17.1 – Ordinary least squares regression

Introduction

Linear regression is a toolkit for developing linear models of cause and effect between a ratio scale data type, response or dependent variable, often labeled “Y,” and one or more ratio scale data type, predictor or independent variables, X. Like ANOVA, linear regression is a special case of the general linear model. Regression and correlation both test linear hypotheses: we state that the relationship between two variables is linear (the alternative hypothesis) or it is not (the null hypothesis). The difference?

Correlation is a test of linear association (are variables correlated, we ask?), but are not tests of causation: we do not imply that one variable causes another to vary, even if the correlation between the two variables is large and positive, for example. Correlations are used in statistics on data sets not collected from explicit experimental designs incorporated to test specific hypotheses of cause and effect.
Linear regression is to cause and effect as correlation is to association. With regression and ANOVA, which again, are special cases of the general linear model (LM), we are indeed making a case for a particular understanding of the cause of variation in a response variable: modeling cause and effect is the goal.

We start our LM model as $Y \sim model$

where “~”, tilda, is an operator used by R in formulas to define the relationship between the response variable and the predictor variable(s).

From R Commander we call the linear model function by Statistics → Fit models → Linear model … , which brings up a menu with several options (Fig 1).

Screenshot rcmdr linear model menu

Figure 1. R commander menu interface for linear model.

Our model was

Matings ~ Body.Mass

R commander will keep track of the models created and enter a name for the object. You can, and probably should, change the object name yourself. The example shown in Figure 1 is a simple linear regression, with Body.Mass as the Y variable and Matings the X variable. No other information need be entered and one would simply click OK to begin the analysis.

Example

The purpose of regression is similar to ANOVA. We want a model, a statistical representation to explain the data sample. The model is used to show what causes variation in a response (dependent) variable using one or more predictors (independent variables). In life history theory, mating success is an important trait or characteristic that varies among individuals in a population. For example we may be interested in determining the effect of age (X₁) and body size (X₂) on mating success for a bird species. We could handle the analysis with ANOVA, but we would lose some information. In a clinical trial, we may predict that increasing Age (X₁) and BMI (X₂) causes increase blood pressure (Y).

Our causal model looks like

$\begin{align*} X_{1}+X_{2}\rightarrow Y \end{align*}$

Let’s review the case for ANOVA first.

The response (dependent variable), the number of successful matings for each individual male bird, would be a quantitative (interval scale) variable. (Reminder: You should be able to tell me what kind of analysis you would be doing if the dependent variable was categorical!) If we use ANOVA, then factors have levels. For example, we could have several adult birds differing in age (factor 1) and of different body sizes. Age and body size are quantitative traits, so, in order to use our standard ANOVA, we would have to assign individuals to a few levels. We could group individuals by age (eg, < 6 months, 6 – 10 months, > 12 months) and for body size (eg, small, medium, large). For the second example, we might group the subjects into age classes (20-30, 30-40, etc), and by AMA recommended BMI levels (underweight < 18.5, normal weight 18.5 – 24.9, overweight 25-29.9, obese > 30).

We have not done anything wrong by doing so, but if you are a bit uneasy by this, then your intuition will be rewarded later when we point out that in most cases you are best to leave it as a quantitative trait. We proceed with the test of ANOVA, but we are aware that we’ve lost some information — continuous variables (age, body size, BMI) were converted to categories — and so we suspect (correctly) that we’ve lost some power to reject the null hypothesis. By the way, when you have a “factor” that is a continuous variable, we call it a “covariate.” Factor typically refers to a categorical explanatory (independent) variable.

We might be tempted to use correlation — at least to test if there’s a relationship between Body Mass and Number of Matings. Correlation analysis is used to measure the intensity of association between a pair of variables. Correlation is also used to to test whether the association is greater than that expected by chance alone. We do not express one as causing variation in the other variable, but instead, we ask if the two variables are related (covary). We’ve already talked about some properties of correlation: it ranges from -1 to +1 and the null hypothesis is that the true association between two variables is equal to zero. We will formalize the correlation next time to complete our discussion of the linear relationship between two variables.

But regression is appropriate here because we are indeed making a causal claim: we selected Age and Body Size, and we selected Age and BMI in the second example wish to develop a model so we can predict and maybe even advise.

Least squares regression explained

Regression is part of the general linear model family of tests. If there is one linear predictor variable, then that is a simple linear regression (SLR), also called ordinary least squares (OLS), if there are two or more linear predictor variables, then that is a multiple linear regression (MLR, Chapter 18).

First, consider one predictor variable. We begin by looking at how we might summarize the data by fitting a line to the data; we see that there’s a relationship between mass and mating success in both young and old females (and maybe in older males).

The data set was

Table 1. Our data set of number of matings by male bird by body mass (g).

Body.Mass	Matings
29	0
29	2
29	4
32	4
32	2
35	6
36	3
38	3
38	5
38	8
40	6

Note 1: Unfortunately, I’ve lost track where Table 1 data set came from! It’s likely a simulated set inspired by my readings back in 2005.

And a scatterplot of the data (Fig 2)

Scatterplot, number of matings by body mass (g) of the male bird.

Figure 2. Number of matings by body mass (g) of the male bird.

There’s some scatter, but our eyes tell us that as body size increases, the number of matings also increases. We can go so far as to say that we can predict (imperfectly) that larger birds will have more matings. We fit the best fit line to the data and added the line to our scatterplot (Fig 3). The best fit line meets the requirements that the error about the line is minimized (see below). Thus, we would predict about six matings for a 40 g bird, but only two matings for a 28 g bird. And this is a good feature of regression, prediction, as long as used with some caution.

Same scatterplot, but with fitted simple linear regression line

Figure 3. Same data as in Fig 2, but with the “best fit” line.

The prediction works best for the range of data for which the regression model was built. Outside the range of values, we predict with caution.

The simplest linear relationship between two variables is the SLR. This would be the parameter version (population, not samples), where

$\begin{align*} Y_{i} = \beta_{0} + \beta_{1} X_{i} + \epsilon_{i} \end{align*}$

$\beta_{0}$ = the Y-intercept coefficient and it is defined as

$\begin{align*} \beta_{0} = \bar{Y} -b\bar{X} \end{align*}$

solve for intercept by setting X = 0.

$\beta_{1}$ = the regression coefficient (slope)

$\begin{align*} \beta_{1} = \frac{\sum \left (X_{i} - \bar{X}\right ) \cdot \left (Y_{i} - \bar{Y} \right ) }{\sum \left (X_{i} - \bar{X} \right )^2} \end{align*}$

Note 2: The denominator is just our corrected sums of squares that we’ve seen many times before. The numerator is the cross-product and is referred to as the covariance.

ε = the error or “residual”

$\begin{align*} \epsilon_{i} = Y_{i} - \hat{Y}_{i} \end{align*}$

The residual is an important concept in regression. We briefly discussed “what’s left over,” in ANOVA, where an observation Yi is equal to the population mean plus the factor effect of level i plus the remainder or “error”.

In regression, we speak of residuals as the departure (difference) of an actual Y_i (observation) from the predicted Y ( $\hat{Y}$ , say “why-hat”).

the linear regression predicts Y

$\begin{align*} \sum_{i=1}^{n} \left (Y_{i} - \hat{Y}_{i} \right )^2 \end{align*}$

and what remains unexplained by the regression equation is called the residual

$\begin{align*} \epsilon_{i} = Y_{i} - \hat{Y}_{i} \end{align*}$

There will be as many residuals as there were observations.

But why THIS particular line? We could draw lines anywhere through the points. Well, this line is termed the “best fit“ because it is the only line that minimizes the sum of the squared deviations for all values of Y (the observations) and the predicted $\hat{Y}$ . The best fit line minimizes the sum of the squared residuals.

$\begin{align*} sum_{i=1}^{n} \left (Y_{i} - \hat{Y}_{i} \right )^2 \end{align*}$

Note 3: More technically we justify the OLS as best fit by saying” Following the Gauss–Markov theorem, OLS estimators are the best linear unbiased estimators (BLUE). Gauss (1821) proved that the least squares method produces unbiased estimates with the smallest variance, a key result for regression analysis, while Markov (1900) rediscovered Gauss’ work and showed the conclusions held under less restrictive conditions.

Thus, like ANOVA, we can account for the total sums of squares (SS_Tot) as equal to the sums of squares (variation), explained by the regression model, SS_reg, plus what’s not explained, what’s left over, the residual sums of squares, SS_res, aka SS_error.

$\begin{align*} SS_{Tot} = SS_{reg} + SS+{res} \end{align*}$

How well does our model explain — fit — our data? We cover this topic in Chapter 17.3 — Estimation of linear regression coefficients, but for now, we introduce $R^2$ , the coefficient of determination.

$\begin{align*} R^2= \frac{SS_{reg}}{SS_{Tot}} \end{align*}$

$R^2$ ranges from zero (0%), the linear model fails completely to explain the data, to one (100%), the linear model explains perfectly every datum.

Models used to predict new values

Once a line has been estimated, one use is to predict new observations not previously measured!

Figure 3 redrawn to extend the line to the Y intercept.

Figure 4. Figure 3 redrawn to extend the line to the Y intercept.

This is an important use of models in statistics: use an equation to fit to some data, then predict Y values from new values of X. To use the equation, simply insert new values of X into the equation, because the slope and intercept are already “known.” Then for any X_i we can determine $\hat{y}$ (predicted Y value that is on the best fit regression line).

This is what people do when they say

“if you are a certain weight (or BMI) you have this increased risk of heart disease”

“if you have this number of black rats in the forest you will have this many nestlings survive to leave the nest”

“if you have this much run-off pollution into the ocean you have this many corals dying”

“if you add this much enzyme to the solution you will have this much resulting product”.

R Code

We can use the drop down menu in Rcmdr to do the bulk of the work, supplemented with a little R code entered and run from the script window. Scrape data from Table 1 and save to R as bird.matings.

LinearModel.3 <- lm(Matings ~ Body.Mass, data=bird.matings)
summary(LinearModel.3)

Output from R — we’ll pull apart the important parts, color-highlights used to help orient the reader as we go.

Call:
lm(formula = Matings ~ Body.Mass, data = bird.matings)

Residuals:
     Min       1Q   Median      3Q     Max 
-2.29237 -1.34322 -0.03178 1.33792 2.70763

Coefficients:
            Estimate Std. Error  t value  Pr(>|t|) 
(Intercept)  -8.4746     4.6641   -1.817    0.1026 
Body.Mass     0.3623     0.1355    2.673    0.0255 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.776 on 9 degrees of freedom
Multiple R-squared: 0.4425, Adjusted R-squared: 0.3806 
F-statistic: 7.144 on 1 and 9 DF, p-value: 0.02551

What are the values for the coefficients? See Estimate .

Table 2. Coefficient estimates.

	Value
$b_{0}$ , Y-intercept	-8.4746
$b_{1}$ , Slope	0.3623

Get the sum of squares from the ANOVA table

myAOV.full <- anova(LinearModel.3); myAOV.full

Output from R, the ANOVA table

Analysis of Variance Table

Response: Matings
           Df  Sum Sq  Mean Sq  F value     Pr(>F)
Body.Mass   1  22.528  22.5277   7.1438   0.02551 *
Residuals   9  28.381   3.1535
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Looking for $R^2$ , the coefficient of determination? From the R output identify Multiple R-squared. We find 0.4425. Grab the SS_reg and SS_Tot from the ANOVA table and confirm

$\begin{align*} R^2= \frac{22.528}{22.528+28.381} \end{align*}$

Note 2: The value of $R^2$ increases with each additional predictor variable included in the model, whether statistically significant contributor or not, regardless of whether the association between the predictor is simply due to chance. The Adjusted R-squared is an attempt to correct for this bias.

The residual standard error, RSE, is another model fit statistic.

$\begin{align*} RSE = \sqrt{MSE} \end{align*}$

The smaller the residual standard error, the better the model fit to the data. See Residual standard error: 1.776 and MSE = 3.1535 above.

And the “statistical significance?” See Pr(>|t|) .There were tests for each coefficient, $H_{O}:b_{0}=0}$ and $H_{O}:b_{1}=0}$ :

Table 3. Coefficient estimates.

	P-value
$b_{0}$ , Y-intercept	0.1026
$b_{1}$ , Slope	0.0255

And, at last, note that p-value, Pr(>F), reported in the Analysis of Variance Table, is the same as the p-value for the slope reported as Pr(>|t|).

Make a prediction:

How many matings expected for a 40 g male?

predict(LinearModel.3, data.frame(Body.Mass=40))

R output

1 
6.016949

Answer: we predict 6 matings for a 40g male.

See below for predicting confidence limits.

How to extract information from an R object

We can do more — the lm() returnd a great deal of information about our regression model. Explore all that is available in the objects LinearModel.3 and myAOV.full with the str() command.

Note 3: str() command lets us look at an object created in R. Type ?str or help(str) to bring up the R documentation. Here, we use str() to look at the structure of the objects we created. “Classes” refers to the R programming class attribute inherited by the object. We first used str() in Chapter 8.3.

str(LinearModel.3)

Output from R

List of 12
$ coefficients : Named num [1:2] -8.475 0.362
..- attr(*, "names")= chr [1:2] "(Intercept)" "Body.Mass"
$ residuals : Named num [1:11] -2.0318 -0.0318 1.9682 0.8814 -1.1186 ...
..- attr(*, "names")= chr [1:11] "1" "2" "3" "4" ...
$ effects : Named num [1:11] -12.965 4.746 2.337 1.327 -0.673 ...
..- attr(*, "names")= chr [1:11] "(Intercept)" "Body.Mass" "" "" ...
$ rank : int 2
$ fitted.values: Named num [1:11] 2.03 2.03 2.03 3.12 3.12 ...
..- attr(*, "names")= chr [1:11] "1" "2" "3" "4" ...
$ assign : int [1:2] 0 1
$ qr :List of 5
..$ qr : num [1:11, 1:2] -3.317 0.302 0.302 0.302 0.302 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:11] "1" "2" "3" "4" ...
.. .. ..$ : chr [1:2] "(Intercept)" "Body.Mass"
.. ..- attr(*, "assign")= int [1:2] 0 1
..$ qraux: num [1:2] 1.3 1.3
..$ pivot: int [1:2] 1 2
..$ tol : num 0.0000001
..$ rank : int 2
..- attr(*, "class")= chr "qr"
$ df.residual : int 9
$ xlevels : Named list()
$ call : language lm(formula = Matings ~ Body.Mass, data = bird_matings)
$ terms :Classes 'terms', 'formula' language Matings ~ Body.Mass
.. ..- attr(*, "variables")= language list(Matings, Body.Mass)
.. ..- attr(*, "factors")= int [1:2, 1] 0 1
.. .. ..- attr(*, "dimnames")=List of 2
.. .. .. ..$ : chr [1:2] "Matings" "Body.Mass"
.. .. .. ..$ : chr "Body.Mass"
.. ..- attr(*, "term.labels")= chr "Body.Mass"
.. ..- attr(*, "order")= int 1
.. ..- attr(*, "intercept")= int 1
.. ..- attr(*, "response")= int 1
.. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
.. ..- attr(*, "predvars")= language list(Matings, Body.Mass)
.. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
.. .. ..- attr(*, "names")= chr [1:2] "Matings" "Body.Mass"
$ model :'data.frame': 11 obs. of 2 variables:
..$ Matings : num [1:11] 0 2 4 4 2 6 3 3 5 8 ...
..$ Body.Mass: num [1:11] 29 29 29 32 32 35 36 38 38 38 ...
..- attr(*, "terms")=Classes 'terms', 'formula' language Matings ~ Body.Mass
.. .. ..- attr(*, "variables")= language list(Matings, Body.Mass)
.. .. ..- attr(*, "factors")= int [1:2, 1] 0 1
.. .. .. ..- attr(*, "dimnames")=List of 2
.. .. .. .. ..$ : chr [1:2] "Matings" "Body.Mass"
.. .. .. .. ..$ : chr "Body.Mass"
.. .. ..- attr(*, "term.labels")= chr "Body.Mass"
.. .. ..- attr(*, "order")= int 1
.. .. ..- attr(*, "intercept")= int 1
.. .. ..- attr(*, "response")= int 1
.. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
.. .. ..- attr(*, "predvars")= language list(Matings, Body.Mass)
.. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
.. .. .. ..- attr(*, "names")= chr [1:2] "Matings" "Body.Mass"
- attr(*, "class")= chr "lm"

Yikes! A lot of the output is simply identifying how R handled our data.frame and applied the lm() function.

