More multiple populations!
In the last chapter we introduced the idea of comparing parameters among populations. Now we will extend those ideas to instances when continuous data is collected.
When we introduced NHST for continuous data, we focused on sepal length from I. virginica.
set.seed(42)
library(ggplot2)
ggplot(iris[iris$Species == "virginica",],
aes(x=Sepal.Length)) +
geom_histogram( fill="blue", color="black") +
labs(title=expression(paste("Sepal lengths of ",italic("I. virginica"))),
x= "Sepal length (cm)",
y= "Frequency")`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
and tested if it was equal to a given value (7.0 cm)
\[ \begin{split} H_O: \mu_{sepal \ length} = 7 \ cm \\ H_A: \mu_{sepal \ length} \neq 7 \ cm \end{split} \]
We then considered how to assess these types of hypotheses using z, t, Wilcoxon, and sign tests.
However, as we noted in the last chapter, we often instead have data from multiple populations. For example, we may have data from 3 species of flowers. We commonly see this data plotted as a bar chart with error bars
What do we test now and how?
As previously noted, we can’t compare heights among group. Height is a random variable, and it’s highly unlikely they will be exactly the same. Seeing the actual data may help us remember this.
ggplot(iris, aes(y=Sepal.Length, x=Species, color=Species)) +
geom_jitter() +
labs(title=expression(paste("Sepal lengths of ",italic("I. species"))),
y= "Sepal length (cm)",
x= "Species")
So we typically focus on a parameter that describes the distribution of a focal trait, such as the mean. Just like binomial data, our hypotheses can then be written several ways:
\\\[ \begin{split} H_O: \mu_{sepal \ length, \ setosa} = \mu_{sepal \ length, \ virginica} = \mu_{sepal \ length, \ versicolor},\\ \textrm{there is no difference among the population means}\\ \textrm{all means are the same}\\ H_A: \textrm{all means are not equal to each other;}\\ \textrm{at least one mean is different than the others}\\ \textrm{(a little odd to write mathematically!)}\\ \end{split} \]
Given that, our overall idea is to consider if the data are better explained by an overall group average or by species-specific averages. To visualize this, we could use this image.
library(viridis)
ggplot(iris, aes(Species,Sepal.Length)) +
geom_jitter(size = 3) +
geom_errorbar(aes(ymin=Sepal.Length, ymax=Sepal.Length, color="group means"), size=1.5,
data = function_output) +
geom_hline(aes(yintercept=mean(Sepal.Length), color = "overall average"), size=1.5)+
scale_color_manual(values = c("#619CFF","#F8766D"))+
labs(title=expression(paste("Sepal lengths of ",italic("I. species"))),
y= "Sepal length (cm)",
x= "Species",
color= "Focus")Warning: Using `size` aesthetic for lines was deprecated in ggplot2 3.4.0.
ℹ Please use `linewidth` instead.
Like our earlier considerations of variance and SSE, the data will obviously be fit better by species-specific averages. Its impossible for them to do worse than the overall average, and at worst they all are the group average. However, we should also remember that these are samples, so we know sampling error is an issue. Therefore, we have to consider if the species-specific averages do enough of a better job explaining the data to warrant using them. To put this in our SSE and hypothesis framework, we need to consider if a more complicated view of the world is worth it.
To test this, we can (as always) carry out a sampling experiment. The general idea is that species does snot matter (just like we saw in contingency analysis). Given that, we can draw samples that match our respective sample sizes for each population from a single population. The mean can be set as the pooled mean for the data (since under the null tested factors don’t matter). Since we don’t know or set sigma, we can again estimate it from the data. One such sample might look like
set.seed(7)
variance_estimate <- sum((function_output$N -1) * (function_output$sd)^2)/(sum(function_output$N)-length(function_output$N))
mean_sepal <- mean(iris$Sepal.Length)
simulated_data <- data.frame(Species=c(rep("setosa", 50),
rep("versicolor", 50),
rep("virginica",50)),
Sepal.Length=rnorm(150, mean_sepal,
sd= sqrt(variance_estimate)))
function_output <- summarySE(simulated_data, measurevar="Sepal.Length", groupvars =
c("Species"))
ggplot(simulated_data, aes(Species,Sepal.Length)) +
geom_jitter(size = 3) +
geom_errorbar(aes(ymin=Sepal.Length, ymax=Sepal.Length, color="group means"), size=1.5,
data = function_output) +
geom_hline(aes(yintercept=mean(Sepal.Length), color = "overall average"), size=1.5)+
scale_color_manual(values = c("#619CFF","#F8766D"))+
labs(title=expression(paste("Sepal lengths of ",italic("I. species"))),
y= "Sepal length (cm)",
x= "Species",
color= "Focus")
As expected, under the null hypothesis the overall and species-specific averages are closer. Now that we have the data, however, we are stuck with a new question: What is our test statistic?
