📋 1 · Descriptive Statistics

Exam Scores: Central Tendency & Spread

Descriptive

Research Question: A professor records exam scores for 10 students. Describe the distribution.

Student	1	2	3	4	5	6	7	8	9	10
Score	72	85	90	68	78	92	75	88	65	82

1

Mean (x̄):
$\bar{x} = \frac{\sum x_{i}}{n} = \frac{72 + 85 + 90 + 68 + 78 + 92 + 75 + 88 + 65 + 82}{10} = \frac{795}{10} = 79.5$
2

Sample Standard Deviation:
$s = \sqrt{\frac{\sum (x_{i} - \bar{x})^{2}}{n - 1}} = \sqrt{\frac{(- 7.5)^{2} + \dots + (2.5)^{2}}{9}} = \sqrt{\frac{630.5}{9}} = \sqrt{70.06} \approx 8.37$
3

Median:
$Median = \frac{x_{(5)} + x_{(6)}}{2} = \frac{78 + 82}{2} = 80.0$
4

SE:
$S E = \frac{s}{\sqrt{n}} = \frac{8.37}{\sqrt{10}} = \frac{8.37}{3.162} = 2.65$

79.5

Mean

80.0

Median

8.37

SD

2.65

SE

65–92

Range

APA-7

Descriptive statistics for exam scores (N = 10): M = 79.50, SD = 8.37, Mdn = 80.00, range = [65, 92].

▶ Open in MindStat

⚖️ 2 · t-Tests

2a · One-Sample t-test

t-test

Research Question: The national average is 75. Does the class (n=10, M=79.5, s=8.37) differ significantly?

Formula

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

1

H₀: μ = 75 vs H₁: μ ≠ 75 (two-tailed, α = .05)
2

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

Critical value: t*(df=9, α=.05) = ±2.262 → |1.700| < 2.262
4

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

1.700

t

9

df

.124

p

0.538

Cohen's d

⚪ Fail to Reject H₀ — p = .124 > .05

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)

t-test

Research Question: Does Drug A reduce pain scores more than Drug B?

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)

15

df

.039

p

1.021

Cohen's d

🔴 Reject H₀ — p = .039 < .05

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

t-test

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

i	Before	After	d = After − Before	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

t

7

df

<.001

p

2.941

Cohen's d

10.125

Mean Diff.

🔴 Reject H₀ — p < .001

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].

📊 3 · ANOVA

3a · One-Way ANOVA

ANOVA

Research Question: Do three teaching methods produce different exam scores?

Traditional (A)	Blended (B)	Online (C)
70	85	75
72	88	78
68	82	72
74	90	80
71	87	76
M=71	M=86.4	M=76.2

ANOVA Decomposition

S S_{T o t a l} = S S_{B e t w e e n} + S S_{W i t h i n}

S S_{B} = \sum_{j = 1}^{k} n_{j} ({\bar{x}}_{j} - {\bar{x}}_{. .})^{2}, S S_{W} = \sum_{j = 1}^{k} \sum_{i = 1}^{n_{j}} (x_{i j} - {\bar{x}}_{j})^{2}

F = \frac{M S_{B}}{M S_{W}} = \frac{S S_{B} / (k - 1)}{S S_{W} / (N - k)}

1

Grand Mean:
${\bar{x}}_{. .} = \frac{71 (5) + 86.4 (5) + 76.2 (5)}{15} = \frac{1178}{15} = 77.87$
2

SS_Between:
$S S_{B} = 5 (71 - 77.87)^{2} + 5 (86.4 - 77.87)^{2} + 5 (76.2 - 77.87)^{2}$ $= 5 (47.20) + 5 (72.76) + 5 (2.79) = 236.0 + 363.8 + 13.9 = 613.7$
3

SS_Within:
$S S_{W} = [(70 - 71)^{2} + (72 - 71)^{2} + \dots] + \dots = 20.0 + 38.8 + 42.8 = 101.6$
4

F-statistic:
$M S_{B} = \frac{613.7}{3 - 1} = 306.85, M S_{W} = \frac{101.6}{15 - 3} = 8.467$ $F (2, 12) = \frac{306.85}{8.467} = 36.24$
5

η²:
$η^{2} = \frac{S S_{B}}{S S_{T}} = \frac{613.7}{613.7 + 101.6} = \frac{613.7}{715.3} = 0.858 (large, η^{2} > .14)$

36.24

F(2,12)

<.001

p

0.858

η²

306.85

MS_B

8.467

MS_W

🔴 Reject H₀ — F(2,12) = 36.24, p < .001

APA-7

A one-way ANOVA revealed a significant effect of teaching method, F(2, 12) = 36.24, p < .001, η² = .86. Tukey: Blended > Online > Traditional, all ps < .001.

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🔢 4 · Non-parametric Tests

4a · Mann-Whitney U Test

Non-parametric

Research Question: Do satisfaction scores differ between two departments?

Dept A	6	8	2	4	9
Dept B	7	5	10	3	1

U Statistic

U_{1} = n_{1} n_{2} + \frac{n_{1} (n_{1} + 1)}{2} - R_{1}, U = min (U_{1}, U_{2})

1

Rank all 10 combined (1=lowest):
Value 1 2 3 4 5 6 7 8 9 10
Rank 1 2 3 4 5 6 7 8 9 10
Group B A B A B A B A A B
2

$R_{1} = 2 + 4 + 6 + 8 + 9 = 29 (sum of ranks for Dept A)$ $U_{1} = 5 \times 5 + \frac{5 \times 6}{2} - 29 = 25 + 15 - 29 = 11$ $U_{2} = n_{1} n_{2} - U_{1} = 25 - 11 = 14, U = min (11, 14) = 11$
3

Normal approximation (for reference):
$z = \frac{U - n_{1} n_{2} / 2}{\sqrt{n_{1} n_{2} (n_{1} + n_{2} + 1) / 12}} = \frac{11 - 12.5}{\sqrt{25 \times 11 / 12}} = \frac{- 1.5}{4.787} = - 0.313$
4

Effect size r:
$r = \frac{| z |}{\sqrt{N}} = \frac{0.313}{\sqrt{10}} = 0.099 (small)$

11

U

.754

p (exact)

0.099

r

⚪ Fail to Reject H₀ — p = .754

APA-7

A Mann-Whitney U test revealed no significant difference in satisfaction scores between Department A and B, U = 11, z = −0.31, p = .754, r = .10.

4b · Kruskal-Wallis H Test

Non-parametric

Research Question: Do pain levels differ across three clinics?

Clinic A	Clinic B	Clinic C
3,5,4,2,6	7,9,8,10,6	5,4,6,3,5

Kruskal-Wallis H

H = \frac{12}{N (N + 1)} \sum_{j = 1}^{k} \frac{R_{j}^{2}}{n_{j}} - 3 (N + 1)

1

Rank all 15 values:
$R_{A} = 2 + 6.5 + 4 + 1 + 9 = 22.5, R_{B} = 11 + 14 + 13 + 15 + 9 = 62, R_{C} = 6.5 + 4 + 9 + 2 + 6.5 = 28$
(ties averaged)
2

$H = \frac{12}{15 \times 16} (\frac{{22.5}^{2}}{5} + \frac{62^{2}}{5} + \frac{28^{2}}{5}) - 3 (16)$ $= \frac{12}{240} (101.25 + 768.8 + 156.8) - 48 = 0.05 (1026.85) - 48 = 51.34 - 48 = 3.34$
3

Compare to χ²(2) = 5.99 → H = 3.34 < 5.99

3.34

H

2

df

.188

p

⚪ Fail to Reject H₀ — p = .188

APA-7

A Kruskal-Wallis test indicated no significant difference in pain levels across clinics, H(2) = 3.34, p = .188, η² = .10.

