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)

14.7 – Rcmdr Multiway ANOVA

Introduction

We have been talking about the two-way randomized, balanced, replicated design.  Here, we take you step by step through use of R to conduct the multiway ANOVA.

R code: Multiway ANOVA

Rcmdr: Statistics → Means → Multiway ANOVA… we will review this as Option 1

or

Rcmdr: Statistics → Fit Models → Linear model… we will review this as Option 2

In either case, as a reminder, your data set must be stacked worksheet, like the data in this worksheet

Table 1. Data set, example.14.7†

Option 1

Your first option is to use the ANOVA menus via “Means.” This is a perfectly good way to handle a standard two-way, fully-crossed, fixed effects model.  However, other designs will not run with this command and R will return a report of errors for ANOVA models that do not conform to the replicated, balanced, crossed design.

Rcmdr: Statistics → Means → Multiway Analysis of variance …

Factors: highlight “Diet” AND “Population”

Response variable: pick one (in this window, all we see is “Response”)

Figure 1. Screenshot Rcmdr multi-way ANOVA.

Interpret the output

AnovaModel.2 <- (lm(Response ~ Diet*Population, data=example.14.7))

Anova(AnovaModel.2)
 
Anova Table (Type II tests)
Response: Response
                  Sum Sq    Df    F value      Pr(>F) 
Diet              36.750     1    12.2500    0.008079 **
Population        10.083     1     3.3611    0.104104 
Diet:Population   52.083     1    17.3611    0.003136 **
Residuals         24.000     8
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
tapply(example.14.7 Response, list(Diet=example.14.7 Diet, 
+ Population=example.14.7 Population), mean, na.rm=TRUE) # means
Population
Diet 1 2
A 5.00000 7.333333
B 12.66667 6.666667

tapply(example.14.7 Response, list(Diet=example.14.7 Diet, 
+ Population=example.14.7 Population), sd, na.rm=TRUE) # std. deviations
Population
Diet 1 2
A 1.000000 2.081666
B 2.081666 1.527525

tapply(example.14.7 Response, list(Diet=example.14.71 Diet, 
+ Population=example.14.7 Population), function(x) sum(!is.na(x))) # counts
Population
Diet 1 2
A 3 3
B 3 3

End R output.

Summary of multi-way ANOVA command

The multi-way ANOVA command returns our ANOVA table plus the adjusted means, along with standard deviations and number of observations (counts).  The adjusted means would then be good to put into a chart to present group comparisons following adjustments from the effects of levels within groups.

Rcmdr: Models → Graphs → Predictor effect plots …

Here’s the chart (hint: )

Figure 2. Predictor effect plots, Diet and Population on Response variable.

Option 2

A more general approach is to use the General linear model. This approach can handle the standard 2-way fixed effects ANOVA (above), but any other model as well. The model is Response ~ Diet*Population.

Rcmdr: Statistics → Fit Models → Linear model… 

Figure 3. Screenshot Rcmdr linear model menu.

Interpret the output

LinearModel.1 <- lm(Response ~ Diet * Population, data=example.14.7)

summary(LinearModel.1)

Call:
lm(formula = Response ~ Diet * Population, data = example.14.7)

Residuals:
Min 1Q Median 3Q Max 
-2.3333 -1.1667 0.1667 1.0833 2.3333 

Coefficients:
                        Estimate    Std. Error ..t value .  Pr(>|t|) 
(Intercept) .              5.000         1.000 .   5.000    .0.00105 ** 
Diet[T.B] .                7.667       ..1.414 .   5.421    .0.00063 ***
Population[T.2]           .2.333       ..1.414 .   1.650    .0.13757 
Diet[T.B]:Population[T.2] -8.333       ..2.000    -4.167    .0.00314 ** 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 1.732 on 8 degrees of freedom
Multiple R-squared: 0.8047, Adjusted R-squared: 0.7315 
F-statistic: 10.99 on 3 and 8 DF, p-value: 0.003285 

End R output

Lots to sort through, so let’s begin with what is in common between the two approaches, the Multi-way ANOVA command versus the linear model command.