In general, using a difference among means makes sense for cases when we have only 2 populations. This will be used when we introduce t-tests. However, it does not work for 3+ populations, so we will need a different approach.
ANOVAs offer an approach that can be used for any group of 2+ populations when certain assumptions are met. ANOVAs stands for analysis of variance which may seem odd given our hypotheses are focused on means. However, the idea (not fully developed here) is that we can get an overall estimate of variance by
calculating variance for each point around its respective group-specific mean
\[ \begin{split} \textrm{Remember, }s^2 = \frac{\sum_{i=1}^{n} (Y_{i}-\overline{Y})^2}{n-1} \sim \sigma^2\\ \textrm{So if we have j groups, for each group we can see}\\ s_j^2 = \frac{\sum_{i=1}^{n_j} (Y_{ij}-\overline{Y_j})^2}{n_j-1} \sim \sigma_j^2\\ \textrm{which we can combine to estimate }\sigma^2_{overall}\\ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2 ...(n_j-1)s_j^2}{n_1+n_2+...n_j-j} = s_{overall}^2\\ \end{split} \]
calculating variance for each group mean around the overall mean
this can be multiplied by \(n\) to get \(\sigma^2\)
this is referred to as mean square treatment
\[ \begin{split} \frac{\sum_{j=1}^{j} (\overline{Y_j}-\overline{\overline{Y}})^2}{j-1} = \frac{s^2}{n} \\ \textrm{where j is the number of groups and }\ \overline{\overline{Y}}\ \textrm{is the overall mean.}\\ \end{split} \]
In other words, variance among groups should be equal to variance within groups. Under the null hypothesis, the ratio of these estimates should tend towards 1. We can see this using a sampling experiment.
#sample
ratio <- data.frame(rep = 1:10000, mse = rep(NA,10000),
msg = rep(NA,10000), ratio = rep(NA,10000))
for(i in 1:10000){
setosa <- rnorm(50, mean_sepal, sd= sqrt(variance_estimate))
versicolor <- rnorm(50, mean_sepal, sd= sqrt(variance_estimate))
virginica <- rnorm(50, mean_sepal, sd= sqrt(variance_estimate))
mean_overall <- mean(c(setosa, versicolor, virginica))
ratio$mse[i] <- (49 * var(setosa) + 49 * var(versicolor) + 49 * var(virginica))/(150 - 3)
ratio$msg[i] <- (50 * (mean(setosa)-mean_overall)^2 +
50 * (mean(versicolor)-mean_overall)^2 +
50 * (mean (virginica)-mean_overall)^2)/2
ratio$ratio[i] <- ratio$msg[i]/ratio$mse[i]
}
ggplot(ratio, aes(ratio)) +
geom_histogram(aes(y=..count../sum(..count..)), fill = "orange", bins=15)+
labs(main = "Ratio under null hypothesis", y= "Probability")Warning: The dot-dot notation (`..count..`) was deprecated in ggplot2 3.4.0.
ℹ Please use `after_stat(count)` instead.
Using this approach, we could determine how unusual our data were (get a p-value!). In our example, we found a signal of 119.265. Looking at the x-axis above, it’s obvious getting a sample this or more extreme is going to be very rare by chance!
However, finding a distribution that approximates this shape would make future work easier. It turns out all the squared terms above lead to this being a rato of \(\chi^2\) distributions, where the numerator has degrees of freedom j-1 (# of groups - 1) and the denominator has degrees of freedom n-j-1 (the “lost” degrees of freedom are used to estimate group and overall means). We call this the F distribution.
ggplot(ratio, aes(ratio)) +
geom_histogram(aes(y=..count../sum(..count..)), fill = "orange", bins = 12) +
labs(main = "Ratio under null hypothesis", y= "Probability")+
stat_function(fun = df, args = list(df1 =2, df2 = 147), color = "green") 
This allows us to estimate the p-value using distributional assumptions.
You should also note this means we can partition the variance for any given observation as its distance from its group mean and its group means distance from the overall. It turns out when we do this, we can decompose the total sums of squares (similar to variance but not divided!) into those attributed to the distance of points from the group mean and the distance of group means from the grand mean (not proven here). These are additive (equal to) to the total sums of squares. . For a proof (and more links!) see Soch et al. (2024); main part reproduced below:
Theorem: Given one-way analysis of variance,
\[ y_{ij} = \mu_i + \varepsilon_{ij}, \; \varepsilon_{ij} \overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0, \sigma^2) \]
sums of squares can be partitioned as follows
\[ \mathrm{SS}_\mathrm{tot} = \mathrm{SS}_\mathrm{treat} + \mathrm{SS}_\mathrm{res} \]
where \(\mathrm{SS} _\mathrm{tot}\) is the **total sum of squares](/D/tss), \(\mathrm{SS} _\mathrm{treat}\) is the treatment sum of squares (sometimes call explained sum of squares, and \(\mathrm{SS} _\mathrm{res}\) is the residual sum of squares (sometimes called sum squares error).
