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)

18.1 – Multiple Linear Regression

Introduction

Last time we introduced simple linear regression:

• one independent X variable
• one dependent Y variable.

The linear relationship between Y and X was estimated by the method of Ordinary Least Squares (OLS). OLS minimizes the sum of squared distances between the observed responses, and responses predicted by the line. Simple linear regression is analogous to our one-way ANOVA — one outcome or response variable and one factor or predictor variable (Chapter 12.2).

But the world is complicated and so, our one-way ANOVA was extended to the more general case of two or more predictor (factor) variables (Chapter 14). As you might have guessed by now, we can extend simple regression to include more than one predictor variable. In fact, combining ANOVA and regression gives you the general linear model! And, you should not be surprised that statistics has extended this logic to include not only multiple predictor variables, but also multiple response variables. Multiple response variables falls into a category of statistics called multivariate statistics.

Like multi-way ANOVA, multiple regression is the extension of simple linear regression from one independent predictor variable to include two or more predictors. The benefit of this extension is obvious — our models gain realism. All else being equal, the more predictors, the better the model will be at describing and/or predicting the response. Things are not all equal, of course, and we’ll consider two complications of this basic premise, that more predictors are best; in some cases they are not.

However, before discussing the exceptions or even the complications of a multiple linear regression model, we begin by obtaining estimates of the full model, then introduce aspects of how to evaluate the model. We also introduce  and whether a reduced model may be the preferred model.

R code

Multiple regression is easy to do in Rcmdr — recall that we used the general linear model function, lm(), to analyze one-way ANOVA and simple linear regression. In R Commander, we access lm() by

Rcmdr: Statistics → Fit model → Linear model

You may, however, access linear regression through R Commander

We use the same general linear model function for cases of multi-way ANOVA and for multiple regression problems. Simply enter more than one ratio-scale predictor variable and boom!

You now have yourself a multiple regression. You would then proceed to generate the ANOVA table for hypothesis testing

Rcmdr: Models → Hypothesis testing → ANOVA tables

From the output of the regression command, estimates of the coefficients along with standard errors for the estimate and results of t-tests for each coefficient against the respective null hypotheses for each coefficient are also provided. In our discussion of simple linear regression we introduced the components: the intercept, the slope, as well as the concept of model fit, as evidenced by R², the coefficient of determination. These components exist for the multiple regression problem, too, but now we call the slopes partial regression slopes because there are more than one.

Our full multiple regression model becomes

where the coefficients β₁, β₂, … β_n are the partial regression slopes and β₀ is the Y-intercept for a model with 1 – n predictor variables. Each coefficient has a null hypothesis, each has a standard error, and therefore, each coefficient can be tested by t-test.

Now, regression, like ANOVA, is an enormous subject and we cannot do it justice in the few days we will devote to it. We can, however, walk you through a fairly typical example. I’ve posted a small data set diabetesCholStatin at the end of this page. Scroll down or click here. View the data set and complete your basic data exploration routine: make scatterplots and box plots.  We think (predict) that body size and drug dose cause variation in serum cholesterol levels in adult men. But do both predict cholesterol levels?

Selecting the best model

We have two predictor variables, and we can start to see whether none, one, or both of the predictors contribute to differences in cholesterol levels. In this case, both contribute significantly. The power of multiple regression approaches is that it provides a simultaneous test of a model which may have many explanatory variables deemed appropriate to describe a particular response. More generally, it is sometimes advisable to think more philosophically about how to select a best model.

In model selection, some would invoke Occam’s razor — given a set of explanations, the simplest should be selected — to justify seeking simpler models. There are a number of approaches (forward selection, backward selection, or stepwise selection), and the whole effort of deciding among competing models is complicated with a number of different assumptions, strengths and weaknesses. I refer you to the discussion below, which of course is just a very brief introduction to a very large subject in (bio)statistics!

Let’s get the full regression model

The statistical model is

As written in R format, our model is ChLDL ~ BMI + Dose + Statin.

We can now proceed in R Commander to fit the model.

Rmdr: Statistics → Fit models → Linear model

How the model is inputted into linear model menu is shown in Figure 1.