Compare R output from ANOVA and linear model

As a direct output, the linear model option does not provide an ANOVA summary table. Instead of our ANOVA table the linear model returns estimates of coefficients along with t-test results for each coefficient of the model from the lm() command output

Recall that we can get ANOVA tables through the following R commands via Rcmdr.

Rcmdr: Models → Hypothesis tests → ANOVA Table.

Let’s do so for this linear model (accept the default for type of tests = “Type II”).

And the output from lm() stored in the object LinearModel.1 is

Anova(LinearModel.1, type="II")
Anova Table (Type II tests)

Response: Response
                Sum Sq Df F value   Pr(>F)   
Diet            36.750  1 12.2500 0.008079 **
Population      10.083  1  3.3611 0.104104   
Diet:Population 52.083  1 17.3611 0.003136 **
Residuals       24.000  8                    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End R output

Now, we’re in business, and, using the lm() function, we have the estimates for each model coefficient plus our ANOVA table.

Same answer! Of course. Which to choose, Option 1 or Option 2? Use the lm() options; more flexible, covers more designs than the multiway ANOVA which is strictly for the crossed fully replicated design.

Questions

1. Write out the two-way model described for the data in Table 1.
2. Write the null hypotheses and provide a summary of the statistical significance of the model.

Quiz Chapter 14.7

Rcmdr Multiway ANOVA

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Chapter 14 contents

• Introduction
• Crossed, balanced, fully replicated
• Sources of variation
• Fixed effects, Random effects
• Randomized block design
• Nested designs
• Some other ANOVA designs
• Rcmdr Multiway ANOVA
• More on the linear model in Rcmdr
• References and suggested readings

9.2 – Chi-square contingency tables

Introduction

This page covers a lot of ground to extend our discussion of implementations of the chi-square test to experimental data and introduces several related statistics, including a correlation coefficient for categorical data and an introduction to agreement and reliability measurement statistics.

We begin with contingency tables.

We just completed a discussion about goodness of fit tests, inferences on categorical traits for which a theoretical distribution or expectation is available. We offered Mendelian ratios as an example in which theory provides clear-cut expectations for the distribution of phenotypes in the F₂ offspring generation. This is an extrinsic model — theory external to the study guides in the calculation of expected values — and it reflects a common analytical task in epidemiology.

But for many other kinds of categorical outcomes, no theory is available. Today, most of us accept the link between tobacco cigarette smoking and lung cancer. The evidence for this link was reported in many studies, some better designed than others.

Consider the famous Doll and Hill (1950) report on smoking and lung cancer. They reported (Table IV Doll and Hill 1950), for men with lung cancer, only two out of 649 were non-smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were smokers (Table 1).

Table 1. Doll and Hill (1950).

A more recent example from the study of the efficacy of St John’s Wort (Hypericum perforatum) as a treatment for major depression (Shelton et al 2001, Apaydin et al. 2016). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve (Table 2).

Table 2. St. John’s Wort and depression

These kinds of problems are direct extensions of our risk analysis work in Chapter 7. Now, instead of simply describing the differences between case and control groups by Relative Risk Reduction or odds ratios, we instruct how to do inference on the risk analysis problems.

Thus, faced with absence of coherent theory as guide, the data themselves can be used (intrinsic model), and at least for now, we can employ the test.

In this lesson we will learn how to extend the analyses of categorical data to cases where we do not have a prior expectation — we use the data to generate the tests of hypotheses. This would be an intrinsic model. One of the most common two-variable categorical analysis involves 2X2 contingency tables. We may have one variable that is the treatment variable and the second variable is the outcome variable. Some examples provided in Table 3.

Table 3. Some examples of treatment and outcome variables suitable for contingency table analysis.