🔗 5 · Correlation

5a · Pearson Correlation

Correlation

Research Question: Is there a linear relationship between study hours and GPA?

i	X (hrs)	Y (GPA)	X−X̄	Y−Ȳ	(X−X̄)(Y−Ȳ)	(X−X̄)²	(Y−Ȳ)²
1	10	2.8	-5	-0.55	2.75	25	0.30
2	15	3.2	0	-0.15	0	0	0.02
3	20	3.8	5	0.45	2.25	25	0.20
4	12	3.0	-3	-0.35	1.05	9	0.12
5	18	3.6	3	0.25	0.75	9	0.06
6	8	2.5	-7	-0.85	5.95	49	0.72
7	22	4.0	7	0.65	4.55	49	0.42
8	15	3.3	0	-0.05	0	0	0.003
Σ	120	27.2	0	0	17.30	166	1.843

r = \frac{\sum (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sqrt{\sum (x_{i} - \bar{x})^{2} \cdot \sum (y_{i} - \bar{y})^{2}}} = \frac{17.30}{\sqrt{166 \times 1.843}} = \frac{17.30}{\sqrt{305.9}} = \frac{17.30}{17.49} = 0.989

Significance test (t)

t = \frac{r \sqrt{n - 2}}{\sqrt{1 - r^{2}}} = \frac{0.989 \sqrt{6}}{\sqrt{1 - 0.978}} = \frac{0.989 \times 2.449}{\sqrt{0.022}} = \frac{2.422}{0.148} = 16.36, d f = 6

95% CI (Fisher-z)

z_{r} = \tanh^{- 1} (r) = \frac{1}{2} \ln \frac{1 + r}{1 - r} = \frac{1}{2} \ln \frac{1.989}{0.011} = 2.693

C I_{z} : 2.693 \pm \frac{1.96}{\sqrt{n - 3}} = 2.693 \pm 0.981 \Rightarrow [1.712, 3.674] \Rightarrow r \in [.936, .998]

.989

r

.978

r²

16.36

t(6)

<.001

p

🔴 Strong positive correlation, r = .989, p < .001

APA-7

Study hours and GPA were strongly positively correlated, r(6) = .99, p < .001, 95% CI [.94, 1.00]. Study hours accounted for 97.8% of the variance in GPA.

📈 6 · Simple Linear Regression

Predict GPA from Study Hours

Regression

Research Question: Predict GPA from study hours (same data as correlation example).

OLS Coefficients

b_{1} = \frac{\sum (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum (x_{i} - \bar{x})^{2}} = \frac{17.30}{166} = 0.1042

b_{0} = \bar{y} - b_{1} \bar{x} = 3.40 - 0.1042 \times 15 = 3.40 - 1.563 = 1.837

\hat{y} = 1.837 + 0.1042 x

R², F, SE

S S_{R e s} = \sum (y_{i} - {\hat{y}}_{i})^{2} = 0.040, S S_{T o t} = 1.843

R^{2} = 1 - \frac{S S_{R e s}}{S S_{T o t}} = 1 - \frac{0.040}{1.843} = 0.978

F (1, 6) = \frac{R^{2} / (p - 1)}{(1 - R^{2}) / (n - p)} = \frac{0.978 / 1}{0.022 / 6} = \frac{0.978}{0.00367} = 267.0

S E_{b_{1}} = \sqrt{\frac{S S_{R e s} / (n - 2)}{\sum (x - \bar{x})^{2}}} = \sqrt{\frac{0.040 / 6}{166}} = \sqrt{0.0000402} = 0.00634

Coefficient	B	SE	t	p	95% CI
b₀ (Intercept)	1.837	0.107	17.17	<.001	[1.575, 2.099]
b₁ (Hours)	0.1042	0.00634	16.44	<.001	[0.0887, 0.1197]

0.978

R²

267.0

F(1,6)

<.001

p

0.1042

β (slope)

✅ Every additional hour → GPA +0.104. R² = 97.8%.

APA-7

Simple linear regression: GPA = 1.84 + 0.10 × Hours, F(1, 6) = 267.0, p < .001, R² = .978.

🗂️ 7 · Chi-Square Test

Smoking vs Cancer

Categorical

Research Question: Is smoking associated with cancer diagnosis (N=200)?

	Cancer: Yes	Cancer: No	Row Total
Smoker	50 (E=32)	30 (E=48)	80
Non-Smoker	30 (E=48)	90 (E=72)	120
Col Total	80	120	200

Expected: E = (R × C) / N

E_{11} = \frac{80 \times 80}{200} = 32, E_{12} = \frac{80 \times 120}{200} = 48

E_{21} = \frac{120 \times 80}{200} = 48, E_{22} = \frac{120 \times 120}{200} = 72

χ²

χ^{2} = \sum \frac{(O - E)^{2}}{E} = \frac{(50 - 32)^{2}}{32} + \frac{(30 - 48)^{2}}{48} + \frac{(30 - 48)^{2}}{48} + \frac{(90 - 72)^{2}}{72}

= \frac{324}{32} + \frac{324}{48} + \frac{324}{48} + \frac{324}{72} = 10.125 + 6.75 + 6.75 + 4.5 = 28.125

Cramér's V

V = \sqrt{\frac{χ^{2}}{N \cdot min (r - 1, c - 1)}} = \sqrt{\frac{28.125}{200 \times 1}} = \sqrt{0.1406} = 0.375 (medium-large)

28.125

χ²

1

df

<.001

p

0.375

Cramér's V

🔴 Reject H₀ — p < .001. Smoking and cancer are significantly associated.

APA-7

A chi-square test of independence indicated a significant association between smoking and cancer, χ²(1, N = 200) = 28.13, p < .001, V = .375.

🔒 8 · Cronbach's Alpha

5-Item Job Satisfaction Scale

Reliability

Research Question: Is this 5-item Likert satisfaction scale internally consistent?

Resp.	Q1	Q2	Q3	Q4	Q5	Sum
1	4	3	5	4	3	19
2	2	2	3	2	2	11
3	5	4	5	5	4	23
4	3	3	4	3	3	16
5	1	2	2	1	2	8
6	4	5	4	4	5	22
s²	2.07	1.27	1.27	2.27	1.07	σ²_T=28.56

Cronbach's α

α = \frac{k}{k - 1} (1 - \frac{\sum s_{i}^{2}}{σ_{T}^{2}})

\sum s_{i}^{2} = 2.07 + 1.27 + 1.27 + 2.27 + 1.07 = 7.95

α = \frac{5}{4} (1 - \frac{7.95}{28.56}) = 1.25 \times (1 - 0.2783) = 1.25 \times 0.7217 = 0.902

0.902

α

5

Items

Excellent

Rating

✅ α = .902 — Excellent internal consistency (α > .90)

APA-7

The internal consistency of the 5-item job satisfaction scale was excellent, α = .90, exceeding the recommended threshold of .70.

🔁 9 · Repeated-Measures ANOVA

Memory Scores: 3 Time Points

RM-ANOVA

Research Question: Do memory scores change over 3 time points in 5 participants?

Subj.	T1	T2	T3	P̄
1	4	6	8	6.0
2	5	7	9	7.0
3	3	5	7	5.0
4	6	8	10	8.0
5	2	4	6	4.0
T̄ⱼ	4.0	6.0	8.0	6.0 (Grand)

SS Decomposition

S S_{T i m e} = n \sum_{j} ({\bar{T}}_{j} - {\bar{x}}_{. .})^{2} = 5 [(4 - 6)^{2} + (6 - 6)^{2} + (8 - 6)^{2}] = 5 [4 + 0 + 4] = 40

S S_{S u b j e c t s} = k \sum_{i} ({\bar{P}}_{i} - {\bar{x}}_{. .})^{2} = 3 [(6 - 6)^{2} + (7 - 6)^{2} + (5 - 6)^{2} + (8 - 6)^{2} + (4 - 6)^{2}] = 3 [0 + 1 + 1 + 4 + 4] = 30

S S_{E r r o r} = S S_{T o t a l} - S S_{T i m e} - S S_{S u b j e c t s}

S S_{T o t a l} = \sum (x_{i j} - {\bar{x}}_{. .})^{2} = 70, S S_{E r r o r} = 70 - 40 - 30 = 0

F = \frac{M S_{T i m e}}{M S_{E r r o r}} = \frac{40 / 2}{0 / 8} \to \infty (perfect linear trend — theoretical)

η_{p a r t i a l}^{2} = \frac{S S_{T i m e}}{S S_{T i m e} + S S_{E r r o r}} = \frac{40}{40 + 0} = 1.000

📌 In real data: use MindStat for Mauchly's test + GG/HF corrections.

APA-7 (template)

A one-way RM-ANOVA indicated a significant effect of time, F(2, 8) = XX, p < .001, η²_p = .XX. Pairwise Bonferroni comparisons: all periods differ significantly.

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⚡ 10 · Power Analysis & Sample Size

Sample Size for Independent t-test

Power

Scenario: Two-group RCT, d = 0.5 (medium), α = .05, power = .80.

Cohen's n per group

n = \frac{(z_{α / 2} + z_{β})^{2} \times 2}{d^{2}}

z_{α / 2} = z_{.025} = 1.96, z_{β} = z_{.20} = 0.842

n = \frac{(1.96 + 0.842)^{2} \times 2}{{0.5}^{2}} = \frac{(2.802)^{2} \times 2}{0.25} = \frac{7.851 \times 2}{0.25} = \frac{15.70}{0.25} = 62.8 \approx 63 per group

d	80% Power	90% Power	95% Power
0.2 (small)	197	265	327
0.5 (medium)	63	85	105
0.8 (large)	26	34	42
1.0 (very large)	17	22	27

63

n per group

126

Total N

80%

Power

APA-7

An a priori power analysis indicated that 63 participants per group (N = 126) were required to detect d = 0.50 with 80% power at α = .05 (two-tailed).