Proof: The total sum of squares for one-way ANOVA is given by
\[ \mathrm{SS}_\mathrm{tot} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y})^2 \]
where \(\bar{y}\) is the mean across all values \(y_{ij}\). This can be rewritten as
\[ \begin{split} \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y})^2 &= \sum_{i=1}^{k} \sum_{j=1}^{n_i} \left[ (y_{ij} - \bar{y}_i) + (\bar{y}_i - \bar{y}) \right]^2 \\ &= \sum_{i=1}^{k} \sum_{j=1}^{n_i} \left[ (y_{ij} - \bar{y}_i)^2 + (\bar{y}_i - \bar{y})^2 + 2 (y_{ij} - \bar{y}_i) (\bar{y}_i - \bar{y}) \right] \\ &= \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y}_i)^2 + \sum_{i=1}^{k} \sum_{j=1}^{n_i} (\bar{y}_i - \bar{y})^2 + 2 \sum_{i=1}^{k} (\bar{y}_i - \bar{y}) \sum_{j=1}^{n_i} (y_{ij} - \bar{y}_i) \; . \end{split} \]
Note that the following sum is zero
\[ \sum_{j=1}^{n_i} (y_{ij} - \bar{y}_i) = \sum_{j=1}^{n_i} y_{ij} - n_i \cdot \bar{y}_i = \sum_{j=1}^{n_i} y_{ij} - n_i \cdot \frac{1}{n_i} \sum_{j=1}^{n_i} y_{ij} \; , \]
so that the sum in in the original equation reduces to
\[ \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y})^2 = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (\bar{y}_i - \bar{y})^2 + \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y}_i)^2 \; . \]
With the treatment sum of square for one-way ANOVA
\[ \mathrm{SS}_\mathrm{treat} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (\bar{y}_i - \bar{y})^2 \]
and the residual sum of squares for one-way ANOVA
\[ \mathrm{SS}_\mathrm{res} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (y_{ij} - \bar{y}_i)^2 \; , \]
we finally have:
\[ \mathrm{SS}_\mathrm{tot} = \mathrm{SS}_\mathrm{treat} + \mathrm{SS}_\mathrm{res} \; \]
This fact, and some clever math tricks, made them easy to work with. Eventually, however, we are focusing on the fact we are estimating the variance in multiple ways.
Remember from the notes on contingency analysis that a \(\chi^2\) distribution is a Z variate squared.It’s defined by degrees of freedom equal to the number of free variates it is summed over.
An ANOVA is just one case of a linear model. We will fully explore these later, but noting this now is useful in that all linear models have the same sets of assumptions. In general, linear models assume the residuals of the model are are independent, identically distributed, and follow a normal distribution. You’ll sometime see this written as
\[ \epsilon \approx i.i.d.\ N(\mu,\sigma) \]
But what does this mean?
Residuals are the distance between a measurement and its model-predicted value. A closely related term, error, is actually the distance between a measurement and the unknown population mean for a group. Linear models assume the residuals are independent of each other (this follows from independent data points) and that their spread (around their predicted values) is normally distributed and similar for all points.
Understanding this explains two key points. The residuals, not the data, need to be normally distributed. Also, we have to build the model to get the residuals, then we check the assumptions.
We can do this in R using the lm function. This approach also lets us use a single set of functions to build many model types. As always, there are many ways to do anything in R, so there are specific ANOVA functions that we will not introduce here.
For our data, we can build an lm object
iris_anova <- lm(Sepal.Length~Species, iris)then use plot to check the assumptions.
plot(iris_anova)
These 4 plots focus on the residuals (not the data).
If our assumptions are met, we can look at the output. One way to do this is using the summary function.
summary(iris_anova)
Call:
lm(formula = Sepal.Length ~ Species, data = iris)
Residuals:
Min 1Q Median 3Q Max
-1.6880 -0.3285 -0.0060 0.3120 1.3120
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.0060 0.0728 68.762 < 2e-16 ***
Speciesversicolor 0.9300 0.1030 9.033 8.77e-16 ***
Speciesvirginica 1.5820 0.1030 15.366 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.5148 on 147 degrees of freedom
Multiple R-squared: 0.6187, Adjusted R-squared: 0.6135
F-statistic: 119.3 on 2 and 147 DF, p-value: < 2.2e-16This output may be confusing, however. The overall p value shown in the bottom right is for the entire model - that works for now, but soon won’t. We also see individual p values for 2 levels of species, plus an odd intercept term.
These are model artifacts and may be confusing. By default (based on contrasts method), R lets the first factor level (typically alphabetical) be an intercept for the linear model, and then considers the other factor levels as deviations from that. It also shows if all the resulting Estimates are significantly different from 0. Note this means a significant intercept term does not mean your groups actually differ. Other approaches to these will be noted later.