Table 4 were examples of published studies returned from a quick PubMed search; there are many examples (meta-analysis opportunities!). However, while they all can be analyzed by contingency tables analysis, they are not exactly the same. In some cases, the treatments are fixed effects, where the researcher selects the levels of the treatments (see Ch12.3 and Ch14.3 for discussion about fixed vs random effects in experimental design).

In other cases, each of these treatment variables need not be actual treatments (in the sense that an experiment was conducted), but it may be easier to think about these as types of experiments. These types of experiments can be distinguished by how the sampling from the reference population were conducted. Before we move on to our main purpose, to discuss how to calculate contingency tables, I wanted to provide some experimental design context by introducing two kinds of sampling (Chapter 5.5).

Our first kind of sampling scheme is called unrestricted sampling. In unrestricted sampling, you collect as many subjects (observations) as possible, then assign subjects to groups. A common approach would be to sample with a grand total in mind; for example, your grant is limited and so you only have enough money to make 1000 copies of your survey and you therefore approach 1000 people. If you categorize the subjects into just two categories (eg, Liberal, Conservative), then you have a binomial sample. If instead you classify the subjects a number of variables, eg, Liberal or Conservative, Education levels, income levels, homeowners or renters, etc., then this approach is called multinomial sampling. The point in either case is that you have just utilized a multinomial sampling approach. The aim is to classify your subjects to their appropriate groups after you have collected the sample.

Logically what must follow unrestricted sampling would be restricted sampling. Sampling would be conducted with one set of “marginal totals” fixed. Margins refers to either the row totals or to the column totals. The sampling scheme is referred to as compound multinomial. The important distinction from the other two types is that the number of individuals in one of the categories is fixed ahead of time. For example, in order to determine if smoking influences respiratory health, you approach as many people as possible to obtain 100 smokers.

The contingency table analysis

We introduced the 2×2 contingency table (Table 4) in Chapter 7.

Table 4. 2X2 contingency table

Contingency tables are all of the form (Table 5)

Table 5.  Basic format of a 2X2 table

Regardless of how the sampling occurred, the analysis of the contingency table is the same; mechanically, we have a chi-square type of problem. In both contingency table and the “goodness of fit” Chi-Square analyses the data types are discrete categories. The difference between GOF and contingency table problems is that we have no theory or external model to tell us about what to expect. Thus, we calculate expected values and the degrees of freedom differently in contingency table problems. To learn about how to perform contingency tables we will work through an example, first the long way, and then using a formula and the a, b, c, and d 2X2 table format. Of course, R can easily do contingency tables calculations for us.

Doctors noticed that a new form of Hepatitis (HG, now referred to as GB virus C), was common in some HIV+ populations (Xiang et al 2001). Accompanying the co-infection, doctors also observed an inverse relationship between HG loads and HIV viral loads: patients with high HG titers often had low HIV levels. Thus, the question was whether co-infection alters the outcome of patients with HIV — do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? I’ve represented the Xiang et al (2001) data in Table 6.

Table 6. Progression of AIDS for patients co-infected with HG [GB virus C], Xiang et al data (2001)

Our question was: Do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? A typical approach to analyze such data sets is to view the data as discrete categories and analyze with a contingency table. So we proceed.

Setup the table for analysis

Rules of contingency tables (see Kroonenberg and Verbeek 2018). What follows is a detailed walk through setting up and interpreting a 2X2 table. Note that I maintain our a, b, c, d cell order, with row 1 referencing subjects exposed or part of treatment group and row 2 referencing subjects not exposed (or part of the control group), as introduced in Chapter 7.

The Hepatitis G data from Xiang et al 2001 data, arranged in 2X2 format (Table 7).

Table 7.  Format of 2X2 table Xiang et al (2001) dataset.

The data are placed into the cells labeled a, b, c, and d.

Cell a : The number of HIV+ individuals infected with HG that lived beyond the end of the study.

Cell b : The number of HIV+ individuals infected with HG that died during the study.

Cell c : The number of HIV+ individuals not infected with HG that lived.

Cell d : The number of HIV+ individuals not infected with HG that died.

