# Student’s t-test explained with R and Pokemon

R
Statistics
Gotta Test ’Em All!
Author

Mauricio “Pachá” Vargas S.

Published

June 24, 2023

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# Motivation

I got more than 3 questions about the use of Student’s t-test in the last week. I think it is a good idea to write a blog post about it.

# What is Student’s t-test?

Student’s t-test is a statistical test that is used to compare the mean of a group against a specified value and to compare the means of two groups against each other. For example, to evaluate if the mean speed of electric and water Pokemon are statistically different. Student wasn’t the name of the test’s creator. The name of the creator was William Gosset, but he published his work under the pseudonym “Student”, like Madonna or Prince.

When we use the test we aim at finding differences that are statistically different. What do we mean by that? Let’s consider the next averages for the different Pokemon type:

# run one and one time only
# install.packages("d3po")
# install.packages("dplyr")

library(d3po)
library(dplyr)

pokemon %>%
select(type_1, attack, defense, speed) %>%
group_by(type_1) %>%
summarise_if(is.numeric, mean, na.rm = TRUE) %>%
arrange(type_1) %>%
print(n = 15)
# A tibble: 15 × 4
type_1   attack defense speed
<fct>     <dbl>   <dbl> <dbl>
1 bug        63.8    57.1  57.1
2 dragon     94      68.3  66.7
3 electric   62      64.7  98.9
4 fairy      57.5    60.5  47.5
5 fighting  103.     61    66.1
6 fire       83.9    62.6  84
7 ghost      50      45    95
8 grass      70.7    69.6  52.1
9 ground     81.9    86.2  58.1
10 ice        67.5    67.5  90
11 normal     67.7    53.5  69.3
12 poison     74.4    67    58.8
13 psychic    60.1    57.5  93
14 rock       82.2   110    58.3
15 water      70.2    77.5  67.7

In the table, the means speed for electric and water pokemon are 98.9 and 67.7 pokemon measurement points (pmp), and we know that those two numbers are different. The question becomes: *Is the difference between 98.9 and 67.7 pmp statistically different from zero? Or in other words, is the difference due to chance or is it due to a real difference between the groups?

We define a null hypothesis, such as “the means for electric an water pokemon are equal” and an alternative hypothesis, such as “the means for electric and water pokemon are different”. The observations support evidence to reject or fail to reject the null hypothesis. In Statistics we never accept the null hypothesis, we just fail to reject it. To read more about inference and hypothesis testing, I recommend Introduction to Modern Statistics by Mine Çetinkaya-Rundel and Johanna Hardin.

# Comparing the means of two groups

From the previous example, we can define $$H_0: \mu^{\text{electric}} = \mu^{\text{water}}$$ and $$H_1: \mu^{\text{electric}} \neq \mu^{\text{water}}$$. Before proceeding to the formal test, let’s explore the quantiles and the box and whiskers for both groups.

pokemon %>%
filter(type_1 == "electric") %>%
pull(speed) %>%
quantile(na.rm = TRUE)
  0%  25%  50%  75% 100%
45   90  100  110  140 
pokemon %>%
filter(type_1 == "water") %>%
pull(speed) %>%
quantile(na.rm = TRUE)
    0%    25%    50%    75%   100%
15.00  57.25  70.00  82.00 115.00 
# run one and one time only
# install.packages("ggplot2")

library(ggplot2)

pokemon %>%
filter(type_1 %in% c("electric", "water")) %>%
ggplot() +
geom_boxplot(aes(x = type_1, y = speed)) +
theme_minimal() From the quantiles and the plot we already have an intution. If we move one box on top of the other, the central quantiles do not overlap, suggesting that there is a statistically significant difference.

The the t.test function in R by default returns the p-value of a two sided test.

electric <- pokemon %>%
filter(type_1 == "electric") %>%
pull(speed) %>%
na.omit()

water <- pokemon %>%
filter(type_1 == "water") %>%
pull(speed) %>%
na.omit()

t.test(electric, water)

Welch Two Sample t-test

data:  electric and water
t = 2.9897, df = 11.019, p-value = 0.01228
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
8.22914 54.12007
sample estimates:
mean of x mean of y
98.88889  67.71429 

The calculated p-value is 0.01228 which is less than the critical p-value 0.05. We reject the null hypothesis that the means of the two groups are equal. We conclude that the means of the two groups are different.

What if we are interested in the sign of the alternative hypothesis? We can use the alternative argument in the t.test function. For example, if we want to specify the alternative $$H_1: \mu^{\text{electric}} > \mu^{\text{water}}$$ for the same $$H_0$$, we specify the alternative = "greater" argument.

t.test(electric, water, alternative = "greater")

Welch Two Sample t-test

data:  electric and water
t = 2.9897, df = 11.019, p-value = 0.006141
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
12.45138      Inf
sample estimates:
mean of x mean of y
98.88889  67.71429 

The calculated p-value is 0.00614 which is less than the critical p-value 0.05. We reject the null hypothesis that the means of the two groups are equal. We conclude that the mean of the electric group is statistically greater than the mean of the water group.

As another example, if we want to specify the alternative $$H_1: \mu^{\text{electric}} < \mu^{\text{water}}$$ for the same $$H_0$$, we specify the alternative = "less" argument.

t.test(electric, water, alternative = "less")

Welch Two Sample t-test

data:  electric and water
t = 2.9897, df = 11.019, p-value = 0.9939
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf 49.89783
sample estimates:
mean of x mean of y
98.88889  67.71429 

The calculated p-value is 0.9939 which is greater than the critical p-value 0.05. We fail to reject the null hypothesis that the means of the two groups are equal. We conclude that the mean of the electric group is statistically equal than the mean of the water group.

# Exercises

Repeat the previous analysis for the attack and defense variables for two Pokemon types of your choice.

Repeat the previous analysis for a critical p-value 0.01 and 0.1.

Find a clinical dataset and perform a Student’s t-test for the trial and control groups. Would you be interested in a particular type of alternative hypothesis? Why?