17.6 – ANCOVA – analysis of covariance

Introduction

Analysis of covariance (ANCOVA) is intended to help with analysis of designs with categorical treatment variables on some response (dependent) variable, but a known confounding variable is also present. Thus, the researcher is also likely to know of additional ratio scale variables that covary with the response variable and, moreover, must be included in the experimental design in some way.

Take for example the well-known relationship between body size and whole-animal metabolic rate as measured by rates of carbon dioxide production or rates of oxygen consumption for aerobic organisms. We may be interested in how addition or blocking of stress hormones affects resting metabolism; we may be interested in comparing men and women for activity metabolism, and so on. We’d like to know if the regressions were the same (eg, metabolic rate covaried with body mass in the same way — that is, the slope of the relationship was the same).

This situation arises frequently in biology. For example, we might want to know if male and female birds have different mean field metabolic rates, in which case we might be tempted to use a one-way ANOVA or t-test (since one factor with two levels). However, if males and females also differ for body size, then any differences we might see in metabolic rate could be due to differences in metabolic rate are confounded by differences in the covariable body size. We already discussed one approach to correction: calculate a ratio. Thus, a logical approach to correcting or normalizing for the covariation would be to divide body mass (units of kilograms) into metabolic rate (eg, volume of oxygen, O₂, consumed), and make comparisons, say, among different species, on mass-specific trait (). However, because the regression between mass and metabolic rate is allometric, i.e., not equal to one, the ratio does not, in fact normalize for body mass.  We made this point in Chapter 6.2, and remarked that analysis of covariance ANCOVA was a solution.

ANCOVA allows you to test for mean differences in traits like metabolic rate between two or more groups, but only after first accounting for covariation due to another variable (eg, body size). However, ANCOVA makes the assumption that relationship between the covariable and the response variable is the same in the two groups. This is the same as saying that the regression slopes are the same. We discussed how to use t-test to test hypothesis of equal slopes between regression models in Chapter 17.5, but a more elegant way is to include this in your model.

Example

We return to our sample of 13 tadpoles (Rana pipiens), hatched in the laboratory. I’ve repeated the data set in this page, scroll or click here.

Our linear model was and a scatterplot of the data set is shown in Figure 1 (a repeat of Figure 3 from Chapter 17.5, but now points identified to developmental group).

Figure 1. Scatterplot of oxygen consumption by R. pipiens tadpoles vs body mass (g) by developmental group (Gosner stages I or II).

The project looked at whether metabolism as measured by oxygen consumption was consistent across two developmental stages. Metamorphosis in frogs and other amphibians represents profound reorganization of the organism as the tadpole moves from water to air. Thus, we would predict some cost as evidenced by change in metabolism associated with later stages of development. Figure 2 shows a box plot of tadpole oxygen consumption by Gosner (1960) developmental stage (Figure 2 is a repeat of Figure 4 Chapter 17.5).

Figure 2. Copy of Figure 4, Chapter 17.5; boxplot of oxygen consumption of R. pipiens tadpoles by Gosner developmental stages.

Looking at Figure 2 we see a trend consistent with our prediction; developmental stage may be associated with increased metabolism. However, older tadpoles also tend to be larger, and the plot in Figure 2 does not account for that. Thus, body mass is a confounding variable in this example. There are several options for analysis here (eg, ANCOVA), but one way to view this is to compare the slopes for the two developmental stages. While this test does not compare the means, it does ask a related question: is there evidence of change in rate of oxygen consumption relative to body size between the two developmental stages? The assumption that the slopes are equal is a necessary step for conducting the ANCOVA.

So, divide the data set (Table 1) into two groups by developmental stage (12 tadpoles could be assigned to one of two developmental stages; one was at a lower Gosner stage than the others and so is dropped from the subset.

Table 1. and of 12 R. pipiens tadpole larvae by Gosner developmental stage.

Gosner stage I

1.76	109.41
1.88	329.06
1.95	82.35
2.13	198
2.26	607.7

Gosner stage II

Body mass
2.28	362.71
2.35	556.6
2.62	612.93
2.77	514.02
2.97	961.01
3.14	892.41
3.79	976.97

The slopes and standard errors we obtained in Chapter 17.5 were

Make a plot (Figure 3).

Figure 3. Scatterplot with best-fit regression lines of by for Gosner State I (closed circle, solid line) and Gosner Stage II (open circle, dashed line) R. pipiens tadpoles.

R code for plot in Figure 3.

#Used Rcmdr scatterplot(), then modified code
scatterplot(VO2~Body.mass | Gosner, regLine=FALSE, smooth=FALSE, 
boxplots=FALSE, xlab="Body mass (g)", ylab="Oxygen consumption (ml/h)", 
main="", cex=1.4, cex.axis=1.5, cex.lab=1.5, pch=c(19,19), by.groups=TRUE,
col=c("blue","red"), grid=FALSE, 
legend=list(coords="bottomright"), data=Tadpoles)
#Get regression equations for groups, subset by Gosner
abline(lm(VO2~Body.mass, data=Stage01), lty=1, lwd=2, col="blue")
abline(lm(VO2~Body.mass, data=Stage02), lty=1, lwd=2, col="red")

Returning to the important question, are the two slopes statistically indistinguishable (Ho: b_I = b_II), where b_I is the slope for the Gosner Stage I subset and b_II is the slope for the Gosner Stage II subset? We look at the plot, and since the lines cross, we tend to see a difference. Of course, we need to consider that our perception of slope differences may simply be chance, especially because the sample size is small. Proceed to test.

R code

The ANCOVA is a new ANOVA model where the factor variables are adjusted or corrected for the effects of the continuous variable.

R code for ANCOVA example, crossed or interaction model.

tadpole.1 <- lm(VO2 ~ Body.mass*Gosner, data=example.Tadpole)
summary(tadpole.1)
Anova(tadpole.1, type="II")

Output

summary(tadpole.1)

Call:
lm(formula = VO2 ~ Body.mass * Gosner, data = example.Tadpole)

Residuals:
    Min      1Q   Median      3Q      Max 
-167.80 -117.93    13.81   94.66   214.65

Coefficients:
                       Estimate   Std. Error   t value    Pr(>|t|) 
(Intercept)             -1231.6       783.0     -1.573    0.1544 
Body.mass                 750.0       390.7      1.919    0.0912 .
Gosner[T.II]              790.4       859.2      0.920    0.3845 
Body.mass:Gosner[T.II]   -350.1       409.5     -0.855    0.4174 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 155.8 on 8 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.821, Adjusted R-squared: 0.7539 
F-statistic: 12.23 on 3 and 8 DF, p-value: 0.002336

The coefficients for the first factor (GII) and then the differences in the coefficient for the second factor. You can just add the second coefficient to the first so they’re on the same scale.

Anova Table (Type II tests)

Response: VO2
                  Sum Sq   Df    F value     Pr(>F) 
Body.mass         330046    1    13.6030   0.006146 **
Gosner              5630    1     0.2321   0.642908 
Body.mass:Gosner   17736    1     0.7310   0.417423 
Residuals 194102 8 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Suggests interaction is not significant, i.e., the slopes are not different.

We can then proceed to check to see if the intercepts are different, now that we’ve confirmed no significant difference in slope.

R code for ANCOVA as additive model

tadpole.2 <- lm(VO2 ~ Body.mass + Gosner, data=example.Tadpole)
summary(tadpole.2)
Anova(tadpole.2, type="II")

Output

> summary(tadpole.2)

Call:
lm(formula = VO2 ~ Body.mass + Gosner, data = example.Tadpole)

Residuals:
    Min       1Q    Median      3Q      Max 
-163.12  -125.53    -20.27   83.71   228.56

Coefficients:
                   Estimate   Std. Error     t value    Pr(>|t|) 
(Intercept)         -595.37       239.87      -2.482     0.03487 * 
Body.mass            431.20       115.15       3.745     0.00459 **
Gosner[T.II]          64.96       132.83       0.489     0.63648 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 153.4 on 9 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.8047, Adjusted R-squared: 0.7613 
F-statistic: 18.54 on 2 and 9 DF, p-value: 0.0006432

Anova(tadpole.2, type="II")
Anova Table (Type II tests)

Response: VO2
              Sum Sq    Df    F value      Pr(>F) 
Body.mass     330046     1    14.0221   0.004593 **
Gosner          5630     1     0.2392   0.636482 
Residuals     211839     9 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Instead of the interaction, or as added, try a nested model, with body mass nested within stage.

tadpole.3 <- lm(VO2 ~ Body.mass/Gosner, data=example.Tadpole)

summary(tadpole.3)

Call:
lm(formula = VO2 ~ Body.mass/Gosner, data = example.Tadpole)

Residuals:
    Min       1Q    Median      3Q      Max 
-168.66  -131.14    -20.28   90.33   225.36

Coefficients:
                 Estimate      Std. Error    t value Pr(>|t|) 
(Intercept)           -575.10      319.51     -1.800   0.1054 
Body.mass              423.65      162.50      2.607   0.0284 *
Body.mass:Gosner[T.II]  21.95       63.73      0.344   0.7384 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 154.4 on 9 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.8021, Adjusted R-squared: 0.7581 
F-statistic: 18.24 on 2 and 9 DF, p-value: 0.0006823

Gets the true coefficient (nested lm() version).

The two test different hypotheses.

lm(VO2 ~ Body.mass * Gosner) tests whether or not the regression has a nonzero slope.

lm(VO2 ~ Body.mass */Gosner) test whether or not the slopes and intercepts from different factors are statistically significant.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. An OLS approach was used for the analysis of tadpole oxygen consumption body mass. Consider the RMA approach — would that be a more appropriate regression model? Explain why or why not.
3. Consider an experiment in which you plan to administer a treatment that has a carry-over effect. For example, Compare and contrast “crossed” and “nested” designs.
4. True or False. The nested design option for the ANCOVA assumes the slopes for the two groups of tadpoles for the regression line of by are equal. Explain your choice.
5. Metabolic rates like oxygen consumption over time are well-known examples of allometric relationships. That is, many biological variables (eg, is related as aM^b, where M is body mass, slope b is scaling exponent), and best evaluated on log-log scale. Repeat the analysis above on log₁₀-transformed and for
   • crossed design (eg, tadpole.1 model)
   • added design (eg, tadpole.2 model)
   • nested design (eg, tadpole.3 model)
6. Create the plot and add the fitted lines from crossed design to the plot.

lgVO2 <- log10(VO2)

lm(log10(VO2) ~ log10(Body.mass ), data=example.Tadpole)

Quiz Chapter 17.6

ANCOVA - analysis of covariance

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Data sets

Oxygen consumption, , of Anuran tadpoles, dataset=example.Tadpole

Gosner refers to Gosner (1960), who developed a criteria for judging metamorphosis staging.