Note 4: The “how” refers to the particular algorithms used by R. For example, qraux, pivot, and tol refer to arguments used in the QR decomposition of the matrix (derived from our data.frame) used to solve our linear least squares problem.

However, it’s from this command we learn what is available to extract in our object. For example, what were the actual fitted values ( $\hat{Y}$ , the predicted Y values) from our linear equation? I marked the relevant str() output above in yellow. We call these from the object with the R code

RegModel.1$fitted.values

and R returns

       1        2        3        4        5        6        7        8        9       10       11 
2.031780 2.031780 2.031780 3.118644 3.118644 4.205508 4.567797 5.292373 5.292373 5.292373 6.016949

Here’s another look at an object, str(myAOV.full), and how to extract useful values.

Output from R

Classes 'anova' and 'data.frame': 2 obs. of 5 variables:
$ Df : int 1 9
$ Sum Sq : num 22.5 28.4
$ Mean Sq: num 22.53 3.15
$ F value: num 7.14 NA
$ Pr(>F) : num 0.0255 NA
- attr(*, "heading")= chr [1:2] "Analysis of Variance Table\n" "Response: Matings"

Extract the sum of squares: type the object name then $ Sum Sq at the R prompt.

myAOV.full $"Sum Sq"

Output from R

[1] 22.52773 28.38136

Get the residual sum of squares

SSE.myAOV.full <- myAOV.full $"Sum Sq"[2]; SSE.myAOV.full

Output from R

[1] 22.52773

Get the regression sum of squares

SSR.myAOV.full <- myAOV.full $"Sum Sq"[1]; SSR.myAOV.full

Output from R

[1] 50.90909

Now, get the total sums of squares for the model

ssTotal.myAOV.full <- SSE.myAOV.full + SSR.myAOV.full; ssTotal.myAOV.full

Calculate the coefficient of determination

myR_2 <- 1 - (SSE.myAOV.full/(ssTotal.myAOV.full)); myR_2

Output from R

[1] 0.4425091

Which matches what we got before, as it should.

Regression equations may be useful to predict new observations

True. However, you should avoid making estimates beyond the range of the X – values that were used to calculate the best fit regression equation! Why? The answer has to do with the shape of the confidence interval around the regression line.

I’ve drawn an exaggerated confidence interval (CI), for a regression line between an X and a Y variable. Note that the CI is narrow in the middle, but wider at the end. Thus, we have more confidence in predicting new Y values for data that fall within the original data because this is the region where we are most confident.

Calculating the CI for the linear model follows from CI calculations for other estimates. It is a simple concept — both the intercept and slope were estimated with error, so we combine these into a way to generalize our confidence in the regression model as a whole given the error in slope and intercept estimation.

$\begin{align*} \% 95 \ CI=b_{1} \pm t_{df}SE_{b_{1}} \end{align*}$

The calculation of confidence interval for the linear regression involves the standard error of the residuals, the sample size, and expressions relating the standard deviation of the predictor variable X — we use the t distribution.

Confidence interval about the best fit line.

Figure 5. 95% confidence interval about the best fit line.

How I got this graph

plot(bird_matings$Body.Mass,bird_matings$Mating,xlim=c(25,45),ylim=c(0,10))
mylm <- lm(bird_matings$Mating~bird_matings$Body.Mass)
predict(mylm, interval = c("confidence"))
abline(mylm, col = "black")
x<-bird_matings$Body.Mass
lines(x, prd[,2], col= "red", lty=2)
lines(x, prd[,3], col= "red", lty=2)

Nothing wrong with my code, but getting all of this to work in R might best be accomplished by adding another package, a plug-in for Rcmdr called RcmdrPlugin.HH. HH refers to “Heiberger and Holland,” and provides software support for their textbook Statistical Analysis and Data Display, 2nd edition.

Regression through the origin

A regression through the origin in OLS is appropriate when theory dictates that the response must be zero when the predictor is zero, that is, the intercept is known to be exactly 0. In population genetics, for example, if you model pairwise genetic distance as a function of some measure of separation, (eg, time or geographic distance), and the definition of the metric ensures that identical individuals have zero distance, then forcing the regression through the origin can be justified. Doing so reduces parameter estimation to the slope alone and can improve efficiency when the assumption is correct. However, if the true relationship includes a nonzero intercept, constraining it to zero biases the slope.

This choice also affects the coefficient of determination (R²): in a no-intercept model, R² is computed relative to zero rather than the mean of Y, so it is not directly comparable to the standard R² from a model with an intercept and can even appear artificially higher or lower. If the origin constraint is correct, the no-intercept model may show a better (more meaningful) fit; if not, the apparent R² can be misleading despite biased estimates.

R code for regression without the Y-intercept:

noOrigin.model <- lm(formula = Matings ~ 0 + Body.Mass, data = bird.matings)
summary(noOrigin.model)

And the R output:

Call:
lm(formula = Matings ~ 0 + Body.Mass, data = bird.matings)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.4112 -1.4406  0.2359  0.9418  3.5301 

Coefficients:
          Estimate Std. Error t value Pr(>|t|)    
Body.Mass  0.11763    0.01726   6.816 4.65e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.97 on 10 degrees of freedom
Multiple R-squared:  0.8229,	Adjusted R-squared:  0.8052 
F-statistic: 46.45 on 1 and 10 DF,  p-value: 4.65e-05

Interpretation and comparison between model with and without fit through the origin.

Without an intercept (regression through the origin), the model estimates a slope of 0.1176 (SE = 0.0173, t = 6.82, p = 4.65×10⁻⁵), indicating a strong, statistically significant positive association between Body Mass and the response. Because the intercept is constrained to zero, this slope is effectively forced to account for all variation in Y, which often inflates its magnitude of significance and can bias the estimate if the true relationship does not pass through the origin.

When the intercept was included, the slope increased to 0.3623 (SE = 0.1355, t = 2.67, p = 0.0255), still statistically significant but less so, while the intercept is estimated at –8.47 (SE = 4.66, p = 0.103), which is not significantly different from zero at conventional levels. The change in slope (from 0.12 to 0.36) shows that allowing a nonzero intercept alters the fitted relationship substantially, suggesting that the origin constraint may not be appropriate in this case. Even though the intercept is not statistically significant, its inclusion redistributes variance between intercept and slope, yielding a more reliable slope estimate and a standard R² that is comparable across models, unlike the no-intercept case.

The negative Y-intercept, while statistically justifiable, is also biologically unrealistic, implying a negative number of matings. The statistical justification is simply that the model is used to predict mating success for organisms in the range of body size included in the data set used to make the model — the negative Y-intercept than is interpreted that we are trying to extend the model beyond its range.

In our example with the birds, a biological justification for a non intercept model can be made. For example, mating success is positively associated with body size; therefore, an organism with a hypothetical body size of zero would logically have zero chance of mating success. That said, even when biology suggests the relationship should pass through the origin, it may still be useful to fit a model with an intercept to test that assumption rather than impose it.

Measurement error, in the case of our birds example, chiefly if observed count of matings differs from the true, actual number of matings that occurred, or in a student employing genetic distance, then a number of errors may occur, including sequencing artifacts, alignment bias, or the way genetic distance is computed (eg, corrections for multiple substitutions, cf discussions in Felsenstein 2003, see Legendre and Desdevises 2009 for justifications for through the origin and phylogenetic independent contrasts). As well documented in standard texts on regression and linear models (eg, Introduction to Linear Regression Analysis, by Montgomery et al 2021; An Introduction to Generalized Linear Models, by Dobson and Barnett 2018), measurement error can introduce a small offset, so the estimated intercept serves as a diagnostic for systematic bias. If the intercept is close to zero and not significant, that provides empirical support for the origin constraint; if it is meaningfully different from zero, forcing the regression through the origin would bias the slope and distort inference. Additionally, including the intercept yields a standard R² and residual structure that are comparable across models, helping assess overall model fit before deciding whether the biologically motivated constraint is justified.

Assumptions of OLS, an introduction

We will cover assumptions of Ordinary Least Squares in detail in 17.8, Assumptions and model diagnostics for Simple Linear Regression. For now, briefly, the assumptions for OLS regression include:

Linear model is appropriate: the data are well described (fit) by a linear model
Independent values of Y and equal variances. Although there can be more than one Y for any value of X, the Y’s cannot be related to each other (that’s what we mean by independent). Since we allow for multiple Y’s for each X, then we assume that the variances of the range of Ys are equal for each X value (this is similar to our ANOVA assumptions for equal variance by groups).
Normality. For each X value there is a normal distribution of Ys (think of doing the experiment over and over)
Error (residuals) are normally distributed with a mean of zero.

Additionally, we assume that measurement of X is done without error (the equivalent, but less restrictive practical application of this assumption is that the error in X is at least negligible compared to the measurements in the dependent variable). Multiple regression makes one more assumption, about the relationship between the predictor variables (the X variables). The assumption is that there is no multicolinearity: the X variables are not related or associated to each other.

In some sense the first assumption is obvious if not trivial — of course a “line” needs to fit the data so why not plow ahead with the OLS regression method, which has desirable statistical properties and let the estimation of slopes, intercept and fit statistics guide us? One of the really good things about statistics is that you can readily test your intuition about a particular method using data simulated to meet, or not to meet, assumptions.

What about assumptions and BLUE? Interestingly, OLS coefficient estimates remain unbiased even if the errors are not normal, as long as other assumptions (like a mean of zero, no correlation, and constant variance) hold. However, the OLS estimators are no longer the “best” linear unbiased estimators (BLUE) because the efficiency property is lost. As we have discussed before, the main consequence is the effects on statistical inference. Without normality, standard errors, t-statistics, confidence intervals, and p-values will be unreliable.

Note 5: The efficiency property means the estimator has the smallest variance among all linear, unbiased estimators. And “unbiased,” again, means the estimator’s expected value is equal to the true parameter value.

Coming up with datasets like these can be tricky for beginners. Thankfully, others have stepped in and provide tools useful for data simulations which greatly facilitate the kinds of testing of assumptions — and more importantly, how assumptions influence our conclusions — statisticians greatly encourage us all to do (see Chatterjee and Firat 2007).

Questions

Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
Predict new values for number of matings for male birds of 20g and 60g.
Anscombe’s quartet (Anscombe 1973) is a famous example of this approach and the fictitious data can be used to illustrate the fallacy of relying solely on fit statistics and coefficient estimates.

Here are the data (modified from Anscombe 1973, p. 19) — I leave it to you to discover the message by using linear regression on Anscombe’s data set. Hint: play naïve and generate the appropriate descriptive statistics, make scatterplots for each X, Y set, then run regression statistics, first on each of the X, Y pairs (there were four sets of X, Y pairs).

Anscombe’s quartet

X	Y1	Y2	Y3	Y4
10	8.04	9.14	7.46	6.58
8	6.95	8.14	6.77	5.76
13	7.58	8.74	12.74	7.71
9	8.81	8.77	7.11	8.84
11	8.33	9.26	7.81	8.47
14	9.96	8.10	8.84	7.04
6	7.24	6.13	6.08	5.25
4	4.26	3.10	5.39	12.50
12	10.84	9.13	8.15	5.56
7	4.82	7.26	6.42	7.91
5	5.68	4.74	5.73	6.89
Mean (±SD)	7.50 (2.032)	7.50 (2.032)	7.50 (2.032)	7.50 (2.032)

Anscombe's data (Anscombe 1973). Note that the data set does not include the column summary statistics shown in the last row of the table.

Quiz Chapter 17.1

Ordinary Least Squares

Chapter 17 contents

15.2 – Wilcoxon Rank Sum Test

Introduction
Wilcoxon test, worked example
Questions
Quiz
Data sets
Chapter 15 contents

Introduction

Wilcoxon rank sum test also called the two sample Wilcoxon test. It is equivalent to another nonparametric tests called the Mann-Whitney test, which was independently derived. We get the Wilcoxon test statistic in Rcmdr through the Statistics submenu.

Rcmdr: Statistics → Nonparametric tests → Two-sample Wilcoxon Test

I’ll show you the test with an example. We’ll use the same data set introduced in chapter 10.3, body mass (g) for four geckos (Hemidactylus frenatus, Fig 1) and four green anolis lizards (Anolis carolinensis, Fig 2).

Male Hemidactylus frenatus, central Oahu, M. Dohm 2018

Figure 1. Female common house gecko, Hemidactylus frenatus, central Oahu, M. Dohm 2018.

Male Anolis carolinensis, ʻAkaka Falls, Big Island of Hawai`i, M. Dohm 2018

Figure 2. Male Anolis carolinensis, ʻAkaka Falls, Hawaiʻi, M. Dohm 2018.

Wilcoxon test, worked example

Geckos: 3.186, 2.427, 4.031, 1.995
Anoles: 5.515, 5.659, 6.739, 3.184

Note 1. This test in Rcmdr requires that data were stacked worksheet and not in unstacked worksheet two columns. If you need help with worksheet format, then see Part07 in Mike’s Workbook for Biostatistics.

We choose from the Rcmdr Nonparametric statistics menu the Two sample Wilcoxon test (Fig 3), then a two-tailed test of the null hypothesis (Fig 4) and elect to use the defaults for the tests and calculations of P-values.

Screenshot Rcmdr 2 Sample Wilcoxn test

Figure 3. Screenshot Rcmdr menu 2 sample Wilcoxon test. Options are selected by clicking on “Options” tab (see Fig 4).

Screenshot Wilcoxon options menu

Figure 4. Screenshot of options tab Rcmdr menu 2 sample Wilcoxon test. Keep defaults to run the “Wilcoxon test.”

Don’t forget to stack the data. Rcmdr won’t produce an error message if the data set is in the unstacked, improper conformation. Instead, Rcmdr menu options will not be available. For example, Fig 5 shows a Two-sample Wilcoxon test… dimmed from view, not available for selection.

Screenshot Rcmdr Statistics menu

Figure 5. Screenshot of Rcmdr menu. Note Two- sample Wilcoxon test… not available.

The results of the test, copied from the Output window, are shown below.

wilcox.test(Mass ~ Lizard, alternative="two.sided", data=LizardStacked)

Wilcoxon rank sum test
data: Mass by Lizard
W = 14, p-value = 0.1143
alternative hypothesis: true location shift is not equal to 0

The calculation of the Wilcoxon test statistic (W) is straightforward, involving summing the ranks. Obtaining the P-value of the test of the null is a bit more involved as it depends on permutations of all possible combinations of differences. For us, R will do nicely with the details, and we just need to check the P-value.

Here, we see that the medians are 5.6 g for the Anolis, and 2.8 g for the geckos. The associated P-value is 0.1143. Thus, we fail to reject the null hypothesis and conclude that there was no difference in median body mass.

Note 2. This is the same general conclusion we got when we ran a independent t-test on this data set: no difference between day one and day two.

