T-Tests: Comparing Means

Overview of T-Tests

T-tests are hypothesis tests for comparing means when the population standard deviation is unknown (which is almost always!).

Test Type	Purpose	Example
One-sample t-test	Compare sample mean to hypothesized value	Is average IQ different from 100?
Independent two-sample t-test	Compare means of two independent groups	Do men and women have different avg heights?
Paired t-test	Compare means of matched/paired observations	Did scores improve from pre-test to post-test?

One-Sample T-Test

Tests whether a population mean differs from a specific value.

Hypotheses

H₀: μ = μ₀
H₁: μ ≠ μ₀ (or < or >)

One-Sample T-Statistic

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

With df = n - 1

Assumptions

Random sample from population
Independence of observations
Normality: Population is normal OR n ≥ 30

One-Sample T-Test

A manufacturer claims light bulbs last 1000 hours. A sample of 25 bulbs has:

Mean: 985 hours
SD: 40 hours

Test at α = 0.05 if the mean differs from the claim.

Hypotheses:

H₀: μ = 1000
H₁: μ ≠ 1000 (two-tailed)

Test statistic: $t = \frac{985 - 1000}{40/\sqrt{25}} = \frac{-15}{8} = -1.875$

Critical value: df = 24, t₀.₀₂₅ = ±2.064

P-value: Between 0.05 and 0.10 (approximately 0.073)

Decision: Since |-1.875| < 2.064 (or p > 0.05), fail to reject H₀.

Conclusion: There is not sufficient evidence at α = 0.05 that the mean lifetime differs from 1000 hours.

Independent Two-Sample T-Test

Compares means of two separate, unrelated groups.

Hypotheses

H₀: μ₁ = μ₂ (or μ₁ - μ₂ = 0)
H₁: μ₁ ≠ μ₂ (or one-tailed alternatives)

Two-Sample T-Statistic (Pooled)

When equal variances are assumed:

$t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

Where the pooled standard deviation is: $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}$

With df = n₁ + n₂ - 2

Two-Sample T-Statistic (Welch's)

When variances are NOT assumed equal:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

With adjusted df (Welch-Satterthwaite formula)

Assumptions

Independent samples from two populations
Random sampling
Normality in each group OR large samples
Equal variances (for pooled t-test) - check with F-test or Levene’s test

Two-Sample T-Test

Compare test scores of two teaching methods:

Method A	Method B
n₁ = 30	n₂ = 35
Mean = 78	Mean = 82
SD = 10	SD = 12

Test at α = 0.05 if there’s a difference.

Hypotheses:

H₀: μ₁ = μ₂
H₁: μ₁ ≠ μ₂

Using Welch’s t-test: Standard Error = $\sqrt{10^2/30 + 12^2/35} = \sqrt{3.33 + 4.11} = \sqrt{7.44} = 2.73$

$t = \frac{78 - 82}{2.73} = \frac{-4}{2.73} = -1.47$

df (Welch): approximately 62

P-value: approximately 0.147 (two-tailed)

Decision: Since p = 0.147 > 0.05, fail to reject H₀.

Conclusion: There is not sufficient evidence of a difference in mean scores between the two methods.

Paired T-Test

Compares two measurements on the same subjects or matched pairs.

When to Use Paired T-Test

Before/after measurements on same subjects
Twin studies
Matched case-control studies
Left vs right measurements on same person

Hypotheses

H₀: μ_d = 0 (no mean difference)
H₁: μ_d ≠ 0 (or one-tailed)

Paired T-Statistic

$t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}$

Where:

$\bar{d}$ = mean of differences
$s_d$ = standard deviation of differences
n = number of pairs

With df = n - 1

Assumptions

Paired data (natural pairing)
Random sample of pairs
Normality of differences OR large n

Paired T-Test

A weight loss program measures 10 participants before and after:

Subject	Before	After	Difference (d)
1	180	175	-5
2	220	212	-8
3	195	190	-5
4	185	183	-2
5	240	231	-9
6	170	168	-2
7	200	191	-9
8	175	172	-3
9	210	205	-5
10	190	184	-6

Summary of differences: $\bar{d}$ = -5.4, $s_d$ = 2.59

Test at α = 0.05 if the program produces weight loss.

Hypotheses:

H₀: μ_d = 0 (no change)
H₁: μ_d < 0 (weight decreased) - left-tailed

Test statistic: $t = \frac{-5.4 - 0}{2.59/\sqrt{10}} = \frac{-5.4}{0.819} = -6.59$

Critical value: df = 9, t₀.₀₅ = -1.833

P-value: < 0.0001

Decision: Since t = -6.59 < -1.833, reject H₀.

Conclusion: There is significant evidence that the weight loss program is effective.