Chapter 17 contents

• Introduction
• Simple Linear Regression
• Relationship between the slope and the correlation
• Estimation of linear regression coefficients
• OLS, RMA, and smoothing functions
• Testing regression coefficients
• ANCOVA – Analysis of covariance
• Regression model fit
• Assumptions and model diagnostics for Simple Linear Regression
• References and suggested readings (Ch17 & 18)

12.7 – Many tests one model

Introduction

In our introduction to parametric tests we so far have covered one and two sample t-tests and now the multiple sample or one-way analysis of variance (ANOVA). In subsequent sections we will cover additional tests, each with their own name. It is time to let you in on a little secret. All of these tests, t-tests, ANOVA, and linear and multiple regression we will work on later in the book belong to one family of statistical models. That model is called the general Linear Model (LM), not to be confused with the Generalized Linear Model (GLM) (Burton et al 1998; Guisan et al 2002). This greatly simplifies our approach to learning how to implement statistical tests in R (or other statistical programs) —  you only need to learn one approach: the general Linear Model (LM) function lm().

Brief overview of linear models

With the inventions of correlation, linear regression, t-tests, and analysis of variance in the period between 1890 and 1920, subsequent work led to the realization that these tests (and many others!) were special cases of a general model, the general linear model, or LM. The LM itself is a special case of the generalized linear model, or GLM; among the differences between LM and GLM, in LM, the dependent variable is ratio scale and errors in the response (dependent) variable(s) are assumed to come from a Gaussian (normal) distribution. In contrast, for GLM, the response variable may be categorical or continuous, and error distributions other than normal (Gaussian), may be applied. The GLM user must specify both the error distribution family (eg, Gaussian) and the link function, which specifies the relationship among the response and predictor variables. While we will use the GLM functions when we attempt to model growth functions and calculate EC₅₀ in dose-response problems, we will not cover GLM this semester.

The general Linear Model, LM

In matrix form, the LM can be written as

where is a matrix of response variables predicted by independent variables contained in matrix  and weighted by linear coefficients in the vector b. Basically, all of the predictor variables are combined to produce a single linear predictor . By adding an error component we have the complete linear model.

In the linear model, the error distribution is assumed to be normally distributed, or “Gaussian.”

R code

The bad news is that LM in R (and in any statistical package, actually) is a fairly involved set of commands; the good news is that once you understand how to use this command, and can work with the Options, you will be able to conduct virtually all of the tests we will use this semester, from two-sample t-tests to multiple linear regression. In the end, all you need is the one Rcmdr command to perform all of these tests.

We begin with a data set, ohia.ch12. Scroll down this page or click here to get the R code.

I found a nice report on a common garden experiment with o`hia (Corn and Hiesey 1973). O`hia (Metrosideros polymorpha) is an endemic, but wide-spread tree in the Hawaiian islands (Fig 1). O`hia exhibits pronounced intraspecific variation: individuals differ from each other. O`hia grows over wide range of environments, from low elevations along the ocean right up the sides of the volcanoes, and takes on many different growth forms, from shrubs to trees. Substantial areas of o`hia trees on the Big Island are dying, attributed to two exotic fungal species of the genus Ceratocystis (Asner et al., 2018).

Figure 1. O’hia, Metrosideros polymorpha. Public domain image from Wikipedia. 

The Biology. Individuals from distinct populations may differ because the populations differ genetically, or because the environments differ, or, and this is more realistic, both. Phenotypic plasticity is the ability of one genotype to produce more than one phenotype when exposed to different environments. Environmental differences are inevitable when populations are from different geographic areas. Thus, in population comparisons, genetic and environmental differences are confounded. A common garden experiment is a crucial genetic experiment to separate variation in phenotypes, P, among populations into causal genetic or environmental components.

If you recall from your genetics course,

where G stands for genetic (alleles) differences among individuals and E stands for environmental differences among individuals. In brief, the common garden experiment begins with individuals from the different populations are brought to the same location to control environmental differences. If the individuals sampled from the populations continue to differ despite the common environment, then the original differences between the populations must have a genetic basis, although the actual genetic scenario may be complicated (the short answer is that if “genotype by environment interaction exists, then results from a common garden experiment cannot be generalized back to the natural populations/locations — this will make more sense when we talk about two-way ANOVA). For more about common garden experiments, see de Villemereuil et al (2016). Nuismer and Gandon (2008) discuss statistical aspects of the common garden approach to studying local adaptation of populations and the more powerful “reciprocal translocation” experimental design.

Managing data for linear models

First, your data must be stacked in the worksheet. That means one column is for group labels (independent variable), the other column is for the response (dependent) variable.

If you have not already downloaded the data set, ohia.ch12, do so now. Scroll down this page or click here to get the R code.

Confirm that the worksheet is stacked. If it is not, then you would rearrange your data set using Rcmdr: Data → Active data set → Stack variables in data set…

The data set contains one factor, “Site” with three levels (M-1, 2, 3). M stands for Maui, and collection sites were noted in Figure 2 of Corn and Hiesey (1973). Once the dataset is in Rcmdr, click on View to see the data (Fig 2). There are two response variables, Height (shown in red below) and Width (shown in blue below).

Figure 2. The o`hia dataset as viewed in R Commander.

The data are from Table 5 of Corn and Hiesey 1973.  (I simulated data based on their mean/SD reported in Table 5). This was a very cool experiment: they collected o`hia  seeds from three elevations on Maui, then grew the seeds in a common garden in Honolulu. Thus, the researchers controlled the environment; what varied, then were the genotypes.

As always, you should look at the data. Box plots are good to compare central tendency (Fig 3).

Figure 3. Box plots of growth responses of o`hia seedlings collected from three Maui sites, M-1 (elevation 750 ft), M-2 (elevation 1100 ft), and M-3 (elevation 6600 ft). Data adapted from Table 5 of Corn and Hiersey 1973.

R code to make Figure 4 plots

par(mfrow=c(1,2))
Boxplot(Height ~ Site, data = ohia, id = list(method = "y"))
Boxplot(Width ~ Site, data = ohia, id = list(method = "y"))

This dataset would typically be described as a one-way ANOVA problem. There was one treatment variable (population source) with three levels (M-1, M-2, M-3). From Rcmdr we select the one-way ANOVA: Statistics → Means → One-way ANOVA… and after selecting the Groups (from the Site variable) and the Response variable (eg, Height), we have

AnovaModel.1 <- aov(Height ~ Site, data = ohia.ch12)
summary(AnovaModel.1)
           Df Sum Sq   Mean Sq  F value          Pr(>F) 
Site        2   4070    2034.8    22.63 0.000000131 ***
Residuals  47   4227      89.9 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Let us proceed to test the null hypothesis (what was it???) using instead the lm() function. Four steps in all.

Step 1. Rcmdr: Statistics → Fit models → Linear model … (Fig 4).

Figure 4. R Commander, select to fit a Linear model.

Step 2. The pop up menu below (Fig 5) follows.

First, What is our response (dependent) variable? What is our predictor (independent) variable? We then input our model. In this case, with only the one predictor variable, Sites, our model formula is simple to enter (Fig 5): Height ~ Site.

Figure 5. Input linear model formula, Height ~ Site.

Step 3. Click OK to carry out the command.

Here is the R output and the statistical results from the application of the linear model.

LinearModel.1 <- lm(Height ~ Site, data=ohia.ch12) 
summary(LinearModel.1)

Call:
lm(formula = Height ~ Site, data = ohia.ch12)

Residuals:
    Min     1Q   Median     3Q     Max 
-18.808 -4.761   -1.755  4.758  29.257

Coefficients:
             Estimate  Std. Error  t value      Pr(>|t|)
(Intercept)    15.314       2.121    7.222 0.00000000377 ***
Site[T.M-2]    19.261       2.999    6.423 0.00000006153 ***
Site[T.M-3]     2.924       3.673    0.796          0.43    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 9.483 on 47 degrees of freedom
Multiple R-squared:  0.4905,	Adjusted R-squared:  0.4688 
F-statistic: 22.63 on 2 and 47 DF,  p-value: 0.0000001311

End R output

The linear model has produced a series of estimates of coefficients for the linear model, statistical tests of the significance of each component of the model, and the coefficient of determination, R², which is a descriptive statistic of how well model fits the data. Instead of our single factor variable for Source Population like in ANOVA we have a series of what are called dummy variables or contrasts between the populations. Thus, there is a coefficient for the difference between M-1 and M-2. “Site[T.M-2]” in the output, between M-1 and M-3, and between M-2 and M-3.

It also produced our first statistic that assesses how well the model fits the data called the coefficient of determination, R². An of of 1.0 would indicate that all variation in the data set can be explained by the predictor variable(s) in the model with no residual error remaining. Our value of 49% indicates that nearly 50% of the variation in height of the seedlings grown under common environments are due to the source population (= genetics).

Step 4. But we are not quite there — we want the traditional ANOVA results (recall the ANOVA table).

To get the ANOVA Table we have to ask Rcmdr (and therefore R) to give us this. Select

Rcmdr: Models → Hypothesis tests → ANOVA table … (Fig 6).

Figure 6. To retrieve an ANOVA table, select Models, Hypothesis tests, then ANOVA table…

Here’s the type of tests for the ANOVA table, select the default (Fig 7).

Figure 7. Options for types of tests.

Now, in the future when we work with more complicated experimental designs, we will also need to tell R how to conduct the test. For now, we will accept the default Type II type of test and ignore sandwich estimators. You should confirm that for a one-way ANOVA, Type I and Type II choices give you the same results.

The reason they do is because there is only one factor — when there are more than one factors, and if one or both of the factors are random effects, then Type I, II, and III will give you different answers. We will discuss this more as needed, but see the note below about default choices.

By default, Rcmdr makes Type II. In most of the situations we will find ourselves this semester, this is the correct choice.

Below is the output from the ANOVA table request. Confirm that the information is identical to the output from the call to aov() function.

Anova(LinearModel.1, type = "II")
Anova Table (Type II tests)
Response: Height
          Sum Sq  Df   F value        Pr(>F) 
Site      4069.7   2    22.626  0.0000001311 ***
Residuals 4226.9  47 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

And the other stuff we got from the linear model command? Ignore for now but make note that this is a hint that regression and ANOVA are special cases of the same model, the linear model.

We do have some more work to do with ANOVA, but this is a good start.

Why use the linear model approach?

