Research Methodology Chapter 10.2

Student's t-Test

I. Basic Concepts of Student's t-Test

The Student’s t-test is a statistical test that is used to determine if there is a significant difference between the means of two groups. It is a commonly used test when the sample size is small and the population standard deviation is unknown. The t-test allows researchers to make inferences about the population based on the sample data.

Student’s T-test is a statistical test used to compare the means of two groups of data. It is a parametric test, which means that it assumes that the data is normally distributed. The test compares the means of two groups and determines if the difference between them is statistically significant. The t-test calculates a t-value, which is then compared to a critical value to determine if the difference is significant.

The T-test is used in a variety of fields, including research, medicine, and business. For example, researchers might use the T-test to compare the effectiveness of two different treatments or to compare the average test scores of two different groups of students.

There are three main types of t-tests:

One-sample t-test: This test is used to compare the mean of a sample to a known population mean.
Independent samples t-test: This test is used to compare the means of two independent groups of data, where the groups are not related to each other.
Paired samples t-test: This test is used to compare the means of two related groups of data, where the groups are paired together.

When performing a t-test, we compare the calculated t-value to the critical value at a given level of significance. If the calculated t-value is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the means of the two groups. If the calculated t-value is less than the critical value, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference between the means.

It is important to note that the t-test only provides evidence of a difference between the means, but it does not provide information about the direction of the difference. To determine the direction of the difference, additional analysis or interpretation is required.

The Student’s t-test is a powerful statistical tool that allows researchers to compare the means of two groups and determine if there is a significant difference. By understanding the basic concepts and different types of t-tests, researchers can make informed decisions and draw meaningful conclusions from their data.

II. Types of Student's t-Test

A. One-Sample t-Test

The one-sample t-test is used to compare the mean of a sample to a known population mean. For example, a company might use the one-sample t-test to compare the average weight of its employees to the national average weight, or a researcher might use the one-sample t-test to compare the average IQ of a group of students to the general population.

A company wants to know if the average weight of its employees is different from 170 pounds. The company collects a sample of 100 employees and finds that the average weight of the employees in the sample is 165 pounds.

To perform a one-sample t-test, the company first needs to calculate the t-score:

where:

x̄ is the mean of the sample
μ is the hypothesized population mean
s is the standard deviation of the sample
n is the sample size

In this case, the t-score is:

t = – 2.5

The company then needs to look up the p-value for a t-score of -2.5. At degrees of freedom of 99 and a level of significance of 0.05, the critical t-value is ±1.984.

Since the calculated t-score (-2.5) is outside ±1.984 (or since the calculated t-score (-2.5) is greater than the critical t-value (1.984)), we would reject the null hypothesis.

This means there is enough evidence at the 0.05 significance level to conclude that the average weight of the employees is different from 170 pounds. The negative sign of the t-statistic indicates that the average weight is less than 170 pounds.

B. Independent Samples t-Test

The independent samples t-test is used to compare the means of two independent groups of data, where the groups are not related to each other. For example, a researcher might use the independent samples t-test to compare the average height of males and females or to compare the average test scores of students from two different schools.

Example

Researchers want to compare the average height of males and females. They collect a sample of 100 males and 100 females and find that the average height of the males is 175 cm ± 10 cm and the average height of the females is 165 cm ± 10 cm.

To perform an independent samples t-test, the researchers first need to calculate the t-score:

where:

x̄₁andx̄₂are the means of the two groups
S₁ and S₂ are the standard deviations of the two groups
n₁ and n₂ are the sample sizes of the two groups

In this case, the t-score is:

That simplified will be

t = 7.072

The researcher then needs to look up the t-crtical value for 198 degrees of freedom (df) at 0.05 level of significance. The t-critical value is found to be 1.975. The t-score is greater than the t-critical value, hence we reject the null hypothesis.

Alternatively, we can look up the p-value for a t-score of 7.072 with 198 df and take a decision. The p-value is the probability of obtaining a t-score as extreme or more extreme than the one they observed, assuming that the null hypothesis is true. The p-value for a t-score of 7.072 with 198 df is less than 0.0001. This means that there is a very small chance of obtaining a t-score as extreme or more extreme than the one they observed if the null hypothesis is true.

Therefore, the researchers reject the null hypothesis and conclude that there is a significant difference in the average height of males and females.

Two-Sample t-Test Assuming Equal Variances

The two-sample t-test assuming equal variances is used when the variances of the two groups are assumed to be equal. This assumption is often made when the sample sizes of the two groups are similar and the data is normally distributed. For example, if we are comparing the test scores of students who received different teaching methods, we can use the two-sample t-test assuming equal variances to determine if there is a significant difference in the mean test scores.

