Motivating Scenario: In the previous chapter we compared two groups as a two sample t-test. In this chapter, we basically did the same thing as an ANOVA. You want to know the relationship between these approaches.
Learning Goals: By the end of this subchapter, you should be able to:
Should I conduct an ANOVA or t-test?
This question is commonly asked by students with data in hand, and even by professors on an exam to evaluate students’ understanding. The answer is – for a case with a binary explanatory variable it makes absolutely no difference.
Reassuringly whether conducting and ANOVA or a t-test on a continuous response and a binary explanatory variable we get the exact same p-value. This is reassuring, as we would like our answers to be robust to such an arbitrary choice. It’s up to you whether you prefer to present results in terms of the number of standard errors separating the groups, or as a ratio of mean squares.
Here we go through some connections between these modeling approaches.
This section is fully optional I just think it’s cool.
Let’s consider our example from this chapter – comparing the admixture proportion of pink and white flowers at Squirrel Mountain.
A t-test provides the following results:
t.test(admix_proportion ~ petal_color, data = clarkia_hz, var.equal=TRUE)
Two Sample t-test
data: admix_proportion by petal_color
t = 8.7486, df = 44, p-value = 3.486e-11
alternative hypothesis: true difference in means between group pink and group white is not equal to 0
95 percent confidence interval:
0.009949917 0.015906235
sample estimates:
mean in group pink mean in group white
0.018958210 0.006030134 lm(admix_proportion ~ petal_color, data = clarkia_hz) |>
anova()Analysis of Variance Table
Response: admix_proportion
Df Sum Sq Mean Sq F value Pr(>F)
petal_color 1 0.0018312 0.00183122 76.539 3.486e-11 ***
Residuals 44 0.0010527 0.00002393
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The most obvious similarities here are:
But wait, there’s more:
# finding the pooled variance
clarkia_hz |>
filter(!is.na(petal_color))|>
group_by(petal_color)|>
mutate(mean_admix = mean(admix_proportion ))|>
ungroup()|>
mutate(deviation = admix_proportion - mean_admix) |>
summarise(pooled_var = sum(deviation^2) / (n() - 2))# A tibble: 1 × 1
pooled_var
<dbl>
1 0.0000239This is all to say that the two sample t-test and an ANOVA with two groups are essentially identical. I hope this makes us feel good about moving on to more complex ANOVAs, and about linear models more broadly.
Why does \(F = t^2\)?
t measures the difference between group means in standard error units (i.e., how many standard deviations of the sampling distribution separate the groups).
F compares the variance between groups to the variance within groups, so it’s measured on the variance scale.
Because variance is just the square of a standard deviation, F is simply the square of the corresponding t when there are two groups.
Remember that \(R^2\) is the “proportion of variance explained” (i.e. the proportion of variance in our response variable that we can pin on our explanatory variable).
\[R^2 = \frac{\text{SS}_\text{model}}{\text{SS}_\text{total}}\].
We can find \(R^2\) from the output of our linear model:
library(broom)
aov(admix_proportion ~ petal_color, data = clarkia_hz) |>
glance()|>
kable()| logLik | AIC | BIC | deviance | nobs | r.squared |
|---|---|---|---|---|---|
| 180.4843 | -354.9686 | -349.4827 | 0.0010527 | 46 | 0.6349717 |
You also may remember that \(r\) is the correlation between variables. If we assign a numeric value of zero to the explanatory variable for one category (e.g., white flowers), and one for the other category (e.g., pink flowers), we can calculate a correlation (\(r\)). If we square this correlation \(r\) squared equals \(R^2\)!
\[r = \frac{ \text{cov}_{x,y} }{s_x \times s_y}\]
clarkia_hz |>
filter(!is.na(petal_color))|>
mutate(petal_color_num = ifelse(petal_color == "white", 0 ,1))|>
summarise(r = cor(petal_color_num , admix_proportion),
r.squared = r^2)# A tibble: 1 × 2
r r.squared
<dbl> <dbl>
1 0.797 0.635