Chief among the reasons to use the lm() approach is to emphasize that a model approach is in use. One purpose of developing a model is to provide a formula to predict new values. Prediction from linear models is more fully developed in Chapter 17 and Chapter 18, but for now, we introduce the predict() function with our O`hia example.

myModel <- predict(LinearModel.1, interval = "confidence")
head(myModel, 3)    #print out first 3 rows
#Add the output to the data set
ohiaPred <- data.frame(ohia,myModel)
with(ohiaPred, tapply(fit, list(Site), mean, na.rm = TRUE))   #print out predicted values for the three sites

Output from R

myModel <- predict(LinearModel.1, interval = "confidence")
head(myModel, 3)
       fit      lwr      upr
1 15.31374 11.04775 19.57974
2 15.31374 11.04775 19.57974
3 15.31374 11.04775 19.57974
with(myModel, tapply(fit, list(Site), mean, na.rm = TRUE))
     M-1      M-2      M-3
15.31374 34.57474 18.23796

Questions

1. Revisit ANOVA problems in homework and questions from early parts of this chapter and apply lm() followed by Hypothesis testing (Rcmdr: Models → Hypothesis tests → ANOVA table) approach instead of one-way ANOVA command. Compare results using lm() to results from One-way ANOVA and other ANOVA problems.

Quiz Chapter 12.7

Many tests one model

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Data set and R code used in this page

Corn and Hiesey (1973)

Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

12.2 – One way ANOVA

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 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: ).

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 ( SEM); Below are two possible outcomes of our experiment (Fig 1).

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 . 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.

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

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

or maybe

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).

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

is.factor(Gender)
[1] FALSE

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.

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.

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

and the degrees of freedom, DF, for within groups

where i is the identity of the groups,  is the individual observations within group 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

and the degrees of freedom for among groups

where i is the identity of the groups, is the grand mean as defined in Measures of Central Tendency (Chapter 3.1), 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.

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 , and denominator ).

The Mean Square for (among) groups is

The Mean Square for error is

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

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

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.

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

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.

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 and  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.

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

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Data used in this page

Difference, no difference

Table 12.2 Difference or No Difference

Comet assay, antioxidant properties of tea

Data presented in Chapter 12.1

Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

10.3 – Paired t-test

Introduction

Good experiments include controls. Interested in a new treatment for weight loss? Define a control group to compare the weight loss by a group using the new product. In many cases, the best control is the individual.

Consider now a basic experimental design, the randomized crossover trial (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Crossover Trial.

Subjects randomly selected from population of interest, then again once recruited into one of two treatment arms: arm 1, subjects first receive the experimental treatment, then some time later the subjects receive the control treatment; arm 2, subjects first receive the control treatment, then some time later the subjects receive the experimental treatment. Note the difference between this paired or repeated measures design and the independent sample design (see Chapter 10.1). Repeated measures designs have many advantages; we discuss them further in Chapter 14.6. At the start, repeated measures designs have greater statistical power compared to cross-sectional (independent) sample designs.

Many experiments are designed so that subjects receive all treatments and responses are gauged against the initial values recorded on the subjects. Repeated measures statistical tests, like the paired t-test, are needed however to analyze the data. These types of statistical procedures are similar to the two sample independent t-test that we discussed earlier.

However, there is an important difference between these two types of statistical procedures. For the two independent sample t-test the samples are unpaired: we observed one variable on some individuals assigned to two different groups. These groups might be

• Two locations where we measure plants or animals
• A treatment (or experimental) group with a control group.
• Expression of cytokeratin genes (eg, ΔΔC_T, fold-change) from breast cancer patients compared to healthy donor subjects (Andergassen et al 2016).

The point is that samples in one group are not the same samples in the second group.

In the paired t-test we have two groups but the observations in these two groups are paired. Paired means that there is some relationship between one observation in the first sample and one observation in the second sample (every observation in one sample must be paired with one observation in another sample).

For example, weight change in humans before and after a change in diet could be performed as a paired analysis. Each subject’s weight before the diet was “paired” with the same subject’s weight after the diet.

Another example comes from genetics. Siblings or Monozygotic twins or clones, strains or varieties of plants or animals can be Paired in an experiment.

• You can give one of the twins a particular diet, or the plant or animal clones or strains can be raised in a particular environment (nutrient)
• The other twin or plant or animal clone or variety can serve as a type of control by providing a normal diet or normal environment.

Another example is a study of environmental pollution on cancer rates in many different communities.

• The researchers selects pairs of communities with similar characteristics for many socioeconomic factors.
• Each pair of communities differed with respect to the proximity to a known source of pollution: one of the pair was close to a source of pollution and one of the pair was far from a source of pollution.

The purpose of pairing in this example is to attempt to “control” for all the socioeconomic factors that might contribute to cancer but they did not want to directly measure. These other factors should be similar for each member of the pair.

Example: How repeatable is human running performance?

The subheading is misleading — we leave “repeatability” or, do individuals perform consistently to another time (see Chapter 12.3 – Fixed effects, random effects, and agreement). Here, we focus on a different question — we test whether mean performance differs across repeated trials. This may seem a subtle difference, but consistency asks the question if performance, whether it is answering questions on an exam or running times for a race, is stable across trials. Testing the means of first and second trials asks a different question: we test for a trend in performance, which may reflect learning.

Here’s an example in which a measure was taken twice for the same individuals. The data are running speed or pace during 5K race held annually on Oahu for a selected sample of female runners with repeat race times (20 – 29 years old). The race was run annually on Oahu, and the data reported are the pace for the first race and the second race, which occurred a year later (Jamba Juice – Banana Man Chase, Ala Moana Beach Park, data extracted from source, https://timelinehawaii.com).

Table 1. 5K pace times (kph) for 15 women (20 – 29 years)

Load the data into R as an unstacked data set. Data available at end of this page or click here.

Begin with description and exploration of the data. Start with histograms to get a sense of the sample distributions (hint: we’re looking to see if the data looks like it could come from a normal distribution, see Chapter 13.3 Assumptions) (Fig 2).

Figure 2. Histograms shows the distribution of 5K running times of 15 women who ran the race twice.

R code (stacked data set, then used defaults R Commander to make the histogram, then modified the code and submitted modified code to make Fig 1)

with(stackExCh10.3, Hist(obs, groups=Race, scale="frequency", breaks="Sturges", col="blue", 
xlab="Time (min)", ylab="Frequency")))

Conclusion? The histograms don’t look normally distributed so we keep this in mind as we proceed. (See Chapter 13 – Assumptions of parametric tests for additional information about assumptions.)

A box plot comparing the first and second pace times (Fig 3).

Figure 3. Box plot of  race speed (kph) for 15 women 5K in two successive years.

I added a trend line (linear regression, see Chapter 17, red line) and connected the averages (blue line)  for visual emphasis — no differences between the means — but note that one wouldn’t do this as part of an analysis (see Chapter 4 discussion).

R code for Fig 2

Boxplot(obs~Race, data=stackExCh10.3, id=list(method="y"), xlab="", ylab="Pace (kph)") #boxplot was made in Rcmdr
abline(lm(obs ~ as.numeric(Race), data=stackExCh10.3), col="red", lwd=2)
means <- tapply(obs, Race, mean)
points(1:2, means, pch=7, col="blue")
lines(1:2, means, col="blue", lwd=2)

The box plot works to show the median difference, but loses the paired information. A nice package called PairedData has several functions that work well with paired data, for example a profile plot (Fig 4). A profile plot is useful for showing an association between paired data.

Figure 4. Profile plot, PairedData package.

R commands for Figure 4.

require(PairedData)
attach(example.ch10.3) # remember to attach dataframe so you don't have to call variables like example.ch10.3$Race1
races <- paired(Race1, Race2)
plot(races, type = "profile")

Paired t-test calculation

The paired t-test is a straight-forward extension of the independent sample t-test; the key concept is that the two samples are no longer independent, they are paired. Thus, instead of mean of group 1 minus mean of group two, we test the differences between sample 1 and sample 2 for each paired observation.

1. Compute the differences between the Paired Samples (as in tables above)
2. Calculate the MEAN difference score, : in the previous example = -0.094 kmh
3. Calculate the degrees of freedom df = # pairs – 1 = n – 1, where n is the number of pairs
4. Calculate the standard error of the mean of d.
   
   where
   
5. Calculate the test statistic for paired data
   
6. Compare to the Critical Value in Table C. 4 (Appendix Table 2)
7. Find the Critical Value = t _{α (2), df}

Try as difference instead of paired

Before you answer, take a look at the box plot of the mean difference between the repeat measures of 5K pace for the 15 women (Fig 5).

Figure 5. Box plot of differences, Red dotted lines shows the null hypothesis.

Create a new variable, raceDiff, equal to Race2 minus Race1. Then, use the one sample T-test on raceDiff. I’ll leave you to complete the work (Question 2).

R code

t.test(Race1, Race2, paired = TRUE, alternative = "two.sided")

R output

Paired t-test

data: Race.1 and Race.2
t = 0.19389, df = 14, p-value = 0.849
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.9491017 1.1377521
sample estimates:
mean of the differences 
0.09432517

Rcmdr, paired t-test

Rcmdr: Statistics → Means → Paired t-test…

Note: your two groups must be in two different columns (unstacked!) to run this version of the test.

Figure 6. R Commander Paired t-test menu, Rcmdr version 2.7.

After selecting the variables, set null hypothesis after clicking on Options tab (Fig 6).

Figure 7. R Commander Paired t-Test options, select null hypothesis.

Click OK to proceed. Results will be the same as the output listed above.

the following text needs to be updated

Interpret the results.

So, what can we conclude about the null hypothesis? Interpret the 95% CI, the T-test statistic, and the P-value.

Do not ignore sample dependence

What if we ignored the repeated measures design and treated the first and second races as independent? The important concept here is to ask, what would have happened if we had done a two independent sample t-test instead?

Let’s run the analysis again, this time incorrectly using the independent sample t-test. We need to manipulate the data set before we do.

Manage your data: Stack the data

This is a good time to share how to Stack data in R. If you look at our active data set, the results of the two trials are in two different columns. In order to run the independent sample t-test we need the data in one column (with a label column).

stackExCh10.3 <- stack(example.ch10.3[, c("Race1","Race2")])
names(stackExCh10.3) <- c("obs", "Race")

Rcmdr: Data → Active data set → Stack variables in data set…

Figure 8. R Commander: Stack worksheet. Select the two variables, Race1 and Race2.

I entered values for name of the new data set, the new variable, and the name for the factor (label) column.

Figure 9. R Commander, select independent sample t-Test …

Figure 10. R commander, independent sample t test menu.

Figure 11. R Commander, select options for independent sample t-Test (assume equal variance).