▶ Power Analysis in MindStat

📐 11 · Normality Testing (Shapiro-Wilk)

11a · Data Consistent with Normality

Normality

Research Question: Are BP readings normally distributed before applying a t-test?

i	1	2	3	4	5	6	7	8	9	10
BP	118	120	122	124	124	126	128	130	133	135

Shapiro-Wilk W

W = \frac{b^{2}}{S S}, b = \sum_{i = 1}^{⌊ n / 2 ⌋} a_{i} (x_{(n + 1 - i)} - x_{(i)}), S S = \sum (x_{i} - \bar{x})^{2}

1

Sort ascending, compute x̄ and SS
$\bar{x} = 126.0, S S = (118 - 126)^{2} + \dots + (135 - 126)^{2} = 274$
2

Coefficients (n=10): a₁=0.5739, a₂=0.3291, a₃=0.2141, a₄=0.1224, a₅=0.0399
$x_{(10)} - x_{(1)} = 17, x_{(9)} - x_{(2)} = 13, x_{(8)} - x_{(3)} = 8, x_{(7)} - x_{(4)} = 4, x_{(6)} - x_{(5)} = 2$ $b = 0.5739 (17) + 0.3291 (13) + 0.2141 (8) + 0.1224 (4) + 0.0399 (2) = 9.756 + 4.278 + 1.713 + 0.490 + 0.080 = 16.317$
3

$W = \frac{(16.317)^{2}}{274} = \frac{266.2}{274} = 0.972$

0.972

W

.895

p

10

n

⚪ Fail to Reject H₀ — Normality confirmed, W(10) = 0.97, p = .895

APA-7

Shapiro-Wilk testing confirmed normality, W(10) = 0.97, p = .895.

11b · Detecting Non-Normality

Non-Normal

Research Question: Are ER waiting times normally distributed?

i	1	2	3	4	5	6	7	8	9	10
Min	5	8	10	12	15	20	35	52	78	95

1

$\bar{x} = 33.0, S S = 9,106$ $Differences: x_{(10)} - x_{(1)} = 90, 70, 42, 23, 5$ $b = 0.5739 (90) + 0.3291 (70) + 0.2141 (42) + 0.1224 (23) + 0.0399 (5) = 51.65 + 23.04 + 8.99 + 2.82 + 0.20 = 86.70$ $W = \frac{(86.70)^{2}}{9106} = \frac{7516.9}{9106} = 0.825$

0.825

W

.030

p

<.05

Violation

🔴 Reject H₀ — Data NOT normal, W(10) = 0.83, p = .030 → use non-parametric test

💡 Decision rule: p > .05 → normality assumed; p ≤ .05 → non-parametric or transform data.

▶ Test Normality in MindStat

🔀 12 · Two-Way ANOVA (Factorial Design)

Teaching Method × Class Size Interaction

ANOVA

Research Question: Does teaching method's effect depend on class size? 2×2 factorial, N=24.

	Small	Large	Row Mean
Traditional	68.0	70.0	69.0
Active	82.0	74.0	78.0
Col Mean	75.0	72.0	73.0 (GM)

SS Formulas

S S_{A} = n \cdot k_{B} \cdot \sum ({\bar{M}}_{A_{i}} - GM)^{2}, S S_{B} = n \cdot k_{A} \cdot \sum ({\bar{M}}_{B_{j}} - GM)^{2}

S S_{A \times B} = n \cdot \sum (M_{i j} - {\bar{M}}_{A_{i}} - {\bar{M}}_{B_{j}} + GM)^{2}

1

Main Effect A — Teaching Method
$S S_{A} = 6 \times 2 \times [(69.0 - 73.0)^{2} + (78.0 - 73.0)^{2}] = 12 \times [16 + 25] = 492$ $F_{A} = \frac{492 / 1}{30.25} = 16.27, p = .001$
2

Main Effect B — Class Size
$S S_{B} = 6 \times 2 \times [(75.0 - 73.0)^{2} + (72.0 - 73.0)^{2}] = 12 \times [4 + 1] = 60$ $F_{B} = \frac{60 / 1}{30.25} = 1.98, p = .174$
3

Interaction A×B
$Residuals: (68 - 69 - 75 + 73) = - 3, (70 - 69 - 72 + 73) = 2, (82 - 78 - 75 + 73) = 2, (74 - 78 - 72 + 73) = - 3$ $S S_{A \times B} = 6 \times [(- 3)^{2} + 2^{2} + 2^{2} + (- 3)^{2}] = 6 \times 26 = 156, F_{A \times B} = \frac{156}{30.25} = 5.16, p = .034$
4

Interpret: Active Learning gains +14 pts in small classes but only +4 pts in large classes — the interaction is meaningful.

Source	SS	df	MS	F	p	η²_p
Method (A)	492	1	492	16.27	.001	.448
Class Size (B)	60	1	60	1.98	.174	.090
A × B	156	1	156	5.16	.034	.205
Error	605	20	30.25
Total	1313	23

✅ Significant A×B interaction, F(1,20) = 5.16, p = .034

APA-7

A 2×2 ANOVA showed a significant Method × Class Size interaction, F(1, 20) = 5.16, p = .034, η²p = .21. Active Learning gained +14 points in small classes but only +4 in large classes.

▶ Two-Way ANOVA in MindStat

🏅 13 · Spearman Rank Correlation

Study Hours Rank vs Exam Rank

Spearman

Research Question: Monotonic relationship between study hours rank and exam rank (n=8)?

Student	Hours	Rank X	Exam	Rank Y	d = Rₓ−Rᵧ	d²
A	3	1	62	2	−1	1
B	4	2	55	1	1	1
C	5	3	68	3	0	0
D	6	4	74	5	−1	1
E	7	5	71	4	1	1
F	8	6	85	7	−1	1
G	9	7	80	6	1	1
H	10	8	91	8	0	0
Σ					0	6

Spearman's rₛ

r_{s} = 1 - \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)} = 1 - \frac{6 \times 6}{8 (64 - 1)} = 1 - \frac{36}{504} = 1 - 0.071 = 0.929

t-test for rₛ

t = r_{s} \sqrt{\frac{n - 2}{1 - r_{s}^{2}}} = 0.929 \sqrt{\frac{6}{1 - 0.863}} = 0.929 \times \sqrt{43.8} = 0.929 \times 6.62 = 6.15, d f = 6

.929

rₛ

6.15

t(6)

<.001

p

Large

Effect

🔴 Strong monotonic relationship, rₛ(6) = .929, p < .001

APA-7

A Spearman correlation indicated a strong positive relationship, rₛ(6) = .93, p < .001.

💡 Use Spearman when data is ordinal, non-normal, or contains outliers.

▶ Spearman Correlation in MindStat

📊 14 · Multiple Linear Regression

Exam Score ← Study Hours + Sleep Hours

Multiple Reg.

Research Question: Can study hours and sleep hours together predict exam score (n=10)?

i	X₁ (Study)	X₂ (Sleep)	Y (Score)	Ŷ	e = Y−Ŷ
1	4	6	58	55.3	2.7
2	6	7	64	64.7	−0.7
3	8	8	72	74.0	−2.0
4	5	5	55	56.1	−1.1
5	9	8	78	77.4	0.6
6	7	7	70	68.1	1.9
7	10	9	85	83.4	1.6
8	3	6	52	51.9	0.1
9	8	6	68	68.9	−0.9
10	6	8	65	67.2	−2.2
x̄	6.6	7.0	66.7

OLS solution

S_{x_{1} x_{1}} = 44.4, S_{x_{2} x_{2}} = 14.0, S_{x_{1} x_{2}} = 18.0

S_{y x_{1}} = 196.8, S_{y x_{2}} = 97.0

b_{1} = \frac{S_{y x_{1}} S_{x_{2} x_{2}} - S_{y x_{2}} S_{x_{1} x_{2}}}{S_{x_{1} x_{1}} S_{x_{2} x_{2}} - S_{x_{1} x_{2}}^{2}} = \frac{196.8 \times 14 - 97.0 \times 18}{44.4 \times 14 - 18^{2}} = \frac{1009.2}{297.6} = 3.39

b_{2} = \frac{S_{y x_{2}} S_{x_{1} x_{1}} - S_{y x_{1}} S_{x_{1} x_{2}}}{297.6} = \frac{764.4}{297.6} = 2.57

b_{0} = \bar{Y} - b_{1} {\bar{X}}_{1} - b_{2} {\bar{X}}_{2} = 66.7 - 3.39 (6.6) - 2.57 (7.0) = 26.34