Figure 1. Screenshot of Rcmdr linear model menu with our model elements in place.

The output

summary(LinearModel.1)
Call:
lm(formula = ChLDL ~ BMI + Dose + Statin, data = cholStatins)

Residuals:
    Min      1Q  Median     3Q    Max
-3.7756 -0.5147 -0.0449 0.5038 4.3821

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)
(Intercept)           1.016715   1.178430   0.863 0.39041
BMI                   0.058078   0.047012   1.235 0.21970
Dose                 -0.014197   0.004829  -2.940 0.00411 **
Statin[Statin2]       0.514526   0.262127   1.963 0.05255 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.31 on 96 degrees of freedom
Multiple R-squared: 0.1231, Adjusted R-squared: 0.09565
F-statistic: 4.49 on 3 and 96 DF, p-value: 0.005407

Question. What are the estimates of the model coefficients (rounded)?

• = intercept = 1.017
• = slope for variable BMI = 0.058
• = slope for variable Dose = -0.014
• = slope for variable Statin = -0.515

Question. Which of the three coefficients were statistically different from their null hypothesis (p-value < 5%)?

Answer: Only coefficient was judged statistically significant at the Type I error level of 5% (p = 0.0041).  Of the four null hypotheses we have for the coefficients:

• = 0
• = 0
• = 0; we only reject the null hypothesis for Dose coefficient.
• = 0)

The important take-home concept is about the lack of a direct relationship between the magnitude of the estimate of the coefficient and the likelihood that it will be statistically significant! In absolute value terms , but was not close to statistical significance (p = 0.220).

We generate a 2D scatterplot and include the regression lines (by group=Statin) to convey the relationship between at least one of the predictors (Fig 2).

Figure 2. Scatter plot of predicted LDL against dose of a statin drug. Regression lines represent the different statin drugs (Statin1, Statin2).

Question. Based on the graph, can you explain why there will be no statistical differences between levels of the statin drug type, Statin1 (shown open circles) vs. Statin2 (shown closed red circles).

Because we have two predictors (BMI and Statin Dose), you may also elect to use a 3D-scatterplot. Here’s one possible result (Fig 3).

Figure 3. 3D plot of BMI and dose of Statin drugs on change in LDL levels (green Statin2, blue Statin1).

R code for Figure 3.

Graph made in Rcmdr: Graphs → 3D Graph → 3D scatterplot …

scatter3d(ChLDL~BMI+Dose|Statin, data=diabetesCholStatin, fit="linear", 
residuals=TRUE, parallel=FALSE, bg="white", axis.scales=TRUE, grid=TRUE, 
ellipsoid=FALSE)

Figure 4. An example of a possible interactive 3D plot; the file embedded in this page is not interactive, just an animation.

Diagnostic plots

While visualization concerns are important, let’s return to the statistics. All evaluations of regression equations should involve an inspection of the residuals. Inspection of the residuals allows you to decide if the regression fits the data; if the fit is adequate, you then proceed to evaluate the statistical significance of the coefficients.

The default diagnostic plots (Fig 5) R provides are available from

Rcmdr: Models → Graphs → Basic diagnostics plots

Four plots are returned

Figure 5. R’s default regression diagnostic plots.

Each of these diagnostic plots in Figure 4 gives you clues about the model fit.

1. Plot of residuals vs. fitted helps you identify patterns in the residuals
2. Normal Q-Q plot helps you to see if the residuals are approximately normally distributed
3. Scale-location plot provides a view of the spread of the residuals
4. The residuals vs. leverage plot allows you to identify influential data points.

We introduced these plots in Chapter 17.8 when we discussed fit of simple linear model to data. My conclusion? No obvious trend in residuals, so linear regression is a fit to the data; data not normally distributed Q-Q plot shows S-shape.

Interpreting the diagnostic plots for this problem

The “Normal Q-Q” plot allows us to view our residuals against a normal distribution (the dotted line). Our residuals do no show an ideal distribution: low for the first quartile, about on the line for intermediate values, then high for the 3rd and 4th quartile residuals. If the data were bivariate normal we would see the data fall along a straight line. The “S-shape” suggests log-transformation of the response and or one or more of the predictor variables.

