Time course experiment - which statistical analysis fits best?

Question

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1. What the student is being asked to do (plain language)

The student has measured the amount of protein X on Western blots for two experimental factors

Factor	Levels	Replicates per level
Treatment	Control, Drug	3 wells each
Incubation time	30 min, 1 h, 2 h, 3 h	3 wells each (per treatment)

All wells come from the same organ lysate, so the only source of variation is the technical variability between wells.

The student wants to know how to test three questions:

Overall drug effect – does the drug change protein X irrespective of time?
Overall time effect – does protein X change with incubation time irrespective of drug?
Interaction – does the drug effect depend on the incubation time (i.e., does the drug work differently at 30 min vs 3 h)?

The appropriate statistical tool for simultaneously answering these three questions is a two‑way (factorial) ANOVA (analysis of variance) with the factors Treatment and Time and their interaction.

2. Step‑by‑step solution

Below each step we give the what (action) and the why (reason).

Step 0 – Organise the data

Create a table (or a spreadsheet) with one row per well:

Well	Treatment	Time (h)	Protein X (raw intensity)
1	Control	0.5	…
2	Control	0.5	…
3	Control	0.5	…
4	Drug	0.5	…
…	…	…	…
24	Drug	3	…

There are 24 observations (4 × 2 × 3).

Tip – If the Western‑blot intensities span several orders of magnitude, take a log‑transform (e.g. log10(intensity)) before analysis; this often improves normality and variance homogeneity.

Step 1 – Check ANOVA assumptions

ANOVA assumes that, within each combination of factors, the residuals are

Assumption	How to check (in R)
Normality	`shapiro.test(residuals(lm))` or a Q‑Q plot (`qqnorm`, `qqline`)
Equal variances (homoscedasticity)	`plot(lm, which = 1)` (residuals vs fitted) or Levene’s test: `car::leveneTest(Y ~ Treatment*Time)`
Independence	Design ensures wells are independent (no repeated measures).

If the assumptions are seriously violated (e.g., Shapiro p < 0.01 and strong skew), either:

Transform the data (log, square‑root), or
Use a non‑parametric factorial method such as the Aligned Rank Transform (ART) ANOVA (ARTool::art in R).

For the remainder we assume the assumptions are reasonably met (or that a log‑transform fixed them).

Step 2 – Fit the two‑way ANOVA model

The linear model is

[ Y_{ijk}= \mu \;+\; \alpha_i \;+\; \beta_j \;+\; (\alpha\beta){ij} \;+\; \varepsilon{ijk} ]

(Y_{ijk}) = protein level in well k of treatment i and time j
(\mu) = overall mean
(\alpha_i) = effect of treatment (i = Control, Drug)
(\beta_j) = effect of time (j = 0.5, 1, 2, 3 h)
((\alpha\beta)_{ij}) = interaction term (does drug effect depend on time?)
(\varepsilon_{ijk}) = random error (assumed (N(0,\sigma^2)))

In R:

# assume the data frame is called df and the response column is called X
df$Treatment <- factor(df$Treatment)      # two levels
df$Time      <- factor(df$Time)           # four levels

# optional log‑transform
df$logX <- log10(df$X)

model <- aov(logX ~ Treatment * Time, data = df)   # or aov(X ~ ...) if no transform
summary(model)

The summary table gives three F‑tests:

Source	df (numerator)	df (denominator)	F value	p‑value
Treatment	1	18	…	…
Time	3	18	…	…
Treatment:Time	3	18	…	…
Residual	16	—	—	—

(df for residual = total n – number of parameters = 24 – (1 + 1 + 3 + 1) = 16)

Step 3 – Interpret the three tests

Question	Null hypothesis (H₀)	What a significant p‑value (typically < 0.05) means
Treatment main effect	The average protein level is the same for Control and Drug after averaging over all times.	The drug changes protein X on average (regardless of time).
Time main effect	All four time points have the same mean (after averaging over treatment).	Protein X changes with incubation time (regardless of drug).
Interaction	The difference between Drug and Control is the same at every time point.	The drug effect depends on incubation time – e.g., it may be strong at 2 h but weak at 30 min.

If the interaction is significant, the main‑effect p‑values are less informative; you should explore the simple effects (drug vs control) within each time.

Step 4 – Post‑hoc / simple‑effects analysis (if needed)

Scenario A – Interaction NOT significant
You can report the two main effects and stop there.