This is a contingency table because the probability of living or dying for these patients may have been contingent on coinfection with hepatitis G.

State the examined hypotheses

H_O: There is NO association between the probability of living and coinfection with hepatitis G.

For a contingency table this means that there should be the same proportion of individuals with and without hepatitis G that either lived or died (1:1:1:1).

Now, compute the Expected Values in the Four cells (a, b, c, d)

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total / Total = Expected Proportion of those that lived beyond the study.
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total / Total = Expected Proportion that died during the study.
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (step 1 above) x Row 1 Total = Expected For Cell A.
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died x Row 1 Total = Expected For Cell B
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived x Row 2 Total = Expected For Cell C
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died x Row 2 Total = Expected For Cell D

Yes, this can get a bit repetitive! But, we are now done — remember, we’re working through this to review how the intrinsic model is applied to obtain expected values.

To summarize what we’ve done so far, we now have

Observations arranged in a 2X2 table

Calculated the expected proportions, ie, the Expected values under the Null Hypothesis.

Now, we proceed to conduct the chi-square test of the null hypothesis — The observed values may differ from these expected values, meaning that the Null Hypothesis is False.

Table 8. Copy Table 6 data set

Get the Expected Values

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total (198) / Total (362) = 0.5470
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total (164) / Total (362) = 0.4530
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (0.547) x Row 1 Total (218) = Cell A = 119.25
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died (0.453) x Row 1 Total (218) = Cell B = 98.75.
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived (0.547) x Row 2 Total (144) = Cell C = 78.768
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died 0(.453) X Row 2 Total = Cell D = 65.232

Thus, we have (Table 9)

Table 9. Expected values for Xiang et al (2001) data set (Table 6 and Table 8)

Now we are ready to calculate the Chi-Square Value

Recall the formula for the chi-square test

Table 10. Worked contingency table for Xiang et al (2001) data set (Table 6 and Table 8)

Cell
a	(103 – 87.768)2 = 78.768	7.4547
b	(41 – 65.232)2 = 65.232	9.002
c	(95 – 119.25)2 = 119.25	4.93134
d	(123 – 98.75)2 = 98.75	5.95506
		27.3452

Adding all the parts we have the chi-square test statistic . To proceed with the inference, we test using the chi-square distribution.

Determine the Critical Value of test and evaluate the null hypothesis

Now recall that we can get the critical value of this test in one of two (related!) ways. One, we could run this through our statistical software and get the p-value, the probability of our result and the null hypothesis is true. Second, we look up the critical value from an appropriate statistical table. Here, we present option 2.

We need the Type I error rate α and calculate the degrees of freedom for the problem. By convention, we set α = 0.05. Degrees of freedom for contingency table is calculated as

Degrees of Freedom = (# rows – 1) x (# columns – 1)

and for our example → DF = (2 – 1) x (2 – 1) = 1.

Note. How did we get the 1 degree of freedom? Aren’t there 4 categories and shouldn’t we therefore have k – 1 = 3 df? We do have four categories, yes, but the categories are not independent. We need to take this into account when we evaluate the test and this lack of independence is accounted for by the loss of degrees of freedom.

What exactly is not independent about our study?

1. 1. The total number of observations is set in advance from the data collection.
   2. We also have the column and row totals set in advance. The number of HIV individuals with or without Hepatitis G infection was determined at the beginning of the experiment. The number of individuals that lived or died was also determined before we we conducted the test.
   3. Then the first cell (let us make that cell A) can still vary anywhere from 0 to N _i (sample size of the first drug).
   4. Once the first cell (A) is determined then the next cell in that row (B) will have to add up to N _i.
   5. Also the other cell in the same column as the first cell (C) must add up to the Column 1 Total.
   6. Lastly, the last cell (D) must have the Row 1 Total add up to the correct number.

All this translates into there being only one cell that is FREE TO VARY, hence, only one degree of freedom.

Get the Critical Value from a chi-square distribution table: For our example, look up the critical value for DF = 1, α = 0.05, you should get 3.841. The 3.841 is the value of the chi-square we would expect at 5% and the null hypothesis is in fact true condition in the population.