There also seems to be a pattern in residuals vs the predicted (fitted) values. There is a trend of increasing residuals as cholesterol levels increase, which is particularly evident in the “scale-location” plot. Residuals tended to be positive at low and high doses, but negative at intermediate doses. This suggests that the relationship between predictors and cholesterol levels may not be linear, and it demonstrates what statisticians refer to as a monotonic spread of residuals.

The last diagnostic plot looks for individual points that influence, change, or “leverage” the regression — in other words, if a point is removed, does the general pattern change? If so, then the point had “leverage” and thus we need to decide whether or not to include the datum. diagnostic plots Cook’s distance is a measure of the influence of a point in regression. Points with large Cook’s distance values warrant additional checking.

The multicollinearity problem

Statistical model building is a balancing act by the statistician. While simpler models may be easier to interpret and, perhaps, to use, it is a basic truism that the more predictor variables the model includes, the more realistic the statistical model. However, each additional parameter that is added to the statistical model must be independent of all other parameters already in the model. To the extent that this assumption is violated, the problem is termed multicollinearity. If predictor variables are highly correlated, then they are essentially just linear combinations and do not provide independent evidence. For example, one would naturally not include two core body temperature variables in a statistical model on basal metabolic rate, one in degrees Fahrenheit and the other in degrees Celsius, because  it is a simple linear conversion between the two units. This would be an example of structural collinearity: the collinearity is because of misspecification of the model variables. In contrast, collinearity among predictor variables may because the data are themselves correlated. For example, if multiple measures of body size are included: weight, height, length of arm, etc., then we would expect these to be correlated, ie, data multicollinearity.

Collinearity in statistical models may have a number of undesirable effects on a multiple regression model. These include

• estimates of coefficients not stable: with collinearity, values of coefficients depend on other variables in the model; if collinear predictors, then the assumption of independent predictor variables is violated.
• precision of the estimates decreases (standard error of estimates increase).
• statistical power decreases.
• p-values for individual coefficients not trust worthy.

Tolerance and Variance Inflation Factor

Absence of multicollinearity is important assumption of multiple regression. A partial test is to calculate product moment correlations among predictor variables. For example, when we calculate the correlation between BMI and Dose for our model, we get r = 0.101 (P = 0.3186), and therefore would tentatively conclude that there was little correlation between our predictor variables.

A number of diagnostic statistics have been developed to test for multicollinearity. Tolerance for a particular independent variable (Xi) is defined as 1 minus the proportion of variance it shares with the other independent variables in the regression analysis (1 − R²_i) (O’Brien 2007). Tolerance reports the proportion of total variance explained by adding the X_i–^th predictor variable that is unrelated to the other variables in the model. A small value for tolerance indicates multicollinearity — and that the predictor variable is nearly a perfect combination (linear) of the variables already in the model and therefore should be omitted from the model. Because tolerance is defined in relation to the coefficient of determination you can interpret a tolerance score as the unique variance accounted for by a predictor variable.

A second, related diagnostic of multicollinearity is called the Variance Inflation Factor, VIF. VIF is the inverse of tolerance.

VIF shows how much of the variance of a regression coefficient is increased because of collinearity with the other predictor variables in the model. VIF is easy to interpret: a tolerance of 0.01 has a VIF of 100; a tolerance of 0.1 has a VIF of 10; a tolerance of 0.5 has a VIF of 2, and so on. Thus, small values of tolerance and large values of VIF are taken as evidence of multicollinearity.

Rcmdr: Models → Numerical diagnostics → Variation-inflation factors

vif(RegModel.2)
      BMI     Dose 
1.010256 1.010256

A rule of thumb is that if VIF is greater than 5 then there is multicollinearity; with VIF values close to one we would conclude, like our results from the partial correlation estimate above, that there is little evidence for a problem of collinearity between the two predictor variables. They can therefore remain in the model.

