Statistics 2nd ed

within-subjects

Factorial ANOVA

factorial layout
factorial means interaction
factorial interaction

Goal. Test the effects of Method (Lecture vs. Online) and Time (Early vs. Late) on exam scores, and whether there is an interaction between Method and Time.

Design & Experiment

  • Factor A (Method): Lecture vs. Online
  • Factor B (Time): Early vs. Late
  • Balanced design: \(n=5\) per cell ⇒ total \(N=20\).

Students are randomly assigned to one of four cells (Method × Time). After a short module, all students take the same 100-point exam.

Figure 1: 2 × 2 layout (Method × Time).


Data

Scores by cell (five students per cell):

MethodTimeScoresCell Mean
LectureEarly686870727270.0
LectureLate767678808078.0
OnlineEarly707072747472.0
OnlineLate717173757573.0

Within each cell the sample variance is 4 (SD = 2), so the within-cell sum of squares is \((n-1)s^2 = 4\times4 = 16\) per cell.

Figure 2: Means with SEM by Time, separate lines for Method.

Figure 3: Interaction plot (Lecture rises sharply; Online nearly flat).


Step 1 — Marginal Means and Grand Mean

Cell means: \[ \bar X_{\text{Lecture,Early}}=70,\; \bar X_{\text{Lecture,Late}}=78,\; \bar X_{\text{Online,Early}}=72,\; \bar X_{\text{Online,Late}}=73. \] Marginal means: \[ \bar X_{\text{Lecture}}=\frac{70+78}{2}=74,\quad \bar X_{\text{Online}}=\frac{72+73}{2}=72.5; \qquad \bar X_{\text{Early}}=\frac{70+72}{2}=71,\quad \bar X_{\text{Late}}=\frac{78+73}{2}=75.5. \] Grand mean: \[ \bar X=\frac{70+78+72+73}{4}=73.25. \]


Step 2 — Sums of Squares (Between)

Balanced design formulas (with \(n\) per cell, \(a=b=2\)):

  • \(SS_A = nb \sum_a(\bar X_{a\cdot}-\bar X)^2\), here \(nb=10\).
  • \(SS_B = na \sum_b(\bar X_{\cdot b}-\bar X)^2\), here \(na=10\).
  • \(SS_{AB} = n \sum_{a,b}\big(\bar X_{ab}-\bar X_{a\cdot}-\bar X_{\cdot b}+\bar X\big)^2\), here \(n=5\).

Compute each term:

Factor A (Method): \[ \begin{aligned} SS_A &= 10\Big[(74-73.25)^2 + (72.5-73.25)^2\Big]\\ &= 10\big[0.75^2 + (-0.75)^2\big] = 10(0.5625+0.5625)=\mathbf{11.25}. \end{aligned} \]

Factor B (Time): \[ \begin{aligned} SS_B &= 10\Big[(71-73.25)^2 + (75.5-73.25)^2\Big]\\ &= 10\big[(-2.25)^2 + (2.25)^2\big] = 10(5.0625+5.0625)=\mathbf{101.25}. \end{aligned} \]

Interaction \(A\times B\): For each cell compute \(d_{ab}=\bar X_{ab}-\bar X_{a\cdot}-\bar X_{\cdot b}+\bar X\). Here each \(d_{ab}=\pm1.75\) so \(d_{ab}^2=3.0625\) and there are four cells: \[ SS_{AB}=5\times(4\times3.0625)=\mathbf{61.25}. \]


Step 3 — Within-Group (Error) and Total SS

Within each cell, \((n-1)s^2=16\). With 4 cells: \[ SS_{\text{within}}=\mathbf{64.00}. \]

Total: \[ SS_{\text{total}}=SS_A+SS_B+SS_{AB}+SS_{\text{within}} =11.25+101.25+61.25+64.00=\mathbf{238.75}. \]


Step 4 — Degrees of Freedom & Mean Squares

\[ \begin{aligned} &df_A=a-1=1,\quad df_B=b-1=1,\quad df_{AB}=(a-1)(b-1)=1,\\ &df_{\text{within}}=N-ab=20-4=\mathbf{16},\quad df_{\text{total}}=N-1=19. \end{aligned} \] \[ MS_A=\frac{11.25}{1}=11.25,\quad MS_B=\frac{101.25}{1}=101.25,\quad MS_{AB}=\frac{61.25}{1}=61.25,\quad MS_{\text{within}}=\frac{64.00}{16}=\mathbf{4.00}. \]