Questions

Conduct an independent t-test on the Lizard body mass data.
- Make a box plot to display the two groups and describe the middle and variability.
- Compare results of test of hypothesis. do they agree with the Wilcoxon test? If not, list possible reasons why the two tests disagree.
Using the dataset below, test null hypothesis using independent t-test, Welch’s test, and nonparametric Wilcoxon’s test.
1. Make a box plot to display the two groups and describe the middle and variability.
2. Compare results of test of hypothesis. do they agree with the Wilcoxon test? If not, list possible reasons why the tests disagree.

Quiz Chapter 15.2

Wilcoxon Rank Sum Test

Data set

var1	var2
5.84	5.93
5.72	5.95
5.75	6.02
5.78	5.81
5.81	6.16
5.81	5.95
5.73	6.09
5.77	5.89
5.76	5.99
5.86	5.60
5.84	6.16
5.83	6.16
5.80	6.06
5.78	6.07
5.89	5.66
5.83	6.14
5.79	5.99
5.84	6.15
5.90	5.81
5.86	6.20

Chapter 15 contents,

17 – Linear Regression

Introduction
Components of a statistical model
Regression and correlation test linear hypotheses
Assumptions of linear regression
Build a statistical model, make predictions
Questions
Suggested homework
Quiz
References

Introduction

Regression is a toolkit for developing models of cause and effect between one ratio scale data type dependent response variables, and one (simple linear regression) or more or more (multiple linear regression) ratio scale data type independent predictor variables. By convention the dependent variable(s) is denoted by Y, the independent variable(s) represented by X₁, X₂, $\cdots$ X_n for n independent variables. Like ANOVA, linear regression is simply a special case of the general linear model, first introduced in Chapter 12.7.

Components of a statistical model

Regression statistical methods return model estimates of the intercept and slope coefficients, plus statistics of regression fit (eg, R², aka “R-squared,” the coefficient of determination).

Chapter 17.1 – 17.9 cover the simple linear model

$\begin{align*} Y_{i} = \alpha + \beta X_{i} + \epsilon_{i} \end{align*}$

Chapter 18.1 – 18.5 cover the multiple regression linear model

$\begin{align*} Y_{i} = \beta_{0} + \beta_{1}X_{1} + \beta_{2}X_{2} + \cdots + \beta_{n}X_{n} + \epsilon_{i} \end{align*}$

where α or β₀ represent the Y-intercept and β or β₁, β₂, … β_n represent the regression slopes.

Regression and correlation test linear hypotheses

We state that the relationship between two variables is linear (the alternative hypothesis) or it is not (the null hypothesis). The difference? Correlation is a test of linear association (are variables correlated, we ask?), imply possible causation, but are not sufficient evidence for causation: we do not imply that one variable causes another to vary, even if the correlation between the two variables is large and positive, for example. Correlations are used in statistics on data sets not collected from explicit experimental designs incorporated to test specific hypotheses of cause and effect.

Linear regression, however, is to cause and effect as correlation is to association. With regression and ANOVA, we are indeed making a case for a particular understanding of the cause of variation in a response variable: modeling cause and effect is the goal. Regression, ANOVA, and other general linear models are designed to permit the statistician to control for the effects of confounding variables provided the causal variables themselves are uncorrelated.

When to use correlation and when to apply linear regression to data set? For two ratio scale variables, one can always apply either approach. Correlation is always appropriate if both variables are measured in the course of the study whereas a regression modeling approach would be implied where one variable was the result of manipulation by the researcher.

Assumptions of linear regression

The key assumption in linear regression is that a straight line indeed is the best fit of the relationship between dependent and independent variables. The additional assumptions of parametric tests (Chapter 13) also hold. In Chapter 18 we conclude with an extension of regression from one to many predictor variables and the special and important topic of correlated predictor variables or multicollinearity.

Build a statistical model, make predictions

In our exploration of linear regression we begin with simple linear regression, also called ordinary least squares regression, starting with one predictor variable. Practical aspects of model diagnostics are presented. Regression may be used to describe or to provide a predictive statistical framework. In Chapter 18 we conclude with an extension of regression from one to many predictor variables. We conclude with a discussion of model selection. Throughout, use of Rcmdr and R have multiple ways to analyze linear regression models are presented; we will continue to emphasize the general linear model approach, but note that use of linear model in Rcmdr provides a number of default features that are conveniently available.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.

Suggested homework

Homework 11: Simple linear regression.

Quiz

In addition to the quiz on this page, a total of 81 questions among the several subchapters, a mix of true or false and multiple choice question format.

Quiz Chapter 17

Linear Regression

References

Linear regression is a huge topic; references I include are among my favorite on the subject, but are only a small and incomplete sampling. For simplicity, I merged references for Chapter 17 and Chapter 18 into one page at References and suggested readings (Ch17 & 18)

Chapter 17 contains

14.2 – Sources of variation

Introduction
Unbalanced designs
Questions
Quiz
Chapter 14 contents

Introduction

Sources of variation, or components of the two-way ANOVA include two factors, each with two or more levels (groups), and collectively, factors are often referred to as the main effects in these types of ANOVA. The other source of variation in a two-way ANOVA is the interaction between the two factors. Below, I have listed the important components, although I have not included how the sum of squares are calculated. You are expected to know the sources of variation for this most basic two-way ANOVA table (Fig 1). You should also be able to solve any missing elements in one of these tables by utilizing any included information.

Sources of varation 2-way ANOVA fixed random effects balanced replicated design

Figure 1. ANOVA table for two-way, balanced, replicated design.

Taking each row from Figure 1 one at a time we have

Source	DF	Mean Squares	F-statistic
First Factor	$a - 1$	$\frac{factor \ A \ SS}{factor \ A \ DF}$	$\frac{factor \ A \ MS}{error \ MS}$

where Source refers to the source of variation, DF refers to Degrees of Freedom, a is the number of levels (groups) of the first factor, SS refers to Sum of Squares, and MS refers to the Mean Squares.

Next is the second factor

Source	DF	Mean Squares	F-statistic
First Factor	$b - 1$	$\frac{factor \ B \ SS}{factor \ B \ DF}$	$\frac{factor \ B \ MS}{error \ MS}$

where b is the number of levels (groups) of the second factor. Next is the interaction between the first and second factors.

Source	DF	Mean Squares	F-statistic
Interaction	$(a - 1)(b - 1)$	$\frac{AXB \ SS}{AXB \ DF}$	$\frac{AXB \ MS}{error \ MS}$

and lastly the Within-cell Error or residual source of variation

Source	DF	Mean Squares	F-statistic
Error	$ab(n - 1)$	$\frac{Error \ SS}{error \ DF}$	$N/A$

where n is the number of experimental units for each group. Note that if the sample size differs for one or more groups (levels),then the design would be unbalanced and this formula does not work to determine the degrees of freedom. The total degrees of freedom for the two-way ANOVA is simply N – 1, where N is the sample size for the entire problem; a little algebra shows that N may be calculated as

$\begin{align*} N = abn \end{align*}$

Unbalanced designs

An unbalanced design implies that observations are missing value for one or more groups. What to do if data are missing? Decision depends on how the data are missing (see Chapter 5). For example, if data are missing at random with respect to treatment, then this should not affect inference. If data are missing not at random, then inference, logically, must be impacted. Calculating the ANOVA, moreover, becomes a different matter. In the one-way ANOVA, no real problem arises although setting up contrasts among the levels requires a weighting term to be factored into the calculations. For higher-level ANOVA involving two or more factors the sums of squares for treatment effects are no longer simple partitioning into the different sources of variation. The sources overlap and the order by which the Factors enter into the statistical model now affects the calculations. Thus, while setting up the calculations for the balanced design is straight-forward, perhaps surprisingly, if group sizes differ, this simple relationship for calculating the degrees of freedom, sums of squares, and Mean squares become an unsolvable problem. This problem is largely solved by the general linear model.

Questions

Based on the results of a two-way ANOVA, the error sums of squares (SSE) was computed to be 160. If we ignore one of the factors and perform a one-way ANOVA using the same data, will the SSE be the same as in the two-way ANOVA, or will it increase? Decrease? Explain your choice.
While conducting a two-way ANOVA, you conclude that a statistically significant interaction exists between factor 1 and factor 2. What should be your next step? Do you drop the interaction term from the model and redo the analysis or do you report the results of factor effects including the non-significant interaction?

Quiz Chapter 14.2

Sources of variation

Chapter 14 contents

13.4 – Tests for Equal Variances

Introduction
More about the F-test
R code
Levene’s test of equal variances
Cahoy’s bootstrap method
Questions
Quiz
R code reminders
Chapter 13 contents

Introduction

In order to carry out statistical tests correctly, we must test our data first to see if our sample conforms to the assumptions of the statistical test. If our data do not meet these assumptions, then inferences drawn may be incorrect. How far off our inferences may be depends on a number of factors, but mostly it depends on how far from the expectations our data are.

One assumption we make with parametric tests involving ratio scale data is that the data could be from a normally distributed population. The other key assumption introduced, but not described in detail for the two-sample t-test, was that the variability in the two groups must be the same, ie, homoscedasticity. Thus, in order to carry out the independent sample t-test, we must assume that the variances are equal.

There are two general reasons we may want to concern ourselves with a test for the differences between two variances

The t-test (and other tests like one-way ANOVA) requires that the two samples compared have the same variances. If the Variances are Not Equal we need to perform a modified t-test (see Welch’s formula).
We may also be interested in the differences between the variances in two populations.

Example 1: In genetics we might be interested in the difference between the variability of response of inbred lines (little genetic variation but environmental variation) versus an outbred populations (lots of genetic and environmental variation).

Example 2: Environmental stress can cause organisms to have developmental instability. This might cause organisms to be more variable in morphology or the two sides (right & left) of an organism may develop non-symmetrically. Therefore, polluted environments might cause organisms to have greater variability compared to non-polluted environments.

The first way to test the variances is to use the F test. This works for two groups.

$\begin{align*} F = \frac{s_{1}^{2}}{s_{2}^{2}} \end{align*}$

For more than two groups, we’ll use different tests (eg, Bartlett’s test, Levene’s test).

Remember that the formula for the sample variance is

The Null Hypothesis is that the two samples have the same variances:

$\begin{align*} H_{0}: s_{1}^{2}=s_{1}^{2} \end{align*}$

The Alternative Hypothesis is that the two samples do not have the same variances:

$\begin{align*} H_{0}: s_{1}^{2}\ne s_{1}^{2} \end{align*}$

Note: I prefer to evaluate this as a one-tailed test: identify the larger of the two variances and take that as the numerator Then, the null hypothesis is that

$\begin{align*} H_{0}: s_{1}^{2}\leq s_{1}^{2} \end{align*}$

and therefore, the alternative hypothesis is that (ie, a one-tailed test).

$\begin{align*} H_{0}: s_{1}^{2} > s_{1}^{2} \end{align*}$

Another way to state equal variance test is that we are testing for homogeneity of variances. You may run across the term homoscedasticity; it is the same thing, just a different three dollar word for “equal variances.”

Stated yet another way, if we reject the null hypothesis, then the variances are unequal or show heterogeneity. An additional and equivalent $3 word for inequality of variances is called heteroscedasticity.

More about the F-test

For the F-test, the null hypothesis is that the variances are equal. This means that the “expected” F value will be one: F = 1.0. (The F-distribution differs from t-distribution because it requires 2 values for DF, and ranges from 1 to infinity for every possible combination of v₁ and v₂).

To evaluate the null hypothesis we need the degrees of freedom. For the F test we need two different degrees of freedom, one set for each group): from Table 2, Appendix – F distribution, look up 5% Type I error line in this table because we make it one tailed.

I need the F-test statistic at $F_{0.05_{1},v_{1},v_{2}}$

Examples of difference between two variances, Table 1.

Sample 1: Aggressiveness of Inbred Mice (number of bites in 30 minutes)

Sample 2: Aggressiveness of Outbred Mice (number of bites in 30 minutes)

Table 1. Aggression by inbred and outbred mice.

Sample 1 Aggressiveness of Inbred Mice (number of bites in 30 minutes)	Sample 2 Aggressiveness of Outbred Mice (number of bites in 30 minutes)
3	4
5	10
4	4
3	7
4	7
5	10
4 = mean	7 = mean

Identify the null and alternative hypotheses
Calculate Variances
Calculate F-test
The “test statistic” for this hypothesis test was F = 7.2 / 0.8 = 9.0
Determine Critical Value of the F table (Table 2, Appendix – F distribution)

Example of how to find the critical values of the F distribution for $F_{0.05_{2}}$ and numerator $DF_{v_{1}}=5$ and denominator $DF_{v_{2}}=5$ .

Table 2. Portion of F distribution, see Appendix – F distribution.

α = 0.05		v1
α = 0.05		1	2	3	4	5	6	7	8
v2	1					230
	2					19.3
	3					9.01
	4					6.26
	5	6.61	5.79	5.41	5.19	5.05	4.95	4.88	4.82
	6

Or, instead of using tables, use R

Rcmdr: Distributions → F distribution → F probabilities

and enter the numbers as shown below (Fig 1).

Screenshot Rcmdr menu F distribution

Figure 1. Screenshot R Commander F distribution probabilities.

This will return the p-value, and you would interpret this against your Type I error rate of 5% as you do for other tests.

From Table 2, Appendix – F distribution we find $F_{0.05\left ( 1 \right ),5,5}=5.05$

And the p-value = 0.015

pf(c(9), df1=5, df2=5, lower.tail=FALSE)
[1] 0.01537472

Question: Reject or Accept null hypothesis?

Question: What is the biological interpretation or conclusion from this result?

R code

Rather than play around with the tables of critical values, which are awkward and old-school (and I am showing you these stats tables so that you get a feel for the process, not so you’d actually use them in real practice), use Rcmdr to generate the F test and therefore return the F distribution probability value. As you may expect, R provides a number of options for testing the equal variances assumption, including the F test. The F test is limited to only two groups and, because it is a parametric test, it also makes the assumption of normality, so the F test should not be viewed as necessarily the best test for the equal variances assumption among groups. We present it here because it is a logical test to understand and because of its relevance to the Mean Square ratios in the ANOVA procedures.

So, without further justification, here is the presentation on how to get the F test in Rcmdr. At the end of this section I present a better procedure than the F test for evaluating the equal variance assumption called the Levene test.

Return to the bite data in the table above and enter the data into an R data frame. Note that the data in the table above are unstacked; R expects the data to be stacked, so either create a stacked worksheet and transcribe the data appropriately into the cells of the worksheet, or, go ahead and enter the values into two separate columns then use the Stack variables in active data set… command from the Data menu in Rcmdr.

Then, proceed to perform the F test.

Rcmdr: Statistics → Variances → Two variances F-test…

The first context menu popup is where you enter the variables (Fig 2).

Screenshot Rcmdr variances F-test menu

Figure 2. Screenshot data options R Commander F test

Because there are only two variables in the data set and because Strain contains the text labels of inbred or outbred whereas the other variable is numeric data type, R will correctly select the variables for you by default. Select the “Options” tab to set the parameters of the F test (Fig 3).

Screenshot Rcmdr F-test menu

Figure 3. Screenshot menu options R Commander F test.

When you are finished setting the alternative hypothesis and confidence levels, proceed with the F test by clicking the OK button.

    F test to compare two variances

data:  mice.aggression
F = 0.1111, num df = 5, denom df = 5, p-value = 0.03075
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.01554788 0.79404243
sample estimates:
ratio of variances 
         0.1111111

End of R output.

Levene’s test of equal variances

We will discuss this test in more detail following our presentation on ANOVA. For now, we note that the test works on two or more groups and is a conservative test of the equal variance assumption. Nonparametric tests in general make fewer assumptions about the data and in particular make no assumption of normality like the F test. It is in this context that the Levene’s test would be preferable over the F test. Below we present only how to calculate the statistic in R and Rcmdr and provide the output for the same mouse data set.

Assuming the data are stacked, obtain the Levene’s test in Rcmdr by clicking on (Fig 4).