\hat{Y} = 26.34 + 3.39 X_{1} + 2.57 X_{2}

Predictor	b	SE	t	p	95% CI	β (std)
Intercept	26.34	4.18	6.30	<.001	[17.1, 35.6]	—
Study Hours	3.39	0.41	8.27	<.001	[2.4, 4.4]	.74
Sleep Hours	2.57	0.74	3.47	.010	[0.9, 4.2]	.32

R², Adjusted R², F

S S_{t o t} = 942.1, S S_{r e s} = 25.6, S S_{r e g} = 916.5

R^{2} = \frac{916.5}{942.1} = .973, R_{a d j}^{2} = 1 - \frac{25.6 / 7}{942.1 / 9} = .965

F (2, 7) = \frac{916.5 / 2}{25.6 / 7} = \frac{458.3}{3.66} = 125.2, p < .001

.973

R²

.965

Adj R²

125.2

F(2,7)

<.001

p

✅ Model explains 97.3% of variance — study hours (β=.74) and sleep hours (β=.32) both significant.

APA-7

Multiple regression: F(2, 7) = 125.2, p < .001, R² = .973. Study hours (β=.74, p<.001) and sleep hours (β=.32, p=.010) both predicted exam scores.

▶ Multiple Regression in MindStat

⚙️ 15 · Logistic Regression

Predict Pass/Fail from Study Hours

Logistic

Research Question: Does study hours predict pass/fail probability (n=20)?

Logistic Model

P (pass) = \frac{1}{1 + e^{- (b_{0} + b_{1} X)}} = \frac{e^{b_{0} + b_{1} X}}{1 + e^{b_{0} + b_{1} X}}

1

Maximum likelihood estimates
$b_{0} = - 8.76 (S E = 3.42), b_{1} = 1.12 (S E = 0.48)$ $\hat{Y} = \frac{1}{1 + e^{- (- 8.76 + 1.12 X)}}$
2

Predicted probabilities

Study hrs 6 7 8 9 10

P(pass) .115 .283 .550 .798 .922

$At X = 8 : P (pass) = \frac{1}{1 + e^{- (- 8.76 + 8.96)}} = \frac{1}{1 + e^{- 0.20}} = \frac{1}{1.819} = 0.55$
3

Odds Ratio (OR)
$O R = e^{b_{1}} = e^{1.12} = 3.06 95 [1.19, 7.87]$ $Each additional study hour multiplies the odds of passing by 3.06×$
4

Model fit
$Overall model: χ^{2} (1) = 14.5, p < .001, Nagelkerke R^{2} = .72$ $Wald test: χ^{2} (1) = 5.44, p = .020 (for b_{1})$ $Correctly classified: 85 %$

3.06

OR

<.001

Model p

.72

Nagelkerke R²

85%

Classified

✅ Study hours predicts pass/fail, OR = 3.06, p < .001

APA-7

Logistic regression: Study hours predicted pass/fail, χ²(1) = 14.5, p < .001, OR = 3.06, 95% CI [1.19, 7.87].

▶ Logistic Regression in MindStat

🎯 16 · Effect Sizes

Four Effect Size Measures with Worked Examples

Effect Size

Effect size measures the practical importance of a finding, independent of sample size.

① Cohen's d — for t-tests

Cohen's d

d = \frac{M_{1} - M_{2}}{S D_{p o o l e d}}, S D_{p o o l e d} = \sqrt{\frac{(n_{1} - 1) S D_{1}^{2} + (n_{2} - 1) S D_{2}^{2}}{n_{1} + n_{2} - 2}}

Worked example

S D_{p o o l e d} = \sqrt{\frac{14 (6.5)^{2} + 14 (5.9)^{2}}{28}} = \sqrt{\frac{592 + 487}{28}} = \sqrt{38.5} = 6.21

d = \frac{82 - 76}{6.21} = \frac{6}{6.21} = 0.97 (large)

② Eta-squared η² — for ANOVA

η^{2} = \frac{S S_{b e t w e e n}}{S S_{t o t a l}}, ω^{2} = \frac{S S_{b e t w e e n} - d f_{b e t w e e n} \cdot M S_{w i t h i n}}{S S_{t o t a l} + M S_{w i t h i n}} (less biased)

Example: S S_{b e t w e e n} = 180, S S_{t o t a l} = 520 \Rightarrow η^{2} = \frac{180}{520} = .346 (large)

③ Cramér's V — for χ²

V = \sqrt{\frac{χ^{2}}{n \cdot min (r - 1, c - 1)}}

Example: χ^{2} = 8.64, n = 45, 2 \times 2 table \Rightarrow V = \sqrt{\frac{8.64}{45 \times 1}} = \sqrt{.192} = .438 (large)

④ r (non-parametric effect size)

r = \frac{| z |}{\sqrt{N}}

Example: z = - 2.45, N = 30 \Rightarrow r = \frac{2.45}{\sqrt{30}} = \frac{2.45}{5.48} = .447 (medium-large)

Measure	Use with	Small	Medium	Large
Cohen's d	t-tests	0.20	0.50	0.80
η² (eta-squared)	ANOVA	.01	.06	.14
η²p (partial)	Factorial ANOVA	.01	.06	.14
Cramér's V	Chi-square	.10	.30	.50
r (Pearson/Spearman)	Correlation	.10	.30	.50
r (z/√N)	Mann-Whitney	.10	.30	.50

💡 Cohen's thresholds are benchmarks, not rules — interpret effect sizes in context.

▶ Compute Effect Sizes in MindStat

📏 17 · Confidence Intervals

Three CI Types — Mean, Proportion, Difference

CI

① 95% CI for a Single Mean

Scenario: n=30 BP readings, M=128.4, SD=14.2 → 95% CI

C I = \bar{X} \pm t^{*} \frac{S D}{\sqrt{n}}, S E = \frac{14.2}{\sqrt{30}} = \frac{14.2}{5.477} = 2.593, t_{(29, 0.025)}^{*} = 2.045

C I = 128.4 \pm 2.045 \times 2.593 = 128.4 \pm 5.30 = [123.1, 133.7] mmHg

128.4

Mean

±5.3

Margin

[123.1, 133.7]

95% CI

② 95% CI for a Proportion

Scenario: 142/200 patients satisfied (p̂=0.71) → 95% CI

C I = \hat{p} \pm z^{*} \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}, S E = \sqrt{\frac{0.71 \times 0.29}{200}} = \sqrt{0.001030} = 0.0321, z^{*} = 1.96

C I = 0.71 \pm 1.96 \times 0.0321 = 0.71 \pm 0.063 = [0.647, 0.773]

71.0%

p̂

±6.3%

Margin

[64.7%, 77.3%]

95% CI

③ 95% CI for Difference Between Two Means

Drug (M=85.2, SD=7.4) vs Placebo (M=79.6, SD=8.1), n=20 each → 95% CI for M₁−M₂

S D_{p} = \sqrt{\frac{19 (7.4)^{2} + 19 (8.1)^{2}}{38}} = \sqrt{\frac{1038 + 1246}{38}} = \sqrt{60.1} = 7.75

S E_{d i f f} = S D_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} = 7.75 \sqrt{\frac{2}{20}} = 7.75 \times 0.316 = 2.45

C I = (85.2 - 79.6) \pm t_{(38)}^{*} \times 2.45 = 6.0 - 2.024 \times 2.45 \Rightarrow [1.04, 10.96 points]

6.0

Δ Mean

±4.96

Margin

[1.04, 10.96]

95% CI

✅ CI excludes 0 → significant difference. Drug improves score by 1–11 points.

APA-7

95% CI for the mean difference = [1.04, 10.96]; excludes zero → significant at α = .05.

▶ Confidence Intervals in MindStat

🔄 18 · Mediation Analysis

Mindfulness → Stress → Anxiety (Simple Mediation)

Mediation

Research Question: Does mindfulness reduce anxiety indirectly through stress? N=90.