Here are the results of the independent sample t-test from R.

t.test(obs~Race, alternative='two.sided', conf.level=.95, var.equal=TRUE, + data=stackCh10.3)

Two Sample t-test
data: obs by Race
t = 0.070645, df = 28, p-value = 0.9442
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.640719 2.829369
sample estimates:
mean in group Race1 mean in group Race2 
 9.886342 9.792017

End R output

In this case, we would have reached the same general conclusion, but the p-values are different. The p-value from the paired t-test was about 0.85 whereas the p-value from the independent sample t-test was higher, nearly 0.95, suggesting little difference between the two trials.

While the general conclusion holds, this time, that there were no statistically significant difference between the means for first and second trials. However, it won’t always work out that way. And besides, if you treated the paired data as independent, you’ve clearly violated one of the assumptions of the test.

Take a look at the degrees of freedom for the two analyses. By ignoring the pairing of samples we gain twice the number of degrees of freedom … that can’t both be right. The way to distinguish between the two is to go back to the experimental units.

Question: What are the sampling units in the case of repeat measures on individuals: the individuals themselves? the pairs of burst speed trials? something else?

it is important to note that the paired t-test is still the best for this situation because it accurately reflects the experiment — individuals were measured twice, therefore the two groups (trial 1 and trial 2) are not independent! Thus, the p-value from the paired t-test correctly reflect our best analyses of the test of the null hypothesis because the correct degrees of freedom were 14 and not 28.

In the case of the independent sample t-test we necessarily make the assumption that the two groups are independent — that is, that they are measured on different sampling units (eg, different individuals or subjects). In statistical terms, that means that you assume that the correlation between trial results is equal to zero. By incorrectly choosing an independent sample test in these repeated measures cases, I would make two null hypotheses: (1) that the means are the same and (2) that the correlation between repeat measures is zero. The problem? The t-test only evaluates the first hypothesis (means).

Questions

1. Refer to Figure 6 again and related data set. Were runners faster the second year or the first year running the 5k? What about the points labeled 4 and 14? What was the average difference between first and second races?
2. Complete the test of the null hypothesis of no difference between race 1 and race 2 (raceDiff) with the one sample t-test. Set up a table to compare the test statistic, df, and p-values for results from paired t test, one sample t-test, and independent sample t-test. How do these results compare?
3. I’ve called the observed value “pace,” but runners would know that pace is actually amount of time per kilometer, not the total time over 5k, which is what I called pace.
   • Create a new variable and report average pace for Race1 and Race2.
   • Redo the paired analysis, including box plot, on your new variable.
   • What is the null hypothesis for your new variable?
   • Summarize your results and add to the table you created for question 2.
4. Redo the two plots in Figure 2 so that the two histograms are on the sample plot. Instructions and examples were provided in Chapter 4.2.

Quiz Chapter 10.3

Paired t-test

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Data set

example.ch10.3 <- read.table(header=TRUE, text = "
ID Race1 Race2
1 15.28 15.61
2 11.22 11.19
3 8.80 9.14
4 8.88 5.46
5 9.81 10.50
6 6.12 5.69
7 8.31 8.71
8 6.26 7.42
9 17.16 16.41
10 16.23 15.82
11 5.90 7.12
12 8.31 10.48
13 5.93 8.64
14 10.54 5.99
15 9.53 8.69
")

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

10.1 – Compare two independent sample means

Introduction

We introduced the concept of comparing a sample statistic (mean) against a population parameter (Chapter 6.7, Normal deviate) or one-sample t-test against a specified mean (eg, from published data or from theory, Chapter 8.5).

Consider now a basic experimental design, the randomized control trial, or RCT (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Control Trial.

Subjects randomly selected from population of interest, then again — random assignment — once recruited into one of two treatment groups. Importantly, subjects belong to one treatment arm only: no subject simultaneously receives the treatment and the control. This is in contrast to the paired design, in which subjects receive both treatments (see Chapter 10.3).

In inferential statistics about an experiment, we are more likely trying to test if sample means are different. For example

• two species grown in common garden, do they differ in growth rate?
• human subjects given a new treatment have better outcomes compare to those receiving a control treatment (eg, placebo).

The equivalent null hypothesis is that two samples are pulled from the same population. We write the null hypothesis as

and the corresponding alternative hypothesis, H_A, then must be

Question: Is this a one tailed or two-tailed hypothesis?

Answer: Two-tailed (review Chapter 8.4)

Worked example

We introduce the two-sample t-test, or better, the independent sample t-test.

where the numerator is the difference between the two sample means and the denominator is the standard error of the differences between the two groups standard errors. The formula for

The choice of independent sample over two-sample is best because it emphasizes that the two groups (the two samples), must be comprise of independent sampling units. This is a pretty straight-forward requirement; you have randomly assigned twenty individuals to two groups, a control group (n = 10) and a treatment group (n = 10). Individuals are either in the control group or they are in the treatment group — they cannot simultaneously appear in both groups.

We will work our way through this test by example. For starters, let’s use the same lizard dataset (see Example data set, below), four body mass recordings (grams) each for house geckos (Hemidactylus frenatus, Fig 2) and the Carolina anole (Anolis carolensis, Fig 3), two of many lizard species introduced to Hawaiʻi.

Figure 2. Male Hemidactylus frenatus, central Oahu, M. Dohm.

Figure 3. Male Anolis carolinensis, ʻAkaka Falls, Big Island of Hawaiʻi, M. Dohm.

Example data set

Geckos: 3.186, 2.427, 4.031, 1.995

Anoles: 5.515, 5.659, 6.739, 3.184

Question: How would you go about creating a data frame with the values in long form (stacked worksheet), including a label variable and the body mass?

Answer: At the R prompt, type

Geckos <- c(3.186, 2.427, 4.031, 1.995); Anoles = c(5.515, 5.659, 6.739, 3.184) #create two vectors
bmass <- c(Geckos, Anoles)    #combine the two vectors into a single vector holding all of the body mass records
species <- c("gecko", "gecko", "gecko", "gecko", "anole", "anole", "anole", "anole")   #create your label variable
lizards <- data.frame(species, bmass)  #create your data frame
lizards   #print your data frame  species bmass
1   gecko 3.186
2   gecko 2.427
3   gecko 4.031
4   gecko 1.995
5   anole 5.515
6   anole 5.659
7   anole 6.739
8   anole 3.184

Note also that you can enter data into the Data editor by creating the data frame first then adding values. To edit the data frame “lizards” type fix(lizards) at the R prompt, then close the data frame when you have added or changed values as needed.

As always, begin with an exploration of the data, including a graph (Fig 4).

Figure 4. Box plot of lizard body mass.

We can see already that there’s greater spread of data for the Anoles compared to the Geckos, but the median values differ. Small sample sizes can be a problem for analyses as we can only have reduced confidence in our conclusions. However, we press on for the sake of demonstration.

Let’s test the null hypothesis, H₀, ie, the two species of lizards have the same mean body mass.

Rcmdr: Statistics → Means → Independent-samples t-test…

In this next image I posted the Rcmdr menu popup for the Independent Samples t-test (Fig 5). Later versions of Rcmdr split the settings for this command into two tabs; the first tab allows for the selection of the variables and setting the hypotheses whereas the second tab, labeled Options, permits additional choices. The default selections need your attention: to actually conduct the t-test you need to answer “No” to the question, “Assume equal variance?” (Fig 6).

Figure 5. Rcmdr menu for Independent sample t-test.

Select the Options tab (Fig 6) to select null hypothesis and to select the t-test and not the Welch-test (which is the default, ie, No to the prompt “Assume equal variances?”).

Figure 6. Rcmdr Options menu for Independent sample t-test.

Let’s look at the results and break down the parts of the test.

t.test(Body.mass~Lizard, alternative='two.sided', conf.level=.95, 
+ var.equal=TRUE, data=lizards)

Two Sample t-test

data: Body.mass by Lizard 
t = 2.7117, df = 6, p-value = 0.03503
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
0.2308685 4.4981315 
sample estimates:
mean in group Anolis mean in group Gecko 
5.27425 2.90975

Consider the R session output above and answer the following questions.

Questions for the worked example

1. Which lizard group had the greater mean value, Anolis or Gecko?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

Try another example

DNA damage, changes in the chemical structure of nucleotide bases or breakage of the DNA chains, occurs in cells under many circumstances. The comet assay, or single-cell gel electrophoresis, is one method for visualizing and measuring DNA strand breaks in cells. Exposed cells are mixed with a low-melting temperature agarose and placed onto a microscope slide. The cells are then lysed with an alkaline detergent and high salts. When current is applied across the slide, undamaged DNA remains in the nucleus, whereas damaged DNA extends towards the anode to form a comet-like tail, with imaging assisted by including a fluorescent dye like Sybr-Green. Examples of comets are shown below (Fig 7).

Figure 7. Comet examples. A Intact cell, no DNA damage, B Cell with some DNA damage, a slight tail to the right is evident, C  Cell with significant DNA damage, a large tail is evident. M. Dohm.

In an experiment, immortalized lung epithelial cells were exposed to dilute copper solutions for 30 minutes then washed with PBS. The comet assay was applied to these cells and for comparison, to cells without copper exposure but otherwise treated the same way (controls). The data are available at the bottom of this page (scroll down or click here).

Again, you should begin all analyses with an exploration of the data, including a graph (Fig 8).

Figure 8. Boxplot of comet tail lengths for cells with and without (control) exposure to copper in the cell medium for 30 minutes.

Let’s look at the R output for the t-test analysis.

    Two Sample t-test

data:  CometTail by Treatment
t = -5.8502, df = 38, p-value = 9.139e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -22.39865 -10.88213
sample estimates:
mean in group Control  mean in group Copper 
             11.14533              27.78571

Consider the R session output above and answer the following questions.

Questions for Comet assay data set

1. Which cell group had the greater mean value, Copper-exposed or Control-exposed cells?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

T test from summary statistics

In some cases you may only have to summary statistics for data, eg, the means and the standard deviations. We can use the equations of the t test to write a simple formula, where the user provides the known means, standard deviations, and sample size. For example, create a simple function with readline for user input.

myTtest <- function() {
   mnx <- as.numeric(readline(prompt="Enter mean of x: "))
   stdevx <- as.numeric(readline(prompt="Enter sd of x: "))
   nx <- as.numeric(readline(prompt="Enter n of x: "))
   mny <- as.numeric(readline(prompt="Enter mean of y: "))
   stdevy <- as.numeric(readline("Enter sd of y: "))
   ny <- as.numeric(readline(prompt="Enter n of y: "))
   myTvalue <- abs(((mnx-mny)-0)/sqrt(((stdevx^2)/nx)+(stdevy^2)/ny))
   myDF <- as.integer(nx+ny-2)
   myPvalue <- pt(myTvalue,myDF,lower.tail=FALSE)*2
   myResults <- c(myTvalue, myDF, myPvalue)
   report <- c("T-test: ", "df: ", "two-tailed p-value: ")
   cat(sprintf("%s %3.3f, ", report, myResults))
}

then run the function by typing myTest() at the R prompt and entering the means, standard deviations, and sample size when prompted.

myTtest()
Enter mean of x: 2.91
Enter sd of x: .895
Enter n of x: 4
Enter mean of y: 5.27
Enter sd of y: 1.497
Enter n of y: 4
T-test: 2.706, df: 6.000, two-tailed p-value: 0.035

P-values from confidence intervals

While we expect certain reporting criteria for published, it is not uncommon that one or more elements are missing. For example, while means plus or minus standard deviations for two groups may be reported, the confidence interval of the difference may not be reported, or, an inexact p-value is reported like “< 0.05,” but we “need” an exact p-value for our meta-analysis.  A little math, and these missing statistics can be calculated.