Solutions for multicollinearity

If there is substantial multicollinearity then you cannot simply trust the estimates of the coefficients. Assuming that there hasn’t been some kind of coding error on your part, then you may need to find a solution. One solution is to drop one of the predictor variables and redo the regression model. Another option is to run what is called a Principal Components Regression (PCR).  One takes the predictor variables and runs a Principal Component Analysis to reduce the number of variables, then the regression is run on the PCA components. By definition, the PCA components are independent of each other. Another option is to use ridge regression approach.

Like any diagnostic rule, however, one should not blindly apply a rule of thumb. A VIF of 10 or more may indicate multicollinearity, but it does not necessarily lead to the conclusion that the linear regression model requires that the researcher reduce the number of predictor variables or analyze the problem using a different statistical method to address multicollinearity as the sole criteria of a poor statistical model. Rather, the researcher needs to address all of the other issues about model and parameter estimate stability, including sample size. Unless the collinearity is extreme (like a correlation of 1.0 between predictor variables!), larger sample sizes alone will work in favor of better model stability (by lowering the sample error) (O’Brien 2007).

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. Please explain why the magnitude of the slope is not the key to statistical significance of a slope? Hint: look at the equation of the t-test for statistical significance of the slope.
3. Consider the following scenario. A researcher measures his subjects multiple times for blood pressure over several weeks, then plots all of the values over time. In all, the data set consists of thousands of readings. He then proceeds to develop a model to explain blood pressure changes over time. What kind of collinearity is present in his data set? Explain your choice.
4. We noted that Dose could be viewed as categorical variable. Convert Dose to factor variable (fDose) and redo the linear model. Compare the summary output and discuss the additional coefficients.
   • Use Rcmdr: Data → Manage variables in active data set →  Convert numeric Variables to Factors to create a new factor variable fDose. It’s ok to use the numbers as factor levels.
5. We flagged the change in LDL as likely to be not normally distributed. Create a log10-transformed variable for ChLDL and perform the multiple regression again.
   1. Write the new statistical model
   2. Obtain the regression coefficients — are they statistically significant?
   3. Run basic diagnostic plots and evaluate for

Quiz Chapter 18.1

Multiple Linear Regression

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Data set

Chapter 18 contents

• Introduction
• Multiple linear regression
• Nonlinear regression
• Logistic regression
• Generalized Linear Squares
• Selecting the best model
• Compare two linear models
• References and suggested readings (Ch17 & 18)

17.8 – Assumptions and model diagnostics for Simple Linear Regression

Introduction

The assumptions for all linear regression:

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

Note the mnemonic device: Linear, Independent, Normal, Error or LINE.

Each of the four elements will be discussed below in the context of Model Diagnostics. These assumptions apply to how the model fits the data. There are other assumptions that, if violated, imply you should use a different method for estimating the parameters of the model.

Ordinary least squares makes the additional assumption about the quality of the independent variable that e that measurement of X is done without error. Measurement error is a fact of life in science, but the influence of error on regression differs if the error is associated with the dependent or independent variable. Measurement error in the dependent variable increases the dispersion of the residuals but will not affect the estimates of the coefficients; error associated with the independent variables, however, will affect estimates of the slope. In short, error in X leads to biased estimates of the slope.

The equivalent, but less restrictive practical application of this assumption is that the error in X is at least negligible compared to the measurements in the dependent variable.

Multiple regression makes one more assumption, about the relationship between the predictor variables (the X variables). The assumption is that there is no multicollinearity, a subject we will bring up next time (see Chapter 18).

Model diagnostics

We just reviewed how to evaluate the estimates of the coefficients of the model. Now we need to address a deeper meaning — how well the model explains the data. Consider a simple linear regression first. If H_o: b = 0 is not rejected — then the slope of the regression equation is taken to not differ from zero. We would conclude that if repeated samples were drawn from the population, on average, the regression equation would not fit the data well (lots of scatter) and it would not yield useful prediction.

However, recall that we assume that the fit is linear. One assumption we make in regression is that a line can, in fact, be used to describe the relationship between X and Y.

Here are two very different situations where the slope = 0.