Path	Description	b	SE	t	p
a	Training → Stress	−6.40	1.30	−4.92	<.001
b	Stress → Anxiety	0.62	0.11	5.64	<.001
c	Total effect (c)	−7.80	1.90	−4.11	<.001
c'	Direct effect (c')	−3.83	1.82	−2.11	.037

Indirect Effect = a × b

Indirect effect = a \times b = (- 6.40) \times (0.62) = - 3.97

Bootstrap 95% CI for a b : [- 6.12, - 1.94] (excludes 0 → significant mediation)

Proportion mediated = \frac{| a b |}{| c |} = \frac{3.97}{7.80} = 50.9 %

−3.97

Indirect

[−6.12, −1.94]

Boot CI

50.9%

Mediated

✅ Significant partial mediation: 51% of training's effect on anxiety is through stress reduction.

APA-7

Simple mediation: ab = −3.97, 95% bootstrap CI [−6.12, −1.94]; 51% of total effect mediated through stress. Partial mediation (direct effect c' = −3.83, p = .037 remains).

💡 Full mediation: c' non-significant. Partial: c' remains significant. Use bootstrap CIs, not Sobel test.

▶ Mediation Analysis in MindStat

📉 19 · Survival Analysis (Kaplan-Meier)

Time-to-Relapse: Drug A vs Drug B

Survival

Research Question: Does Drug A prolong relapse-free survival vs Drug B? (†=censored)

Drug A (n=12)	3	5	8	11	15	20†	22†	24†	24†	26†	28†	30†
Drug B (n=12)	2	4	6	8	10	14	18	20†	22†	24†	26†	28†

1

KM formula: S(tⱼ) = S(tⱼ₋₁) × (1 − dⱼ/nⱼ)

t (months)	nⱼ (at risk)	dⱼ (events)	1−dⱼ/nⱼ	S(t) Drug A
0	12	0	—	1.000
3	12	1	11/12	0.917
5	11	1	10/11	0.833
8	10	1	9/10	0.750
11	9	1	8/9	0.667
15	7	1	6/7	0.571

2

Summary statistics

Median Survival S(12 mo) Events/N

Drug A 22 months 0.750 5/12

Drug B 14 months 0.500 7/12
3

Log-rank test
$χ_{l o g - r a n k}^{2} = \frac{(O_{1} - E_{1})^{2}}{E_{1}} + \frac{(O_{2} - E_{2})^{2}}{E_{2}} = 5.02, d f = 1, p = .025$

22 mo

Median A

14 mo

Median B

5.02

χ²(1)

.025

p

✅ Drug A significantly prolongs survival: 22 vs 14 months, p = .025

APA-7

Kaplan-Meier analysis: Drug A (median = 22 mo) vs Drug B (median = 14 mo), log-rank χ²(1) = 5.02, p = .025. At 12 months, 75% vs 50% remained relapse-free.

💡 Censored observations (†) contribute data up to their last follow-up. KM correctly handles censoring.

▶ Survival Analysis in MindStat

📐 20 · Sample Size

How Many Participants Do I Need?

Sample Size

Research Question: Master's student comparing two teaching methods.

Parameter	Value	Why
d	0.5	متوسط (Cohen 1988)
α	0.05	معياري
Power	0.80	الحد الأدنى الموصى به

Formula

n = \frac{2 (z_{α / 2} + z_{β})^{2}}{d^{2}}

1

z-values:
$z_{α / 2} = z_{0.025} = 1.96 z_{β} = z_{0.20} = 0.842$
2

n per group:
$n = \frac{2 (1.96 + 0.842)^{2}}{(0.5)^{2}} = \frac{2 \times 7.857}{0.25} = \frac{15.713}{0.25} \approx 63$
3

Attrition buffer:
$n_{a d j} = \frac{63}{1 - 0.15} = \frac{63}{0.85} \approx 75 per group$
4

Total N:
$N_{t o t a l} = 75 \times 2 = 150 participants$

63

Min/group

75

Recruit/group

150

Total N

80%

Power

APA-7

APA-7 power statement.

▶ Calculate in MindStat

📋 Open your SPSS file

From SPSS to MindStat

Migration

Why:

How:

Open MindStat →

💻 21 · SPSS Output

Interpreting SPSS t-test Output

SPSS

Research Question: How to read the SPSS Independent Samples Test table.

	Levene F	Sig.	t	df	p	Δ̄	SE	CI₋	CI₊
Equal var.	2.14	.149	−3.42	48	.001	−4.80	1.40	−7.63	−1.97
Unequal var.			−3.42	45.7	.001	−4.80	1.40	−7.64	−1.96

1

Levene's Test: F = 2.14, p = .149 → use Equal variances row.
2

t, df, p: t(48) = −3.42, p = .001 → significant.
3

Cohen's d:
$d = \frac{| t |}{\sqrt{n_{1} + n_{2}}} \times \sqrt{2} = \frac{3.42}{\sqrt{50}} \times \sqrt{2} = \frac{3.42}{7.07} \times 1.414 = 0.68 (large)$
4

95% CI [−7.63, −1.97]: CI excludes zero → significant.

−3.42

t(48)

.001

p

−4.80

Δ̄

0.68

Cohen's d

🔴 Reject H₀ — p = .001 < .05

APA-7

APA-7 result.

▶ Run t-test in MindStat

✍️ 22 · APA-7 Writing

APA-7 Result Sentences

APA-7

Why: APA-7 rules for reporting statistics.

الخطأ	الصحيح
p = 0.043	p = .043
p = 0.000	p < .001
t = 3.42 (df=48)	t(48) = 3.42
No effect size	Add d / η² / r / R²
R² = 0.47	R² = .47

t

t-test:

t-test APA sentence.
F

ANOVA:

ANOVA APA sentence.
r

Correlation:

Correlation APA sentence.
R²

Regression:

Regression APA sentence.

p = .043

✓

p < .001

✓

t(48)

✓

η² / d

✓

APA-7

APA-7 methods sentence.

▶ Generate APA in MindStat

📋 23 · Likert Scale

Likert Scale — Complete Analysis

Likert

Research Question: 5-item Likert scale for 8 employees.

م	ف1	ف2	ف3	ف4	ف5	∑
1	4	3	4	5	4	20
2	2	2	3	2	3	12
3	5	4	5	4	5	23
4	3	3	3	4	3	16
5	4	5	4	5	4	22
6	2	3	2	3	2	12
7	5	4	5	5	4	23
8	3	4	3	3	4	17

1

Descriptives:
${\bar{x}}_{t o t a l} = \frac{20 + 12 + 23 + 16 + 22 + 12 + 23 + 17}{8} = \frac{145}{8} = 18.13, S D = 4.36$
2

Item mean:
${\bar{x}}_{i t e m} = \frac{18.13}{5} = 3.63 out of 5$
3

Cronbach's α:
$α = \frac{k}{k - 1} (1 - \frac{\sum s_{i}^{2}}{s_{T}^{2}}) = \frac{5}{4} (1 - \frac{5.33}{19.0}) = \frac{5}{4} (0.720) = 0.90$
4

Interpretation: 3.63 falls in High range (3.41–5.00).

18.13

M total

3.63

M item

0.90

α

HIGH

Level

APA-7

APA-7 Likert result.

▶ Calculate Alpha

📊 24 · Multiple Regression

Multiple Regression — Step by Step

Multiple Reg.

Research Question: Predict job performance from 3 predictors.

م	Y	X₁	X₂	X₃
1	72	3	40	3.5
2	85	7	60	4.2
3	90	10	80	4.8
4	68	2	30	3.0
5	78	5	55	3.8
6	92	12	90	5.0
7	75	4	45	3.6
8	88	8	70	4.5
9	65	1	25	2.8
10	82	6	65	4.0

Model

\hat{Y} = b_{0} + b_{1} X_{1} + b_{2} X_{2} + b_{3} X_{3}

\hat{Y} = 48.2 + 1.83 X_{1} + 0.21 X_{2} + 3.94 X_{3}

1

Coefficients:

المتنبئ	b	SE	β	t	p
b₀	48.2	4.1	—	11.76	<.001
X₁	1.83	0.42	.52	4.36	.003
X₂	0.21	0.09	.28	2.33	.052
X₃	3.94	1.21	.38	3.26	.014

2

Model fit:
$R^{2} = .973, R_{a d j}^{2} = .959, F (3, 6) = 71.9, p < .001$
3

Strongest predictor: X₁ β = .52 → strongest predictor.
4

Interpret b₁: +1 year experience → +1.83 performance points.

.973

R²

71.9

F(3,6)

<.001

p

β=.52

X₁ best

🟢 Model significant — p < .001

APA-7

APA-7 multiple regression result.

▶ Run Regression

🌀 26 · Mauchly's Sphericity Test

Mauchly's Test of Sphericity

RM-ANOVA

Research Question: Do we meet the sphericity assumption before an RM-ANOVA?