Scenario B – Interaction significant

Simple‑effect t‑tests (or one‑way ANOVAs) for each time point:

library(emmeans)
em <- emmeans(model, ~ Treatment | Time)   # means per time
pairs(em)                                   # drug vs control at each time

Adjust for multiple testing (e.g., Tukey’s HSD) because you are making four comparisons.
```
contrast(em, method = "pairwise", adjust = "tukey")
```

Optionally plot the interaction:

library(ggplot2)
ggplot(df, aes(x = Time, y = logX, colour = Treatment, group = Treatment)) +
  stat_summary(fun = mean, geom = "point", size = 3) +
  stat_summary(fun = mean, geom = "line") +
  stat_summary(fun.data = mean_se, geom = "errorbar", width = .2)

The plot visually shows whether the two lines are parallel (no interaction) or cross/diverge (interaction).

Step 5 – Report the results

A typical write‑up (using log‑transformed data) might look like:

A two‑way ANOVA was performed on log₁₀‑transformed protein X levels with Treatment (Control vs Drug) and Incubation time (0.5, 1, 2, 3 h) as fixed factors. The interaction was significant, F(3,16) = 5.42, p = 0.009, indicating that the drug effect differed across time points. Simple‑effect comparisons (Tukey‑adjusted) showed that the drug significantly increased protein X at 2 h (p = 0.003) and 3 h (p = 0.021) but not at 30 min or 1 h (p > 0.10). The main effect of Time was also significant, F(3,16) = 8.73, p = 0.001, reflecting a general increase of protein X with longer incubation. The main effect of Treatment was not significant after accounting for the interaction (F(1,16) = 2.01, p = 0.17).

If the interaction were non‑significant, you would instead report the two main effects and their p‑values.

Step 6 – Caveats about the experimental design

All wells originate from a single organ; therefore the data represent technical replicates, not biological replicates. The inference you can draw is limited to “the assay can detect a change under these laboratory conditions.”
If you later obtain lysates from different animals, you would treat Animal as a random factor and use a mixed‑effects model (e.g., lmer(logX ~ Treatment*Time + (1|Animal))).
With only n = 3 per cell, the ANOVA has low power; a non‑significant result does not prove there is no effect – it may simply be undetectable with this sample size.

3. Final answer (concise)

Question	Recommended test	What the test tells you
Overall drug effect (ignoring time)	Two‑way ANOVA – main effect of Treatment	Whether the drug changes protein X on average.
Overall time effect (ignoring drug)	Two‑way ANOVA – main effect of Time	Whether protein X changes with incubation time.
Does drug effect vary with time?	Two‑way ANOVA – Treatment × Time interaction	Whether the drug’s impact is different at the various time points.
If the interaction is significant	Post‑hoc simple‑effects (pairwise comparisons of Drug vs Control at each time, Tukey‑adjusted)	Which specific times show a drug effect.

The complete workflow:

Log‑transform the intensity (if needed).
Verify normality and equal variances.
Fit aov(Y ~ Treatment * Time).
Inspect the three F‑tests (Treatment, Time, Interaction).
If Interaction = significant → run emmeans/Tukey post‑hoc for each time.
Summarise results with F‑values, p‑values, and effect‑size estimates (e.g., η²).

4. Common mistakes (and how to avoid them)

Mistake	Why it’s wrong	How to fix it
Treating the 24 wells as independent biological replicates	All wells come from the same organ, so they only reflect assay variability, not animal‑to‑animal variation.	State clearly that the analysis is on technical replicates; plan a follow‑up with true biological replicates.
Running three separate one‑way ANOVAs (Treatment, Time, Interaction)	Inflates Type I error and ignores the factorial structure; you lose the ability to test interaction properly.	Use a single two‑way ANOVA that simultaneously evaluates both main effects and the interaction.
Ignoring the interaction when it is significant	The main‑effect p‑values become misleading; you might claim a “drug effect” that only exists at certain times.	When the interaction term is significant, focus on simple‑effects (drug vs control at each time) and report those.
Not checking assumptions	ANOVA’s p‑values are unreliable if residuals are highly non‑normal or variances are unequal.	Plot residuals, run Shapiro‑Wilk and Levene’s tests; transform data or use a non‑parametric ART‑ANOVA if assumptions fail.
Using the raw Western‑blot intensities without normalization	Loading differences or background can create systematic bias.	Normalize each lane to a loading control (e.g., β‑actin) and use the normalized values as the response variable.
Reporting only “p < 0.05” without effect size	Statistical significance does not convey biological relevance, especially with small n.	Report effect sizes (η² for

Original question: Time course experiment - which statistical analysis fits best? on Cross Validated (Stats Stack Exchange), licensed CC BY-SA.

1. What the student is being asked to do (plain language)

2. Step‑by‑step solution