We can use the lizard study as an example; while all of the expected elements were reported, including the p-value (0.03503), let’s say the p value was reported as “statistically significant at p < 0.05.”

We need

lower and upper confidence intervals: 0.2309 and 4.498, respectively.

critical value from t distribution, df = 6 (2 tailed, therefore 0.05/2). A little R code we have:

the standard error, which can be calculated as

and then the p-value,

which is pretty close to the result from R (p = 0.03503). The difference is likely due to the small sample size.

Questions

1. Don’t forget to work through the Questions for the Comet tail data set (scroll up or click here).
2. Microsoft Excel, LibreOffice Calc, and Google sheets spreadsheet software all include t-test functions and return the p-value. Consider two variables big (100, 110, 120, 100, 110, 210, 200) and small (0,1,1,2,0,1,0). (Note — these two groups are obviously very different, calculating a t-test on their difference is silly, just for this question.)  If formatting is set to the default two decimal places for Number cell category, the p-value will return as “0.00.” How should you report the p-value in this case?

Quiz Chapter 10.1

Compare two independent sample means

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Comet assay data set

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

4.10 – Graph software

Introduction

You may already have experience with use of spreadsheet programs to create bar charts and scatter plots. Microsoft Office Excel, Google Sheets, Numbers for Mac, and LibreOffice Calc are good at these kinds of graphs — although arguably, even the finished graphics from these products are not suitable for most journal publications. 

For bar charts, pie charts, and scatter plots, if a spreadsheet app is your preference, go for it, at least for your statistics class. This choice will work for you, at least it will meet the minimum requirements asked of you.

However, you will find spreadsheet apps are typically inadequate for generating the kinds of graphics one would use in even routine statistical analyses (e.g., box plots, dot plots, histograms, scatter plots with trend lines and confidence intervals, etc.). And, without considerable effort, most of the interesting graphics (e.g., box plots, heat maps, mosaic plots, ternary plots, violin plots), are impossible to make with spreadsheet programs.

At this point, you can probably discern that, while I’m not a fan of spreadsheet graphics, I’m also not a purist — you’ll find spreadsheet graphics scattered throughout Mike’s Biostatistics Book. Beyond my personal bias, I can make the positive case for switching from spreadsheet app to R for graphics is that the learning curve for making good graphs with Excel and other spreadsheet apps is as steep as learning how to make graphs in R (see Why do we use R Software?). In fact, for the common graphs, R and graphics packages like lattice, ggplot2, make it easier to create publishing-quality graphics.

Alternatives to base R plot

This is a good point to discuss your choice of graphic software — I will show you how to generate simple graphs in R and R Commander which primarily rely on plotting functions available in the base R package. These will do for most of the homework. R provides many ways to produce elegant, publication-quality graphs. However, because of its power, R graphics requires lots of process iterations in order to get the graph just right. Thus, while R is our software of choice, other apps may be worth looking at for special graphics work. 

My list emphasizes open source and or free software available both on Windows and macOS personal computers. Data set used for comparison from Veusz (Table).

1. GrapheR — R package that provides a basic GUI (Fig 1) that relies on Tcl/Tk — like R Commander — that helps you generate good scatter plots, histograms, and bar charts.  Box plot with confidence intervals of medians (Fig 2).

Figure 1. Screenshot of GrapheR GUI menu, box plot options.

Figure 2. Box plot made with GrapheR.

2. RcmdrPlugin.KMggplot2 — a plugin for R Commander that provides extensive graph manipulation via the ggplot2 package, part of the Tidyverse environment (Fig 3). Box plot with data point, jitter (Fig 4).

Figure 3. Screenshot of KMggplot2 GUI menu, box plot options.

Figure 4. Box plot graph made with GrapheR with jitter applied to avoid overplotting of points.

Figure 5. Box plot graph made with GrapheR with beeswarm applied to avoid overplotting of points.

3. A bit more work, but worth a look. Use plotly library to create interactive web application to display your data. 

install.packages("plotly")
library(plotly)
fig <- plot_ly(y = Bees, type = "box", name="Bees")
fig <- fig %>% add_trace(y = Butterflies, name="Butterflies")
fig

code modified from example at https://plotly.com/r/box-plots/

Figure 6. Screenshot of plotly box plot. Live version, data points visible when mouse pointer hover.

4. Veusz, at https://veusz.github.io/. Includes a tutorial to get started. Mac users will need to download the dmg file with the curl command in the terminal app instead of via browser, as explained here.

Figure 7. Screenshot of box plot example in Veusz GUI.

5. SciDAVis, “Scientific Data Analysis and Visualization,” is a package capable of generating lots of kinds of graphs along with curve fitting routines and other mathematical processing, https://scidavis.sourceforge.net/. SciDAVis is very similar to QtiPlot and OriginLab, and runs on both macOS and WinPC.

Figure 8. Screenshot SciDAVis app with default box plot.

More sophisticated graphics can, and when you gain confidence in R, you’ll find that there are many more sophisticated packages that you could add to R to make really impressive graphs. However, the point is to get the best graph, and there are many tools out there that can serve this end.

Chapter 4 contents

• Graphs and tables
• Bar (column) charts
• Histograms
• Box plot
• Mosaic plots
• Scatter plots
• Adding a second Y axis
• Q-Q plot
• Ternary plots
• Heat maps
• Graph software
• References and suggested readings

4.3 – Box plot

Introduction

Box plots, also called whisker plots, should be your routine choice for exploring ratio scale data. Like bar charts, box plots are used to compare ratio scale data collected for two or more groups. Box plots serve the same purpose as bar charts with error bars, but box plots provide more information.

Purpose and design criteria

Box plots are useful tool for getting a sense of central tendency and spread of data. These types of plots are useful diagnostic plots. Use them during initial stages of data analyses. All summary features of box plots are based on ranks (not sums). So, they are less sensitive to extreme values (outliers). Box plots reveal asymmetry. Standard deviations are symmetric.

The median splits each batch of numbers in half (center line). The “hinge” (median value) splits the remaining halves in half again (the quartiles). The first, second (median), and third quartiles describes the interquartile range, or IQR, 75% of the data (Fig 1). Outlier points can be identified, for example, with an asterisk or by id number (Fig 1).

Figure 1. A box plot. Elements of box plot labelled. 

We’ll use the data set described in the previous section, so if you have not already done so, get the data from Table 1, Chapter 4.2 into your R software.

R Code

Command line

We’ll provide code for the base graph shown in Figure 2A. At the R prompt, type

boxplot(OliveMoment~Treatment)

Figure 2A. Box plot, default graph in base package

Boxplot is a common function offered in several packages. In the base installation of R, the function is boxplot(). The car package, which is installed as part of R Commander installation, includes Boxplot(), which is a “wrapper function” for boxplot(). Note the difference: base package is all lower case, car package the “B” is uppercase. One difference, base boxplot() permits horizontal orientation of the plot (Fig 2B).

boxplot(OliveMoment ~ Treatment,  horizontal=TRUE, col="steelblue")

Figure 2B. Same graph, but with color and made horizontal; boxplot(), default graph in base package

Base package boxplot() has additional features and options compared to Note 3: Boxplot() in the car package. i.e., not all barcode() options are wrapped. For example, I had more success adding original points to boxplot() graph (Fig 2C) following the function call with stripchart().

stripchart(OliveMoment ~ Treatment, method = "overplot", pch = 19, add = TRUE)

Figure 2C. Same graph, added original points; boxplot(), default graph in base package.

Rcmdr: Graph → Boxplot…

Select the response variable, then click on the Plot by: button

Figure 3. Popup menu in R Commander: Select the response variable and set the Plot by: option.

Next, select the Groups (Factor) variables (Fig 4). Click OK to proceed

Figure 4. Select the group variable

Back to the Box Plot menu, click “Options” tab to add details to the plot, including a graph title and how outliers are noted (Fig 5), 

Figure 5. Options tab, enter labels for axes and a title.

And here is the resulting box plot (Fig 6)

Figure 6. Resulting box plot from car package implemented in R Commander. Outliers are identified by row id number.

The graph is functional, if not particularly compelling. The data set was “olive moments” from Comet Assays of an immortalized rat lung cell line exposed to dilute copper solution (Cu), Hazel tea (Hazel), or Hazel & Copper solution. 

Apply Tidyverse-view to enhance look of boxplot graphic

Load the ggplot2 package via the Rcmdr plugin to add options to your graph. As a reminder, to install Rcmdr plugins you must first download and install them from an R mirror like any other package, then load the plugin via Rcmdr Tools → Load Rcmdr plug-in(s)… (Fig 7, Fig 8).

Figure 7. Screen shot of Load Rcmdr plug-ins menu, Click OK to proceed (see Fig 8).

Figure 8. To complete installation of the plug-in, restart R Commander.

Significant improvement, albeit with an “eye of the beholder” caveat, can be made over the base package. For example, ggplot2 provides additional themes to improve on the basic box plot. Figure 8 shows the options available in the Rcmdr plugin KMggplot2, and the default box plot is shown in Fig 9.

Figure 9. Menu of KMggplot2. A title was added, all else remained set to defaults.

The next series of plots, Fig 10 – 12,  explore available formats for the charts.

Figure 10. Default box plot from KMggplot.

Figure 11. “Economist” theme box plot from KMggplot2.

And finally, since the box plot is often used to explore data sets, some recommend including the actual data points on a box plot to facilitate pattern recognition. This can be accomplished in the KMggplot2 plugin by checking “Jitter” under the Add data points option (see Fig 8). Jitter helps to visualize overlapping points at the expense of accurate representation. I also selected the Tufte theme, which results in the image displayed in Figure 12. 

Figure 12. Tufte theme and data points added to the box plot.

Conclusions

As part of your move from the world of Microsoft Excel graphics to recommended graphs by statisticians, the box plot is used to replace the bar charts plus error bars that you may have learned in previous classes. The second conclusion? I presented a number of versions of the same graph, differing only by style. Pick a style of graphics and be consistent.