P	A	B	C	D
1	8	7	1	6
2	9	5	2	5
3	6	2	3	8
4	5	3	1	9
5	8	4	5	8
6	7	5	6	7
7	10	2	7	2
8	12	6	8	1

Mauchly's W

W = \frac{det (S)}{{[\frac{tr (S)}{k - 1}]}^{k - 1}}

χ^{2} = - [n - 1 - \frac{2 (k - 1)^{2} + (k - 1) + 2}{6 (k - 1)}] \ln W, d f = \frac{(k - 1) k}{2} - 1

1

Orthonormal contrasts: Transform to k−1 = 3 orthonormal contrast scores.
2

Covariance matrix S:
$tr (S) = 21.91, det (S) = 53.1$
3

W:
$W = \frac{53.1}{{(21.91 / 3)}^{3}} = \frac{53.1}{389.6} = 0.1362$
4

χ² approximation:
$χ^{2} = - [8 - 1 - \frac{2 (9) + 3 + 2}{18}] \ln (0.1362) = - 6.72 \times (- 1.994) = 11.41, d f = 5$
5

Decision: p = .044 → violated → apply GG/HF correction.

0.1362

W

11.41

χ²

5

df

.044

p

🔴 Sphericity violated — W = 0.136, χ²(5) = 11.41, p = .044.

APA-7

Mauchly's test indicated the sphericity assumption was violated, χ²(5) = 11.41, p = .044; degrees of freedom were corrected using Greenhouse-Geisser estimates (ε = .62).

▶ Open in MindStat

💻 27 · Reading SPSS ANOVA Output

Interpreting SPSS ANOVA Output

SPSS

Research Question: How to read the SPSS ANOVA and homogeneity tables.

Source	SS	df	MS	F	Sig.
Between	613.73	2	306.87	36.24	.000
Within	101.60	12	8.47
Total	715.33	14

1

Levene's Test first: Levene Sig. > .05 → standard ANOVA valid.
2

F and Sig.: F(2,12) = 36.24, and .000 means p < .001, not p = 0.
3

Effect size η²:
$η^{2} = \frac{S S_{B e t w e e n}}{S S_{T o t a l}} = \frac{613.73}{715.33} = 0.858 (large)$
4

Post-hoc: Tukey 'Multiple Comparisons': pairs with Sig. < .05 differ.

36.24

F(2,12)

<.001

p (from .000)

0.858

η²

🔴 Significant — F(2,12) = 36.24, p < .001, η² = .86.

APA-7

A one-way ANOVA showed a significant effect, F(2, 12) = 36.24, p < .001, η² = .86. (SPSS's .000 is reported as p < .001.)

▶ Open in MindStat

📊 28 · Welch's ANOVA

Welch's ANOVA — Unequal Variances

ANOVA

Research Question: Does course format affect scores despite wildly unequal variances?

A	B	C
58	50	78
62	100	82
59	55	79
61	95	81
60	75	80

Standard F

F = \frac{M S_{B}}{M S_{W}} = \frac{S S_{B} / (k - 1)}{S S_{W} / (N - k)}

Welch's F

F_{W} = \frac{\sum w_{i} ({\bar{x}}_{i} - {\bar{x}}_{w})^{2} / (k - 1)}{1 + \frac{2 (k - 2)}{k^{2} - 1} \sum \frac{(1 - w_{i} / \sum w)^{2}}{n_{i} - 1}}, w_{i} = \frac{n_{i}}{s_{i}^{2}}

1

Means and variances:
${\bar{x}}_{A} = 60.00, s_{A}^{2} = 2.50 {\bar{x}}_{B} = 75.00, s_{B}^{2} = 512.50 {\bar{x}}_{C} = 80.00, s_{C}^{2} = 2.50$
B's variance is 200× A's or C's.
2

Levene's test first:
$F_{L e v e n e} (2, 12) = 12.96, p = .001$
Variances significantly unequal → prefer Welch.
3

Standard ANOVA (for contrast):
$S S_{B} = 1083.33, S S_{W} = 2070.00, F (2, 12) = \frac{541.67}{172.50} = 3.14, p = .080$
Misleading 'not significant' — B's variance inflates the pooled error.
4

Welch's ANOVA:
$w_{A} = 2.00, w_{B} = 0.0098, w_{C} = 2.00, {\bar{x}}_{w} = 70.01$ $F_{W} (2, 7.12) = \frac{200.12}{1.0936} = 182.99, p < .001$
B is down-weighted to near-zero; F jumps from 3.14 to 182.99.
5

Decision: Report Welch's F — the standard F would be a Type II error here.

182.99

F_Welch(2,7.12)

<.001

p (Welch)

3.14

F standard

.080

p standard

🔴 Significant by Welch's ANOVA — F(2, 7.12) = 182.99, p < .001.

APA-7

Because Levene's test indicated unequal variances, a Welch's ANOVA was conducted, F(2, 7.12) = 182.99, p < .001; the standard ANOVA alone would have misleadingly suggested no effect, F(2, 12) = 3.14, p = .080.

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🧪 29 · Levene's Test

Levene's Test — Variance Homogeneity

Assumption

Research Question: Do novice and expert typists have equal reaction-time variance?

Novice	Expert
812	430
940	465
705	410
1050	455
880	398
690	448
965	421

Levene's F

z_{i j} = | x_{i j} - {\bar{x}}_{i} |, F = \frac{\sum_{i} n_{i} ({\bar{z}}_{i} - \bar{z})^{2} / (k - 1)}{\sum_{i} \sum_{j} (z_{i j} - {\bar{z}}_{i})^{2} / (N - k)}

1

Means and spread:
${\bar{x}}_{N o v i c e} = 863.1, S D_{N o v i c e} = 134.8 {\bar{x}}_{E x p e r t} = 432.4, S D_{E x p e r t} = 24.6$
Experts are faster AND far more consistent.
2

Transform then ANOVA on z: Levene's test = an ANOVA run on |deviations|.
3

F:
$F (1, 12) = 12.67, p = .0039$
4

Decision: Violated → use Welch's t-test (MindStat's unconditional default).

12.67

F(1,12)

.0039

p

134.8 / 24.6

SD ratio

🔴 Variances significantly unequal — F(1,12) = 12.67, p = .004.

APA-7

Levene's test indicated the homogeneity-of-variance assumption was violated, F(1, 12) = 12.67, p = .004; a Welch's t-test was used for the comparison.

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🔢 30 · Wilcoxon Signed-Rank Test

Wilcoxon Signed-Rank Test

Non-parametric

Research Question: Did coping scores change after the program, without assuming normality?

#	Before	After	Diff
1	60	66	+6
2	58	62	+4
3	65	74	+9
4	72	70	−2
5	55	62	+7
6	63	66	+3
7	59	64	+5
8	68	76	+8
9	61	62	+1

W+ / W−

W^{+} = \sum rank (| d_{i} |) for d_{i} > 0, W^{-} = \sum rank (| d_{i} |) for d_{i} < 0, W = min (W^{+}, W^{-})

1

Rank |d|:
$| d | : 1, 2, 3, 4, 5, 6, 7, 8, 9 \to ranks 1, 2, 3, 4, 5, 6, 7, 8, 9 (each unique)$
2

Sum ranks by sign:
$W^{+} = 6 + 4 + 9 + 7 + 3 + 5 + 8 + 1 = 43, W^{-} = 2, W = min (43, 2) = 2$
Only employee #4 regressed.
3

Exact test: Exact distribution used (n<50, no ties).
$p_{exact} (W = 2, n = 9) = .0117 (two-tailed)$
4

Decision: p = .012 → reject H₀ → significant change.

2

W

9

n

.012

p

🔴 Significant — W = 2, p = .012 (exact).

APA-7

A Wilcoxon signed-rank test showed a significant increase in coping scores, W = 2, p = .012 (exact, two-tailed).

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🔢 31 · Friedman Test

Friedman Test — Non-parametric RM

Non-parametric

Research Question: Do the three study techniques differ in recall, without assuming normality?

P	Re-reading	Practice Testing	Spaced Practice
1	12	15	17
2	14	16	18
3	10	14	16
4	16	18	19
5	13	15	17
6	17	14	16
7	15	17	14
8	12	15	17

Friedman χ²

χ^{2} = \frac{12}{N k (k + 1)} \sum_{j = 1}^{k} R_{j}^{2} - 3 N (k + 1), d f = k - 1

1

Rank within each row: Not every participant follows the overall pattern.
2

Rank sums:
$R_{Re-reading} = 11, R_{Practice Testing} = 16, R_{Spaced Practice} = 21$
3

χ²:
$χ^{2} = \frac{12}{8 (3) (4)} (11^{2} + 16^{2} + 21^{2}) - 3 (8) (4) = 0.125 (818) - 96 = 6.25, d f = 2$
4

Decision + post-hoc: Reject H₀; Spaced Practice > Re-reading survives post-hoc correction.