Step 5 — F Tests & p-values

\[ F_A=\frac{MS_A}{MS_{\text{within}}}=\frac{11.25}{4}= \mathbf{2.8125},\qquad F_B=\frac{MS_B}{MS_{\text{within}}}=\frac{101.25}{4}= \mathbf{25.3125},\qquad F_{AB}=\frac{MS_{AB}}{MS_{\text{within}}}=\frac{61.25}{4}= \mathbf{15.3125}. \] With \(df_1=1\), \(df_2=16\): \[ p_A \approx 0.11\;(\text{n.s.}),\quad p_B < 0.001,\quad p_{AB} \approx 0.001. \]


ANOVA Summary Table

SourceSSdfMSFp
Method (A)11.25111.252.8125≈ 0.11
Time (B)101.251101.2525.3125< 0.001
A × B61.25161.2515.3125≈ 0.001
Within (Error)64.00164.00
Total238.7519

Interpretation

Main effect of Time (B) is significant: Late > Early on average. Main effect of Method (A) is not significant at conventional levels. The interaction (A × B) is significant: Lecture improves markedly from Early→Late, while Online changes little—non-parallel lines in the interaction plot.

Figure 4: Interaction plot highlighting non-parallel lines.

Assumptions (checklist)

  • Independence of observations within and across cells.
  • Approximately normal scores within each cell.
  • Homogeneity of variances across cells (here, each cell variance ≈ 4).

Practice self-test quiz

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Identify the Design


Case 1

Scenario: A teacher compares test scores of students in two different classrooms (Class A vs. Class B).
Question: Are the two groups significantly different in mean score?
Answer: Independent-samples t-test.


Case 2

Scenario: A researcher tests the same group of students before and after tutoring.
Question: Did their scores improve after the program?
Answer: Paired-samples t-test (dependent t-test).


Case 3

Scenario: Three groups of students use different study methods: flashcards, highlighting, and practice tests.
Question: Do the study methods lead to different mean scores?
Answer: One-way ANOVA.


Case 4

Scenario: A psychologist measures anxiety scores in patients given three different drugs.
Question: Do the drugs produce different mean anxiety scores?
Answer: One-way ANOVA.


Case 5

Scenario: A study compares two groups of athletes: runners vs. swimmers, on reaction time.
Question: Are the two sports groups different in mean reaction time?
Answer: Independent-samples t-test.


Case 6

Scenario: Students are tested at three times: beginning, middle, and end of the semester.
Question: Did their scores change over time?
Answer: Repeated-measures ANOVA.


Case 7

Scenario: Two teaching methods (Lecture, Online) are tested across two times of day (Morning, Afternoon).
Question: What are the effects of method, time, and their interaction?
Answer: Two-way (factorial) ANOVA.


Case 8

Scenario: A company compares productivity of three work shifts (Day, Evening, Night) across two departments (Sales, Service).
Question: Are there main effects of shift and department, and is there an interaction?
Answer: Two-way (factorial) ANOVA.


Case 9

Scenario: Students are randomly assigned to a control or experimental group, and both groups are measured three times (Weeks 1, 2, 3).
Question: Is there an effect of group, time, and interaction?
Answer: Mixed (split-plot) ANOVA.


Case 10

Scenario: A survey asks students to choose their favorite subject: Math, Science, or English.
Question: Is the distribution of responses different from chance?
Answer: Chi-square goodness-of-fit test.


Case 11

Scenario: A researcher studies whether gender (Male, Female) is related to preference for sports (Soccer, Basketball, Tennis).
Question: Is there an association between gender and sport preference?
Answer: Chi-square test of independence.


Case 12

Scenario: Students are ranked by teacher ratings: 1st, 2nd, 3rd, etc. Two different teaching methods are compared on these ranks.
Question: Do the groups differ in median ranks?
Answer: Mann–Whitney U test (non-parametric).


Case 13

Scenario: The same students are ranked before and after a training program.
Question: Did the ranks change after training?
Answer: Wilcoxon signed-rank test (non-parametric).

Practice self-test quiz

In the space below, please find practice problems and self-test quizzes. For full access, please signup free.