To perform this t-test, we calculate the mean and standard deviation of each group. The pooled standard deviation is then calculated by combining the standard deviations of the two groups. The t-value is calculated by dividing the difference in means by the standard error of the difference. This t-value is then compared to the critical value to determine if the difference is statistically significant.

The formula for calculating the t-statistic in this case is as follows:

where:

x̄₁andx̄₂ are the means of the two groups
S_pis the pooled standard deviation (see below for calculation)
n₁ and n₂are the sample sizes of the two groups

The Pooled Standard Deviation is calculated using the following formula:

S₁ and S₂ are the standard deviations of the two groups
n₁ and n₂ are the sample sizes of the two groups

Example:

Suppose we want to compare the average scores of students from two different tutoring programs. We have a sample of 23 students from Program A and another sample of 30 students from Program B. The average score for Program A is 75 with a standard deviation of 8, and the average score for Program B is 72 with a standard deviation of 7. Assume equal variances.

We know that for Program A, n = 25, sample mean (x̄₁) is 75, and standard deviation (s1) is 8, while for Program B, n = 30, sample mean (x̄₂) is 72 and standard deviation (s2) is 7.

First, we calculate the Pooled Standard Deviation using the formula:

S₁ and S₂ are the standard deviations of the two groups
n₁ and n₂ are the sample sizes of the two groups

By substituting values in the formula, we get

S_p= 7.46

Now, we calculate the t-statistic using the formula:

where:

x̄₁andx̄₂ are the means of the two groups
S_pis the pooled standard deviation
n₁ and n₂are the sample sizes of the two groups

By substituting values in the formula, we get:

The degree of freedom for this problem is n1 + n2 – 2, which is 25 + 30 – 2 = 53.

To conclude, we need to compare the calculated t-statistic (1.48) with the critical t-value at a specified significance level (0.05) and degrees of freedom (df = 53).

Remember, to follow these steps:

Step 1. Find the Critical t-value: Using a t-distribution table or statistical software, find the critical t-value for a two-tailed test with a significance level of 0.05 and degrees of freedom (df) equal to 53.

Step 2. Compare the t-statistic and Critical t-value: If the calculated t-statistic (1.48) is greater than the critical t-value, we would reject the null hypothesis. If it’s less than the critical t-value, we would fail to reject the null hypothesis.

Step 3. Draw a Conclusion: Based on the following assumptions,

If calculated |t| > t critical, reject the null hypothesis. This would suggest that there is a significant difference between the average scores of the two tutoring programs.

If calculated |t| < t critical, fail to reject the null hypothesis. This would suggest that there is not enough evidence to conclude a significant difference between the average scores of the two tutoring programs.

(Also note that if the p-value is less than the significance level (usually 0.05), the null hypothesis is rejected. This means that there is a statistically significant difference in the average scores between Program A and Program B).

Conclusion

Using a t-distribution table, the critical t-value for a two-tailed test with a significance level of 0.05 and degrees of freedom (df) equal to 53 was found to be 2.0. The calculated t-statistic (1.48) is lesser than the critical t-value (2.0), we fail to reject the null hypothesis, suggesting that there is not enough evidence to conclude a significant difference between the average scores of the two tutoring programs.

Two-Sample t-Test Assuming Unequal Variances

The two-sample t-test assuming unequal variances is used when the variances of the two groups are not assumed to be equal. This assumption is often made when the sample sizes of the two groups are different or when the data is not normally distributed. For example, if we are comparing the heights of men and women, we can use the two-sample t-test assuming unequal variances to determine if there is a significant difference in the mean heights.

To perform this t-test, we calculate the differences between the means of the groups, and and variances of the groups. The t-value is then calculated by dividing the difference of means by the square root of the sum of the ratios of variance of the first group and the number of observations in that group and ratios of variance of the second group and the number of observations in that group. This t-value is then compared to the critical value to determine if the difference is statistically significant. The two-sample test assuming unequal variances is also known as Welch’s t-test.

To perform a two-sample t-test assuming unequal variances, the researchers first need to calculate the t-score by using the formula:

where:

x̄₁ is the mean of the first group
x̄₂ is the mean of the second group
σ₁²is the variance of the first group
σ₂²is the variance of the second group
n₁ is the sample size of the first group
n₂ is the sample size of the second group

The variance for both groups is calculated using the following formulae:

and

The degrees of freedom is calculated using the following formula:

Example:

A researcher wants to compare the average blood pressure of patients before and after a new treatment. The researcher knows that blood pressure can vary widely from person to person, so he assumes that the variances of the two groups are unequal.

To perform a paired two-sample t-test with unequal variances, the researcher first needs to calculate the means of both groups. Then needs to calculate the variances of both the groups, followed by the degrees of freedom. Then substitute the values in the t-test formula to calculate the t-statistic.