Given these issues, why use the summary command? It does present some other useful information. For example, the R2 values is a measure of how variation the model explains. It can range from 0 (the model explains nothing) to 1 (all residuals are zero, in this case meaning all members of a given species have the exact same height). The adjusted-R2 value is similar, but it adjusts the measure to account for the fact more complex models will always explain more variation.
You can also remove the intercept to get group estimates for all groups
summary(lm(Sepal.Length~Species-1, iris))
Call:
lm(formula = Sepal.Length ~ Species - 1, data = iris)
Residuals:
Min 1Q Median 3Q Max
-1.6880 -0.3285 -0.0060 0.3120 1.3120
Coefficients:
Estimate Std. Error t value Pr(>|t|)
Speciessetosa 5.0060 0.0728 68.76 <2e-16 ***
Speciesversicolor 5.9360 0.0728 81.54 <2e-16 ***
Speciesvirginica 6.5880 0.0728 90.49 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.5148 on 147 degrees of freedom
Multiple R-squared: 0.9925, Adjusted R-squared: 0.9924
F-statistic: 6522 on 3 and 147 DF, p-value: < 2.2e-16Note these match our group means, which is good, but the overall p value is now less useful (it compares our data to a null that assumes everything is equal to 0), and the output is still confusing.
Since we are doing an omnibus test, what we typically want is a single p value associated with our factor of interest (species in this case). To get that, we’ll use the Anova function from the car package.
library(car)
Anova(iris_anova, type="III")Anova Table (Type III tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
(Intercept) 1253.00 1 4728.16 < 2.2e-16 ***
Species 63.21 2 119.26 < 2.2e-16 ***
Residuals 38.96 147
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Here, the type argument is noting how residuals, and thus sums of squares, should be calculated. Residuals can be calculated in multiple ways. For simple models (those with one variable, which we have now) and those with balanced designs (the same number of observations for each grouping, coming back to this next chapter), these approaches will lead to the same answer. However, we’ll introduce these ideas here and return to them later.
Let’s show that here and briefly look at the approaches.
Type I sums of squares calculate residuals (and thus sums of squares) based on the order in which factors are added to a model. This is why they are also known as sequential sums of squares. Since we only have one factor now, this distinction isn’t important, but we’ll come back to it. The Anova function won’t calculate these, but we can use another approach (don’t use this otherwise!)
summary(aov(iris_anova)) Df Sum Sq Mean Sq F value Pr(>F)
Species 2 63.21 31.606 119.3 <2e-16 ***
Residuals 147 38.96 0.265
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Type 2 sums of squares do not consider interactions (which we haven’t either yet, so again, we’ll come back to it).
Anova(iris_anova, type="II")Anova Table (Type II tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
Species 63.212 2 119.26 < 2.2e-16 ***
Residuals 38.956 147
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1So for our simple design, these all look the same. However, if we have an unbalanced design, things change a little:
#from https://jennybc.github.io/purrr-tutorial/ls12_different-sized-samples.html
library(purrr)
library(tidyr)
set.seed(19)
iris_unbalanced <- iris %>%
group_by(Species) %>%
nest() %>%
ungroup() %>%
mutate(n = c(50, 30, 20)) %>%
mutate(samp = map2(data, n, sample_n)) %>%
dplyr::select(-data) %>%
unnest(samp)
iris_unbalanced_anova <- lm(Sepal.Length~Species, iris_unbalanced)Now’s lets compare our different sums of squares.
summary(aov(iris_unbalanced_anova)) Df Sum Sq Mean Sq F value Pr(>F)
Species 2 35.07 17.54 76.11 <2e-16 ***
Residuals 97 22.35 0.23
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova(iris_unbalanced_anova, type="II")Anova Table (Type II tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
Species 35.070 2 76.114 < 2.2e-16 ***
Residuals 22.347 97
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova(iris_unbalanced_anova, type="III")Anova Table (Type III tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
(Intercept) 1253.00 1 5438.846 < 2.2e-16 ***
Species 35.07 2 76.114 < 2.2e-16 ***
Residuals 22.35 97
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Wait..they are the same. Let’s try even more unbalanced data
set.seed(19)
iris_really_unbalanced <- iris %>%
group_by(Species) %>%
nest() %>%
ungroup() %>%
mutate(n = c(50, 4, 2)) %>%
mutate(samp = map2(data, n, sample_n)) %>%
dplyr::select(-data) %>%
unnest(samp)
iris_really_unbalanced_anova <- lm(Sepal.Length~Species, iris_really_unbalanced)summary(aov(iris_really_unbalanced_anova)) Df Sum Sq Mean Sq F value Pr(>F)
Species 2 7.551 3.776 26.74 9.35e-09 ***
Residuals 53 7.483 0.141
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova(iris_really_unbalanced_anova, type="II")Anova Table (Type II tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
Species 7.5511 2 26.74 9.346e-09 ***
Residuals 7.4832 53
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Anova(iris_really_unbalanced_anova, type="III")Anova Table (Type III tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
(Intercept) 1253.00 1 8874.42 < 2.2e-16 ***
Species 7.55 2 26.74 9.346e-09 ***
Residuals 7.48 53
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1That’s because R is assuming variance is the same for each of the groups, which means unequal sample sizes isn’t an issue. However, we didn’t test this assumption, and it would be really hard to given the small sizes of some groups. In the next chapter, we’ll discuss these issues more. For now, we’ll get in the habit of using type=“III” for this class.