Rcmdr: Statistics → Variances → Levene’s test…

Screenshot menu options R Commander Levene's test

Figure 4. Screenshot menu options R Commander Levene’s test.

Select the median and the factor variable (in our case “Strain”) and the numeric outcome variable (“Bites”), then click OK button.

Tapply(bites ~ strain, var, na.action=na.omit, data=mice.aggression) # variances by group
inbred outbred
   0.8     7.2

leveneTest(bites ~ strain, data=mice.aggression, center="median")
Levene's Test for Homogeneity of Variance (center = median)
      Df F value  Pr(>F)
group  1       4 0.07339 .
      10
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End of R output

Compare the p-values. In both cases — F test and Levene’s test — we’re testing the same hypothesis (equal variances), but the p-values do not agree between the parametric F test and the nonparametric Levene’s test! If we were to go by the results of the F test, p-value was 0.031, less than out Type I error of 5% and we would tentatively conclude that the assumption of equal variances may not apply. On the other hand, if we go with the Levene’s test the p-value was 0.074, which is greater than our Type I error rate of 5% and we would therefore conclude the opposite, that the assumption of equal variances might apply! Both conclusions can’t hold, so which test result of equal variances do we prefer, the parametric F test or the nonparametric Levene’s test? Answer — we’d go with Levene’s because the F test is parametric and, therefore, also assumes normality.

Cahoy’s bootstrap method

Draft. Cahoy (2010). Variance-based statistic bootstrap test of heterogeneity of variances. we discuss bootstrap methods in Chapter 19.2; in brief, bootstrapping involves resampling the dataset and computing a statistic on each sample. This method may be more powerful, that is, more likely to correctly reject the null hypothesis when warranted, compared to Levene’s test.

for now, install package testequavar.

Function for testing two samples.

equa2vartest(inbred, outbred, 0.05, 999)

R output

[[1]]
[1] "Decision: Reject the Null"
$Alpha
[1] 0.05
$NumberOfBootSamples
[1] 999
$BootPvalue
[1] 0.006

The output “Decision: Reject the Null” reflects output from a box-type acceptance region.

Compare results from Levene’s test: p-value 0.07339 suggests accept hypothesis of equal variances, whereas bootstrap method indicates variances heterogenous, ie, reject equal variance hypothesis. However, re-running the test without setting the seed for R’s pseudorandom number generator will result in different p-values. For example, I re-ran Cahoy’s test five times with the following results:

Questions

Test assumption of equal variances by Bartlett’s method and by Levene’s test on OliveMoment variable from the Comet tea data set introduced in Chapter 12.1. Do the methods agree? If not, which test result would you choose?
- BONUS. Retest homogeneous variance hypothesis by equa3vartest(Copper.Hazel, Copper, Hazel, 0.05, 999). Reject or fail to reject the null hypothesis by bootstrap method?
Test assumption of equal variances by Bartlett’s method and by Levene’s test on Height from the O’hia data set introduced in Chapter 12.7. Do the methods agree? If not, which result would you choose?
- BONUS. Retest homogeneous variance hypothesis by equa3vartest(M.1, M.2, M.3, 0.05, 999). Reject or fail to reject the null hypothesis by bootstrap method?

Quiz Chapter 13.4

Tests for Equal Variances

R code reminders

The bootstrap method expects the variables in unstacked format. A simple method to extract the variables from the stacked data is to use a command like the following. For example, extract OliveMoment values for Copper-Hazel treatment.

Copper.Hazel <- cometTea$OliveMoment[1:10]

For Copper, Hazel, replace above with [11:20], and [21-30] respectively. Note: changed variable name from Copper-Hazel to Copper.Hazel. The hyphen is a reserved character in R.

Chapter 13 contents

13.3 – Test assumption of normality

Introduction
Goodness of fit tests of normality hypothesis
Example
Which test of normality?
Questions
Quiz
Data set
Chapter 13 contents

Introduction

I’ve commented numerous times that your conclusions from statistical inference are only as good as the validity of making the and applying the correct procedures. This implies that we know the assumptions that go into the various statistical tests, and where possible, we critically test the assumptions. From time to time then I will provide you with “tests of assumptions.”

Here’s one. The assumption of normality, that your data were sampled from a population with a normal distribution for the variable of interest, is key and there are a number of ways to test this assumption. Hopefully as you read through the next section you can extend the logic to any distribution; if the data are presumed to come from a binomial, or a Poisson, or a uniform distribution, then the logic of goodness of fit tests would apply.

How to test normality assumption

It’s not a statistical test per se, but the best is to simply plot (histogram) the data. You can do these graphics by hand, or, install the data mining package rattle which will generate nice plots useful for diagnosing normality issues.

Note 1: rattle (R Analytic Tool To Learn Easily) is a great “data-mining” package. We introduced the package in Chapter 3 — Exploring data.

The rattle histogram plot (Fig 1) superimposes a normal curve over the data, which allows you to “eyeball” the data.

First, the eye test. I used the R-package rattle for this on a data set of comet tail lengths of rat lung cells exposed to different amounts of copper in growth media (scroll to bottom of page or click here to get the data).

In addition to the histogram (Fig 1 top image), I plotted the cumulative function (Fig 1 bottom image). In short, if the data set were normal, then the cumulative frequency plot should look like a straight line.

rattle normal

cumulative frequency

Figure 1. Rattle descriptive graphics on Comet Copper dataset. Dotted line (top image) and red line (bottom image) follow the combined observations regardless of treatment.

So, just looking at the data set, we don’t see clear evidence for a normal-like data set. The top image (Fig 1) looks stacked to the left and the cumulative plot (bottom image) is bumped in the middle, not falling on a straight line. We’ll need to investigate the assumption of normality more for this data set. We’ll begin by discussing some hypothetical points first, then return to the data set.

Goodness of fit tests of normality hypothesis

While graphics and the “eyeball test” are very helpful, you should understand that whether or not your data fits a normal distribution, that’s a testable hypothesis. The null hypothesis is that your sample distribution fits are normal distribution. In general terms, these “fit” hypotheses are viewed as “goodness of fit” tests. Often times, the test is some variation of a $\chi^{2}$ problem: you have your observed (sample distribution) and you compare it to some theoretical expected value (eg, in this situation, the normal distribution). If your data fit a normal curve, then the test statistics will be close to zero.

We have discussed before that the data should be from a normal distributed population. To the extent this is true, then we can trust our statistical tests are performing the way they are expected to do. We can test the assumption of normality by comparing our sample distribution against a theoretical distribution, the normal curve. I’ve shown several graphs in the past that “looked normal”. What are the alternatives for unimodal (single peak) distributions (Fig 2)?

Kurtosis describes the shape of the distribution, whether it is stacked up in the middle (leptokurtosis), or more spread out and flattened (platykurtosis).

Skewness describes differences from symmetry about the middle. Left skew the tail of the distribution extends to the left, ie, smaller values.

“Leptokurtosis”

Leptokurtosis

“Platykurtosis”

Platykurtosis

Negative skew, left skewed

Negative skew, left skewed

Positive skew, right skewed

Positive skew, right skewed

Figure 2. Graphs describing different distributions. From top to bottom: Leptokurtosis, platykurtosis, negative skew, positive skew.

The easiest procedures for goodness of fit tests of normality are based on the $\chi^{2}$ distribution and yield a “goodness of fit” test for normal distribution. We discussed the $\chi^{2}$ distribution in Chapter 6.9, and used the test in Chapter 9.1.

$\begin{align*} \chi^2 = \sum_{i = 1}^{k}\frac{\left ( O_{i}-E{i} \right )^2}{E_{i}} \end{align*}$

where O refers to the observed data (what we’ve got) and E refers to the expected (eg, data from a normal curve with same mean and standard deviation as our sample).

To illustrate, I simulated a data set in R.

Rcmdr: Distributions → Continuous distributions → Normal distribution → Sample from normal distribution

I created 100 values, stored them in column 1, and set the population mean = 125 and population standard deviation = 10. And therefore the population standard error of the mean was 1.0.

The resulting MIN/MAX was from 99.558 to 146.16; the sample mean was 124.59 with a sample standard deviation of 9.9164. And therefore the sample standard error of the mean was 0.9916.

Question: After completing the steps above, your data will be slightly different from mine… Why?

But getting back to the main concept, Does our data agree with a normal curve? We have discussed how to construct histograms and frequency distributions.

Let’s try six categories (why six? we discussed this when we talked about histograms).

All chi-square tests are based on categorical data, so we use the counts per category to get our data. Group the data, then count the number of OBSERVED in each group. To get the EXPECTED values, use the Z-score (normal deviate) with population mean and standard deviation as entered above.

Table 1. Tabulated values for test of normality.

Number of observations	Weight	Normal deviate (Z)	Expected Proportion	Expected number	(Obs – Exp)² / Exp
105	3	less than or equal to -2	0.0228	2.28	0.227368421
105 < 115	17	between -1 & -2 = 0.1587 – 0.0228	0.1359	13.59	0.855636497
115 < 125	34	between -0 & -1 = 0.5 – 0.1587	0.3413	34.13	0.000495166
125 < 135	30	between +0 & +1 = 0.5 – 0.1587	0.3413	34.13	0.499762672
135 < 145	15	between +1 & +2 = 0.9772 – 0.8413	0.1359	13.59	0.146291391
> 145	1	greater than or equal to +2 = 1 – 0.9772	0.0228	2.28	0.718596491
					$\chi^2 = 2.44815064$

Then obtain the critical value of the $\chi^{2}$ with df = 6 – 1 = 5 (see Appendix, Table of chi-square critical values , critical $\chi^{2}$ = 11.1 with df = 5).

Thus, we would not reject the null hypothesis and would proceed with the assumption that our data could have come from a normally distributed population.

This would be an OK test, but different approaches, although based on a chi-square-like goodness of fit, have been introduced and are generally preferred. We have just shown how one could use the chi-square goodness of fit approach to testing whether your data fit a normal distribution. A number of modified tests based on this procedure have been proposed; we have already introduced and used one (Wilks-Shapiro), which is easily accessed in R.

Rcmdr: Summaries → Wilks-Shapiro

Like Wilks-Shapiro, another common “goodness-of-fit” test of normality is the Anderson-Darling test. This test is now included with Rcmdr, but it’s also available in the package nortest. After the package is installed and you run the library (you should by now be able to do this!), then at the R prompt (>) type:

ad.test(dataset$variable)

replacing “dataset$variable” with the name of your data set and variable name.

In the context of goodness of fit, a perfect fit means the data are exactly distributed as a normal curve, and the test statistics would be zero. Differences away from normality increase the value of the test statistic.

How do these tests perform on data?

The histogram of our simulated normal data of 100 observations with mean = 125 and standard deviation = 10 (Fig 3).

Histogram of simulated normal dataset, μ = 125, σ = 10.

Figure 3. Histogram of simulated normal dataset, μ = 125, σ = 10.

and here’s the cumulative plot (Fig 4).

Cumulated frequency of simulated normal dataset, μ = 125, σ = 10

Figure 4. Cumulated frequency of simulated normal dataset, μ = 125, σ = 10.

Results from Anderson-Darling test were A = 0.2491, p-value = 0.7412, where A is the Anderson-Darling test statistic.

Results of the Shapiro-Wilks test on the same data. W = 0.9927, p-value = 0.8716, where W is the Shapiro-Wilks test statistic. We would not reject the null hypothesis in either case because p > 0.05.

Example

Histogram of a data set, highly skewed to the right. 90 observations in three groups of 30 each, with mean = 0 and standard deviation = 1 (Fig 5).

Histogram of simulated normal dataset, μ = 0, σ = 1.

Figure 5. Histogram of simulated normal dataset, μ = 0, σ = 1.

and the cumulative frequency plot (Fig 6).

Cumulated frequency of simulated normal dataset, μ = 0, σ = 1.

Figure 6. Cumulated frequency of simulated normal dataset, μ = 0, σ = 1.

Results from Anderson-Darling test were A = 9.0662, p-value < 2.2e^-16. Results of the Shapiro-Wilks test on the same data. W = 0.6192, p-value = 6.248e^-14. Therefore, we would reject the null hypothesis because p < 0.05.

The Shapiro-Wilk test in Rcmdr

Let’s go back to our data set and try tests of normality on the entire data set, ie, not by treatment groups.

Rcmdr: Statistics → Summaries → Test of normality…

normalityTest(~CometTail, test="shapiro.test", data=CometCopper)

 Shapiro-Wilk normality test

data: CometCopper$CometTail
W = 0.91662, p-value = 0.006038

End of R output

Another test, built on the basic idea of a chi-square goodness of fit is the Anderson Darling test. Some statisticians prefer this test and it is one built-in to some commercial statistical packages (eg, Minitab). To obtain the Anderson-Darling test in R, you need to install a package. After installing nortest package, run the AD test at the command prompt.

require(nortest)
ad.test(CometCopper$CometTail)

Anderson-Darling normality test

data: CometCopper$CometTail
A = 1.0833, p-value = 0.006787

End of R output.

Note 2: Because you’ve installed Rcmdr the Anderson-Darling test is available (since version 2.6) without installing the nortest package. You can call Anderson-Darling via the menu system, Statistics → Summaries → Test of normality … or type in the script the following command

normalityTest(~CometTail, test="ad.test", data=CometCopper)

In contrast, Shapiro-Wilks test is part of the base R package, so can be called directly

shapiro.test(CometCopper$CometTail)

P-values are both much less than 0.05, so we would reject the assumption of normality for this data set.

Which test of normality?

Why show you two tests for normality, the Shapiro-Wilks and Anderson-Darling? The simple answer is that both are good as general tests of normality, both are widely used in scientific papers, so just pick one and go with it as your general test of normality.

The more complicated answer is that each is designed to be sensitive to different kinds of departure from normality. By some measures, the Shapiro-Wilks test is somewhat better (ie, more statistical power to test the null hypothesis), than other tests, but this is not something you want to get into as a beginner. So, I show both of them to you so that you are at least introduced to the concept that there are often more than one way to test a hypothesis. The bottom-line is that plots may be best!

Questions

Work describe in this chapter involves statistical tests of the assumption of normality. It is just as important, maybe more so, to also apply graphics to take advantage of our built-in pattern recognition functions. What graphic techniques, besides histogram, should be used to view the distribution of the data?
In R, what command would you use so that you can call the variable name, CometTail, directly instead of having to refer to the variable as CometCopper$CometTail?
Why are Anderson-Darling, Shapiro-Wilks and other related tests referred to as “goodness of fit” tests? You may wish to review discussion in Chapter 9.1.
The example tests presented for the Comet Copper data set were conducted on the whole set, not by treatment groups. Re-run tests of normality via Rcmdr, but this time, select the By groups option and select Treatment.

Quiz Chapter 13.3

Test assumption of normality

Data set used in this page

Treatment	CometTail
Control	17.86
Control	16.52
Control	14.93
Control	14.03
Control	13.33
Control	8.81
Control	14.70
Control	9.26
Control	21.78
Control	6.18
Control	9.20
Control	5.54
Control	6.72
Control	2.63
Control	7.19
Control	5.39
Control	11.29
Control	15.44
Control	17.86
Contro	l4.25
Copper	53.21
Copper	38.93
Copper	18.93
Copper	30.00
Copper	28.93
Copper	15.36
Copper	17.86
Copper	17.50
Copper	21.07
Copper	29.29
Copper	28.21
Copper	16.79
Copper	21.07
Copper	37.50
Copper	38.22
Copper	17.86
Copper	29.64
Copper	11.07
Copper	35.00
Copper	49.29

Chapter 13 contents

13.1 – ANOVA Assumptions

Introduction
Impact of assumptions
Statistical outlier
ANOVA is robust to modest deviations from assumptions
Example
Lack of independence among data
Questions
Quiz
Data set
Chapter 13 contents

Introduction

Like all parametric tests, assumptions are made about the data in order to justify and trust estimates and inferences drawn from ANOVA. While acknowledging the “experimental design” message suggested by the xkcd panel in Figure 1 — the presence of a large effect size between groups mitigates the technical necessities of inferential statistics — of course, not all of our projects will have such obvious results. Understanding the impact of assumptions on inference is a key concept in (bio)statistics: when the p-value is close to the Type I error rate, deviations from assumptions may reduce our confidence in the “statistical significance” of the result.

