T-Test Assumptions : Excellent Tutorial You Will Love - Datanovia (2024)

Demo data

Demo dataset: mice [in datarium package]. Contains the weight of 10 mice:

# Load and inspect the datadata(mice, package = "datarium")head(mice, 3)

## # A tibble: 3 x 2## name weight## <chr> <dbl>## 1 M_1 18.9## 2 M_2 19.5## 3 M_3 23.1

Identify outliers

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

mice %>% identify_outliers(weight)

## [1] name weight is.outlier is.extreme## <0 rows> (or 0-length row.names)

There were no extreme outliers.

Check normality assumption

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

mice %>% shapiro_test(weight)

## # A tibble: 1 x 3## variable statistic p## <chr> <dbl> <dbl>## 1 weight 0.923 0.382

From the output, the p-value is greater than the significance level 0.05 indicating that the distribution of the data are not significantly different from the normal distribution. In other words, we can assume the normality.

You can also create a QQ plot of the weight data. QQ plot draws the correlation between a given data and the normal distribution.

ggqqplot(mice, x = "weight")

All the points fall approximately along the (45-degree) reference line, for each group. So we can assume normality of the data.

Demo data

Demo dataset: genderweight [in datarium package] containing the weight of 40 individuals (20 women and 20 men).

Load the data and show some random rows by groups:

# Load the datadata("genderweight", package = "datarium")# Show a sample of the data by groupset.seed(123)genderweight %>% sample_n_by(group, size = 2)

## # A tibble: 4 x 3## id group weight## <fct> <fct> <dbl>## 1 6 F 65.0## 2 15 F 65.9## 3 29 M 88.9## 4 37 M 77.0

Identify outliers by groups

genderweight %>% group_by(group) %>% identify_outliers(weight)

## # A tibble: 2 x 5## group id weight is.outlier is.extreme## <fct> <fct> <dbl> <lgl> <lgl> ## 1 F 20 68.8 TRUE FALSE ## 2 M 31 95.1 TRUE FALSE

There were no extreme outliers.

Check normality by groups

# Compute Shapiro wilk test by goupsdata(genderweight, package = "datarium")genderweight %>% group_by(group) %>% shapiro_test(weight)

## # A tibble: 2 x 4## group variable statistic p## <fct> <chr> <dbl> <dbl>## 1 F weight 0.938 0.224## 2 M weight 0.986 0.989

# Draw a qq plot by groupggqqplot(genderweight, x = "weight", facet.by = "group")

From the output above, we can conclude that the data of the two groups are normally distributed.

Check the equality of variances

This can be done using the Levene’s test. If the variances of groups are equal, the p-value should be greater than 0.05.

genderweight %>% levene_test(weight ~ group)

## # A tibble: 1 x 4## df1 df2 statistic p## <int> <int> <dbl> <dbl>## 1 1 38 6.12 0.0180

The p-value of the Levene’s test is significant, suggesting that there is a significant difference between the variances of the two groups. Therefore, we’ll use the Welch t-test, which doesn’t assume the equality of the two variances.

Demo data

Here, we’ll use a demo dataset mice2 [datarium package], which contains the weight of 10 mice before and after the treatment.

# Wide formatdata("mice2", package = "datarium")head(mice2, 3)

## id before after## 1 1 187 430## 2 2 194 404## 3 3 232 406

# Transform into long data: # gather the before and after values in the same columnmice2.long <- mice2 %>% gather(key = "group", value = "weight", before, after)head(mice2.long, 3)

## id group weight## 1 1 before 187## 2 2 before 194## 3 3 before 232

First, start by computing the difference between groups:

mice2 <- mice2 %>% mutate(differences = before - after)head(mice2, 3)

## id before after differences## 1 1 187 430 -242## 2 2 194 404 -210## 3 3 232 406 -174

Identify outliers

mice2 %>% identify_outliers(differences)

## [1] id before after differences is.outlier is.extreme ## <0 rows> (or 0-length row.names)

There were no extreme outliers.

Check normality assumption

# Shapiro-Wilk normality test for the differencesmice2 %>% shapiro_test(differences) 

## # A tibble: 1 x 3## variable statistic p## <chr> <dbl> <dbl>## 1 differences 0.968 0.867

# QQ plot for the differenceggqqplot(mice2, "differences")

From the output above, it can be assumed that the differences are normally distributed.