Choosing the Right T-Test

Situation	Test
One sample, compare to specific value	One-sample t
Two separate groups	Independent two-sample t
Same subjects measured twice	Paired t
Before/after on same subjects	Paired t
Treatment vs control (different people)	Independent two-sample t

Effect Size: Cohen’s d

P-values don’t tell you how big the effect is. Use Cohen’s d for effect size.

Cohen's d

For one-sample: $d = \frac{\bar{x} - \mu_0}{s}$

For two-sample: $d = \frac{\bar{x}_1 - \bar{x}_2}{s_p}$

For paired: $d = \frac{\bar{d}}{s_d}$

Cohen’s d	Interpretation
0.2	Small effect
0.5	Medium effect
0.8	Large effect

Effect Size for Weight Loss

From the paired example: $d = \frac{-5.4}{2.59} = -2.08$

This is a very large effect (|d| > 0.8).

Both statistically significant AND practically meaningful!

Confidence Intervals from T-Tests

Every t-test can produce a confidence interval:

CI for Difference in Means

$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \times SE$

95% CI from Two-Sample Test

From the teaching methods example:

Difference = 78 - 82 = -4 SE = 2.73 t₀.₀₂₅,₆₂ ≈ 2.00

95% CI: -4 ± 2.00(2.73) = -4 ± 5.46 = (-9.46, 1.46)

Since this interval includes 0, we can’t conclude the means differ (consistent with failing to reject H₀).

Assumptions Violations

Summary

In this lesson, you learned:

One-sample t-test: Compare sample mean to hypothesized value
Two-sample t-test: Compare means of independent groups
Paired t-test: Compare paired/matched observations
Use Welch’s t-test when unsure about equal variances
Effect size (Cohen’s d) measures practical significance
Confidence intervals complement hypothesis tests
Always check assumptions: normality, independence, (equal variances)

Practice Problems

1. A sample of 16 students has mean GPA 3.2 with SD 0.5. Test at α = 0.05 if the mean differs from 3.0.

2. Compare two groups:

Group A: n = 20, mean = 45, SD = 8
Group B: n = 25, mean = 50, SD = 10

Test at α = 0.05 if there’s a significant difference.

3. Eight patients’ blood pressure before and after medication:

Before: 145, 150, 138, 155, 142, 148, 152, 140
After: 140, 142, 135, 148, 138, 145, 147, 136

Test at α = 0.05 if the medication lowered blood pressure.

4. For problem 3, calculate Cohen’s d and interpret the effect size.

Click to see answers

1. One-sample t-test

t = (3.2 - 3.0)/(0.5/√16) = 0.2/0.125 = 1.6
df = 15, critical t = ±2.131
|1.6| < 2.131, fail to reject H₀
Not enough evidence mean differs from 3.0

2. Two-sample t-test (Welch’s)

SE = √(64/20 + 100/25) = √(3.2 + 4) = √7.2 = 2.68
t = (45 - 50)/2.68 = -1.87
df ≈ 42, p ≈ 0.068
Fail to reject H₀ (p > 0.05)

3. Paired t-test

Differences: -5, -8, -3, -7, -4, -3, -5, -4
Mean d = -4.875, SD_d = 1.73
t = -4.875/(1.73/√8) = -4.875/0.612 = -7.97
df = 7, critical t = -1.895 (one-tailed)
Reject H₀, medication significantly lowered BP

d = -4.875/1.73 = -2.82
This is a very large effect (|d| > 0.8)
Both statistically and practically significant

Next Steps

Continue with hypothesis testing:

ANOVA - Comparing more than two groups
Chi-Square Tests - Categorical data analysis
T-Test Calculator - Practice t-test calculations

Subject	Before	After	Difference (d)
1	180	175	-5
2	220	212	-8
3	195	190	-5
4	185	183	-2
5	240	231	-9
6	170	168	-2
7	200	191	-9
8	175	172	-3
9	210	205	-5
10	190	184	-6

Subject	Before	After	Difference (d)
1	180	175	-5
2	220	212	-8
3	195	190	-5
4	185	183	-2
5	240	231	-9
6	170	168	-2
7	200	191	-9
8	175	172	-3
9	210	205	-5
10	190	184	-6

Overview of T-Tests

One-Sample T-Test

Hypotheses

Assumptions

Independent Two-Sample T-Test

Hypotheses

Assumptions

Paired T-Test

When to Use Paired T-Test

Hypotheses

Assumptions

Choosing the Right T-Test

Effect Size: Cohen’s d

Confidence Intervals from T-Tests

Assumptions Violations

Summary

Practice Problems

Next Steps

Was this lesson helpful?

Subject	Before	After	Difference (d)
1	180	175	-5
2	220	212	-8
3	195	190	-5
4	185	183	-2
5	240	231	-9
6	170	168	-2
7	200	191	-9
8	175	172	-3
9	210	205	-5
10	190	184	-6