Questions

1. Why is a box plot preferred over a bar chart for ratio scale data, even if an appropriate error bar is included?
2. With your comet data (Table 1, Chapter 4.2), explore the different themes available in the box plot commands available to you in Rcmdr. Which theme do you prefer and why?

Quiz Chapter 4.3

Box plots

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Chapter 4 contents

• Graphs and tables (How to report statistics)
• Bar (column) charts
• Histograms
• Box plot
• Mosaic plots
• Scatter plots
• Add a second Y axis
• Q-Q plot
• Ternary plots
• Heat maps
• Graph software
• References and suggested readings

4.1 – Bar (column) charts

Introduction

Bar chart, also called a bar graph, or column chart in many spreadsheet apps (e.g., Google Sheets, Microsoft Excel, Libreoffice Calc — “bar charts” is reserved for horizontal bar plots),  are used to compare counts among two or more categories, i.e., an alternative to pie charts (Fig. 1). 

Figure 1. Single nucleotide variants for human gene ACTB by DNA and functional element (data collected 19 May 2022 from NCBI SNP database with Advanced search query). A. Pie chart. Note that slide for “exon – nonsense” is not visible. B. Bar chart – color coded bars to facilitate comparison with pie chart.

Although bar charts are common in the literature (Cumming et al 2007; Streit and Gehlenborg 2014), bar charts may not be a good choice for comparisons of ratio scale data (Streit and Gehlenbor 2014). Bar charts for ratio data are misleading. Parts of the range implied by the bar may never have been observed: the bars of the chart always start at zero. Box (whisker) plots are better for comparisons of ratio scale data and are presented in the next section of this chapter. That said, I will go ahead and present how to create bar charts for both count — generally considered acceptable — and ratio scale data, for which their use is controversial.

Purpose of the bar chart

Like all graphics, a bar chart should tell a story. The purpose of displaying data is to give your readers a quick impression of the general differences among two or more groups of the data. For counts, that’s where the bar chart comes in. The bar chart is preferred over the pie chart because differences are represented by lengths of the bars in the bar chart. Differences among categories in a pie chart are reflected by angles, and it seems that humans are much better at judging lengths than angles.

Bar chart with counts

Figure 2. A simple bar chart

myCombo <- seq(0,10, by=1) 
myCounts <- choose(10, myCombo)    #combinations
barplot(myCounts, names.arg = myCombo, xlab = "Number correct", ylab = "Count",col = "darkblue")

 A stacked bar chart is used to 

myCombo <- seq(0,10, by=1) 
myCounts <- choose(10, myCombo)    #combinations
barplot(myCounts, names.arg = myCombo, xlab = "Number correct", ylab = "Count",col = "darkblue")

Figure 3. The luxury ship RMS Titanic, which sunk 15 April 1912, More than 1500 souls were lost. Public domain image, Wikipedia. Click image to view full sized image.

Stacked bar chart, data set TitanicSurvival in package carData.

Figure 4. A stacked bar chart, survival Titanic

Barplot(passengerClass, by=survived, style="divided", legend.pos="above", xlab="passengerClass", ylab="Frequency")

Bar charts with error bars

Although many data visualization specialists argue against the bar chart — they tend to obscure data — their use is well established. For the familiar bar chart with ratio scale data, the X (horizontal) axis displays the categories of one variable (e.g., location, or treatment group). You plot groups to emphasize comparisons. The Y (vertical) axis then is the mean for each group.

You need error bars with your bar chart to convey variability. A bar chart plus error bars is sometimes called (pejoratively) a dynamite plot. If the mean is displayed, some measure of precision should be (must be?) displayed (Cumming et al 2007). And, as you should recall by now, your choices are standard deviation (Chapter 3.2), standard error of the mean (SEM) (Chapters 3.2, 3.4), or confidence interval (see Chapter 3.4). It is strongly advised, that without a representation of precision, one should not interpret trends or group differences from representations of means (i.e., height of bars) alone.

The bar charts on this page are means plus or minus the standard error of the mean, + SEM. I’m not advocating here for the bar chart — box plots with efforts to show the actual data are better. However, bar charts remain common and may be more familiar to certain audiences. We’ll discuss which choice to make.

Examples

The Copper_rats_PMID3357063 dataset will be used for the next series of graphs (Data set). Refer to Mike’s Workbook for Biostatistics Part07 to review how to import the data.

A portion of the data set is shown below

head(Copper)
         Diet     Body    Heart    Liver
1 Adequate-Cu 320.1381 1.125037 10.259657
2 Adequate-Cu 329.6879 1.158982 9.843295
3 Adequate-Cu 327.9838 1.090374 9.855975
4 Adequate-Cu 334.6669 1.118183 9.942997
5 Adequate-Cu 338.3134 1.172636 9.860971
6 Adequate-Cu 345.4608 1.056183 8.885820

The data set consists of organ weights (heart, liver) from rats fed a diet adequate in copper, deficient in copper, and then a third group who received the adequate diet from perspective of amount of copper, but calorie restricted to match the decreased feeding rates of the rats fed the copper deficient diet. Copper is an essential trace element in our diet. The data set was simulated from descriptive statistics (means, standard deviation, number of subjects) of published data by sampling from a normal distribution. (Table 1, Ovecka et al 1988).

On we go with some graphs.

But, I’m still learning about R graphics, and for bar charts with error bars, I find other packages more straight-forward. So, I’ll use this moment to point out that making a good graph is more about the end product then the particular tools. I have been using other tools for years to make my graphics, so I tend to default on these options first. Graphs presented here are mostly from Veusz software program (pronounced “views”). I like it because the software allows me to edit any of the elements of the graph.

Here’s a typical looking bar chart with ratio scale data: means ± SEM. Let’s look at them more critically.

Figure 5. A bar chart with error bars (standard error of the mean).

Figure 6. Another bar chart. with standard errors of mean

When making comparisons, make sure the axes have the same scale, or consider putting the graphs together.

Figure 7. Bar chart that allows for a comparison among levels of a a factor (organs, liver vs. heart).

Here’s the organ weights again, but taken as ratio of body mass.

Figure 8. Same chart as in Figure 6, but on ratios.

How to make a bar chart with error bars in R

Option 1. First, let’s try a work-around. Instead of an error bar option for bar chart menu, Rcmdr provides a plot of means that allows you to plot with error bars. These are equivalent graphs, the “bar chart” and the “plot of means”, though you should favor the bar chart format for publishing.

Rcmdr: Graphs → Plot of means…

Figure 9. Rcmdr menu popup for Plots Means

Here, Rcmdr takes the data and calculates the mean and your choice of standard errors or deviations, confidence intervals, or no error bars. the resulting graph is below.

Figure 10. Plot of means

That’s an ugly graph (Fig. 10). Functional, good enough for data exploration and preliminary results, and certainly good enough for a Biostatistics homework or report. Additionally, connecting the dots here is a no-no. It implies that if we had measured categories between “adequate copper” and “deficient copper,” then the points would fall on those lines. That would be a complete guess. So, why did I include the connecting lines? That was the default setting for the command, and it makes the point — think before you click. One argument for connecting points in a graph is that it makes it easier for the reader to visualize trends.

This graph (Fig. 10) is fine for exploring data, but you will want to do better for publication.

Let’s make some better graphs with R

Once you are ready to go beyond the default settings available in Rcmdr, there is tremendous functionality in R for graphics. To access R’s potential, you’ll need to get into the commands a bit. I’m going to continue to try and shield you from the programming aspects of R, but from time to time you really need to see what is possible with R. Graphics is one such area. I use the package gplots, with 23 different graphing functions (type at R prompt ?gplots to call up the manual pages).

gplots should be among the packages on your R installation; if not, then install the package and run library(gplots) to complete the installation. We’ll try the barchart2 function.

But first, we need to get means for each of our groups.

At the R command prompt:

hrtWt <- tapply(Dataset$HeartWt, list(Group=Dataset$Group), mean, na.rm=TRUE) 

This code extracts means from our HeartWt variable for each Group, then  stores the three (in this case, because our data set has 3 groups) in the place holder I had called hrtWt. To verify that the three means are there, type “hrtWt” without the quotes then enter.

You should see

hrtWt Group Cu adequate Cu deficient Pair-fed 
               1.200000     1.566667 0.900000

R functions used: tapply, list, mean; na.rm was not needed but would be used to remove all missing values (recall during our import phase we were asked how missing observations were noted in our file; the default is NA).

Next, I want to apply standard error bars

stdDEV <- tapply(Dataset$HeartWt, list(Group=Dataset$Group), sd, na.rm=TRUE)
cil <- hrtWt-(stdDEV/sqrt(3))
ciu <- hrtWt+(stdDEV/sqrt(3))

I used cil and ciu to designate the lower cil and upper ciu values for my ± SEM (standard error of the mean). ciu stands for “confidence interval lower;” ciu stands for “confidence interval upper.”

Finally, here’s the plot command

barplot2(hrtWt, beside = TRUE, main=c("Mice fed different amounts of copper in diet"), col = c("blue", "red", "green"), xlab="Copper in diet", ylab="Heart mass (g)", ylim = c(0, 2), plot.ci=TRUE, ci.l=cil, ci.u=ciu)

Now, draw a box around the graphic

box()

Whew!

What does your new graph look like? My graph is below (Fig. 11).

Figure 11. A bar chart using barplot2.

This works and the point is that once the script is written it easy to make small changes as you need in the future to make nice graphs.

