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t-test

Independent-Samples t-test (Welch) — Drug A vs. Drug B

Compare pain scores between two drug groups using Welch's t-test. Welch-Satterthwaite df, Cohen's d=1.02, 95% CI.

📊 t(15) = −2.26, p = .039, d = 1.02

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⚖️ 2 · t-Tests

2a · One-Sample t-test: Is the class average = 75?

t-test

Research Question: The national average exam score is 75. Does a sample of 10 students differ from this standard?

Formula

t = \frac{\bar{x} - μ_{0}}{s / \sqrt{n}}, d f = n - 1

1

State hypotheses: H₀: μ = 75 vs H₁: μ ≠ 75 (two-tailed)
2

$t = \frac{79.5 - 75}{8.37 / \sqrt{10}} = \frac{4.5}{2.646} = 1.700, d f = 9$
3

Critical value at α=.05, df=9: t* = ±2.262. Since |1.700| < 2.262 → Fail to reject H₀.
4

Cohen's d (effect size):
$d = \frac{\bar{x} - μ_{0}}{s} = \frac{79.5 - 75}{8.37} = 0.538 (medium)$

1.700

.124

0.538

Cohen's d

⚪ Fail to Reject H₀ — The class mean (M=79.5) does not differ significantly from 75, t(9)=1.70, p=.124.

APA-7

A one-sample t-test indicated that the class mean (M = 79.50, SD = 8.37) did not significantly differ from the national standard of 75, t(9) = 1.70, p = .124, d = 0.54, 95% CI [−1.48, 10.48].

2b · Independent-Samples t-test (Welch): Drug A vs Drug B

t-test

Research Question: Does Drug A reduce pain scores more than Drug B? (Higher score = more pain)

Group	n	Mean	SD
Drug A	12	4.2	1.3
Drug B	10	5.8	1.9

Welch t-statistic

t_{W} = \frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}

1

$S E = \sqrt{\frac{{1.3}^{2}}{12} + \frac{{1.9}^{2}}{10}} = \sqrt{\frac{1.69}{12} + \frac{3.61}{10}} = \sqrt{0.1408 + 0.361} = \sqrt{0.5018} = 0.7084$
2

$t_{W} = \frac{4.2 - 5.8}{0.7084} = \frac{- 1.6}{0.7084} = - 2.258$
3

Welch-Satterthwaite df:
$d f = \frac{(s_{1}^{2} / n_{1} + s_{2}^{2} / n_{2})^{2}}{\frac{(s_{1}^{2} / n_{1})^{2}}{n_{1} - 1} + \frac{(s_{2}^{2} / n_{2})^{2}}{n_{2} - 1}} = \frac{(0.5018)^{2}}{\frac{(0.1408)^{2}}{11} + \frac{(0.361)^{2}}{9}} \approx 15.7 \approx 15$
4

Cohen's d:
$d = \frac{| {\bar{x}}_{1} - {\bar{x}}_{2} |}{s_{p}}, s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{11 (1.69) + 9 (3.61)}{20}} = \sqrt{1.554} = 1.567$
$d = \frac{1.6}{1.567} = 1.021 (large)$

−2.258

t(Welch)

.039

1.021

Cohen's d

🔴 Reject H₀ — Drug A produces significantly lower pain scores than Drug B, t(15)=−2.26, p=.039.

APA-7

An independent-samples Welch t-test revealed that Drug A (M = 4.20, SD = 1.30) produced significantly lower pain scores than Drug B (M = 5.80, SD = 1.90), t(15.00) = −2.26, p = .039, d = 1.02, 95% CI [−3.11, −0.09].

2c · Paired-Samples t-test: Before vs After Training

t-test

Research Question: Did a training program improve performance scores for the same 8 employees?

i	Before	After	Diff (d)	d − d̄	(d − d̄)²
1	60	72	12	2.75	7.56
2	55	68	13	3.75	14.06
3	70	75	5	-4.25	18.06
4	65	74	9	-0.25	0.06
5	58	71	13	3.75	14.06
6	72	80	8	-1.25	1.56
7	63	70	7	-2.25	5.06
8	68	82	14	4.75	22.56
Σ			81	0	83.00

t = \frac{\bar{d}}{s_{d} / \sqrt{n}}, \bar{d} = \frac{81}{8} = 10.125, s_{d} = \sqrt{\frac{83.00}{7}} = \sqrt{11.857} = 3.443

t = \frac{10.125}{3.443 / \sqrt{8}} = \frac{10.125}{1.217} = 8.320, d f = n - 1 = 7

Cohen's d (paired)

d = \frac{\bar{d}}{s_{d}} = \frac{10.125}{3.443} = 2.941 (very large)

8.320

<.001

2.941

Cohen's d

10.125

Mean Diff.

🔴 Reject H₀ — Training significantly improved performance, t(7) = 8.32, p < .001, d = 2.94.

APA-7

A paired-samples t-test indicated a statistically significant improvement in performance following training (M_diff = 10.13, SD_diff = 3.44), t(7) = 8.32, p < .001, d = 2.94, 95% CI [7.25, 13.00].

⚠️ Example not found

Independent-Samples t-test (Welch) — Drug A vs. Drug B

⚖️ 2 · t-Tests

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