Example 1. Linear Slope = 0, No relationship between X and Y

Example 2. Linear Slope = 0, A significant relationship between X and Y

But even if H_o: b = 0 is rejected (and we conclude that a linear relationship between X and Y is present), we still need to be concerned about the fit of the line to the data — the relationship may be more nonlinear than linear, for example. Here are two very different situations where the slope is not equal to 0.

Example 3. Linear Slope > 0, a linear relationship between X and Y

Example 4. Linear Slope > 0, curve-linear relationship between X and Y

How can you tell the difference? There are many regression diagnostic tests, many more than we can cover, but you can start with looking at the coefficient of determination (low R² means low fit to the line), and we can look at the pattern of residuals plotted against the either the predicted values or the X variables (my favorite). The important points are:

1. In linear regression, you fit a model (the slope + intercept) to the data;
2. We want the usual hypothesis tests (are the coefficients different from zero?) and
3. We need to check to see if the model fits the data well. Just like in our discussions of chi-square, a “perfect fit would mean that the difference between our model and the data would be zero.

Graph options

Using residual plots to diagnose regression equations

Yes, we need to test the coefficients (intercept H_o = 0; slope H_o = 0) of a regression equation, but we also must decide if a regression is an appropriate description of the data. This topic includes the use of diagnostic tests in regression. We address this question chiefly by looking at

1. scatterplots of the independent (predictor) variable(s) vs. dependent (response) variable(s).
   what patterns appear between X and Y? Do your eyes tell you “Line”? “Curve”? “No relation”?
2. coefficient of determination
   closer to zero than to one?
3. patterns of residuals plotted against the X variables (other types of residual plots are used to, this is one of my favorites)

Our approach is to utilize graphics along with statistical tests designed to address the assumptions.

One typical choice is to see if there are patterns in the residual values plotted against the predictor variable. If the LINE assumptions hold for your data set, then the residuals should have a mean of zero with scatter about the mean. Deviations from LINE assumptions will show up in residual plots.

Here are examples of POSSIBLE outcomes:

Figure 1. An ideal plot of residuals

Solution: Proceed! Assumptions of linear regression met.

Compare to plots of residuals that differ from the ideal.

Figure 2. We have a problem. Residual plot shows a “funnel” shape — unequal variance (aka heteroscedasticity).

Solution. Try a transform like the log₁₀-transform.

Figure 3. Problem. Residual plot shows systematic trend.

Solution. Linear model a poor fit; May be related to measurement errors for one or more predictor variables. Try adding an additional predictor variable or model the error in your general linear model.

Figure 4. Problem. Residual plot shows nonlinear trend.

Solution. Transform data or use more complex model

This is a good time to mention that in statistical analyses, one often needs to do multiple rounds of analyses, involving description and plots, tests of assumptions, tests of inference. With regression, in particular, we also need to decide if our model (e.g., linear equation) is a good description of the data.

Diagnostic plot examples

Return to our example.Tadpole dataset. To obtain residual plots, Rcmdr: Models → Graphs → Basic diagnostic plots yields four graphs. Graph A is

Figure 5. Basic diagnostic plots. A: residual plot; B: Q-Q plot of residuals; C: Scale-location (aka spread-location) plot; D: leverage residual plot.

In brief, we look at plots

A, the residual plot, to see if there are trends in the residuals. We are looking for a spread of points equally above and below the mean of zero. In Figure 5 we count seven points above and six points below zero so there’s no indication of a trend in the residuals vs the fitted VO2 (Y) values.

B, the Q-Q plot is used to see if normality holds. As discussed before, if our data are more or less normally distributed, then points will fall along a straight line in a Q-Q plot.

C, the Scale- or spread-location plot is used to verify equal variances of errors.

D, Leverage plot — looks to see if an outlier has leverage on the fit of the line to the data, i.e., changes the slope. Additionally, provides location of Cook’s distance measure (dashed red lines). Cook’s distance measures the effect on the regression by removing one point at a time and then fitting a line to the data. Points outside the dashed lines have influence.

Statistical test options

After building linear models, run statistical diagnostic tests that compliment graphics approaches. Here, simply to introduce the tests, I included the Gosner factor as potential predictor — many of these numerical diagnostics are more appropriate for models with more than one predictor variable, the subject of the next chapter.

