T-test

After learning about the Z-test we now move on to another important statistical test called the t-test. While the Z-test is useful when we know the population variance. The t-test is used to compare the averages of two groups to see if they are significantly different from each other. Suppose you want to compare the test scores of two groups of students:

• Group 1: 30 students who studied with Method A.
• Group 2: 30 students who studied with Method B.

We use a t-test to check if there is a significant difference in the average test scores between the two.

The t-test is part of Hypothesis testing where you start with an assumption the null hypothesis that the two group means are the same. Then the test helps you decide if there’s enough evidence to reject that assumption and conclude that the groups are different.

Assumptions in T-test

• Independence: The observations within each group must be independent of each other means that the value of one observation should not influence the value of another observation.
• Normality: The data within each group should be approximately normally distributed i.e., the data within each group being compared should resemble a normal bell-shaped distribution.
• Homogeneity of Variances: The variances of the two groups being compared should be equal. This assumption ensures that the groups have a similar spread of values.
• Absence of Outliers: There should be no outliers in the data as outliers can influence the results especially when sample sizes are small.

Types of T-tests

There are three types of t-tests and they are categorized as dependent and independent t-tests.

One sample T-test

One sample t-test is used for comparison of the sample mean of the data to a particularly given value. We can use this when the sample size is small. (under 30) data is collected randomly and it is approximately normally distributed. It can be calculated as:

Example: The weights of 25 obese people were taken before enrolling them into the nutrition camp. The population mean weight is found to be 45 kg before starting the camp. After finishing the camp for the same 25 people the sample mean was found to be 75 with a standard deviation of 25. Did the fitness camp work?

Python Implementation

import scipy.stats as stats
import numpy as np

popu_mean = 45
s_mean = 75
s_std = 25
s_size = 25

t_stat = (s_mean - popu_mean) / (s_std / np.sqrt(s_size))
df = s_size - 1
alpha = 0.05
cr_t = stats.t.ppf(1 - alpha, df)

p_v = 1 - stats.t.cdf(t_stat, df)

print("T-Statistic:", t_stat)
print("Critical t-value:", cr_t)
print("P-Value:", p_v)

print('With T-value :')
if t_stat > cr_t:
    print("Significant difference. Camp had effect.")
else:
    print("No significant difference. Camp had no effect.")

print('With P-value :')
if p_v >alpha:
    print("Significant difference. Camp had effect.")
else:
    print("No significant difference. Camp had no effect.")

Output:

T-Statistic: 6.0
Critical t-value: 1.7108820799094275
P-Value: 1.703654035845048e-06
With T-value :
Significant difference. Camp had effect.
With P-value :
No significant difference. Camp had no effect.

Interpretation

• T-value (6.0) is much greater than the critical t-value (1.71), so we reject the null hypothesis.
• The P-value (0.0000017) is much smaller than 0.05, also indicating a significant result.

The fitness camp had a significant effect on participants weights, causing a measurable change.

Independent sample T-test

An Independent sample t-test commonly known as an unpaired sample t-test is used to find out if the differences found between two groups is actually significant or just a random occurrence. We can use this when:

• the population mean or standard deviation is unknown. (information about the population is unknown)
• the two samples are separate/independent. For i.e. boys and girls (the two are independent of each other)

It can be calculated using:

Let's Take a example to understand

Researchers want to see if two teaching methods, A and B, produce different exam scores. Samples for both methods are collected independently.

Sample A (Teaching Method A): 78,84,92,88,75,80,85,90,87,7978,84,92,88,75,80,85,90,87,79
Sample B (Teaching Method B): 82,88,75,90,78,85,88,77,92,8082,88,75,90,78,85,88,77,92,80

Python Implementation

from scipy import stats
import numpy as np

A = np.array([78,84,92,88,75,80,85,90,87,7978,84,92,88,75,80,85,90,87,79])
B = np.array([82,88,75,90,78,85,88,77,92,8082,88,75,90,78,85,88,77,92,80])

t_val, p_val = stats.ttest_ind(A, B)

alpha = 0.05
df = len(A)+len(B)-2

crit_t = stats.t.ppf(1 - alpha/2, df)

print("T-value:", t_val)
print("P-Value:", p_val)
print("Critical t-value:", crit_t)

print('T-test Result:')
if np.abs(t_val) >crit_t:
    print('Significant difference found.')
else:
    print('No significant difference.')

print('P-test Result:')
if p_val >alpha:
    print('Fail to reject H0. No strong evidence of difference.')
else:
    print('Reject H0. Significant difference found.')

Output:

T-value: -0.008275847896130646
P-Value: 0.9934425963209128
Critical t-value: 2.0280940009804502
T-test Result:
No significant difference.
P-test Result:
Fail to reject H0. No strong evidence of difference.

Interpretation

• T-value (0.989) is less than the critical t-value (2.1009), so we fail to reject the null hypothesis.
• P-value (0.336) is greater than 0.05, meaning no significant difference.

There is no statistically significant difference between exam scores of Teaching Method A and Teaching Method B.

Paired Two-sample T-test

Paired sample t-test also known as dependent sample t-test is used to find out if the difference in the mean of two samples is 0. The test is done on dependent samples usually focusing on a particular group of people or things. In this each entity is measured twice resulting in a pair of observations. 

We can use this when:

• Two similar samples are given. [i.e. Scores obtained in English and Math (both subjects)]
• The dependent variable data is continuous.
• The observations are independent of one another.
• The dependent variable is approximately normally distributed.

It can be calculated using

Example Problem

Consider the following example. Scores (out of 25) of the subjects Math1 and Math2 are taken for a sample of 10 students. We have to perform the paired sample t-test for this data. 

Math1: 4, 4, 7, 16, 20, 11, 13, 9, 11, 15
Math2:15, 16, 14, 14, 22, 22, 23, 18, 18, 19

Python Implementation

from scipy import stats
import numpy as np

A = np.array([4, 4, 7, 16, 20, 11, 13, 9, 11, 15])
B = np.array([15, 16, 14, 14, 22, 22, 23, 18, 18, 19])

t_val, p_val = stats.ttest_rel(A, B)

alpha = 0.05
df = len(A)-1

c_t = stats.t.ppf(1 - alpha/2, df)

print("T-value:", t_val)
print("P-Value:", p_val)
print("Critical t-value:", c_t)

print('T-test:')
if np.abs(t_val) >c_t:
    print('Significant difference found.')
else:
    print('No significant difference.')

print('P-test:')
if p_val >alpha:
    print('Reject H0')
else:
    print('Fail to reject H0')

Output:

T-value: -4.953488372093023
P-Value: 0.0007875235561560145
Critical t-value: 2.2621571628540993
T-test:
Significant difference found.
P-test:
Fail to reject H0

Interpretation

• T-value (-4.95) is less than the negative critical t-value (-2.2622), so we reject the null hypothesis.
• P-value (0.00079) is less than 0.05, indicating significance.

There is a significant difference between Math1 and Math2 scores. The difference is unlikely due to chance.

Related Articles

• Degrees of Freedom
• Statistical Significance
• Student's t-distribution in Statistics