Doing this we see Species has a significant impact on explaining variation in the data (and our very high F value). So we reject the null hypothesis that mean sepal length does not differ among species. Now what?
Just like for a multi-population \(\chi^2\) tests, we need to do follow-up tests to compare groups while controlling for the FWER. For linear models (and more), we will use the glht function from the multcomp package to conduce these tests.
library(multcomp)
compare_cont_tukey <- glht(iris_anova, linfct = mcp(Species = "Tukey"))
summary(compare_cont_tukey)
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: lm(formula = Sepal.Length ~ Species, data = iris)
Linear Hypotheses:
Estimate Std. Error t value Pr(>|t|)
versicolor - setosa == 0 0.930 0.103 9.033 <1e-08 ***
virginica - setosa == 0 1.582 0.103 15.366 <1e-08 ***
virginica - versicolor == 0 0.652 0.103 6.333 <1e-08 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- single-step method)Our first approach uses a new (to us) method. It is not exactly Tukey’s method (very confusing) but is closely related to it. Tukey’s method is focuses on all possible pair-wise comparisons and controls for the FWER using a studentized range approach (not fully developed here, but similar to z-transform but focused on the range of means and using the estimated standard deviation; similar to t-statistic in this aspect and also developed by Student). It is also called Tukey’s Honestly Significant Difference/HSD, Tukey-Kramer method, and many other names. In glht, specifying “Tukey” tells the program to do all possible pairs comparison (like Tukey’s method). The post-hoc control for FWER, however, uses a slightly different approach that can handle interactions (still to be explained) and some other things a little bit easier.
Using this approach we see that all species differ significantly from all others; the output also provides estimates of the differences, which match up with our model summary output.
We can also use the methods we previously discussed such as Bonferroni and FDR. These requre us to set up the comparisons, which also means we can limit our number of tests if so desired. For example, we could focus only on differences with I. virginica.
compare_virginica_only <- glht(iris_anova, linfct = mcp(Species =
c("virginica - versicolor = 0",
"virginica - setosa = 0")))After setting up the comparison, we can specify the method to use to correct for FWER,
summary(compare_virginica_only, test=adjusted("holm"))
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: User-defined Contrasts
Fit: lm(formula = Sepal.Length ~ Species, data = iris)
Linear Hypotheses:
Estimate Std. Error t value Pr(>|t|)
virginica - versicolor == 0 0.652 0.103 6.333 2.77e-09 ***
virginica - setosa == 0 1.582 0.103 15.366 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- holm method)summary(compare_virginica_only, test=adjusted("fdr"))
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: User-defined Contrasts
Fit: lm(formula = Sepal.Length ~ Species, data = iris)
Linear Hypotheses:
Estimate Std. Error t value Pr(>|t|)
virginica - versicolor == 0 0.652 0.103 6.333 2.77e-09 ***
virginica - setosa == 0 1.582 0.103 15.366 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- fdr method)Note for our small number of tests and relatively large differences in means and large sample sizes, differences in p values are minimal.
There are instances when FWER are not an issue and thus p values do not need to be adjusted. This occurs when we explore orthogonal contrasts.
Let’s return to linear models to help explain orthogonal contrasts (and some other things). Linear models are a sysem of equations (a matrix), where
\[ \begin{split} Y=X\beta+\epsilon, \textrm{ where }\\ \textrm{Y is our observations (an nx1 matrix)}\\ \textrm{X is a matrix showing our explanatory variables (an nxk matrix)}\\ \beta \textrm{ is our coefficient matrix(an kx1 matrix)}\\ \epsilon \textrm{is a matrix (an nx1) of residuals)}\\ \end{split} \]
In our case, \(\beta\) is simply a 3x1 matrix where each entry is a species average (or one is an intercept and other two are distances from it - same thing) and X is a matrix with dummy variables (1 or 0) indicating which group each observation belongs too. X is sometimes called a model or design matrix. We can see this using R. First, we can pull the design matrix from our model object
library(rmarkdown)
paged_table(data.frame(model.matrix(iris_anova)))We can also pull the model coefficients, which form our \(\beta\) matrix, and place them in the correct orientation.