"Statistics," by xkcd at https://xkcd.com/2400/

Figure 1. xkcd.com “Statistics,” https://xkcd.com/2400/.

ANOVA assumptions include:

Data come from normal distributed population. View with a histogram or Q-Q plot. Test with Shapiro-Wilks or other appropriate goodness of fit test (see Question 1 below). Normality tests are the subject of Chapter 13.3.
Sample size equal among groups.
- This is an example of a potentially confounding factor — If sample sizes differ, then any difference in means could be simply because of differences in sample size! This gets us into weighted versus unweighted means.
- You shouldn’t be surprised that modern implementations of ANOVA in software easily handle (adjust for) these known confounding factors. Depending on the program, you’ll see “Adjusted means,” “least squares means,” “Marginal means” etc. This just implies that the group means are compared after accounting for confounding factors.
- Importantly, as long as sample sizes among the groups are roughly equivalent, normality assumption is not a big deal (low impact on risk of Type I error).
Independence of errors. One consequence of this assumption is that you would not view 100 repeated observations of a trait on the same subject as 100 independent data points. We’ll return to this concept more in the next two lectures. Some examples:

- Colorimetric assay where the signal changes over time, and you measure in order (eg, samples from group 1 first, samples from group 2 second, etc.) — this confounds group with time.
  - The consequence is that you are far more likely to reject the null hypothesis, committing a Type I error.
- Let’s say you are observing running speeds of ten mongoose. However, it turns out that five of your subjects are actually from the same family, identical quintuplets! Do you really have ten subjects?
- Compare brain-body mass ratio among different species; this is a classic comparative method problem (Fig 3). Since 1985 (Felsenstein 1985), it was recognized that the hierarchical evolutionary relationships among the species must be accounted for to control for lack of independence among the taxa tested. See Phylogenetically independent contrasts, Chapter 20.12.

Equal variances (homogeneity) among groups. See Chapter 13.4 for how to test this for multiple groups.

Impact of assumptions

R — and pretty much all statistics packages — will calculate the ANOVA table and the p-value, but it is up to you to recognize that the P-value is accurate only if the assumptions are met. Violation of the assumption of normality can lead to Type I errors occurring more often than the 5% level. What to do if the assumptions are violated?

If the violation is due to only a handful of the data, you might proceed anyway. But following a significant test for normality, we could avoid the ANOVA in favor of nonparametric alternatives (Chapter 15), or, we might try to transform the data.

Consider a histogram of brains weight measures in grams for a variety of mammals (Fig 2).

Histogram of brain mass (g) for 24 mammals (data from Boddy et al 2012).

Figure 2. Histogram of body mass (g) for 24 mammals (data from Boddy et al 2012).

We will introduce a variety of statistical tests of the assumption of normality in Chapter 13.3, but looking at a histogram as part of our data exploration, we clearly see the data are right-skewed (Fig 2). Is this an example of normal distributed sample of observations? Clearly not. If we proceed with statistical tests on the raw data set, then we are more likely to commit a Type I error (ie, we will reject the null hypothesis more often than we should).

A comment on normality and biology. It is very possible that data may not be normally distributed or have equal variances on the original scale, but a simple mathematical manipulation may take care of that. In fact, in many case in biology that involve growth, many types of variables are expected to not be normal on the original scale. For example, while the relationship between body mass, $M$ , and metabolic rate, $MR$ , in many groups of organisms is allometric and increases positively, the relationship

$\begin{align*} MR = a\cdot M^b \end{align*}$

is not directly proportional (linear) on the original scale. By taking the logarithm of both body mass and metabolic rate, however, the relationship is linear

$\begin{align*} log\left ( MR \right ) = log\left ( a \right ) + b\cdot log\left ( M \right ) \end{align*}$

In fact, taking the logarithm (base 10, base 2, or base e) is often a common solution to both non-normal data (Fig 2) and unequal variances.

Histogram of log-transformed body mass observations from Figure 1

Figure 3. Histogram of log₁₀-transformed body mass observations from Figure 2.

Other common transformations include taking the square root or the inverse of the square-root for skewed or kurtotic sample distributions, and the arcsine for frequencies (since frequencies can only be from 0 to 1 — need to “stretch the data” to make frequencies fit procedures like ANOVA). There are many issues about data transformation, but keep in mind three points. After completing the transformation, you should check the assumptions (normality, equal variances) again.

You may need to recode the data before applying a transform. For example, you cannot take the square root or logarithm of negative numbers. If you do not recode the data, then you will lose these observations in your subsequent analyses. In many cases, this problem is easily solved by adding 1 to all data points before making the transform. I prefer to make the minimum value 1 and go from there. The justification for data transformation is basically to realize that there is no necessity to use the common arithmetic or linear scale: many relationships are multiplicative or nonadditive (eg, rates in biology and medicine).

Note 1: Instead of transformations or other ad hoc manipulations of the data, modern statistical approaches favor modeling the error structure of the data within a Generalized Linear Model framework (St.-Pierre et al 2018). The advantage of the model approach is that parameter estimation occurs on the raw data. Use of transformations may, however, remain a better choice. While statistically justified, the generalized linear model approach may also tend to have higher rates of Type II error compared to simple transformations.

Statistical outlier

Another topic we should at least mention here is the concept of outliers. While most observations tend to cluster around the mean or middle of the sample distribution, occasionally one or more observations may differ substantially from the others. Such values are called outliers, and we note that there are two possible explanations for an outlier

the value could be an error.
it is a true value.

Note 2: “it is a true value,” and there may be an interesting biological explanation for its cause, ie, extraordinary individuals.

We encountered a clear outlier in the BMI homework. If the reason is (1), then we go back and either fix the error or delete it from the worksheet. If (2), however, then we have no objective reason to exclude the point from our analyses.

We worry if the outlier influences our conclusions — so it is a good idea to run your analyses with and without the outlier. If your conclusions remain the same, then no worries. If your conclusions change based on one observation, then this is problematic. For the most part you are then obligated to include the outlier and the more conservative interpretation of your statistical tests.

ANOVA is robust to modest deviations from assumptions

A comment about ANOVA assumptions ANOVA turns out to be robust to violations of item (1) or (2). That means unless the data are really skewed or the group sizes are very different, ANOVA will perform well (Type I error rate stays close to the specified 5% level). We worry about this however when p-value is very close to alpha!!

The third assumption is more important in ANOVA.

Like the t-test, ANOVA makes the assumption of equal variances among the groups, so it will be helpful to review why this assumption is important to both the t-test and ANOVA. In the two-sample independent t-test, the pooled sample variance, $s_{p}^2$ , is taken as an estimate of the population variance, $\sigma^{2}$ . If you recall,

where $SS_{1}$ refers to the sum of squares for the first group and $SS_{2}$ refers to the second group sum of squares (see our discussion on measures of dispersion) and $v_{1}$ refers to the degrees of freedom for the first group and $v_{2}$ refers to the second group degrees of freedom. We make a similar assumption in ANOVA. We assume that the variances for each sample are the same and therefore that they all estimate the population variance $\sigma^{2}$ . To say it in another way, we are assuming that all of our samples have identical variability

Once we make this assumption, we may pool (or combine) all of the SSs and DFs for all groups as our best estimate of the population variance, $\sigma^{2}$ . The trick to understanding ANOVA is to realize that there can be two types of variability: there is variability due to being part of a group (eg, even though ten human subjects receive the same calorie-restricted diet, not all ten will loose the same amount of weight) and there is variability among or between groups (eg, on average, all subjects who received the calorie-restricted diet lost more weight than did those subjects who were on the non-restricted diet).

Example

The encephalization index (or encephalization quotient) is defined as the ratio of size the brain compared to body size. While there is a well-recognized increase in brain size given increased body size, encephalization describes a shift of function to cortex (frontal, occipital, parietal, temporal) from noncortical parts of the brain (cerebellum, brainstem). Increased cortex is associated with increased complexity of brain function; for some researchers, the index is taken as a crude estimate of intelligence. Figure 4A shows plot of brain mass in grams versus body size (grams) for 24 mammal species (data sampled from Boddy et al 2012); Figure 4B shows the same data, but following log₁₀-transform of both variables.

encephalization index

Figure 4. Plot of brain and body weights (A) and log₁₀-log₁₀ transform (B) for a variety of species (data from Boddy et al 2012). The ratio is called encephalization index.

Looking at the two figures the linear relationship between the two variables is obvious in Figure 4B, less so for Figure 4A. Thus, one biological justification for transformation of the raw data is exemplified with the brain-body mass dataset: the association is allometric not additive. The other reason to apply a transform is statistical; the log₁₀-transform improves the normality of the variables. Take a look at the Q-Q plot for the raw data (Fig 5) and for the log10-transformed data (Fig 6).

Q-Q plot, raw body mass

Figure 5. Q-Q plot, body mass, raw data. Compare to Figure 2.

Figure 5 is a good test of your observation skills — what are you supposed to look for in a Q-Q plot? Follow the dots — the data don’t fall on the straight line. A few even fall outside of the confidence interval (the curved dashed lines), which suggests the data were not normally distributed (see histogram, Fig 2). And for the transformed data, the Q-Q plot is shown in Figure 6.

Q-Q plot, log-transformed body mass

Figure 6. Q-Q plot same data, log₁₀-transformed, compare to Figure 2.

Compared to the raw data, the transformed data now fall on the line and none are outside of the confidence interval. We would conclude that the transformed data are more normal, thus, better meeting the assumptions of our parametric tests.

Lack of independence among data

Species comparisons are common in evolutionary biology and related fields. As noted earlier, comparative data should not be treated as independent data points (discussed in 20.12 – Phylogenetically independent contrasts). For our 24 species, I plotted the estimate of the phylogeny (timetree.org), represented in Figure 7. Some species are closely related (eg the two elephant shrews, Rhynchocyon genus). Thus, because of the hierarchical, nested nature of evolution — first articulated by Charles Darwin — instead of 24 independent data points, we have eight clusters of points (see Question 4).

phylogeny

Figure 7. Phylogenetic tree of 24 species used in this report.

Note 3: The relevant quote from Charles Darwin: “… the inevitable result is that the modified descendants proceeding from one progenitor become broken up into groups subordinate to groups,” and was illustrated by the only figure in Origin of Species (both quote and illustration are in Chapter 4).

The conclusion? We don’t have 24 data points, more like 8 points. Because the species are more or less related, there are fewer than 24 independent data points. Statistically, this would mean that the errors are correlated. Various approaches to account for this lack of independence have been developed; perhaps the most common approach is to apply phylogenetically independent contrasts, a topic discussed in Chapter 20.12. (Boddy et al 2012 used this approach.)

Note 4: See 20.11 – Plot a Newick tree for how to plot phylogeny trees in R.

Questions

Make a histogram for the brain mass data set.
Shapiro-Wilks is one test of normality.
1. Can you recall the name of the other normality tests we named?
2. Run tests of normality on both body mass and brain mass data. Interpret the results, frame your response with keywords: “central tendency,” “outliers,” “shape,” and “symmetry.”
Calculate moment statistics for the body mass and brain mass data set, both raw and log₁₀-transformed data. Interpret your results, taking care to frame your answer with the following keywords: “shape,” “symmetry,” and “peakedness” (aka “tailedness”).
We introduced the lack of independence for comparative data attributed to the evolutionary process. Statistically, the impact is on degrees of freedom; for example, if we ignore the lack of independence in the data set, df = 22 for the slope coefficient in a linear regression. However, given the degrees of freedom would be closer to six! What are the effects on Type I error rate of lack of independence?

Quiz Chapter 13.1

ANOVA assumptions

Data set and R code used in this page

species, order, body, brain
'Herpestes ichneumon', Carnivora, 1764, 24.1
'Potos flavus', Carnivora, 2620, 31.05
'Vulpes vulpes', Carnivora, 3080, 49.5
'Madoqua kirkii', Cetartiodactyla, 4570, 37
'Sus scrofa', Cetartiodactyla, 1000, 47.7
'Tragulus napu', Cetartiodactyla, 2510, 18.5
'Casinycteris argynnis', Chiroptera, 40.5, 0.92
'Cynopterus brachyotis', Chiroptera, 29, 0.88
'Potorous platyops', Chiroptera, 718, 6.5
'Cercartetus nanus', Diprotodontia, 12, 0.44548
'Dactylopsila palpator', Diprotodontia, 474, 7.15876
'Vombatus ursinus', Diprotodontia, 1902, 11.396
'Crocidura fuscomurina', Eulipotyphla, 5.6, 0.13
'Scalopus aquaticus', Eulipotyphla, 39.6, 1.48
'Sorex hoyi', Eulipotyphla, 2.6, 0.107
'Elephantulus fuscipes', Macroscelidea, 57, 1.33
'Rhynchocyon cirnei', Macroscelidea, 490, 6.1
'Rhynchocyon petersi', Macroscelidea, 717.3, 4.46
'Aotus trivirgatus', Primates, 480, 15.5
'Cebus capucinus', Primates, 590, 53.28
'Leontopithecus rosalia', Primates, 502.5, 11.7
'Dipodomys agilis', Rodentia, 61.4, 1.34
'Peromyscus melanocarpus', Rodentia, 58.8, 1.03
'Sciurus carolinensis', Rodentia, 367, 6.49

The Newick code for the tree in Figure 6.

((Vombatus_ursinus:48.94499077,((Potorous_platyops:47.59556667,Cercartetus_nanus:47.59556667)'14':0.66887333,Dactylopsila_palpator:48.26444000)'13':0.68055077)'11':109.65259681,(((((Crocidura_fuscomurina:33.74066667,Sorex_hoyi:33.74066667)'10':33.03022424,Scalopus_aquaticus:66.77089091)'19':22.55292749,(((Herpestes_brachyurus:54.32144118,(Vulpes_vulpes:45.52834967,Potos_flavus:45.52834967)'9':8.79309151)'22':23.43351523,((Tragulus_napu:43.96862857,Madoqua_kirkii:43.96862857)'8':17.99735995,Sus_scrofa:61.96598852)'6':15.78896789)'30':0.77378567,(Casinycteris_argynnis:35.20000000,Cynopterus_brachyotis:35.20000000)'29':43.32874208)'27':10.79507632)'35':7.13857077,(((Peromyscus_melanocarpus:69.89837667,Dipodomys_agilis:69.89837667)'43':0.64655123,Sciurus_carolinensis:70.54492789)'42':19.27825952,((Leontopithecus_rosalia:18.38385647,Aotus_trivirgatus:18.38385647)'40':1.29720005,Cebus_capucinus:19.68105652)'48':70.14213090)'51':6.63920175)'47':8.99739162,(Elephantulus_fuscipes:39.23366667,(Rhynchocyon_cirnei:15.34500000,Rhynchocyon_petersi:15.34500000)'39':23.88866667)'56':66.22611412)'55':53.13780679);

Chapter 13 contents

12.6 – ANOVA posthoc tests

Introduction
When to conduct posthoc tests
Posthoc tests
Performing multiple comparisons and the one-way ANOVA
Questions
Quiz
Data set
Chapter 12 contents

Introduction

Tests of the null hypothesis in a one-way ANOVA yields one answer: either you reject the null, or you do not reject the null hypothesis — with the usual caveats about “provisionally reject,” etc (Ch8.2 and Ch8.3). But while there was only one factor (population, drug treatment, etc) in a one-way ANOVA, there are usually many treatments (eg, multiple levels, four different populations, three doses of a drug plus a placebo).