Before Treatment: 120, 122, 118, 130, 125, 128, 121, 123, 126, 129

After Treatment: 118, 120, 115, 128, 123, 126, 120, 122, 124, 127.

Step 1: Calculate the means of both groups.

x̄₁ (mean of the first group) = 1232/10

= 123.2

x̄₂ (mean of the second group) = 1213/10

=121.3

Step 2: Calculate the variances for both groups.

σ₁²(variance of the first group) = 160.2/9

=17.8

σ₂²(variance of the second group) = 154.8/9

= 17.2

Step 3: Calculate degrees of freedom.

df = 17.9

Step 4: Calculate the t-statistic by substituting the above values in the formula:

t = 1.02

Using a t-distribution table, the critical t-value for a two-tailed test with a significance level of 0.05 and degrees of freedom (df) equal to 9 is found. If the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis, suggesting that there is a significant difference between the averages compared.

For the present example, the t-statistic (1.02) is lesser than the t-critical (2.262) indicating that the average of Group 1’s and Group 2’s blood pressure are not different.

The researcher can conclude that the new treatment is not effective in reducing blood pressure.

C. Paired Samples t-Test

The paired samples t-test is used to compare the means of two related groups of data, where the groups are paired together. For example, a teacher might use the paired samples t-test to compare the average pre-test and post-test scores of her students, or a doctor might use the paired samples t-test to compare the blood pressure of patients before and after a new treatment.

The paired two-sample t-test is used when the data is paired or matched. This means that each observation in one group is related to a specific observation in the other group. For example, if we are comparing the effectiveness of a new drug, we might measure the blood pressure of each patient before and after taking the drug. The paired t-test allows us to determine if there is a significant difference in blood pressure before and after taking the drug.

To perform a paired t-test, we calculate the difference between the paired observations, and then calculate the mean and standard deviation of these differences. The t-value is then calculated by dividing the mean difference by ratio of the standard deviation of the differences and square root of number of pairs. This t-value is then compared to the critical value to determine if the difference is statistically significant.

To perform the paired t-test, the researchers first need to calculate the t-score by using the formula:

where:

md is the mean difference between the groups
SDd is the standard deviation of the differences
n is the number of pairs

The mean difference between the groups is calculated by the following formula:

where:

md is the mean difference between the groups
Σd is sum of the differences
nis the number of pairs

The standard deviation of differences is calculated by the following formula:

where:

di is the individual difference
md is the mean difference
nis the number of pairs

Example:

To perform a paired two-sample t-test, the researcher first needs to calculate the differences between After- and Before-treatment blood pressures. Then needs to calculate the mean of differences, followed by the standard deviation of differences. Then substitute the values in the paired t-test formula to calculate the t-statistic.

Before Treatment: 120, 122, 118, 130, 125, 128, 121, 123, 126, 129

After Treatment: 118, 120, 115, 128, 123, 126, 120, 122, 124, 127.

Step 1: Calculate differences, that is, differences is After-treatment minus Before-treatment values.

Differences =

Differences = (118 – 120, 120 – 122, 115 – 118, 128 – 130, 123 – 125, 126 – 128, 120 – 121, 122 – 123, 124 – 126, 127 – 129)

Differences = (-2, -2, -3, -2, -2, -2, -1, -1, -2, -2)

Step 2: Calculate mean of differences.

Mean of Differences = (-2 + -2 + -3 + -2 + -2 + -2 + -1 + -1 + -2 + -2) / 10

Mean of Differences = -19 / 10

Mean of Differences = -1.9

Step 3: Calculate the standard deviation of differences.

Squared deviations = ((-2 – (-1.9))² + (-2 – (-1.9))² + (-3 – (-1.9))²+ (-2 – (-1.9))²+ (-2 – (-1.9))²+ (-2 – (-1.9))²+ (-1 – (-1.9))² + (-1 – (-1.9))² + (-2 – (-1.9))²+ (-2 – (-1.9))²)

Squared deviations = (0.01 + 0.01 + 1.21+ 0.01+ 0.01+ 0.01+ 0.81+ 0.81 + 0.01+ 0.01)

Squared deviations = 2.9

Now calculate the Standard Deviation of differences by substituting the values in the formula.

Squared Deviation of differences =

SDd = 0.567

Step 4: Calculate the t-statistic by substituting the values in the formula.

t = -10.596

For the present example, the t-statistic (|-10.596|) is greater than the t-critical (2.262) indicating that the average of Before Treatement and After Treatment blood pressure are significantly different.

The researcher can conclude that the new treatment is very effective in reducing blood pressure.

Student's t-Test

I. Basic Concepts of Student's t-Test

II. Types of Student's t-Test

A. One-Sample t-Test

B. Independent Samples t-Test

Two-Sample t-Test Assuming Equal Variances

Two-Sample t-Test Assuming Unequal Variances

C. Paired Samples t-Test

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