6.25

χ²(2)

.044

p

8

N

🔴 Significant — χ²(2) = 6.25, p = .044.

APA-7

A Friedman test showed a significant difference across techniques, χ²(2) = 6.25, p = .044, N = 8; spaced practice beat re-reading on post-hoc (p_adj < .001).

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🗂️ 32 · McNemar's Test

McNemar's Test — Paired Proportions

Categorical

Research Question: Did training change the pass rate for the same 20 candidates?

	After: Pass	After: Fail
Before: Pass	6 (a)	2 (b)
Before: Fail	9 (c)	3 (d)

McNemar χ²

χ^{2} = \frac{(b - c)^{2}}{b + c} (uncorrected), χ_{c c}^{2} = \frac{(| b - c | - 1)^{2}}{b + c} (continuity-corrected)

1

Discordant pairs: b=2 (regressed), c=9 (improved).
2

Uncorrected:
$χ^{2} = \frac{(2 - 9)^{2}}{2 + 9} = \frac{49}{11} = 4.455, p = .035$
3

Continuity-corrected:
$χ_{c c}^{2} = \frac{(| 2 - 9 | - 1)^{2}}{2 + 9} = \frac{36}{11} = 3.273, p = .070$
The correction changes the conclusion here.
4

Decision: Corrected result is not significant — the uncorrected test overstates it.

3.27

χ² (cc)

.070

p (cc)

4.5

OR

20

N

🟢 Not significant (corrected) — χ²(1) = 3.27, p = .070.

APA-7

A McNemar's test with continuity correction showed no significant change in pass rate, χ²(1) = 3.27, p = .070 (OR = 4.50); the uncorrected test would have overstated the evidence.

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🔗 33 · Partial Correlation

Partial Correlation

Correlation

Research Question: Does exercise still relate to blood pressure once age is controlled for?

P	X	Y	Z
1	1	148	58
2	2	145	62
3	2	150	45
4	3	138	60
5	4	142	50
6	4	130	40
7	5	128	48
8	6	135	35
9	6	120	42
10	7	118	30
11	8	115	33
12	9	110	28

Partial r

r_{X Y . Z} = \frac{r_{X Y} - r_{X Z} r_{Y Z}}{\sqrt{(1 - r_{X Z}^{2}) (1 - r_{Y Z}^{2})}}

1

Zero-order correlations:
$r_{X Y} = - .939, r_{X Z} = - .869, r_{Y Z} = .778$
Classic confound pattern: all three pairs are correlated.
2

Partial r:
$r_{X Y . Z} = \frac{- .939 - (- .869) (.778)}{\sqrt{(1 - .755) (1 - .605)}} = \frac{- .263}{\sqrt{.245 \times .395}} = \frac{- .263}{.311} = - .846$
3

Significance test (df=n−3):
$t = r \sqrt{\frac{d f}{1 - r^{2}}} = - .846 \sqrt{\frac{9}{1 - .716}} = - 4.765, d f = 9, p = .001$
4

Decision: Relationship survives controlling for age, though attenuated.

−.846

Partial r

−4.765

t(9)

.001

p

−.939

Zero-order r

🔴 Significant — partial r(9) = −.846, p = .001.

APA-7

A partial correlation controlling for age showed exercise remained significantly related to blood pressure, r(9) = −.846, p = .001 (zero-order r = −.939).

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🔗 34 · Point-Biserial Correlation

Point-Biserial Correlation

Correlation

Research Question: How strong is the association between group and outcome, expressed as a correlation?

P	Group	Score
1	0	72
2	0	68
3	0	75
4	0	70
5	0	74
6	0	69
7	1	82
8	1	88
9	1	79
10	1	85
11	1	90
12	1	84

Point-biserial r

r_{p b} = \frac{\sum (g_{i} - \bar{g}) (y_{i} - \bar{y})}{\sqrt{\sum (g_{i} - \bar{g})^{2} \sum (y_{i} - \bar{y})^{2}}}

1

Group means:
${\bar{y}}_{C o n t r o l} = 71.33, {\bar{y}}_{T r e a t m e n t} = 84.67$
2

Compute r:
$r_{p b} = .904$
3

Significance test:
$t = r_{p b} \sqrt{\frac{d f}{1 - r_{p b}^{2}}} = .904 \sqrt{\frac{10}{1 - .818}} = 6.704, d f = 10, p < .001$
4

Same test as t-test: Identical t/df to the independent t-test on the same data.

.904

r_pb

6.704

t(10)

<.001

p

.818

r²_pb

🔴 Significant — r_pb(10) = .904, p < .001.

APA-7

A point-biserial correlation showed a strong association, r_pb(10) = .904, p < .001 (Treatment M = 84.67 vs. Control M = 71.33).

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🔬 35 · Exploratory Factor Analysis

Exploratory Factor Analysis

Reliability

Research Question: Do the 6 items form one factor or two?

	P1	P2	P3	C1	C2	C3
P1	1.00	.65	.62	.20	.18	.22
P2	.65	1.00	.68	.15	.20	.17
P3	.62	.68	1.00	.22	.19	.21
C1	.20	.15	.22	1.00	.64	.60
C2	.18	.20	.19	.64	1.00	.66
C3	.22	.17	.21	.60	.66	1.00

KMO & Bartlett

K M O_{i} = \frac{\sum_{j \neq i} r_{i j}^{2}}{\sum_{j \neq i} r_{i j}^{2} + \sum_{j \neq i} a_{i j}^{2}}, χ_{B a r t l e t t}^{2} = - [(N - 1) - \frac{2 p + 5}{6}] \ln | R |

1

Gatekeeper checks:
$K M O_{o v e r a l l} = .744 (items .708 – .780) χ_{B a r t l e t t}^{2} (15) = 555.13, p < .001$
Both gatekeepers pass — proceed to extraction.
2

Eigenvalues:
$λ : 2.864, 1.705, 0.410, 0.390, 0.353, 0.278 (\sum λ = 6)$
Exactly 2 factors survive Kaiser's rule.

3

Varimax rotation:

Item	F1	F2	h²
P1	.124	.853	.742
P2	.082	.887	.793
P3	.130	.866	.766
C1	.851	.109	.735
C2	.878	.105	.781
C3	.858	.121	.750

Clean 2-factor structure — pay and coworker satisfaction are separate.

4

Decision: Score as 2 subscales, not 1 total.

.744

KMO

555.13

Bartlett χ²

2

Factors

75.1%

Var. explained

🔴 Two-factor structure confirmed (KMO=.744, Bartlett p<.001).

APA-7

An EFA (PCA extraction, varimax rotation) confirmed a clean 2-factor structure, KMO=.744, Bartlett χ²(15)=555.13, p<.001, explaining 75.1% of variance.

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📊 36 · ANCOVA

ANCOVA — Adjusting for a Covariate

ANOVA

Research Question: Does method affect post-test scores after adjusting for pre-test?

A: Pre	A: Post	B: Pre	B: Post
60	80	59	68
65	75	63	75
58	85	61	66
70	83	66	80
62	73	60	64
64	87	68	78

Adjusted mean

{\bar{Y}}_{j}^{'} = {\bar{Y}}_{j} - b_{w} ({\bar{X}}_{j} - {\bar{X}}_{g r a n d}), F = \frac{M S_{g r o u p . a d j}}{M S_{e r r o r . a d j}}

1

Group means:
${\bar{X}}_{A, p r e} = 63.17, {\bar{X}}_{B, p r e} = 62.83, (grand \bar{X} = 63.00) {\bar{Y}}_{A, p o s t} = 80.50, {\bar{Y}}_{B, p o s t} = 71.83$
Groups started nearly level; ANCOVA still gives the correct adjustment.
2

Pooled slope:
$b_{w} = 0.6945$
3

Adjusted means:
${\bar{Y}}_{A}^{'} = 80.50 - 0.6945 (63.17 - 63.00) = 80.38 {\bar{Y}}_{B}^{'} = 71.83 - 0.6945 (62.83 - 63.00) = 71.95$
4

F-test:
$S S_{c o v a r i a t e} = 73.16, S S_{g r o u p . a d j} = 225.33, S S_{e r r o r . a d j} = 307.18 (d f = 9)$ $F (1, 9) = \frac{225.33 / 1}{307.18 / 9} = 6.60, p = .030, η_{p}^{2} = .423$
5

Decision: Method A > Method B after adjustment.

80.38 / 71.95

Adj. means

6.60

F(1,9)

.030

p

.423

η²_p

🔴 Significant — F(1,9) = 6.60, p = .030.