Results for the model were:

lm(formula = VO2 ~ Body.mass + Gosner, data = frogs)

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

Question. Note the p-value for the coefficients — do we include a model with both body mass and Gosner Stage? See Chapter 18.5 – Selecting the best model

In R Commander, the diagnostic test options are available via

Rcmdr: Models → Numerical diagnostics

Variance inflation factors (VIF):  used to detect multicollinearity among the predictor variables. If correlations are present among the predictor variables, then you can’t rely on the the coefficient estimates — whether predictor A causes change in the response variable depends on whether the correlated B predictor is also included in the model. If correlation between predictor A and B, the statistical effect is increased variance associated with the error of the coefficient estimates. There are VIF for each predictor variable. A VIF of one means there is no correlation between that predictor and the other predictor variables. A VIF of 10 is taken as evidence of serious multicollinearity in the model.

R code:

vif(RegModel.2)

R output

Body.mass Gosner 
2.186347 2.186347 

round(cov2cor(vcov(RegModel.2)), 3) # Correlations of parameter estimates
         (Intercept) Body.mass Gosner[T.II]
(Intercept)    1.000    -0.958        0.558
Body.mass     -0.958     1.000       -0.737
Gosner[T.II]   0.558    -0.737        1.000

Our interpretation

The output shows a matrix of correlations; the correlation between the two predictor variables in this model was 0.558, large per our discussions in Chapter 16.1 . Not really surprising as older tadpoles (Gosner stage II), are larger tadpoles. We’ll leave the issue of multicollinearity to Chapter 18.1.

Breusch-Pagan test for heteroscedasticity… Recall that heteroscedasticity is another name for unequal variances. This tests pertains to the residuals; we used the scale (spread) location plot (C) above. The test statistic can be calculated as . The function call is part of the car package, which we download as part of the Rcmdr package. R Commander will pop up a menu of choices for this test. For starters, accept the defaults, enter nothing, and click OK.

R code:

library(zoo, pos=16)
library(lmtest, pos=16)
bptest(VO2 ~ Body.mass + Gosner, varformula = ~ fitted.values(RegModel.2), studentize=FALSE, data=frogs)

R output

Breusch-Pagan test

data:  VO2 ~ Body.mass + Gosner
BP = 0.0040766, df = 1, p-value = 0.9491

Our interpretation

A p-value less than 5% would be interpreted as  statistically significant unequal variances among the residuals.

Durbin-Watson for autocorrelation… used to detect serial (auto)correlation among the residuals, specifically, adjacent residuals. Substantial autocorrelation can mean Type I error for one or more model coefficients. The test returns a test statistic that ranges from 0 to 4. A value at midpoint (2) suggests no autocorrelation. Default test for autocorrelation is r > 0.

R code:

dwtest(VO2 ~ Body.mass + Gosner, alternative="greater", data=frogs)

R output

Durbin-Watson test

data:  VO2 ~ Body.mass + Gosner
DW = 2.1336, p-value = 0.4657
alternative hypothesis: true autocorrelation is greater than 0

Our interpretation

The DW statistic was about 2. Per our definition, we conclude no correlations among the residuals.

RESET test for nonlinearity…  The Ramsey Regression Equation Specification Error Test is used to interpret whether or not the equation is properly specified. The idea is that fit of the equation is compared against alternative equations that account for possible nonlinear dependencies among predictor and dependent variables. The test statistic follows an F distribution.

R code:

resettest(VO2 ~ Body.mass + Gosner, power=2:3, type="regressor", data=frogs)

R output

RESET test

data:  VO2 ~ Body.mass + Gosner
RESET = 0.93401, df1 = 2, df2 = 7, p-value = 0.437

Our interpretation

Also a nonsignificant test — adding a power term to account for a nonlinear trend gained no model fit benefits, p-value much greater than 5%.

Bonferroni outlier test. Reports the Bonferroni p-values for testing each observation to see if it can be considered to be a mean-shift outlier (a method for identifying potential outlier values — each value is in turn replaced with the mean of its nearest neighbors, Lehhman et al 2020).  Compare to results of VIF. The objective of outlier and VIF tests is to verify that the regression model is stable, insensitive to potential leverage (outlier) values.