If you are impatient like me, I like a GUI option, at least to start crafting the graph. The Rcmdr plugin KMggplot2 provides a good set of tools to make bar charts with error bars. An even better option I think is to use a software package that is designed for graphics, at least simple graphics like a bar chart. I use SciDAVis and Veusz for simple graphs like pie charts and bar charts; much easier to control.

ggplot2 bar charts with error bars

Nevertheless, here’s how to make a bar chart with error bars using ggplot2 (Fig. 11). First, we need to create a statistics summary. The script printed here was modified from scripts at R Graph Cookbook website.

require(plyr)
summarySE <- function(data=NULL, measurevar, groupvars=NULL, na.rm=FALSE,
     conf.interval=.95, .drop=TRUE) {
length2 <- function (x, na.rm=FALSE) {
     if (na.rm) sum(!is.na(x))
     else length(x)
}

#returns a vector with N, mean, and sd
datac <- ddply(data,groupvars, .drop=.drop,
    .fun = function(xx,col){
      c(N=length2(xx[[col]],na.rm=na.rm),
      mean=mean(xx[[col]],na.rm=na.rm),
      sd=sd(xx[[col]],na.rm=na.rm)
     )
    },
    measurevar
   )

#Rename the "mean" column
datac <- rename(datac,c("mean"=measurevar))

#Calculate the standard error of the mean
datac$se <- datac$sd/sqrt(datac$N)

#Get confidence interval
ciMult <- qt(conf.interval/2 + .5, datac$N-1)
datac$ci <- datac$se*ciMult

return(datac)
}

Applying this function to the BMI dataset yields the following output.

sumBMI <- summarySE(BMI, measurevar="BMI", groupvars=c("Sex", "Smoke"));sumBMI

   Sex Smoke   N      BMI       sd        se       ci 
1    F    No  23 26.83567 7.610271 1.5868511 3.290928 
2    F   Yes  14 25.76133 4.658625 1.2450698 2.689810 
3    M    No  10 26.35731 3.363575 1.0636557 2.406156 
4    M   Yes  27 26.71879 4.675631 0.8998256 1.849618

Now we are ready to make the bar chart with error bars

ggplot(tgc,aes(x=Smoke,y=BMI,fill=Sex)) + 
 geom_bar(position=position_dodge(),stat="identity",color="black") + 
 geom_errorbar(aes(ymin=BMI,ymax=BMI+se),width=.2,position=position_dodge(.9))

Figure 12. A barchart from ggplot2

Questions

1. Why should you use box plots and not bar charts to display comparisons for a ratio scale variable between categories? Obtain a copy of the article by Streit and Gehlenbor 2014 — it’s free! After reading, summarize the pro and cons for box plots over bar charts with error bars.

2. Enter the following data into R. The data are sulfate levels in water, parts per million. 

type = c("Palolo Stream","Chaminade tap water", "Aquafina","Dasani") 
sulfateppm =c(11, 14, 5, 12)
try = data.frame(type,sulfateppm) 
byWater = tapply(try$sulfateppm,list(Group=try$type),mean)

Make a simple bar chart using the boxplot2 function in gplots package. 

3. Change the range of values on the vertical axis to 0, 20

4. Change the color of the bars from gray to blue

5. Add a label to the vertical axis, “Sulfates, ppm” (without the quotes)

6. Add a box around the graph.

Quiz – Chapter 4.1

Bar charts

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Data set

Chapter 4 contents

• Graphs and tables
• Bar (column) charts
• Histograms
• Box plot
• Mosaic plots
• Scatter plots
• Adding a second Y axis
• Q-Q plot
• Ternary plots
• Heat maps
• Graph software
• References and suggested readings

3.2 – Measures of Central Tendency

Introduction to summary statistics.

For a sample of observations we can begin the summary by identifying the “typical” value. Various statistics are used to describe the middle and collectively these are referred to as measures of central tendency. The mean, the median, and the mode are the most common measures of central tendency. In the situation in which we work with data from a population census, we would calculate population descriptive statistics — not inferential statistics; because we more often work with samples from populations, we report sample descriptive statistics.

First we review the familiar arithmetic mean and introduce the weighted mean. Next we introduce “other means,” which you may not be as familiar with.

Means.

There are several means beyond the simple arithmetic average or mean. Here we review a few. We use this topic also to start introducing standard notation we will use throughout the book.

The population mean, μ (pronounced “mu”) is given by

X_i, an observation on the i-th individual

Σ, or “sigma,” which instructs you to add them up (the X’s), from i = 1 (the first observation) to i = N (the last observation).

The sample mean, , pronounced “eks bar,” of a collection of observations is given by 

Weighted arithmetic mean.

In some cases you may have several samples from the same population. If the sample sizes are the same, you can calculate the average of averages without any fuss — just take all of the sample means and add them up, then divide by the total number of samples. If the sample sizes differ, then you needs to weight (W) each sample mean by its sample size. Simply divide each sample mean by its appropriate sample size, then add all of these up. That is the weighted average.

More generally, we can write

For example, consider a variable containing the following observations, for example, over the course of watching birds gathering at a drinking source (eg, recorded over 20 days, a single 5 minute interval per day), we can count how many times zero, one, two, etc birds are observed. 

Table 1. A sample of counts of observations, eg, how many birds

The observation “4” was observed four times; the observation “5” was observed twice, and so on for a total of 15 observations.

What is the arithmetic mean of these 15 observations? To solve this, well, you have a couple of choices. You could copy the numbers down as often as they appear and then calculate the mean in the usual way.

Other “means”.

The arithmetic average (illustrated above) is not the only way to estimate the mean.

The trimmed mean, also called the truncated mean, is a useful approach when data is widely dispersed — data spread away from the middle (see Chapter 3.3). Thus, the trimmed mean will be less influenced compared to the arithmetic mean by outlier data points, ie, data far from other data points in the set.

You would use the trimmed mean to describe the middle of a data set in which a plot shows most of the values are clumped together around a middle – and yet you see a few values that are much smaller or much greater. A specified percentage of the smallest and largest values are removed from the data set and then the simple arithmetic mean is calculated for the trimmed data set. For example, given a data set of daily rainfall for different cities, you might wish to remove the driest 5% and wettest 5% of the days in order to better compare the rainfall trends for the cities.

Calculating the trimmed mean is straight-forward in R: use the same built-in function, mean(), but add some options.

Note 2: This is a good time to remind you how to get help with R commands. Do you recall how to get help in R? 

At the R prompt type

help(mean) 

or

?mean

The R Documentation page for mean() will pop up (assuming you allowed R to install help pages as html). Figure 1 shows a screenshot of a portion of the help page for mean()

Figure 1. A portion of the R help page about the function mean.

From the help page (Fig 1) we can see that we can specify a trimmed mean by adding options to the mean(x, ...) command. For our x variable defined above, get the trimmed mean after 25% of the data are removed.

mean(x, trim=0.25)

[1] 6

From the help page we see that the only required option you need to feed the mean command is the name of the variable, in this case, “x” (it can be, of course, any name provided the data are attached). In this case we removed 25% of the values – 12.5% of the smallest values and 12.5% of the largest values – that’s also called the interquartile mean. We use a dot plot to show these trends (Fig 2).

Figure 2. Dot plot of our x variable with locations of the mean (blue) and the trimmed mean (red).  The Dotplot(x) function in package RcmdrMisc was used in Rcmdr to make this graphic. Arrows were added by hand. Dotplot() example code presented in Chapter 3.4.

If we recalculate a trimmed mean after dropping 10% of the points, or even 40% of the points, we get the same mean value of 6. The trimmed mean is an example of a robust estimator; it’s resistant to the influence of outliers.

Another useful descriptor of the middle is the geometric mean. The geometric mean is useful for calculating the average of ratios.  Geometric mean would be used when you want to compare central tendency for different variables, each differing in scale. For example, gene expression results, reported as fold-changes, for different genes often shows tremendous differences among genes and would be best described by logarithmic scale, not arithmetic scale. Geometric mean expression values would be better choice for central tendency. Other examples are found in economics: for example, calculating compound interest or interest. The geometric mean applies whenever the scale is multiplicative and not additive

The geometric mean is given by the equation

The geometric mean (gm) is equivalent to log-transforming your data, then calculating the arithmetic mean, and transforming the result back (with the antilog exponent.) As you recall, for our simple data set the arithmetic mean was 6.2. The geometric mean for this data was 5.977. Taking the natural log for each of the values from our simple data set, then calculating the arithmetic mean we have 1.788. 

The antilog of this value is

exp(1.788)

[1] 5.977

Another frequently encountered mean is the harmonic mean, which is defined by the equation

Harmonic mean is appropriate for averaging rates. For example, what is the average speed traveled if you travel 30 miles per hour (mph) between point A and B, then on the return trip, your speed was 40 mph? If you think (30 + 40)/2 = 35 mph, then this would be incorrect — after all, the distance covered has not changed, just the time. The harmonic mean returns 34.2 mph. Let

y = c(30,40)

The harmonic mean returns 34.2 mph ( see below “How to calculate these other means”)

Both harmonic and geometric means apply for values greater than zero.

How to calculate these “other” means.

In Microsoft Excel, calculate geometric mean via the function GEOMEAN(); calculate harmonic mean via the function HARMEAN().

The base R (and Rcmdr) doesn’t have built in functions for these, although you could download and install some R packages which do (eg, package psych, geometric.mean(variable), harmonic.mean(variable)). It is quicker to just to calculate these by submitting a snippet of code into the script window

For geometric mean of variable “x” at the R prompt type

exp(mean(log(x)))

For harmonic mean of variable “x” at the R prompt type

1/mean(1/x)

where is the base of the natural logarithm, Euler’s number, and log is the natural logarithm (in R, to get log to other bases you can use log10 for base 10 logarithm or log2 for base 2 logarithm, or  log(x, base = n) for any base n of the variable x, and variable is the name of the variable you wish to do the calculations on.

R code: Do try on your own!

Here’s some numbers to try your hand. For example, create a variable containing a few numbers, any numbers. and write it to the variable named z

myZ = c(3,4,6,7,9,11,4)

Now, calculate the arithmetic mean, the geometric mean, and the harmonic mean for the variable myZ. You should get Table 4.2)}

Table 2. Comparison of different means for myZ.

Try three more. In R (or R Commander script window), create three new variables.

varA = c(3,3,3,3)

varB = c(1,2,3,12)

varC = c(-3,0,1,3)

Now, calculate the arithmetic mean, geometric mean, and harmonic mean for each variable.

For the simple arithmetic mean

mean(varA)

For the geometric mean, use the formula above

exp(mean(log(varA)))

For the harmonic mean, use the formula above

1/mean(1/varA)

What did you get?

Other measures of central tendency

Median 

The median, which lacks an accepted notation — we’ll go with , divides a set of observed numbers into two equal halves. Half the observations are above the median, half of the observations are below the median. Arrange data from lowest to highest, take the middle measurement:

odd number of measurements minus the middle value minus even number of measurements minus average of the 2 middle values

Or, more succinctly, we have

To get the median in R type at the R prompt

median(variable)

and of course, replace variable with the name of the variable containing the numbers. For our x variable created earlier, the function median returns in R

[1] 6

[Note that the median for x was the same as the trimmed mean for x, which is consistent with with our view that the trimmed mean is a robust estimator of the middle of a data set.

Mode

Mode is another way to express the middle and it refers to the most frequent occurring measurement. Use of mode makes most sense for discrete or countable numbers. For a normal distribution, the mean, median and mode will be the same value. Note that a data set may have more than one mode. For example, what is the mode for the variable  we created earlier?

myX = c(4,4,4,4,5,5,6,6,6,7,7,8,8,8,15)

For this small data set we see that “4” is the most frequent with a count of four occurrences in the set.