Once we obtain the critical value we simply use the previous statement regarding the probability of the Null Hypothesis being TRUE. We Reject the Null Hypothesis when the calculated test statistic is greater than the Critical Value. We Accept the Null Hypothesis when the calculated  is less than the Critical Value. In our example we got 27.3452 for our calculated test statistic, thus we reject the null hypothesis and conclude, that, at least for the sample in this study, there was an association between the probability of patients living past the study period and the presence of hepatitis G.

from a, b, c, and d 2X2 table format formula

If a hand calculation is required, a simpler formula to use is

where N is the sum of all observations in the table, r₁ and r₂ are the marginal row totals, and c₁ and c₂ are the marginal column totals from the 2×2 contingency table (eg, Table 4). If you are tempted to try this in a spreadsheet, and assuming your entries for a, b, c, d, and N look like Table 11, then a straightforward interpretation requires referencing five spreadsheet cells no less than 13 times! Not to mention a separate call to the distribution to calculate the p-value (one tailed test).

Table 11. Example spreadsheet with formulas for odds ratio (OR), Pearson’s , and p-value from distribution.

	A	B	C	D	E	F	G
1
2	a	103			OR	=(2*5)/(3*4)
3	b	41
4	c	95
5	d	123			chisq, p	=(((2*5-3*4)^2)*6)/ (SUM(2:3)*SUM(4:5)* SUM(2,4)*SUM(3,5))	=CHIDIST(F5,1)
6	N	632

Now, let’s see how easy this is to do in R.

effect size

Inference on hypotheses about association, there’s statistical significance — evaluate p-value compared to Type I error, eg, 5% — and then there’s biological importance, the strength of the association. The concept of effect size is an attempt to communicate the likely importance of a result. Several statistics are available to communicate effect size: for that’s φ, phi for the φ (phi) coefficient.

where N is the total. Phi for our example is

R code

sqrt(27.3452/362)

and R returns

[1] 0.274844

Effect size statistics typically range from 0 to 1; Cohen (1992) suggested the following interpretation

Table 12. Cohen’s interpretation of d values.

For our example, effect size of the association between co-infection with GB virus C and mortality was weak in the patients with HIV infection. The function phi() in psych package will calculate phi coefficient from a 2X2 table.

Agreement, kappa, and inter-rater reliability

I imagine by now you have made the leap — can we use the contingency table to assess agreement between (inter-) two judges (raters) who evaluate the same object? A naive approach would be to calculate simple agreement, the percent of objects in the list of items evaluated for which both judges agree (concur). For a binomial Yes or No, then agreement (concordance) occurs when both judges report “Yes,” or both report “No,” displayed on the main diagonal of the 2X2 table. The other possible combinations (Yes vs No and No vs Yes) are disagreement (discordance) and would be displayed on the off-diagonal of the 2X2 matrix.

Unfortunately, simple agreement, while intuitive, cannot discriminate between true agreement and agreement by chance. Cohen’s kappa (Cohen 1960, McHugh 2012) is used to assess agreement, aka inter-rater reliability, for binomial and nominal data types for two judges (raters). Cohen’s 1960 paper has been enormously impactful, with more than 53K citations (Google Scholar, Fall 2024), and spawned a number of additional measures of agreement, eg, weighted kappa for more than binomial levels, Fleiss kappa for three or more raters, ICC for ratio scale, discussed in Ch12.3). It’s beyond the scope of this text to review the context of such terms as agreement, and reliability, for just two terms used to evaluate results from two independent observers of the “same” event or object, and leave that to the reader to confirm how to report such endeavors.

Off we go. Cohen’s kappa is defined as

where p_o is the proportion in agreement and p_c is proportion of agreement expected by chance. Essentially, then, interpretable kappa values fall between 0 and 1 (0% to 100%); negative kappa estimates are possible.  Cohen (1960) noted that for kappa to equal 1, the off diagonal elements would all have to be zero.