ANOVA plus posthoc tests solves the multiple comparison problem we discussed: you still get your tests of all group differences, but with adjustments to the procedures so that these tests are conducted without suffering the increase in Type I error = the multiple comparison problem. If the null hypothesis is rejected, you may then proceed to posthoc tests among the groups to identify differences without increasing the chance of rejecting the null hypothesis too often.

Consider the following example of four populations scored for some outcome, sim.ch12 (scroll down the page, or click here to get the R code).

Bring the data frame, sim.ch12, into current memory in Rcmdr by selecting the data set. Next, run the one-way ANOVA.

Rcmdr: Statistics → Means → One-way ANOVA…

which brings up the following menu (Fig 1).

Figure 1. One-way ANOVA menu in R Commander

Figure 1. One-way ANOVA menu in R Commander.

Note 1: If you look carefully in Figure 1, you can see model name was AnovaModel.8. There’s nothing sacred about that name, it just means this was the 8^th model I had run up to that point. For consistency I renamed all instances of model objects reported in the text as AnovaModel.1 — after all, they were the same model run with different commands. As a reminder, Rcmdr will provide names for models for you; it is better practice to provide model names yourself.

Notice that Rcmdr menu correctly identifies the Factor variable, which contains text labels for each group, and the Response variable, which contains the numerical observations.

Note 2: If your factor is numeric, you’ll first have to tell R that the variable is a factor and hence nominal. this can be accomplished within Rcmdr via the Data → Manage variables… options, or simply submit the command

newName <- as.factor(oldVariable)

Similarly, you may need to change a variable from character to factor. For example, check a variable with class(). If it returns anything but “factor,” then use as.factor() as before.

If your data set contains more variables, then you would need to sort through these and select the correct model (Fig 1).

To get the default Tukey posthoc tests simply check the Pairwise comparisons box and then click OK.

For a test of the null that four groups have the same mean, a publishable ANOVA table would look like Table 1.

Note 3: the ANOVA table is something you put together from the output of R (or other statistical programs). After running one-way ANOVA in Rcmdr, go to Models → Hypothesis tests → ANOVA table…, accept the defaults (marginality, sandwich estimators and are explained in the next chapter 12.7).

Table 1. The ANOVA table

	Df	Mean Square	F	P†
Label	3	389627	76.44	< 0.0001
Error	36	61167

† DrD edited the R output for one-tailed p-value from 1.108357e-15 to a threshold < 0.0001.

Note 4: To automatically replace a small p-value with a threshold cutoff requires conditional formatting, to highlight results by their content. This is a great ability of spreadsheet apps, and, unsurprisingly, R can do this too. For example, the pvalue() function in scales package, which can report p-values rounded to a threshold value.

R code

pvalue(1.108357e-15, accuracy=0.0001, decimal.mark = ".", add_p=TRUE)

Output

[1] "p<0.0001"

h/t norbert at Scripts & Statistics

Here’s the R output for ANOVA

AnovaModel.1 <- aov(Values ~ Label, data = sim.Ch12)
summary(AnovaModel.1)
          Df  Sum Sq Mean Sq  F value     Pr(>F) 
Label      3  389627  129876    76.44   1.11e-15 ***
Residuals 36   61167    1699 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

with(sim.Ch12, numSummary(Values, groups = Label, statistics = c("mean", "sd")))
      mean       sd data:n
Pop1 150.3 35.66838 10
Pop2  99.0 43.25634 10
Pop3 130.9 42.36469 10
Pop4 350.7 43.10723 10

End of R output.

When to conduct posthoc tests.

Recall that all we can say is that a difference has been found and we reject the null hypothesis. However, we do not know if group 1 = group 2, but both are different from group 3, or some other combination. So we may want additional tools. We can conduct posthoc tests (also called multiple comparisons tests).

Once a difference has been detected (F test statistic > F critical value, therefore P < 0.05), then posteriori tests, also called unplanned comparisons, can be used to tell which means differ.

There are also cases for which some comparisons were planned ahead of time and these are called a priori or planned comparisons; even though you conduct the tests after the ANOVA, you were always interested in particular comparisons. This is an important distinction: planned comparisons are more powerful, more aligned with what we understand to be the scientific method. An argument — transparency, avoids Type II errors — can be made that if comparisons are planned, and only a few tests are conducted, no correction for “multiple tests” is warranted. Moreover, it is defensible to report all p-values from a hundred tests without correction, acknowledging that if all null hypotheses were “true,” then 5 would be expected to be less than Type I of 5% by chance alone.

Posthoc tests.

ANOVA was introduced by Fisher in 1918, and, as readers know by know, returns whether or not significant differences are present among groups, but no which groups may differ. Unsurprisingly, given the needs of researchers to compare groups, many tests were developed since Fisher’s ANOVA to conduct multiple comparison tests. Collectively, they are often referred to as posthoc tests (Ruxton and Beauchamp 2008).

There are many different flavors of these tests, and R offers several, but I will hold you responsible only for three such comparisons: Tukey’s, Dunnett’s, and Bonferroni (Dunn). These named tests are among the common ones (Table 2, Midway et al 2020), but you should be aware that the problem of multiple comparisons and inflated error rates has received quite a lot of recent attention because the size of data sets has increased in many fields, eg, genome wide-association studies in genetics or data mining in economics or business analytics. A related topic then is the issue of “false positives.” New approaches include Holm-Bonferroni. There are others — it is a regular “cottage industry” in applied statistics to a problem that, while recognized, has not achieved a universal agreed solution. Best we can do is be aware and deal with it and know that the problem is one mostly of big data (eg, microarray and other high-throughput approaches). Equally important, some of these tests are better than others, and some may not be consistent with best practices.

Let’s take a look at these procedures.

Performing multiple comparisons and the one-way ANOVA

Note 5: In order to do most of the posthoc tests you will need to install the multcomp package; after installing the package, load the library(multcomp). Just using the default option from the one-way ANOVA command yields the Tukey’s HSD test.

a. Tukey’s: “honestly (wholly) significant difference test”

Tests $H_{0}:\ \bar{X}_{B}=\bar{X}_{A}$ versus $H_{A}:\ \bar{X}_{B}\neq \bar{X}_{A}$ where A and B can be any pairwise combination of two means you wish to compare. Thus, this is likely a case of unplanned comparisons. There are $\frac{k\left ( k-1 \right )}{2}$ comparisons.

$\begin{align*} q = \frac{\bar{X}_{B}-\bar{X}_{A}}{SE} \end{align*}$

where

$\begin{align*} SE=\sqrt{\frac{MS_{error}}{n}} \end{align*}$

and n is the harmonic mean of the sample sizes of the two groups being compared. If the sample sizes are equal, then the simple arithmetic mean is the same as the harmonic mean.

Note 6: q is like t for when we are testing means from two samples.

The significance level is the probability of encountering at least one Type I error (probability of rejecting H_o when it is true). This is called the experimentwise (familywise) error rate whereas before we talked about the comparisonwise (individual) error rate.

Three options to get the posthoc test Tukey:

(1) TukeyHSD() function is included in base statistics.

(2) use a function called mcp or in Rcmdr, Tukey is the default option in the one-way ANOVA command. The function mcp is part of the multcomp package, which is included in the Rcmdr package.

(3) specify method = "HSD" in PostHocTest function in DescTools package.

In brief, run the anova first, save the output to an object, then call an appropriate command from a package to conduct the posthoc test.

Rcmdr: Statistics → Means → One-way ANOVA
Check “Pairwise comparisons of means” to get the Tukey’s HSD test (Fig 2).

Rcmdr Tukey test

Figure 2. Screenshot Rcmdr menu: Select Tukey posthoc tests with the one-way ANOVA.

Tukey’s honestly significant difference is available without installing additional packages. Simply run

AnovaModel.1 <- aov(formula = Values ~ Label, data = sim.Ch12)
TukeyHSD(AnovaModel.1)

and R output returned is

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = Value ~ Label, data = sim.Ch12)

$Label
           diff        lwr        upr     p adj
Pop2-Pop1 -51.3 -100.94731  -1.652695 0.0406116
Pop3-Pop1 -19.4  -69.04731  30.247305 0.7200796
Pop4-Pop1 200.4  150.75269 250.047305 0.0000000
Pop3-Pop2 31.9   -17.74731  81.547305 0.3232616
Pop4-Pop2 251.7  202.05269 301.347305 0.0000000
Pop4-Pop3 219.8  170.15269 269.447305 0.0000000

Installation of R Commander also includes installation of the multcomp package, which contains several post hoc tests, including Tukey and Dunnett’s tests.

R output follows. There’s a lot, but much of it is repeat information. Take your time, here we go.

.Pairs <- glht(AnovaModel.1, linfct = mcp(Label = "Tukey"))
summary(.Pairs) # pairwise tests
     Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
 Fit: aov(formula = Values ~ Label, data = sim.Ch12) 

 Linear Hypotheses: 
                  Estimate Std. Error t value Pr(>|t|)     
 Pop2 - Pop1 == 0   -51.30      18.43  -2.783   0.0405 *   
 Pop3 - Pop1 == 0   -19.40      18.43  -1.052   0.7201     
 Pop4 - Pop1 == 0   200.40      18.43  10.871   <0.001 *** 
 Pop3 - Pop2 == 0    31.90      18.43   1.730   0.3233     
 Pop4 - Pop2 == 0   251.70      18.43  13.654   <0.001 *** 
 Pop4 - Pop3 == 0   219.80      18.43  11.924   <0.001 *** 
 --- 
 Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
 (Adjusted p values reported -- single-step method) 

 confint(.Pairs) # confidence intervals 
      Simultaneous Confidence Intervals 

 Multiple Comparisons of Means: Tukey Contrasts 

 Fit: aov(formula = Values ~ Label, data = sim.Ch12) 

 Quantile = 2.6927 
 95% family-wise confidence level 
   
 Linear Hypotheses: 
                  Estimate  lwr       upr       
 Pop2 - Pop1 == 0  -51.3000 -100.9382   -1.6618 
 Pop3 - Pop1 == 0  -19.4000  -69.0382   30.2382 
 Pop4 - Pop1 == 0  200.4000  150.7618  250.0382 
 Pop3 - Pop2 == 0   31.9000  -17.7382   81.5382 
 Pop4 - Pop2 == 0  251.7000  202.0618  301.3382 
 Pop4 - Pop3 == 0  219.8000  170.1618  269.4382

Confidence interval plots, Tukey posthoc

Figure 3. Plot of confidence intervals of Tukey HSD.

R Commander includes a default 95% CI plot (Fig 3). From this graph, you can quickly identify the pairwise comparisons for which 0 (zero, dotted vertical line) is included in the interval, ie, there is no difference between the means (eg, Pop1 is different from Pop4, but Pop1 is not different from Pop3).

b. Dunnett’s Test for comparisons against a control group

There are situations where we might want to compare our experimental Populations to one control Population or group. Thus, this is likely a case of planned comparisons.
This is common in medical research where there is a placebo (control pill with no drug) or sham operations (operations where everything but the critical operation is done).
This is also a common research design in ecological or agricultural research where some animal or plant populations are exposed to an environmental factor (eg, fertilizer, pesticide, pollutant, competitors, herbivores) and other animal or plant populations are not exposed to these environmental factor.
The difference in the statistical procedure for analyzing this type of research design is that the experimental groups may only be compared to the control group.
This results in fewer comparisons.
The formula is the same as for the Tukey’s Multiple Comparison test, except for the calculation of the SE.

Standard Error is changed by multiplying the MS_Error by 2.

And n is the harmonic mean of the sample sizes of the two groups being compared.

$\begin{align*} q = \frac{\bar{X}_{control}-\bar{X}_{A}}{SE} \end{align*}$

where

$\begin{align*} SE=\sqrt{\frac{2/ MS_{error}}{n}} \end{align*}$

R Commander doesn’t provide a simple way to get Dunnett, but we can get it simply enough if we are willing to write some script. Fortunately (OK, by design!), Rcmdr prints commands.

Look at the Output window from the one-way ANOVA with pairwise comparisons: it provides clues as to how we can modify the mcp command (mcp stands for multiple comparisons).

First, I had run the one-way ANOVA command and noted the model (AnovaModel.3). Second, I wrote the following script, modified from above.

Pairs <- glht(AnovaModel.1, linfct = mcp(Label = c("Pop2 - Pop1 = 0", "Pop3 - Pop1 = 0", "Pop4 - Pop1 = 0")))

where Label was my name for the Factor variable. Note that I specified the comparisons I wanted R to make. When I submit the script, nothing shows up in the Output window because the results are stored in Pairs.

Next, I then need to ask R to provide confidence intervals

confint(Pairs)

R output window

Pairs <- glht(AnovaModel.1, linfct = mcp(Label = c("Pop2 - Pop1 = 0", "Pop3 - Pop1 = 0", "Pop4 - Pop1 = 0"))) 
confint(Pairs) 

Simultaneous Confidence Intervals 

 Multiple Comparisons of Means: User-defined Contrasts 

 Fit: aov(formula = Values ~ Label, data = sim.Ch12) 

 Estimated Quantile = 2.4524 
 95% family-wise confidence level 

 Linear Hypotheses: 
  ...... lwr ....... upr 
 Pop2 - Pop1 == 0  .. -51.3000  . -96.5080  ... -6.0920 
 Pop3 - Pop1 == 0  .. -19.4000  . -64.6080  ... 25.8080 
 Pop4 - Pop1 == 0  .. 200.4000  . 155.1920  .. 245.6080

Look for intervals that include zero, therefore, the group does not differ from the Control group (Pop1). How many groups differed from the Control group?

Alternatively, I may write

Tryme <- glht(AnovaModel.1, linfct = mcp(Label = "Dunnett"))
confint(Tryme)

It’s the same — in fact, the default mcp test is Dunnett’s test.

Tryme <- glht(AnovaModel.1, linfct = mcp(Label = "Dunnett")) 
confint(Tryme) 

Simultaneous Confidence Intervals 

Multiple Comparisons of Means: Dunnett Contrasts 

Fit: aov(formula = Values ~ Label, data = sim.Ch12) 

Estimated Quantile = 2.4514 
95% family-wise confidence level 

Linear Hypotheses: 
Estimate lwr upr 
Pop2 - Pop1 == 0 -51.3000 -96.4895 -6.1105 
Pop3 - Pop1 == 0 -19.4000 -64.5895 25.7895 
Pop4 - Pop1 == 0 200.4000 155.2105 245.5895

Note 7: Use relevel() command* to set the order of levels in the data set. By default Dunnett procedure in mcp assumes the first level of the factor is the control or reference level. R sorts levels alphabetically, so clearly one option is to always provide a sort-friendly name for the reference level. Naturally, R has you covered if you need to specify a different level as the reference, the relevel() command. For example, instead of Pop1, set reference to Pop2 as follows:

Level 
[1] Pop1 Pop2 Pop3 Pop4
Levels: Pop1 Pop2 Pop3 Pop4

Set new order

Level <- relevel(Level, ref = "Pop2"); Level
[1] Pop1 Pop2 Pop3 Pop4
Levels: Pop2 Pop1 Pop3 Pop4

relevel() works only on unordered factors; if you get this error message it just means you need to specify your character variable as a factor. As you may recall, to confirm a character variable is factor

is.factor(Level)
[1] TRUE

If FALSE, then set character variable to factor.