APA-7

A one-way ANCOVA with pre-test as covariate showed a significant effect of method, F(1, 9) = 6.60, p = .030, η²p = .42 (Method A adjusted M = 80.38 vs. Method B M = 71.95).

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💻 37 · SPSS Regression Output

Interpreting SPSS Regression Output

SPSS

Research Question: How to read the SPSS regression Model Summary, ANOVA, and Coefficients tables.

Model	B	SE	Beta	t	Sig.
(Constant)	15.20	4.10		3.71	.001
Experience	1.85	0.32	.612	5.78	<.001
Education	2.10	0.68	.331	3.09	.005

Model Summary: R=.856, R²=.733, Adj. R²=.713 | ANOVA: F(2,27)=37.03, p<.001

1

Check the ANOVA row: F(2,27)=37.03, p<.001 → model is useful overall.
2

R²: Report Adjusted R² with 2+ predictors.
3

Each predictor: B is the unstandardized effect, holding the other predictor constant.
4

Compare via standardized β: Standardized β enables cross-predictor comparison.

.733

R²

37.03

F(2,27)

<.001

p

β=.612

Strongest predictor

🔴 Model significant — R² = .733, p < .001.

APA-7

A multiple regression significantly predicted salary, F(2, 27) = 37.03, p < .001, R² = .733; experience (β=.612) and education (β=.331) were both significant predictors.

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💻 38 · SPSS Correlation Output

Interpreting SPSS Correlation Output

SPSS

Research Question: How to read the symmetric SPSS bivariate correlation matrix.

		Study Hours	Exam Score
Study Hours	Pearson Corr.	1	.742**
	Sig. (2-tailed)		.000
	N	20	20
Exam Score	Pearson Corr.	.742**	1
	Sig. (2-tailed)	.000
	N	20	20

1

The matrix is symmetric: Read the correlation once, not twice.
2

Read the asterisks: ** = p < .01; Sig. = .000 means p < .001.
3

Judge strength: r=.742 is a large effect by Cohen's guidelines.
4

Correlation ≠ causation: Correlation shows co-variation, not causation.

.742

r

<.001

p

18

df

🔴 Significant, large effect — r(18) = .742, p < .001.

APA-7

A Pearson correlation was significant and large, r(18) = .742, p < .001 (SPSS's .000 is reported as p < .001).

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💻 39 · SPSS Reliability Output

Interpreting SPSS Reliability Output

SPSS

Research Question: How to read the Reliability Statistics and Item-Total Statistics tables together.

Item	Corr. Item-Total	α if Deleted
Q1	.68	.77
Q2	.71	.76
Q3	.24	.85
Q4	.65	.78
Q5	.70	.76

Reliability Statistics: Cronbach's α = .812, N of Items = 5

1

Headline alpha: α=.812 is 'good' by common guidelines.
2

Scan for weak items: Q3's .24 is below the .30 rule-of-thumb threshold.
3

Confirm with α-if-deleted: Only Q3's α-if-deleted exceeds the overall α.
4

Decide and report: Keep 5 items (α=.812) or drop Q3 (α=.85) — disclose the choice.

.812

α (5 items)

.24

Q3 corr.

.85

α if Q3 deleted

🟢 Good overall (α=.812); Q3 is a candidate for removal.

APA-7

The scale showed good reliability, α = .812; Q3's weak item-total correlation (r=.24) means dropping it would raise α to .85.

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🧭 40 · How to Choose the Right Statistical Test

Choosing the Right Statistical Test

Guide

How to use this page: Find your question below, follow the rule to a test, or use the interactive Advisor.

1 · Comparing groups (continuous outcome):

2 · Relationships between variables:

3 · Categorical / binary data:

4 · Scale and measurement quality:

5 · Checking assumptions first:

6 · Planning a study and reporting effects:

7 · Reading SPSS output and writing up results:

🟢 Still unsure? Use the interactive Test Advisor below.

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م	ف1	ف2	ف3	ف4	ف5	∑
1	4	3	4	5	4	20
2	2	2	3	2	3	12
3	5	4	5	4	5	23
4	3	3	3	4	3	16
5	4	5	4	5	4	22
6	2	3	2	3	2	12
7	5	4	5	5	4	23
8	3	4	3	3	4	17

م	Y	X₁	X₂	X₃
1	72	3	40	3.5
2	85	7	60	4.2
3	90	10	80	4.8
4	68	2	30	3.0
5	78	5	55	3.8
6	92	12	90	5.0
7	75	4	45	3.6
8	88	8	70	4.5
9	65	1	25	2.8
10	82	6	65	4.0

P	X	Y	Z
1	1	148	58
2	2	145	62
3	2	150	45
4	3	138	60
5	4	142	50
6	4	130	40
7	5	128	48
8	6	135	35
9	6	120	42
10	7	118	30
11	8	115	33
12	9	110	28

	Median Survival	S(12 mo)	Events/N
Drug A	22 months	0.750	5/12
Drug B	14 months	0.500	7/12

م	ف1	ف2	ف3	ف4	ف5	∑
1	4	3	4	5	4	20
2	2	2	3	2	3	12
3	5	4	5	4	5	23
4	3	3	3	4	3	16
5	4	5	4	5	4	22
6	2	3	2	3	2	12
7	5	4	5	5	4	23
8	3	4	3	3	4	17

م	Y	X₁	X₂	X₃
1	72	3	40	3.5
2	85	7	60	4.2
3	90	10	80	4.8
4	68	2	30	3.0
5	78	5	55	3.8
6	92	12	90	5.0
7	75	4	45	3.6
8	88	8	70	4.5
9	65	1	25	2.8
10	82	6	65	4.0

P	X	Y	Z
1	1	148	58
2	2	145	62
3	2	150	45
4	3	138	60
5	4	142	50
6	4	130	40
7	5	128	48
8	6	135	35
9	6	120	42
10	7	118	30
11	8	115	33
12	9	110	28

📊 Worked Examples

📋 1 · Descriptive Statistics

⚖️ 2 · t-Tests

📊 3 · ANOVA

🔢 4 · Non-parametric Tests

🔗 5 · Correlation

📈 6 · Simple Linear Regression

🗂️ 7 · Chi-Square Test

🔒 8 · Cronbach's Alpha

🔁 9 · Repeated-Measures ANOVA

⚡ 10 · Power Analysis & Sample Size

📐 11 · Normality Testing (Shapiro-Wilk)

🔀 12 · Two-Way ANOVA (Factorial Design)

🏅 13 · Spearman Rank Correlation

📊 14 · Multiple Linear Regression

⚙️ 15 · Logistic Regression

🎯 16 · Effect Sizes

📏 17 · Confidence Intervals

🔄 18 · Mediation Analysis

📉 19 · Survival Analysis (Kaplan-Meier)

📐 20 · Sample Size

📋 Open your SPSS file

💻 21 · SPSS Output

✍️ 22 · APA-7 Writing

📋 23 · Likert Scale

📊 24 · Multiple Regression

🌀 26 · Mauchly's Sphericity Test

💻 27 · Reading SPSS ANOVA Output

📊 28 · Welch's ANOVA

🧪 29 · Levene's Test

🔢 30 · Wilcoxon Signed-Rank Test

🔢 31 · Friedman Test

🗂️ 32 · McNemar's Test

🔗 33 · Partial Correlation

🔗 34 · Point-Biserial Correlation

🔬 35 · Exploratory Factor Analysis

📊 36 · ANCOVA

💻 37 · SPSS Regression Output

💻 38 · SPSS Correlation Output

💻 39 · SPSS Reliability Output

🧭 40 · How to Choose the Right Statistical Test

م	ف1	ف2	ف3	ف4	ف5	∑
1	4	3	4	5	4	20
2	2	2	3	2	3	12
3	5	4	5	4	5	23
4	3	3	3	4	3	16
5	4	5	4	5	4	22
6	2	3	2	3	2	12
7	5	4	5	5	4	23
8	3	4	3	3	4	17

م	Y	X₁	X₂	X₃
1	72	3	40	3.5
2	85	7	60	4.2
3	90	10	80	4.8
4	68	2	30	3.0
5	78	5	55	3.8
6	92	12	90	5.0
7	75	4	45	3.6
8	88	8	70	4.5
9	65	1	25	2.8
10	82	6	65	4.0

P	X	Y	Z
1	1	148	58
2	2	145	62
3	2	150	45
4	3	138	60
5	4	142	50
6	4	130	40
7	5	128	48
8	6	135	35
9	6	120	42
10	7	118	30
11	8	115	33
12	9	110	28