Mode would seem like a straightforward function in R. However, it turns out there is not a mode function in the base package.

A little explanation is in order. In R, typing mode at the R prompt like so

mode(myX)

returns

[1] "numeric" 

Not the answer we were expecting. In R, mode command is used to tell you what the mode (ie, way or manner in which some task is accomplished) of storage is for the variables.

In order to get the statistical mode we want, we either hunt down a package that contains mode estimation (eg, install the package DescTools use the Mode function — see below), or we can write a little code.

A quick Google search found a number of answers at stackoverflow.com (eg, question 2547402). The simplest response was to use names and max commands like so

temp = table(as.vector(myX))

names (temp)[temp==max(temp)]

Why the median may be a better middle than the mean

Comparing the two measures of central tendency can tell you without plotting how your data are distributed about the middle. Sample distributions are discussed in Chapter 6. 

• When the distribution of the data is symmetric or normally distributed (discussed in Chapter 6.7) then the mean and the median will be about the same value
• When data are right-skewed (a few large values), then the mean will be greater than the median.
• When the data are left-skewed (a few small values), then the mean will be less than the median.

Here’s an illustration (Fig 3). I sampled 100 points from a random normal distribution with mean zero and standard deviation 0ne and another 100 points from a log-normal distribution also with mean zero and standard deviation one. In Figure 3, the histograms (see Chapter 4.2 – Histograms), along with summary statistics (see Note 4). Means are indicated with red arrows and medians are indicated with blue arrows.

Figure 3. Normal and lognormal distributions with mean (red) and median (blue) noted for comparison.

So the median is a better descriptor or the central tendency of a sample distribution when the distribution is NOT normally distributed.

Scaling and transformation of data

Sometimes it is useful to standardize your data so that the variables all have the same scale. One algorithm for standardization is called normalization. Normalization implies that you correct the data so that data has a mean, μ, of zero, and a standard deviation, σ, of 1 (unit variance). There are several ways to standardize, each with strengths and limitations. To normalize we use the Z-score equation (see Chapter 6.7 for other uses of Z score).

where X_i is each observation in your data set.

Normalization will make outliers, the few points in a data set that are noticeably different from the central tendency of the rest of the data, smaller and less influential. When you normalize multiple sets of data, then each will have the same mean (0) and variance (unit variance), but the ranges will differ. An example of this is the simple product moment correlation — by standardizing you change the variances for the different variables to have the same unit variance.

As we will see later in class it is also useful to expand or contract the variability of the data or to change the shape of the distribution (if the data is not normally distributed). For example, if you compare individuals of a population for many morphological traits (eg body size, growth rate), the spread of points (called a distribution) will look more like a Poisson distribution (not symmetrical about the mean, a few individuals may be much larger…). This is partly due to the way in which morphological traits are measured. We normally measure body size on a linear scale (inches or centimeters). However, body size is affected by physiological processes that are more related to volume. Therefore, the more appropriate scale of measurement is on a log scale. We can transform the data measured on a linear scale to a log scale. For morphological traits this can produce a distribution that is normally distributed (bell shape). There are many more statistical procedures for data that is normally distributed than there are statistical procedures for Poisson distributions or any other type of distribution. Additional discussion about data transformation is introduced in Chapter 13.3.

You can always uncode or unstandardize your data after performing the statistical procedures and return to the original scale. In fact, when reporting descriptive statistics you should report the untransformed, uncoded data. Moreover, you will find it useful to report means adjusted for other variables (eg, from ANOVA or regression); if the ANOVA or regression equations are performed on transformed or coded data you would want to back calculate to the original scale after applying the ANOVA or regression adjustments. This advise will make more sense after we’ve discussed ANOVA (Chapter 12) and linear regression (Chapter 17).

R operators

The names command can be used to retrieve the names contained in the variable (if text types) or to set the names of the observations, which is what we are using it for here. We set the numbers to text names “4”, “5”, etc. then find the maximum count of named items in the temp table. The double equals operator (==) is used to tell R to find the object that is “equal to” something we specify, in this case, the max value (R Language Definition 2014). Table 3 shows common operators in R.

Table 3. Common arithmetic* and comparison** operators

help(Arithmetic)

help(Comparison)

The R package modeest has a number of algorithms for calculating the mode, depending on the kind of data you are working with. After installing the package and its dependencies, type at the R prompt 

require(modeest)
mfv(myX)
[1] 4

Creating objects in R

Everything in R is an object (Chambers 2008). We provide many worked examples in Part 07. Working with your own data – Mike’s Workbook for Biostatistics. In brief, create a variable in R by assigning the vector x, either directly at the R prompt or in a script window (Rcmdr, RStudio), like so

myX = c(4,4,4,4,5,5,6,6,6,7,7,8,8,8,15)

The function c(), which stands for combine  is used to combine the set of numbers into the object, myX. For small sets like this you may find it convenient to enter the values one by one and let R store it into the vector for you. For R installed on your computer, use the scan() function and your keyboard — scan() doesn’t work consistently in serverless R instances (eg., Google CoLab). Careful! Make sure that you remember to assign the results from scan to a vector.

I’ll create the object tryScan just to distinguish it from x, although I will enter the same values. Until R receives an interrupt signal from you, it will prompt you to enter numbers one row at a time. When you’ve reached the end, use the keyboard shortcut Ctrl + q (Windows) or Cmd + q (macOS) to interrupt keyboard input. 

tryScan = scan()
1: 4
2: 4
3: 4
4: 4
5: 5
6: 5
7: 6
8: 6
9: 6
10: 7
11: 7
12: 8
13: 8
14: 8
15: 15
Read 15 items

tryScan
 [1] 4 4 4 4 5 5 6 6 6 7 7 8 8 8 15

The function tryScan is a very useful command, with many options, and can be used for more than keyboard entry. For example, you can paste from your computer’s clipboard a column of numbers from your spreadsheet.

R code calculate central tendency

Once we have the vectormyX, calculate the mean by entering at the R prompt

mean(myX)

and you should get the answer of 6.466667

And of course, you don’t type in the R prompt >, right?

Or, for the better option, create two variables, one containing the list of observed numbers and the second that contains the frequency for each observed number in the series. You would then use the command for weighted mean.

myY = c(4,5,6,7,8,15)

myW = c(4/15, 1/15, 3/15, 2/15, 4/15, 1/15)

Note — you can check that the frequencies sum to 1 by using the sum command like so

sum(myW)

For the weighted mean, the command is

weighted.mean(myY,myW)

and, the answer returned is 6.466667, the same as before.

There’s an R package for that.

We list code examples for all of the statistics described on this page, but as you can imagine, there are packages that include these and other useful functions, or “wrapper functions” that make writing code easier in R. A good example is the DescTools package. One advantage of using a package like DescTools is that it cuts through the need to remember specific commands for basic data exploration. For example, just call the Desc() function to yield several measures of central tendency and dispersion (see Chapter 3.3). 

The R code and output follows, along with a plot (Fig 4). The mean and median are highlighted in orange.

Desc(myX)
──────────────────────────────────────────────────────────────────────────────── 
myX (numeric)

length      n   NAs unique   0s  mean meanCI'
    15     15     0      6    0  6.47    4.92
       100.0%  0.0%        0.0%          8.02

   .05    .10   .25 median  .75   .90     .95
  4.00   4.00  4.50   6.00 7.50  8.00   10.10

range      sd vcoef    mad  IQR  skew    kurt
11.00    2.80  0.43   2.97 3.00  1.72    2.94

  value freq   perc cumfreq cumperc
1     4    4  26.7%       4   26.7%
2     5    2  13.3%       6   40.0%
3     6    3  20.0%       9   60.0%
4     7    2  13.3%      11   73.3%
5     8    3  20.0%      14   93.3%
6    15    1   6.7%      15  100.0%

' 95%-CI (classic)

Figure 4. Histogram, box plot, and cumulative distribution function plot generated by default by Desc() call.

Lots of output! This is a good example of how to explore data — describe the middle (eg, mean and median), but also the variability about the middle (eg, range, standard deviation — sd, mean absolute deviation — mad, coefficient of variation — vcoef), and distribution shape (quantiles — 0.05, 0.25 etc, kurt, skew). The plots — histogram, box plot, and cumulative distribution plots — are explained in Chapter 4.

Use aggregate() to calculate summary statistics by group.

So far, all code presumes need to calculate descriptive statistics one vector at a time. In general, we need summary statistics for more than one group or strata. For example, we may ask, what are the average body temperature readings measured by IR device on different body areas from several subjects (question 8a, Homework 2A: Descriptive statistics in Mike’s Workbook for Biostatistics).

The aggregate() function in R simplifies calculating means by groups by providing a concise and efficient way to apply a summary function (like mean()) to a variable, grouped by one or more other variables. The format for aggregate is

 aggregate(x = variable_to_aggregate ~ grouping_variable, FUN = mean, data = your_dataframe)

Questio8a had two levels of body regions, “Forehead” and “Throat.” Assuming the variable is named bodyTemp and the grouping variable is called bodyGroup and stored in the attached data.frame question08a, then the following R code provides the requested summary statistics by groups.

aggregate(bodyTemp ~ bodyGroup, data = question08a, 
function(x) c(mean = mean(x)))

Additional statistics can be requested by adding one after the other following mean(x), eg, median = median(x).

The advantage of use of R Commander, it’s easy to get multiple statistics by groups by following the drop down menus, beginning with Statistics > Summaries > Table of statistics… (Fig 5).

Figure 5. Screenshot R Commander, summary statistics by group.

Questions

1. Find the help page in R for the median function. How does the function handle missing values?
2. For a simple data set like the following myY <- c(1,1,3,6) you should now be able to calculate, by hand, the
   • mean • median • mode
3. If the observations for a ratio scale variable are normally (symmetrically) distributed, which statistic of central tendency is best (eg, less sensitive to outlier values)?
4. In the names() command, what do you think the result will be if you replace max in the command with min?
5. If data are right skewed, what will be the order of the mean, median, and mode?
6. Body mass of Rhinella marina (formerly Bufo marinus, Fig 6),
   bufo <- c(71.3, 71.4, 74.1, 85.4, 85.4, 86.6, 97.4, 99.6, 107, 115.7, 135.7, 156.2)

Figure 6. Female Rhinella marina (formerly Bufo marinus), Chaminade University campus. Body length 23.5 cm. Click image to view full sized image.

7. Under what circumstances is it appropriate to transform raw data?
8. What does “standardize” or “normalize” imply as actions to take with data?

Quiz Chapter 3.2

Measures of Central Tendency

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Chapter 3 contents

• Exploring data
• Data types
• Measures of Central Tendency
• Measures of dispersion
• Estimating parameters
• Statistics of error
• References and suggested readings