ABD 3e Chapter 15
Study of how light exposure shifts the body’s internal clock (circadian rhythm)
22 participants randomly assigned to one of three treatments:
Each person received a single 3-hour light exposure
Researchers measured how much each person’s internal clock shifted
Total variation: how much all observations vary around the overall mean
This can be split into two parts:
ANOVA works by comparing these two sources of variation
\[ F = \frac{MS_{groups}}{MS_{error}} \]
\[ H_0: \mu_1 = \mu_2 = \cdots = \mu_k \]
treatment as the (categorical) explanatory variable and shift as the (numerical) response variable.
anova() function to show the ANOVA table
Analysis of Variance Table
Response: shift
Df Sum Sq Mean Sq F value Pr(>F)
treatment 2 7.2245 3.6122 7.2894 0.004472 **
Residuals 19 9.4153 0.4955
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
| Source | df | SS | MS | F | P |
|---|---|---|---|---|---|
| Groups (treatment) | 2 | 7.22 | 3.61 | 7.29 | 0.004 |
| Error (residuals) | 19 | 9.42 | 0.50 |
| Source | df | SS | MS | F | p-value |
|---|---|---|---|---|---|
| Groups (treatment) | 2 | 7.22 | 3.61 | 7.29 | 0.004 |
| Error (residuals) | 19 | 9.42 | 0.50 | NA | NA |
| Source | df | SS | MS | F | P |
|---|---|---|---|---|---|
| Groups (treatment) | 2 | 7.22 | 3.61 | 7.29 | 0.004 |
| Error (residuals) | 19 | 9.42 | 0.50 | NA | NA |
| Source | df | SS | MS | F | P |
|---|---|---|---|---|---|
| Groups (treatment) | 2 | 7.22 | 3.61 | 7.29 | 0.004 |
| Error (residuals) | 19 | 9.42 | 0.50 |
\[ F = \frac{MS_{groups}}{MS_{error}} \]
\[ R^2 = \frac{SS_{groups}}{SS_{total}} \]
Observations are independent
Data are approximately normally distributed within each group
Groups have similar variance
These assumptions affect validity of results
Comparisons specified before analyzing data
Focus on specific, meaningful hypotheses:
Advantages:
Interpreted similarly to:

BIOL 275 Biostatistics | Spring 2026