t-Test and its types - Probability and Statistics

What is t-Test?

A t-test is a statistical test used to compare the means of two groups and determine if the differences between them are statistically significant. It is widely used in hypothesis testing when the data sets follow a normal distribution and the sample size is relatively small. There are different types of t-tests, including the independent t-test, paired t-test, and one-sample t-test.

Need for t-Test

The t-test is essential in various research scenarios:

Comparing Means: To compare the average values of two different groups.

Hypothesis Testing: To determine whether the observed data significantly differ from what is expected under the null hypothesis.

Small Sample Sizes: To analyze data from small sample sizes where other statistical tests (like z-tests) may not be appropriate.

Types of t-Tests

1. Independent t-Test: Compares the means of two independent groups.

2. Paired t-Test: Compares the means from the same group at different times (e.g., before and after a treatment).

3. One-Sample t-Test: Compares the mean of a single group to a known value.

1. Independent t-Test: Compares the means of two independent groups.
$t = \frac{\overline{X_1} - \overline{X_2}}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}$
Where:
$\overline{X_1}$ and $\overline{X_2}$ are the sample means,
$S_1^2$ and $S_2^2$ are the sample variances,
$n_1$ and $n_2$ are the sample sizes.

For Example: A researcher wants to compare the average test scores of two different teaching methods to determine if one is more effective than the other. The test scores from two groups of students, each taught using different methods, are collected.

Data:
Group 1 (Method A): 78, 82, 88, 90, 85
Group 2 (Method B): 85, 87, 89, 91, 90

Solution:

1. Calculate the means of each group:
$\overline{X_1} = \frac{78 + 82 + 88 + 90 + 85}{5} = \frac{423}{5} = 84.6$
$\overline{X_2} = \frac{85 + 87 + 89 + 91 + 90}{5} = \frac{442}{5} = 88.4$

2. Calculate the variances of each group:
$S_1^2 = \frac{(78 - 84.6)^2 + (82 - 84.6)^2 + (88 - 84.6)^2 + (90 - 84.6)^2 + (85 - 84.6)^2}{5 - 1}$
$S_1^2 = \frac{43.56 + 7.84 + 11.56 + 28.96 + 0.16}{4} = \frac{92.08}{4} = 23.02$
$S_2^2 = \frac{(85 - 88.4)^2 + (87 - 88.4)^2 + (89 - 88.4)^2 + (91 - 88.4)^2 + (90 - 88.4)^2}{5 - 1}$
$S_2^2 = \frac{11.56 + 1.96 + 0.36 + 6.76 + 2.56}{4} = \frac{23.2}{4} = 5.8$

3. Calculate the t-statistic:
$t = \frac{84.6 - 88.4}{\sqrt{\frac{23.02}{5} + \frac{5.8}{5}}}$
$t = \frac{-3.8}{\sqrt{\frac{23.02 + 5.8}{5}}}$
$t = \frac{-3.8}{\sqrt{4.604 + 1.16}}$
$t = \frac{-3.8}{\sqrt{5.764}}$
$t = \frac{-3.8}{2.4}$
$t = -1.583$

Therefore, the t-statistic is approximately -1.583. Researchers would compare this t-value to the critical t-value from the t-distribution table at the desired significance level (e.g., 0.05) to determine if the difference is statistically significant.

2. Paired t-Test: Compares the means from the same group at different times (e.g., before and after a treatment).
$t = \frac{\overline{d}}{S_d / \sqrt{n}}$
Where:
$\overline{d}$ is the mean of the differences,
$S_d$ is the standard deviation of the differences,
$n$ is the number of pairs.

For Example : A researcher wants to test the effectiveness of a new diet plan by comparing the weights of participants before and after the diet. The weights (in kg) of 5 participants are recorded before and after following the diet for one month.

Data:
Before: 70, 75, 80, 85, 90
After: 68, 74, 78, 82, 88

Solution:

1. Calculate the differences ($d$) between the before and after weights:
$d = [70-68, 75-74, 80-78, 85-82, 90-88] = [2, 1, 2, 3, 2]$

2. Calculate the mean of the differences ($\overline{d}$):
$\overline{d} = \frac{2 + 1 + 2 + 3 + 2}{5} = \frac{10}{5} = 2$

3. Calculate the standard deviation of the differences ($S_d$):
$S_d = \sqrt{\frac{(2-2)^2 + (1-2)^2 + (2-2)^2 + (3-2)^2 + (2-2)^2}{5-1}}$
$S_d = \sqrt{\frac{0 + 1 + 0 + 1 + 0}{4}} = \sqrt{\frac{2}{4}} = \sqrt{0.5} = 0.707$

4. Calculate the t-statistic:
$t = \frac{\overline{d}}{S_d / \sqrt{n}}$
$t = \frac{2}{0.707 / \sqrt{5}}$
$t = \frac{2}{0.707 / 2.236}$
$t = \frac{2}{0.316}$
$t = 6.325$

Therefore, the t-statistic is approximately 6.325. This value would be compared to the critical t-value from the t-distribution table at the desired significance level (e.g., 0.05) to determine if the weight change is statistically significant.

3. One-Sample t-Test: Compares the mean of a single sample to a known value (population mean).
$t = \frac{\overline{X} - \mu}{S / \sqrt{n}}$
Where:
$\overline{X}$ is the sample mean,
$\mu$ is the known value (population mean),
$S$ is the sample standard deviation,
$n$ is the sample size.

For Example : Suppose a company claims that the mean waiting time for customers is 5 minutes. A random sample of 10 customers is taken, and their waiting times (in minutes) are recorded as follows:

Data:
4, 6, 5, 3, 5, 4, 6, 7, 5, 4

Solution:

1. Calculate the sample mean ($\overline{X}$):
$\overline{X} = \frac{4 + 6 + 5 + 3 + 5 + 4 + 6 + 7 + 5 + 4}{10} = \frac{49}{10} = 4.9$

2. Assume the population mean ($\mu$) is 5 (as claimed by the company).

3. Calculate the sample standard deviation ($S$):
First, calculate the variance:
$s^2 = \frac{\sum (X_i - \overline{X})^2}{n-1}$
$s^2 = \frac{(4-4.9)^2 + (6-4.9)^2 + \ldots + (4-4.9)^2}{9}$
$s^2 = \frac{0.81 + 1.21 + \ldots + 0.81}{9} = \frac{11.7}{9} = 1.3$
Then, calculate $S$:
$S = \sqrt{1.3} = 1.14$

4. Calculate the t-statistic:
$t = \frac{\overline{X} - \mu}{S / \sqrt{n}}$
$t = \frac{4.9 - 5}{1.14 / \sqrt{10}}$
$t = \frac{-0.1}{1.14 / 3.16}$
$t = \frac{-0.1}{0.36}$
$t = -0.278$

Therefore, the t-statistic is approximately -0.278. This value would be compared to the critical t-value from the t-distribution table at the desired significance level (e.g., 0.05) to determine if there is sufficient evidence to reject the company's claim about the mean waiting time.