t -test

 

What is a t-test?

A t-test is a statistical hypothesis test used to determine whether there is a significant difference between the means of two groups.

It is commonly used when:

• Sample sizes are small

• Population standard deviation is unknown

• The data is approximately normally distributed

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🧪 Types of t-tests

There are three main types of t-tests:

1. One-Sample t-test

Compares the mean of a single sample to a known value (often a population mean).

Example: Is the average IQ of a group of students different from 100?

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2. Two-Sample t-test (Independent)

Compares the means of two independent groups.

Example: Are the exam scores of male and female students significantly different?

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3. Paired Sample t-test (Dependent)

Compares means of the same group at two different times (or matched pairs).

Example: Did students improve after a training session?

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📐 Assumptions of the t-test

• Data is continuous

• Data is normally distributed (or approximately normal)

• Observations are independent

• Variances are equal for independent samples (unless using Welch's t-test)

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🔢 Formulas

✅ One-Sample t-test:

$$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$$

• $\bar{x}$ = sample mean

• $\mu_0$ = population mean

• $s$ = sample standard deviation

• $n$ = sample size

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✅ Two-Sample t-test:

$$t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} }}$$​

Where $s_p$= pooled standard deviation:

$$s_p = \sqrt{ \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} }$$

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✅ Paired Sample t-test:

$$t = \frac{\bar{d}}{s_d / \sqrt{n}}$$

• $\bar{d}$ = mean of the differences

• $s_d$ = standard deviation of the differences

• $n$= number of pairs

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Examples

Example 1: One-Sample t-test

Problem: A teacher claims that the average score in her class is 75. A sample of 10 students has a mean score of 72 with a standard deviation of 5. Is her claim valid?

Solution:

$$t = \frac{72 - 75}{5 / \sqrt{10}} = \frac{-3}{1.58} ≈ -1.90$$

Using a t-table with df = 9, at α = 0.05, critical t ≈ ±2.262.
Since -1.90 > -2.262, we fail to reject the null hypothesis.

✅ Conclusion: No significant difference.

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Example 2: Independent Two-Sample t-test

Problem: Two sections of a class scored as follows:

• Section A: mean = 70, sd = 4, n = 10

• Section B: mean = 75, sd = 5, n = 10

Is there a significant difference in scores?

Solution:

1. Calculate pooled standard deviation:

$$s_p = \sqrt{ \frac{(9)(4^2) + (9)(5^2)}{18} } = \sqrt{ \frac{144 + 225}{18} } = \sqrt{20.5} ≈ 4.53$$

2. Compute t-statistic:

$$t = \frac{70 - 75}{4.53 \cdot \sqrt{\frac{1}{10} + \frac{1}{10}}} = \frac{-5}{4.53 \cdot 0.447} ≈ \frac{-5}{2.02} ≈ -2.48$$

Degrees of freedom = 18. Critical t ≈ ±2.101 at α = 0.05.

✅ Since -2.48 < -2.101 → Reject null hypothesis

✔️ Conclusion: Significant difference in scores.

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Example 3: Paired Sample t-test

Problem: Students took a mock test and then a final test. Scores:

Student	Mock	Final
A	65	70
B	68	72
C	70	75
D	60	66
E	62	65

Compute differences: [5, 4, 5, 6, 3]

Mean difference = 4.6
Standard deviation of difference ≈ 1.14
n = 5

$$t = \frac{4.6}{1.14 / \sqrt{5}} ≈ \frac{4.6}{0.51} ≈ 9.02$$

df = 4, critical t ≈ 2.776.
Since 9.02 > 2.776 → Reject null hypothesis

✅ Conclusion: Significant improvement.

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💡 When to use which t-test?

Scenario	Test Type
Compare one sample with known mean	One-sample t-test
Compare two unrelated group means	Two-sample (independent)
Compare two related (paired) samples	Paired sample t-test