matrix(as.numeric(iris_anova$coefficients), ncol=1) [,1]
[1,] 5.006
[2,] 0.930
[3,] 1.582So for our first observation, which is
iris[1,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 5.1 3.5 1.4 0.2 setosaWe would multiply (remember, rows get multiplied by columns in matrices)
1*5.006 + 0*.930+0*1.58[1] 5.006This explains why all our fitted values are one of three values! We can also see our observation minus prediction
iris[1,"Sepal.Length"]-1*5.006 + 0*.930+0*1.58[1] 0.094matches our first model residual
iris_anova$residuals[1] 1
0.094 Understanding this general setup explains a few things. When models get more complicated you may see errors or warnings related to singularity. This occurs when XTX isn’t invertible(linear algebra!), which it needs to be to find \(\beta\). This occurs if columns in your design matrix are not independent and are actually linear combinations of each other. This happens when you have highly related measurements (we’ll discuss correlation eventually so you can actually measure this!). We will use similar manipulations to eventually find the \(\hat{H}\) matrix when we consider Cook’s Distance (in regression chapter). Degrees of freedom are similarly related to the number of estimated coefficients the model required (the number of rows in the \(\beta\) matrix).
Returning to our contrasts, note when we do post-hoc tests we are effectively testing for differences in \(\beta\) values. We can put these “tests” in a similar system of equations/matrix. For these tests, the coefficients have to equal 0. For pair comparisons, that means we have 1 for one coeffcient and -1 for the other (for example, (1,-1,0). However, we can also compare one coefficient to the average of the others (2,-1,-1). We could write these two contrasts as
\[ \begin{bmatrix} 1 & -1 & 0 \\ 2 & -1 & -1 \\ \end{bmatrix} \]
A group of contrasts are orthogonal if the sum of the multiplied coefficients from each column equals zero. In this case
1*2 + -1*1 + 0*1[1] 1does not equal 0, so these are not orthogonal contrasts. This is also because I can add add the first and third column and get the second (columns are not independent). However, if we instead carried out these contrasts
\[ \begin{bmatrix} 2 & -1 & -1 \\ 0 & 1 & -1\\ \end{bmatrix} \]
they would be independent. There are other options as well, as we can always find a number of orthogonal contrasts equal to the number of groups being compared minus one.
Resulting p values would not required correction for FWER. We can specify contrasts like this using glht. Below I do the same matrix, but note I set the maximum entry to 1 so that estimate of mean differences aren’t doubled.
contr <- rbind("setosa - versicolor - virginica" = c(1, -.5,-.5),
"versicolor - virginica" = c(0,1,-1))
summary(glht(iris_anova, linfct = mcp(Species = contr)), test=adjusted("none"))
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: User-defined Contrasts
Fit: lm(formula = Sepal.Length ~ Species, data = iris)
Linear Hypotheses:
Estimate Std. Error t value Pr(>|t|)
setosa - versicolor - virginica == 0 -1.25600 0.08916 -14.086 < 2e-16 ***
versicolor - virginica == 0 -0.65200 0.10296 -6.333 2.77e-09 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- none method)Output from post-hoc tests is often displayed using compact letter display. Groups that are not significantly differently share the same letter (so in this case they all have different letters).
cld_output <- fortify(cld(compare_cont_tukey))
cld_output$Species <- cld_output$lhs
function_output <- summarySE(iris, measurevar="Sepal.Length", groupvars =
c("Species"))
function_output <- merge(function_output, cld_output)
ggplot(function_output, aes(y=Sepal.Length, x=Species, fill=Species)) +
geom_col(aes(fill=Species)) +
geom_errorbar(aes(ymin=Sepal.Length-ci, ymax=Sepal.Length+ci)) +
labs(title=expression(paste("Sepal lengths of ",italic("I. species"))),
y= "Sepal length (cm)",
x= "Species")+
geom_text(aes(label=letters,y=Sepal.Length+3*ci))
Other options include plotting the differences in means
plot(compare_cont_tukey)
Another popular package for conducting posthoc comparison in R is emmeans. The package has a great starters guide here
So far we have focused on comparing means among multiple groups. This can include include comparing means between only 2 groups (which we already do for the post-hoc tests). In doing this we also introduced new post-hoc tests and the ideas of a linear model.
The linear model framework will unify most of the remaining tests we learn in class. In fact, several tests we’ve already learned can be formulated this way. This is extremely useful given we want statistics to be a related set of tests in a comprehensive framework.
There are many ways to teach statistics, however, and a long history of tests. Many textbooks and approaches build up from one sample tests by moving to two-sample t-tests. These tests bridge the logic noted above and the approach we used for single-sample t-test. This is because the t-distribution is a special case of the F distribution. It occurs when the square root of an F distribution with 1 degree of freedom in the numerator is considered. Thus the degrees of freedom associated with a t-test will be equal to the degrees of freedom associated with the denominator of the associated F-test.