Level <- as.factor(Level)

* Yes, you could change the order any way you want with level(). For example,

Level <- factor(Level, levels = c("Pop2", "Pop4", "Pop1", "Pop3")); Level

returns

[1] Pop1 Pop2 Pop3 Pop4
Levels: Pop2 Pop4 Pop1 Pop3

c. Bonferroni t

The Bonferroni t test is a popular tool for conducting multiple comparisons. This, too, is consistent with an unplanned comparisons approach. The rationale for this particular test is that the MS_error is a good estimate of the pooled variances for all groups in the ANOVA.

$\begin{align*} Bonferroni = \frac{\bar{X}_{B}-\bar{X}_{A}}{\sqrt{MS_{error}\left ( \frac{1}{n_{B}}+\frac{1}{n_{A}} \right )}} \end{align*}$

and DF = N – k.

Note 8: In order to achieve a Type I error rate of 5% for all tests, you must divide the 0.05 by the number of comparisons conducted.

Thus, for k = 4 groups,

$\binom{4}{2} = \frac{4!}{2!\left ( k-2 \right )!}$

Here’s a simplified version if you prefer to get all pairwise tests

$\binom{4}{2} = \frac{4!}{2!\left ( k-2 \right )!}$

Use this information then to determine how many total comparisons will be made, then if necessary, use to adjust Type I error rate for one test (the exerimentwise error rate).

For our example, is 0.05/6 = 0.00833. Thus, for a difference between two means to be statistically significant, the P-value must be less than 0.00833.

For Bonferroni, we will use the following script.

Set up one-way ANOVA model (ours has been saved as AnovaModel.1),
Collect all pairwise comparisons with the mcp(~"Tukey") stored in a vector (I called mine Whynot),
and finally, get the Bonferroni adjusted test of the comparisons with the summary command, but addtest = adjusted("bonferroni").

It’s a bit much, but we end up with a very nice output to work with.

Whynot <- glht(AnovaModel.1, linfct = mcp(Label = "Tukey")) 
summary(Whynot, test = adjusted("bonferroni")) 

 Simultaneous Tests for General Linear Hypotheses

Multiple Comparisons of Means: Tukey Contrasts

Fit: aov(formula = Values ~ Label, data = sim.Ch12)

Linear Hypotheses:
Estimate Std. Error t value Pr(>|t|) 
Pop2 - Pop1 == 0 -51.30 18.43 -2.783 0.0512 . 
Pop3 - Pop1 == 0 -19.40 18.43 -1.052 1.0000 
Pop4 - Pop1 == 0 200.40 18.43 10.871 3.82e-12 ***
Pop3 - Pop2 == 0 31.90 18.43 1.730 0.5527 
Pop4 - Pop2 == 0 251.70 18.43 13.654 5.33e-15 ***
Pop4 - Pop3 == 0 219.80 18.43 11.924 2.77e-13 ***
  --- 
 Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
 (Adjusted p values reported -- bonferroni method)

Questions

Be able to define and contrast experiment-wise and family-wise error rates.
Read and interpret R output
1. Refer back to the Tukey HSD output. Among the four populations, which pairwise groups were considered “statistically significant” following use of the Tukey HSD?
2. Refer back to the Dunnett’s output. Among the four populations, which population was taken as the control group for comparison?
3. Refer back to the Dunnett’s output. Which pairwise groups were considered “statistically significant” from the control group?
4. Refer back to the Bonferroni output. Among the four populations, which pairwise groups were considered “statistically significant” following use of the Bonferroni correction?
5. Compare and contrast interpretation of results for posthoc comparisons among the four populations based on the three different posthoc methods
Be able to distinguish when Tukey HSD and Dunnet’s post hoc tests are appropriate.
Some microarray researchers object to use of Bonferroni correction because it is too “conservative.” In the context of statistical testing, what errors are the researchers talking about when they say the correction is “conservative.”

Quiz Chapter 12.6

ANOVA posthoc tests

Data set used in this page

Label	Value
Pop1	105
Pop1	132
Pop1	156
Pop1	198
Pop1	120
Pop1	196
Pop1	175
Pop1	180
Pop1	136
Pop1	105
Pop2	100
Pop2	65
Pop2	60
Pop2	125
Pop2	80
Pop2	140
Pop2	50
Pop2	180
Pop2	60
Pop2	130
Pop3	130
Pop3	95
Pop3	100
Pop3	124
Pop3	120
Pop3	180
Pop3	80
Pop3	210
Pop3	100
Pop3	170
Pop4	310
Pop4	302
Pop4	406
Pop4	325
Pop4	298
Pop4	412
Pop4	385
Pop4	329
Pop4	375
Pop4	365

R code # copy the data from the full table to clipboard and paste between the double quotes in text=””)

sim.ch12 <- read.table(header=TRUE, sep=",",text="")

#check the dataframe
head(sim.ch12)

Chapter 12 contents

12.2 – One way ANOVA

Introduction
Three important terms you’ll need for one-way ANOVA
Note on experimental design
ANOVA sources of variation
Mean squares
Worked example with R
Questions
Quiz
Data set
Chapter 12 contents

Introduction

In ANalysis Of VAriance, or ANOVA for short, we likewise have tools to test the null hypothesis of no difference between between categorical independent variables — often called factors when there’s just a few levels to keep track of — and a single, dependent response variable. But now, the response variable is quantitative, not qualitative like the $\chi^2$ tests.

Analysis of variance, ANOVA, is such an important statistical test in biology that we will take the time to “build it” from scratch. We begin by reminding you of where you’ve been with the material.

We already saw an example of this null hypothesis. When there’s only one factor (but with two or more levels), we call the analysis of means and “one-way ANOVA.” In the independent sample t-test, we tested whether two groups had the same mean. We made the connection between the confidence interval of the difference between the sample means and whether or not it includes zero (ie, no difference between the means). In ANOVA, we extend this idea to a test of whether two or more groups have the same mean. In fact, if you perform an ANOVA on only two groups, you will get the exact same answer as the independent two-sample t-test, although they use different distributions of critical values (t for the t-test, F for the ANOVA — a nice little fact for you square the t-test statistic, you’ll get the F-test statistic: $t^{2} = F$ ).

Let’s say we have an experiment where we’ve looked at the effect of different three different calorie levels on weight change in middle-aged men.

I’ve created a simulated dataset which we will use in our ANOVA discussions. The data set is available at the end of this page (scroll down or click here).

We might graph the mean weight changes ( $\pm$ SEM); Below are two possible outcomes of our experiment (Fig 1).

Figure 1. Hypothetical results, box plots.

Figure 1. Hypothetical results of an experiment, box plots. Left, no difference among groups; Right, large differences among groups.

As statisticians and scientists, we would first calculate an overall or grand mean for the entire sample of observations; we know this as the sample mean whose symbol is $\bar{X}$ . But this overall mean is made up of the average of the sample means. If the null hypothesis is true, then the all of the sample means all estimate the overall mean. Put another way, being a member of a treatment group doesn’t matter, ie, there is no systematic effect, all differences among subjects are due to random chance.

$\begin{align*} H_{0} : \bar{X} = \bar{X}_{A} = \bar{X}_{B} = \bar{X}_{C} \end{align*}$

The hypotheses among are three groups or treatment levels then are:

$\begin{align*} H_{0} : \bar{X}_{A} = \bar{X}_{B} = \bar{X}_{C}&& \end{align*}$

Null that there are no differences among the group means. And the alternative hypotheses include any (or all) of the following possibilities

$\begin{align*} H_{0} : \bar{X}_{A} \ne \bar{X}_{B} = \bar{X}_{C} \end{align*}$

or maybe

$\begin{align*} H_{0} : \bar{X}_{A} = \bar{X}_{B} \ne \bar{X}_{C} \end{align*}$

or… have we covered all possible alternate outcomes among three groups?

In either case, we could use one-way ANOVA to test for “statistically significant differences.”

Three important terms you’ll need for one-way ANOVA

FACTOR: We have one factor of interest. For example, a factor of interest might be

Diet fed to hypertensive subjects (men and women)
Distribution of coral reef sea cucumber species in archipelagos
Antibiotic drug therapy for adolescents with Acne vulgaris (click see Webster 2002 for review).

Note 1: Character vs factor. In R programming, while it is true that a treatment variable can be a character vector like Gender*, such vectors are handled by R as string. In contrast, by claiming Gender as a factor, R handles the variable as a nominal data type. Hence, ANOVA and other GLM functions expect any treatment variable to be declared as a factor:

Gender <- c("Gender non-conforming", "Men", "Non-binary", "Transgender", "Women", "Declined"); Gender
[1] "Gender non-conforming" "Men" "Non-binary" "Transgender" 
[5] "Women" "Declined"

# it’s a string vector, not a factor

is.factor(Gender)
[1] FALSE

Change string vector to a factor.

Gender <- as.factor(Gender); is.factor(Gender) [1] TRUE

* For discussions of ethics of language inclusivity and whether or not such information should be gathered, see Henrickson et al 2020.

LEVELS: We can have multiple levels (2 or more) within the single factor. Some examples of levels for the Factors listed:

Three diets (DASH), diet rich in fruits & vegetables, control diet)
Five archipelagos (Hawaiian Islands, Line Islands, Marshal Islands, Bonin Islands, and Ryukyu Islands)
Five antibiotics (ciprofloxacin, cotrimoxazole, erythromycin, doxycycline, minocycline).

RESPONSE: There is one outcome or measurement variable. This variable must be quantitative (ie, on the ratio or interval scale). Continuing our examples then

Reduction in systolic pressure
Numbers of individual sea cucumbers in a plot
number of microcomedo.

Note 2: A comedo is a clogged pore in the skin; a microcomedo refers to the small plug. Yes, I had to look that up, too.

The response variable can be just about anything we can measure, but because ANOVA is a parametric test, the response variable must be normally distributed!

Note on experimental design

As we discuss ANOVA, keep in mind we are talking about analyzing results from an experiment. Understanding statistics, and in particular ANOVA, informs how to plan an experiment. The basic experimental design is called a completely randomized experimental design, or CDR, where treatments are assigned to experimental units at random.

In this experimental design, subjects (experimental units) must be randomly assigned to each of these levels of the factor. That is, each individual should have had the same probability of being found in any one of the levels of the factor. The design is complete because randomization is conducted for all levels, all factors.

Thinking about how you would describe an experiment with three levels of some treatment, we would have

Table 1. Summary statistics of three levels for some ratio scale response variable.

Level 1

$\bar_{X}_{1}$

sample mean₁

$s_{1}$

sample standard deviation₁

Level 2

$\bar_{X}_{2}$

sample mean₂

$s_{2}$

sample standard deviation₂

Level 3

$\bar_{X}_{3}$

sample mean₃

$s_{3}$

sample standard deviation₃

Note 3: Table 1 is the basis for creating the box plots in Figure 1.

ANOVA sources of variation

ANOVA works by partitioning total variability about the means (the grand mean, the group means). We will discuss the multiple samples and how the ANOVA works in terms of the sources of variation. There are two “sources” of variation that can occur:

Within Group Variation
Among Groups Variation

So let’s look first at the variability within groups, also called the Within Group Variation.

Consider and experiment to see if DASH diet reduces systolic blood pressure in USA middle-aged men and women with hypertension (Moore et al 2001). After eight weeks we have

We get the corrected sum of squares, SS, for within group

$\begin{align*} within-groups \ SS = \sum_{i=1}^{k}\left [ \sum_{j=1}^{n_{i}}\left ( X_{ij}-\bar{X}_{i} \right )^2 \right ] \end{align*}$

and the degrees of freedom, DF, for within groups

$\begin{align*} within-groups \ DF = \sum_{i=1}^{k}\left ( n_{i}-1 \right ) = N - k \end{align*}$

where i is the identity of the groups, $X_{ij}$ is the individual observations within group i, $\bar{X}_{i}$ is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

Importantly, this value is also referred to as “error sums of squares” in ANOVA. It’s importance is as follows — In our example, the within group variability would be zero if and only if all subjects within the same diet lost exactly the same amount of weight. This is hardly ever the case of course in a real experiment. Because there are almost always some other unknown factors or measurement error that affect the response variable, there will be some unknown variation among individuals who received the same treatment (within the same group). Thus, the error variance will generally be larger than zero.

The first point to consider: your ANOVA will never result in statistical differences among the groups if the error variance is greater than the second type of variability, the variability between groups.

The second type of variability in ANOVA is that due to the groups or treatments. Individuals given a calorie-restricted diet will lose some weight; individuals allowed to eat a calorie-rich diet likely will gain weight, therefore there will be variability (a difference) due to the treatment. So we can calculate the variability among groups. We get the corrected sum of squares for among groups

$\begin{align*} among-groups \ SS = \sum_{i=1}^{k}n_{i}\left ( X_{i}-\bar{X}\right )^2 \end{align*}$

and the degrees of freedom for among groups

$\begin{align*} among-groups \ DF = k - 1 \end{align*}$

where i is the identity of the groups, $\bar{X}$ is the grand mean as defined in Measures of Central Tendency (Chapter 3.1), $\bar{X}_{i}$ is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

The sums of squares here is simply subtracting the mean of each population from the overall mean.

If the Factor is not important in explaining the variation among individuals then all the population means will be similar and the sums of squares among populations would be small.
If the Factor is important in explaining some of the variation among the individuals then all the population means will NOT be the same and the sums of squares among populations would be large.

Finally, we can identify the total variation in the entire experiment. We have the total sum of squares.

$\begin{align*} Total \ SS = among-groups \ SS + within-groups \ SS \end{align*}$

Thus, the insight of ANOVA is that variation in the dataset may be attributed to a portion explained by differences among the groups and differences among individual observations within each group. The inference comes from recognizing that if the among group effect is greater than the within group effect, then there will be a difference due to the treatment levels.

Mean squares

To decide whether the variation associated with the among group differences are greater than the within group variation, we calculate ratios of the sums of squares. These are called Mean Squares or MS for short. The ratio of the Mean Squares is called F, the test statistic for ANOVA.

For the one-way ANOVA we will have two Mean Squares and one F, tested with degrees of freedom for both the numerator $MS_{groups}$ , and denominator $MS_{error}$ ).

The Mean Square for (among) groups is

$\begin{align*} MS_{group} = \frac{among-groups \ SS}{among-groups \ DF} \end{align*}$

The Mean Square for error is

$\begin{align*} MS_{error} = \frac{within-groups \ SS}{within-groups \ DF} \end{align*}$

And finally, the value for F, the test statistic for ANOVA, is

$\begin{align*} F = \frac{MS_{group}}{MS_{error}} \end{align*}$

Worked example with R

A factor with three levels, A, B, and C

group <- c("A", "A", "A", "B", "B", "B", "C", "C", "C")

and their responses, simulated

response <- c(10.8, 11.8, 12.8, 6.5, 7, 8, 3.8, 2.8, 3)

We create a data frame

all <- data.frame(group, response)

Of course, you could place the data into a worksheet and then import the worksheet into R. Regardless, we now have our dataset.

Now, call the ANOVA function, aov, and assign the results to an object (eg, Model.1)

Model.1 <- aov(response ~ group, data=all)

Now, visualize the ANOVA table

summary(Model.1)

and the output from R, the ANOVA table, is shown below

Note 4: Rcmdr: Statistics → Means → One-way ANOVA… see figures 2 and 3 below

Table 2. Output from aov() command, the ANOVA table, for the “Difference” outcome variable.

           Df   Sum Sq    Mean Sq   F value       Pr(>F) 
group       2   111.16      55.58     89.49    0.0000341 ***
Residuals   6     3.73       0.62
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Let’s take the ANOVA table one row at a time

The first row has subject headers defining the columns.
The second row of the table “groups” contains statistics due to group, and provides the comparisons among groups.
The third row, “Residuals” is the error, or the differences within groups.

Moving across the columns, then, for each row, we have in turn,

the degrees of freedom (there were 3 groups, therefore 2 DF for group),
the Sums of Squares, the Mean Squares,
the value of F, and finally,
the P-value.

