R中ggqqplot两种调用结果不同的原因及正确用法咨询

Great question! This is a super common point of confusion when checking ANOVA assumptions—let’s break down exactly why these two ggqqplot() calls produce different outputs, and which one you actually need for your ANOVA work.

1. What ggqqplot(PlantGrowth, "weight") does #

This call creates a QQ plot for the raw weight values across all three groups combined. It’s testing whether the entire pool of weight measurements follows a normal distribution.

But here’s the key: ANOVA doesn’t require your raw response variable to be normally distributed. The critical assumption is about the normality of the model residuals—not the raw data itself.

2. What ggqqplot(residuals(model)) does #

When you pull residuals(model) from your lm(weight ~ group, data = PlantGrowth) model, you’re extracting the error terms for each observation:

• For every plant, residual = observed weight - predicted weight (which is just the mean weight of its group, since the linear model predicts group means here)

This QQ plot checks if these within-group error terms are normally distributed—which is the exact assumption ANOVA depends on. Even if raw weights (mixed across groups) aren’t normal (say, because group means are very different, creating a multi-modal combined distribution), the residuals could still be perfectly normal if each group’s weights vary normally around their own mean.

3. A concrete example from PlantGrowth #

In the PlantGrowth dataset, the three treatment groups have distinct average weights. If you plot all weights together, the combined distribution might look "lumpy" (with peaks near each group’s mean), making the raw data QQ plot deviate from the straight reference line. But when you look at residuals, you’re stripping out the group mean effect—leaving only the random variation within each group, which should follow a normal distribution if your ANOVA assumptions hold.

Bottom line #

For validating ANOVA’s normality assumption, always use the model residuals with ggqqplot(residuals(model)). The raw response variable’s normality isn’t the right metric here—ANOVA is surprisingly robust to deviations in raw data normality, but it relies on normally distributed residuals to generate reliable p-values.

内容的提问来源于stack exchange，提问作者user12838638