2-sample t-tests may also an easier approach to first considering differences among groups since with only 2 populations the difference in means may be considered. However, it can be shown (not here) this is simply a rearrangement of our exploration of variances.
To demonstrate this, let’s only consider two species. Note
two_species_subset <- iris[iris$Species!="setosa",]
t.test(Sepal.Length ~ Species, two_species_subset, var.equal=T)
Two Sample t-test
data: Sepal.Length by Species
t = -5.6292, df = 98, p-value = 1.725e-07
alternative hypothesis: true difference in means between group versicolor and group virginica is not equal to 0
95 percent confidence interval:
-0.8818516 -0.4221484
sample estimates:
mean in group versicolor mean in group virginica
5.936 6.588 yields the same p-value as
Anova(lm(Sepal.Length ~ Species, two_species_subset), type="III")Anova Table (Type III tests)
Response: Sepal.Length
Sum Sq Df F value Pr(>F)
(Intercept) 1761.80 1 5253.038 < 2.2e-16 ***
Species 10.63 1 31.688 1.725e-07 ***
Residuals 32.87 98
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Also note the t statistic is the square root of the F statistic.
-5.6292^2[1] -31.68789Note the t.test function can also use columns holding data from each population as arguments as opposed to the formula interface, but we will not use that approach here.
The var.equal=T argument, however, is not the default in R, and this assumption is one of the major differences in 2-sample t-tests and F tests. Remember, ANOVAs and t-tests both require estimates for sigma. If we do not assume the variances are equal for each group, then the best way to estimate the variance is to calculate the variance for each group and take a weighted (by sample size mean). This approach is known as the Behren-Fisher or Welsh t-test.
t.test(Sepal.Length ~ Species, two_species_subset)
Welch Two Sample t-test
data: Sepal.Length by Species
t = -5.6292, df = 94.025, p-value = 1.866e-07
alternative hypothesis: true difference in means between group versicolor and group virginica is not equal to 0
95 percent confidence interval:
-0.8819731 -0.4220269
sample estimates:
mean in group versicolor mean in group virginica
5.936 6.588 The resulting statistics has a distrubtion that can be approximated by a t-distribution, but the associated degrees of freedom can can be non-integer (decimal) and less than (n1+ n2 - 2).
This means the basic assumptions for 2-sample t-tests are independent data points, groups show the same variance, and means are normally distributed. Much like the one-sample t-test, the central limit theorem implies assumptions about the mean distribution are commonly met. However, if they are not met, we have a few common options.
Options for when assumptions of the t- and F-tests (ANOVAs) are presented below. Note given the history of t-tests being considered apart from ANOVAs, some functions only work with less than 2 populations while others work with three or more. However, the overall approaches are similar.
We can extend the Wilcoxon test to 2-samples. To do so, we rank the data points from smallest to largest. The ranks are then used to calculate a U statistic. The statistic sums the ranks for each group (r), then uses them to calculate
\[ U_1 = n_1n_2+\frac{n_1(n_1+1)}{2}-r_1 \] The U statistics is calculated for each group. The larger U value is then taken and used to compute a p value. We can calculate this using the wilcox.test function,
wilcox.test(Sepal.Length ~ Species, two_species_subset)
Wilcoxon rank sum test with continuity correction
data: Sepal.Length by Species
W = 526, p-value = 5.869e-07
alternative hypothesis: true location shift is not equal to 0This test assumes the two distributions being considered have similar shape (not that the resulting means are normally-distributed). If you remove the default continuity correction (applied as we approximate discrete data with a continuous distribution)
wilcox.test(Sepal.Length ~ Species, two_species_subset, correct=F)
Wilcoxon rank sum test
data: Sepal.Length by Species
W = 526, p-value = 5.765e-07
alternative hypothesis: true location shift is not equal to 0We get the same result as the Kruskal-Wallis test
kruskal.test(Sepal.Length ~ Species, two_species_subset)
Kruskal-Wallis rank sum test
data: Sepal.Length by Species
Kruskal-Wallis chi-squared = 24.989, df = 1, p-value = 5.765e-07which is a rank-based test that can be applied to 3+ populations.
kruskal.test(Sepal.Length ~ Species, data = iris)
Kruskal-Wallis rank sum test
data: Sepal.Length by Species
Kruskal-Wallis chi-squared = 96.937, df = 2, p-value < 2.2e-16If we have more than three populations and this omnibus test reveals a significant p-value, we can follow it up with appropriate post-hoc tests.
pairwise.wilcox.test(iris$Sepal.Length,
iris$Species,
p.adjust.method="holm")
Pairwise comparisons using Wilcoxon rank sum test with continuity correction
data: iris$Sepal.Length and iris$Species
setosa versicolor
versicolor 1.7e-13 -
virginica < 2e-16 5.9e-07
P value adjustment method: holm For a single sample, we also considered the sign/binary test. We will return to this test in the next chapter, as it does not work for data from independent samples.