R provides a helpful guide on the last line of the ANOVA summary table, the “Signif[icance] codes,” which highlights the magnitude of the P-value.

Note 5: On the use of “error” and “residuals.” Error refers to the difference between an observed value and the true population value. Residual is the difference between an observed value and a value predicted by a model. In ANOVA, the “error variance” or “residual variance” is the variation within groups unexplained by the tested factor. It is what remains after the effects of the model are accounted for.

What to report? ANOVA problems can be much more complicated than the simple one-way ANOVA introduced here. For complex ANOVA problems, report the ANOVA table itself! But for the one-way ANOVA it would be sufficient to report the test statistic, the degrees of freedom, and the p-value, as we have in previous chapters (eg, t-test, chi-square, etc.). Thus, we would report:

F = 89.49, df = 2 and 6, p = 0.0000341

where F = 89.49 is the test statistic, df = 2 (degrees of freedom for the among group mean square) and 6 (degrees of freedom for the within group mean square), and p = 0.0000341 is the p-value.

In Rcmdr, the appropriate command for the one-way ANOVA is simply

Rcmdr: Statistics → Means

Screenshot Rcmdr select one-way ANOVA

Figure 2. Screenshot Rcmdr select one-way ANOVA.

which brings up a simple dialog. R Commander anticipates factor (Groups) and Response variable. Optional, choose Pairwise comparisons of means for post-hoc test (Tukey’s) and, if you do not want to assume equal variances (see Chapter 13), select Welch F-test.

Screenshot Rcmdr select one-way ANOVA options

Figure 3. Screenshot Rcmdr select one-way ANOVA options.

Questions

1. Review the example ANOVA Table and confirm the following

How many levels of the treatment were there?
How many sampling units were there?
Confirm the calculation of $MS_{group}$ and $MS_{error}$ using the formulas contained in the text.
Confirm the calculation of F using the formula contained in the text.
The degrees of freedom for the F statistic in this example were 2 and 6 (F_2,6). Assuming a two-tailed test with Type I error rate of 5%, What is the critical value of the F distribution (see Appendix 20.4)

2. Repeat the one-way ANOVA using the simulated data, but this time, calculate the ANOVA problem for the “No.difference” response variable.

3. Leaf lengths from three strains of Arabidopsis thaliana plants grown in common garden are shown in Fig 2. Data are provided for you in the following R script.

arabid <- c("wt","wt","wt","AS1","AS1","AS1","AS2","AS2","AS2")
leaf <- c(4.909,5.736,5.108,6.956,5.809,6.888,4.768,4.209,4.065)
leaves <- data.frame(arabid,leaf)

Write out a null and alternative hypotheses
Conduct a test of the null hypothesis by one-way ANOVA
Report the value of the test statistic, the degrees of freedom, and the P-value
Do you accept or reject the null hypothesis? Explain your choice.

Box plot of length of leaves from three strains of Arabidopsis thaliana

Figure 4. Box plot of lengths of leaves of a 10-day old plant from on of three strains of Arabidopsis thaliana.

4. Return to your answer to question 7 from Chapter 12.1 and review your answer and modify as appropriate to correct your language to that presented here about factors and levels.

5. Conduct the one-way ANOVA test on the Comet assay data presented question 7 from Chapter 12.1. Data are copied for your convenience to end of this page. Obtain the ANOVA table and report the value of the test statistics, degrees of freedom, and p-value.
a. Based on the ANOVA results, do you accept or reject the null hypothesis? Explain your choice.

Quiz Chapter 12.2

One-way ANOVA

Data used in this page

Difference, no difference

Table 12.2 Difference or No Difference

Group	No.difference	Difference
A	12.04822	11.336161
A	12.67584	13.476142
A	12.99568	12.96121
A	12.01745	11.746712
A	12.34854	11.275492
B	12.17643	7.167262
B	12.77201	5.136788
B	12.07137	6.820242
B	12.94258	5.318743
B	12.0767	7.153992
C	12.58212	3.344218
C	12.69263	3.792337
C	12.60226	2.444438
C	12.02534	2.576014
C	12.6042	4.575672

Simulated data

Comet assay, antioxidant properties of tea

Data presented in Chapter 12.1

Chapter 12 contents

12.1 – The need for ANOVA

Introduction
The multiple comparison problem
Corrections to account for the multiple comparisons problem
ANOVA is a solution
Questions
Quiz
Data used in this page
Chapter 12 contents

Introduction

Moving from an experiment with two groups to multiple groups is deceptively simple: we move from one comparison to multiple comparisons. Consider an experiment in which we have randomly assigned patients to receive one of three doses of a statin drug (lower cholesterol), including a placebo (eg, Tobert and Newman 2015). Thus, we have three groups or levels of a single treatment factor and we’ll want to test the null hypothesis that the group (level) means are all equal as opposed to the alternative hypothesis in which one or more of the group means, eg, group A, Group B, group C, are different.

$\begin{align*} H_{0}: \bar{X}_{A} = \bar{X}_{B} = \bar{X}_{C} \end{align*}$

The correct procedure is to analyze multiple levels of a single treatment with a one-way analysis of variance followed by a suitable post-hoc (“after this”) test. Two common post-hoc tests are Tukey’s range test (aka Tukey’s HSD [honestly significant difference] test), which is for all pairwise comparisons, or the Dunnett’s test, which compares groups against the control group. Post-hoc tests are discussed in Chapter 12.6.

Note. To get the number of pairwise comparisons $C$ of $k$ observations, let

$\begin{align*} C = \frac{k\left ( k-1 \right )}{2} \end{align*}$

for more on pairwise comparisons, see 10.1 – Compare two independent sample means

For our three group experiment, how many pairwise comparisons can be tested? Therefore, $k = 3$ , and we have

$\begin{align*} C = \frac{3\left ( 3-1 \right )}{2} = 3 \end{align*}$

Thus, for our three groups, A, B, and C, there are three possible pairwise comparisons.

$\begin{align*} H_{1}: \bar{X}_{A} = \bar{X}_{B}\\ H_{2}: \bar{X}_{A} = \bar{X}_{C}\\ H_{3}: \bar{X}_{B} = \bar{X}_{C} \end{align*}$

How many pairwise comparisons for a four group experiment? Check your work, you should get $C = 6$ .

The multiple comparison problem

Let’s say that we’re stubborn. We could do many single two-sample t-tests — certainly, your statistical software won’t stop you — this is a situation that calls for statistical reasoning. Here’s why we should not: we will increase the probability of rejecting a null hypothesis when the null hypothesis is true (eg, discussion in Jafari and Ansari-Pour 2019). That is, the chance we will commit a Type I error increases if we do not account for the lack of independence in these sets of pairwise tests evaluated by t-tests. This is multiplicity or the multiple comparison problem.

Review: when we perform a two-sample t-test we are willing to reject a true null hypothesis 5% of the time. This is what is meant by setting the critical probability value (alpha) = 0.05. By “willing” we mean that we know that our conclusions could be wrong because we ar3 working with samples, not the entire population. (Of course at the time, we have no way of actually knowing WHEN we are wrong, but we do want to know how likely we could be wrong!) However, if we compare three population means we have three separate null hypotheses.

H_O: one or more of the means are different.

But if we conduct these as separate independent sample t-tests, then we are implicitly making the following null hypothesis statements:

$\begin{align*} H_{1}: \bar{X}_{O} = \bar{X}_{B}\\ H_{2}: \bar{X}_{O} = \bar{X}_{C}\\ H_{3}: \bar{X}_{O} = \bar{X}_{C} \end{align*}$

Thus, we have a 5% chance of being wrong for the first hypothesis and/or a 5% chance of being wrong for the second hypothesis and/or a 5% chance of being wrong for the third hypotheses. The chance that we will be wrong for at least one of these hypotheses must now be greater than 5%.

For three separate hypotheses there is a 14% chance of being wrong when we have the probability value for each individual t-test set at α = 0.05. How did we get this result? The point is that these tests are not independent, they are done on the same data set, therefore you can’t simply apply the multiplication probability rule.

Here’s how to figure this: for the set of three hypotheses, the probability of incorrectly rejecting at least one of the null hypotheses is

$\begin{align*} 1 - 0.953 = 0.14 \end{align*}$

So, for three t-tests on the same experiment, the Type I error overall tests (experiment-wise) is actually 14% not 5%. It gets worse as the number of combinations (groups and therefore hypotheses) increases. For four groups, Type I error is actually

$\begin{align*} 1 - 0.956 = 0.265 \end{align*}$

That’s 26.5%, not 5%.

If we have just five populations means to compare the probability of rejecting a null hypothesis when it is true climbs to 60%! How did this happen? The probability or correctly rejecting all of them is now

$\begin{align*} 0.9510 = 0.599 \end{align*}$

So, the probability of incorrectly rejecting one test (Type I error) is now

$\begin{align*} 1 - 0.9510 = 0.401 \end{align*}$

instead of the 5% we think we are testing.

The key argument for why you must use ANOVA to analyze multiple samples instead of a combination of t-tests!! ANOVA guarantees that the overall error rate is the specified 5%.

Why is the Type I error not 5% for each test? Because we conducted ONE experiment, we can conduct only ONE test (we could be right, we could be wrong 5% of the time). If we conduct the experiment over again, on new subjects, each time resulting in new and therefore independent data sets, then Type I error = 5% for each of these independent experiments.

Now, I hope I have introduced you to the issue of Type I error at the level of a single comparison and the idea of an experiment, holding Type I error-rates at 5% across all hypotheses to be evaluated in an experiment. You may wonder why anyone would make this mistake now. Actually, people make this “mistake” all the time and in some fields like evaluating gene expression for microarray data, this error was the norm, not the exception (see for example discussion of this in Jeanmougin et al 2010).

To conclude, if one does multiple tests on the same experiment, whether it is t-tests or some other test for that matter, then our subsequent tests are related. This is what we mean by independence in statistics — and there are many ways that nonindependence may occur in experimental research. For example, we introduced the concept of pseudoreplication, when observations are treated as if they are independent, but they are not (see Chapter 5.2). The “multiple comparison problem” specifically refers to the lack of independence when all the data set from a single experiment is parsed into lots of separate tests. Philosophically, there must be a logical penalty — and that is reflected in the increase in Type I error.

Clearly something must be done about this!

Corrections to account for the multiple comparisons problem

The family-wise (aka experiment-wise) error rate for multiple comparisons is kept at 5%, and each individual-wise comparison is compared against a more strict (ie, smaller Type I error rate). Put another way, the family-wise error rate is the chance of a number of false positives: making a mistake when we consider many tests simultaneously. The simplest correction for the individual-wise error rate is the Bonferroni correction: test each individual comparison at Type I error equal to $\alpha \slash C$ , where C is the number of comparisons. The Šidák correction, $1-(1-\alpha)^\frac{1}{C}$ , and the Holm-Bonferroni method are alternatives with reportedly better statistical properties (eg, reduced Type II error rates; Ludbrook 1998, Sauder and DeMars 2019).

ANOVA is a solution

One possible solution for getting the correct experiment-wise error rate: adjust for differences in probability for multiple comparisons with the t-test. We used post-hoc tests presented above: you could evaluate the tests after accounting for the change in Type I error. This is what is done in many cases. For example, in genomics. In the early days of gene expression profiling by microarray, it was common to see researchers conduct t-tests for each gene. Since microarrays can have thousands of genes represented on the chip, then these researchers were conducting thousands of t-tests, arranging the t-tests by P-value and counting the number of P-values less than 5% and declaring that the differences were statistically significant.

This error didn’t stand long, and there are now many options available to researchers to handle the “multiple comparisons” problem (some probably better than others, research on this very much an ongoing endeavor in biostatistics). The Bonferroni correction was an available solution, largely replaced by the Holm (aka Holm-Bonferroni) method (Holm 1979). Recall that the Bonferroni correction judged individual p-values statistically significant only if they were less than $\alpha \slash C$ , where, again, α is the family-wise error rate (eg, α = 5%) and C was the number of comparisons. The Holm method orders the C p-values from lowest to highest rank. The method then evaluates lowest p-value, if less than $\alpha \slash C$ , then reject hypothesis for that comparison. Proceed to next p-value, if less than $\alpha \slash C$ , then reject hypothesis for that comparison, and so on until no more comparisons are less than $\alpha \slash C$ .

A MUCH better alternative is to perform a single analysis that takes the multiple-comparisons problem into account: Single Factor ANOVA, also called the one-way ANOVA, plus the post-hoc tests with error correction. We introduce one-way ANOVA in the next section. Post-hoc tests are discussed in Chapter 12.6.

Questions

You should be able to define and distinguish how Bonferoni correction, Dunnett’s test, and Tukey’s test methods protect against inflation of Type I error.
What will be the experiment-wise error rate for an experiment in which there are only two treatment groups?
Experiment-wise error rate may also be called _______________ error rate.
List and compare the three described posthoc approaches to correct for multiple comparison problem,
Glycophosphate-tolerant soybean is the number one GMO (genetically modified organism) crop plant worldwide. Glycophosphate is the chief active ingredient in Roundup, the most widely used herbicide. A recent paper examined “food quality” of the nutrient and elemental composition of plants drawn from fields which grow soy by organic methods (no herbicides or pesticides) and GMO plants subject to herbicides and pesticides. A total of 28 individual t-tests were used to compare the treatment groups for different levels of nutrients and elements (eg, vitamins, amino acids, etc.,); the authors concluded that 10 of these t-tests were statistically significant at Type I error rate of 5%. Discuss the approach to statistical inference by the authors of this report; include correct use of the terms experiment-wise and individual-wise in your response and suggest an alternative testing approach if it is appropriate in your view.
In Chapter 4.2 we introduced a data set from an experiment. The experiment looked at DNA damage quantified by measuring qualities in a Comet Assay including the Tail length, the percent of DNA in the tail, and olive moment. The data set is copied to end of this page. In the next chapter I’ll ask you to conduct the ANOVA on this experiment. For now, answer the following questions.
a. What is the response variable? Defend your choice(s).
b. Why is this an ANOVA problem? Include as part of your explanation a statement of the null hypothesis.

Quiz Chapter 12.1

The need for ANOVA

Data used in this page

comet assay dataset

Treatment	Tail	TailPercent	OliveMoment*
Copper-Hazel	10	9.7732	2.1501
Copper-Hazel	6	4.8381	0.9676
Copper-Hazel	6	3.981	0.836
Copper-Hazel	16	12.0911	2.9019
Copper-Hazel	20	15.3543	3.9921
Copper-Hazel	33	33.5207	10.7266
Copper-Hazel	13	13.0936	2.8806
Copper-Hazel	17	26.8697	4.5679
Copper-Hazel	30	53.8844	10.238
Copper-Hazel	19	14.983	3.7458
Copper	11	10.5293	2.1059
Copper	13	12.5298	2.506
Copper	27	38.7357	6.9724
Copper	10	10.0238	1.9045
Copper	12	12.8428	2.5686
Copper	22	32.9746	5.2759
Copper	14	13.7666	2.6157
Copper	15	18.2663	3.8359
Copper	7	10.2393	1.9455
Copper	29	22.6612	7.9314
Hazel	8	5.6897	1.3086
Hazel	15	23.3931	2.8072
Hazel	5	2.7021	0.5674
Hazel	16	22.519	3.1527
Hazel	3	1.9354	0.271
Hazel	10	5.6947	1.3098
Hazel	2	1.4199	0.2272
Hazel	20	29.9353	4.4903
Hazel	6	3.357	0.6714
Hazel	3	1.2528	0.2506

Data set, comet assay. Rat lung cells treated with Hazel tea extract and exposed to copper metal. Tail refers to length of the comet tail, TailPercent is percent DNA damage in tail, and OliveMoment refers to Olive's (1990), defined as the fraction of DNA in the tail multiplied by tail length