.05).
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.
The data set score was significantly different at the different time points during the diet, F(1.42, 33) = 24.4, p 0.05), except for treatment B at time point T2, as assessed by Shapiro-Wilk’s test.
Based on Shapiro-Wilk’s analysis at time point T2 non-normality observed, so further you need to check the QQ plot for the verification.
From the plot above, as all the points fall approximately along the reference line, we can assume normality.
res.aovObserved statistically significant two-way interactions between treatment and time p
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