Another option is to extend the bootstrapping option. Although we could again develop a simulation using the boot function again, here we again use the MKinfer package.
library(MKinfer)
boot.t.test(Sepal.Length ~ Species, two_species_subset)
Bootstrap Welch Two Sample t-test
data: Sepal.Length by Species
bootstrap p-value < 2.2e-16
bootstrap difference of means (SE) = -0.6533133 (0.1145254)
95 percent bootstrap percentile confidence interval:
-0.878 -0.426
Results without bootstrap:
t = -5.6292, df = 94.025, p-value = 1.866e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.8819731 -0.4220269
sample estimates:
mean in group versicolor mean in group virginica
5.936 6.588 note we can also run this test without assuming variances are different.
boot.t.test(Sepal.Length ~ Species, two_species_subset, var.equal=T)
Bootstrap Two Sample t-test
data: Sepal.Length by Species
bootstrap p-value < 2.2e-16
bootstrap difference of means (SE) = -0.6529151 (0.1314769)
95 percent bootstrap percentile confidence interval:
-0.914 -0.388
Results without bootstrap:
t = -5.6292, df = 98, p-value = 1.725e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.8818516 -0.4221484
sample estimates:
mean in group versicolor mean in group virginica
5.936 6.588 Both these approaches also show the corresponding t-test results, but note you should choose which test you plan to use before seeing the results!
For more than 2 groups, the t1waybt function in the WRS2 package can allow comparison.
library(WRS2)
t1waybt(Sepal.Length~Species, iris)Call:
t1waybt(formula = Sepal.Length ~ Species, data = iris)
Effective number of bootstrap samples was 599.
Test statistic: 111.9502
p-value: 0
Variance explained: 0.716
Effect size: 0.846 If needed, the mcppb20 package allows for appropriate post-hoc comparisons.
bootstrap_post_hoc <- mcppb20(Sepal.Length~Species, iris)
bootstrap_post_hoc_df <-data.frame(bootstrap_post_hoc$comp)
bootstrap_post_hoc_df$adjusted_p <- p.adjust(as.numeric(bootstrap_post_hoc$comp[,6]), "holm")
bootstrap_post_hoc_df$Group <- factor(bootstrap_post_hoc_df$Group)
library(plyr)
bootstrap_post_hoc_df$Group <- revalue(bootstrap_post_hoc_df$Group,
setNames( bootstrap_post_hoc$fnames,as.character(1:length(bootstrap_post_hoc$fnames))))The following `from` values were not present in `x`: 3bootstrap_post_hoc_df$Group.1 <- factor(bootstrap_post_hoc_df$Group.1)
bootstrap_post_hoc_df$Group.1 <- revalue(bootstrap_post_hoc_df$Group.1,
setNames( bootstrap_post_hoc$fnames,as.character(1:length(bootstrap_post_hoc$fnames))))The following `from` values were not present in `x`: 1bootstrap_post_hoc_df Group Group.1 psihat ci.lower ci.upper p.value adjusted_p
1 setosa versicolor -0.9100000 -1.1400000 -0.7100000 0 0
2 setosa virginica -1.5466667 -1.7866667 -1.3000000 0 0
3 versicolor virginica -0.6366667 -0.9166667 -0.3366667 0 0A new option when comparing groups (2 or more) is known as the permutation test. We encountered a similar approach when we learned about the Fisher’s test for binomial data. Using this approach, we can move the measurements between measured populations, calculate test statistics, and consider how unusual our observed statistic was (a p value!). We can do this for 2
library(coin)
independence_test(Sepal.Length ~ Species, data = two_species_subset)
Asymptotic General Independence Test
data: Sepal.Length by Species (versicolor, virginica)
Z = -4.9183, p-value = 8.731e-07
alternative hypothesis: two.sidedor 3+ populations
independence_test(Sepal.Length ~ Species, data = iris)
Asymptotic General Independence Test
data: Sepal.Length by Species (setosa, versicolor, virginica)
maxT = 8.7572, p-value < 2.2e-16
alternative hypothesis: two.sidedPost-hoc test options are available in the rcompanion package.
library(rcompanion)
pairwisePermutationTest(Sepal.Length ~ Species,
data = iris,
method="holm") Comparison Stat p.value p.adjust
1 setosa - versicolor = 0 -7.246 4.28e-13 8.560e-13
2 setosa - virginica = 0 -8.368 5.883e-17 1.765e-16
3 versicolor - virginica = 0 -4.918 8.731e-07 8.731e-07Our following chapters will extend ANOVAs to consider the impact of multiple measured categories. In doing so, we will also explain paired t-tests and sign tests for paired data.