Cohen provided an estimate of the standard error for kappa as

and 95% confidence intervals are then .

Here, we review calculation of simple agreement and Cohen’s kappa, with just the two levels. Both simple agreement and Cohen’s kappa (κ), are available in R via the icc package. Cohen’s kappa is also available in the psych package, with the cohen.kappa() function providing a confidence interval estimate directly.

Use of both is illustrated here with a simulated coin toss example modified from a nice 2018 summary of Cohen’s kappa at r-bloggers. The R code

library(irr)
# simulate some data
N = 24
flips1 <- rbinom(n = N, size = 1, prob = 0.5)
flips2 <- rbinom(n = N, size = 1, prob = 0.5)
coins<-cbind(flips1, flips2)

Confirm the data set.

head(coins)  
flips1 flips2 
[1,] 0 1 
[2,] 1 0 
[3,] 1 0 
[4,] 1 1 
[5,] 0 0 
[6,] 1 0

and create a 2X2 table.

I stacked the data

head(StackedCoin)
Side Coins    HT
1 0 flips1 tails
2 1 flips1 heads
3 1 flips1 heads
4 1 flips1 heads
5 0 flips1 tails
6 1 flips1 heads

then created a simple 2×2 table (see Note 6 below for a simple explanation for why “.Table” as a named object).

.Table <- xtabs(~Coins+HT, data=StackedCoin)

Table 13. coins as 2X2 table.

We can calculate the phi coefficient for Table 13.

phi(.Table, digits = 3)
[1] -0.167

and would conclude no correlation between the two sets of coin tosses.

Now, agreement statistics. Both agree() and kappa2() expect an unstacked (wide) data format. It’s not necessary, but we’ll call a value of 1 as “heads up” and a value of 0 as “tails up” of the tossed coin.

Next, simple agreement.

agree(coins, tolerance=0)

R returns

Subjects = 24 
Raters = 2 
%-agree = 41.7

Now, a proper agreement statistic.

kappa2(coins) 

and R returns

Subjects = 24 
Kappa = -0.135
z = -0.7 
p-value = 0.484

Results with psych:cohen.kappa, which provides the 95% confidence interval as default output.

Cohen Kappa and Weighted Kappa correlation coefficients and confidence boundaries 
                 lower estimate upper
unweighted kappa -0.51    -0.14  0.24
weighted kappa   -0.51    -0.14  0.24
Number of subjects = 24

By chance, the two independent toss of 24 coins had 10 rows of “agreement” (H:H or T:T), simply due to chance. However, results of kappa find no agreement between the two sets of coin tosses.

In summary, we see that simple agreement and the kappa statistic don’t return similar results. Percent agreement reported 41%, whereas kappa was zero. We know percent agreement can’t reflect anything in the coin toss case because each toss is independent from the next. We’ll have you explore repeated trials of coin tosses to help convince you that simple agreement is not a reliable measure of agreement (see question 6 below).

Interpreting agreement values is not straightforward, and some argue agreement statistics should be avoided in favor of disagreement measures (for example, see discussion in Pontius and Millones 2011). Cohen suggested (Table 14)

Table 14. Landis and Koch’s (1977) interpretation of Cohen’s kappa.

Even a cursory look at Table 14 should lead to questions about scenarios in which a range of agreement “substantial” would provide little comfort about the agreement between the judges or the reliability of the instrument used, for example forensic cases or histology of tissue biopsies.

Contingency table analyses in Rcmdr

Assuming you have already summarized the data, you can enter the data directly in the Rcmdr contingency table form. If your data are not summarized, then you would use Rcmdr's Two-way table… for this. We will proceed under the assumption that you have already obtained the frequencies for each cell in the table.

Rcmdr: Statistics → Contingency tables → Enter and analyze two-way table…

Here you can tell R how many rows and how many columns (Fig 1).

Figure 1. Screenshot R Commander menu for 2X2 data entry with counts.

The default is a 2X2 table. For larger tables, use the sliders to increase the number of rows, columns, or both.

Next, enter the counts. You can edit every cell in this table, including the headers. In the next panel, I will show you the data entry and options. The actual calculation of the test statistic is done when you select the “Chi-square test of independence” check-box (Fig 2).

Figure 2. Display of Xiang et al data entered into R Commander menu.

After entering the data, click on the Statistics tab (Fig 3).

Figure 3. Screenshot Statistics options for contingency table.

If you also select “Components of chi-square statistic” option, then R will show you the contributions of each cell (O-E) towards the chi-square test statistic value. This is helpful to determine if rejection of the null is due to a subset of the categories, and it also forms the basis of the heterogeneity tests, a subject we will pick up in the next section.

Here’s the R output from R Commander. Note the “2, 2, byrow=TRUE” instructions (check out R help to confirm what

.Table <- matrix(c(103,41,95,123), 2, 2, byrow=TRUE)
dimnames(.Table) <- list("rows"=c("HG+", "HG-"), "columns"=c("Lived", "Died"))

.Table # Counts
     columns
rows Lived Died
 HG+   103   41
 HG-   95   123

.Test <- chisq.test(.Table, correct=FALSE)
.Test

Pearson's Chi-squared test

data: .Table
X-squared = 27.339, df = 1, p-value = 0.0000001708

round(.Test$residuals^2, 2) # Chi-square Components
    columns
rows Lived Died
HG+   7.46 9.00
HG-   4.93 5.95

remove(.Test)
remove(.Table)

end R output.

Lots of output, take a deep breath and remember… The minimum output we need to look at is… (Table 15)?

Table 15. Minimally expected outcome for report from a statistical test.

We can see that the p-value 1.7^-7 is much less than Type 1 α = 0.05, thus, by our decision criterion we reject the null hypothesis (provisionally of course, as science goes).

myNumber = 0.0000001708
format(myNumber, scientific = TRUE, digits=2)

[1] "1.7e-07"

Questions

1. Instead of R Commander, try the contingency table problem in R directly.

   myData <- matrix(c(103,41,95,123))
   chisq.test(myData)

   Is this the correct contingency table analysis? Why or why not?

2. For many years National Football League games that ended in ties went to “sudden death,” where the winner was determined by the first score in the extra period of play, regardless of whether or not the other team got an opportunity to possess the ball on offense. Thus, in more than 100 games (140), the team that won the coin toss and therefore got the ball first in overtime, won the game following either a kicked field goal or after a touchdown was scored. In 337 other games under this system, the outcome was not determined by who got the ball first. Many complained that the “sudden death” format was unfair and in 2013 the NFL changed its overtime rules. Beginning 2014 season, both teams got a chance to possess the ball in overtime, unless the team that won the coin toss also went on to score a touchdown at which time that team would be declared the winner. In this new era of over-time rules ten teams that won the coin toss went on to score a touchdown in their first possession and therefore win the game, whereas in 54 other overtime games, the outcome was decided after both teams had a chance on offense (data as of 1 December 2015). These data may be summarized in the table
   

   	First possession win?
   	Yes	No
   Coin flip years	140	337
   New era	10	54

   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?

3. Return to the Doll and Hill (1950) data: 2 men with lung cancer were nonsmokers, 647 men with lung cancer were cigarette smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were cigarette smokers.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
4. A more recent example from the study of the efficacy of St John’s Wort as a treatment for major depression (Shelton et al 2001). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
5. Return to the coin toss simulation and repeat comparisons between simple agreement and Cohen’s kappa for different values of N. Do you see any trend(s) for the two measures as sample size increases or decreases?
6. Repeat the coin toss simulation for N = 24 up to five times, each time calculate simple agreement and Cohen’s kappa. Provide a table of your five results plus the example, and generalize simple agreement vs Cohen’s kappa as a reliable measure of agreement. Hint: you may want to use Descriptive statistics to help make your case.

Quiz Chapter 9.2

Chi-square contingency tables

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• 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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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