R code:

outlierTest(LinearModel.3)

R output

No Studentized residuals with Bonferroni p < 0.05
Largest |rstudent|:
  rstudent unadjusted p-value Bonferroni p
7 1.958858        0.085809              NA

Our interpretation

We ran the outlierTest() function on LinearModel.3 to check for influential outliers using Studentized residuals with Bonferroni correction.

The output shows “No Studentized residuals with Bonferroni p < 0.05”, which means no data points were statistically flagged as outliers after adjusting for multiple comparisons. The largest Studentized residual was 1.96 (point 7), with an unadjusted p-value of 0.086. Although this is the largest deviation from the regression line, it is not statistically significant. Therefore, we conclude that no single observation is exerting undue influence on the model fit, and we can be confident that our regression results are not driven by outliers.

Response transformation… The concept of transformations of data to improve adherence to statistical assumptions was introduced in Chapters 3.2, 13.1, and 15.

R code:

summary(powerTransform(LinearModel.3, family="bcPower"))

R output

bcPower Transformation to Normality 
    Est Power Rounded Pwr Wald Lwr Bnd Wald Upr Bnd
Y1     0.9987           1       0.2755       1.7218

Likelihood ratio test that transformation parameter is equal to 0
   (log transformation)
                           LRT df      pval
LR test, lambda = (0) 7.374316  1 0.0066162

Likelihood ratio test that no transformation is needed
                                LRT df    pval
LR test, lambda = (1) 0.00001289284  1 0.99714

Our interpretation

The Box-Cox analysis suggests that the response variable for LinearModel.3 can remain on its original scale. No power transformation was needed, and any standard linear regression inference is appropriate. The estimated transformation parameter was close to one, which suggests that the original response variable did not need a transformation. The LRT tests whether a log transformation (lambda = 0) would improve normality. The significant p-value, 0.0066, for the LRT indicates that a log transformation would not be the best choice, consistent with lambda ≈ 1. And last, the very high p-value, 0.99714, confirms that no transformation is necessary; the model residuals were close to normal.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. Referring to Figures 1 – 4 on this page, which plot best suggests a regression line fits the data?
3. Return to the electoral college data set and your linear models of Electoral vs. POP_2010 and POP_2019. Obtain the four basic diagnostic plots and comment on the fit of the regression line to the electoral college data.
   • residual plot
   • Q-Q plot
   • Scale-location plot
   • Leverage plot
4. With respect to your answers in question 2, how well does the electoral college system reflect the principle of one person one vote?

Quiz Chapter 17.8

Assumptions and model diagnostics for Simple Linear Regression

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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
• Regression model fit
• Assumptions and model diagnostics for Simple Linear Regression
• References and suggested readings (Ch17 & 18)

17.4 – OLS, RMA, and smoothing functions

Introduction

OLS or ordinary least squares is the most commonly used estimation procedure for fitting a line to the data. For both simple and multiple regression, OLS works by minimizing the sum of the squared residuals. OLS is appropriate when the linear regression assumptions LINE apply. In addition, further restrictions apply to OLS including that the predictor variables are fixed and without error. OLS is appropriate when the goal of the analysis is to retrieve a predictive model. OLS describes an asymmetric association between the predictor and the response variable: the slope b_X for Y ~ b_XX will generally not be the same as the slope b_Y for X ~ b_YY.

OLS is appropriate for assessing functional relationships (i.e., inference about the coefficients) as long as the assumptions hold. In some literature, OLS is referred to as a Model I regression.

Generalized Least Squares

Generalized linear regression is an estimation procedure related to OLS but can be used either when variances are unequal or multicollinearity is present among the error terms.

Weighted Least Squares

A conceptually straightforward extension of OLS can be made to account for situation where the variances in the error terms are not equal. If the variance of Yi varies for each Xi, then a weighting function based on the reciprocal of the estimated variance may be used.

then, instead of minimizing the squared residuals as in OLS, the regression equation estimates in weighted least squares minimizes the squared residuals summed over the weights.

Weighted least squares is a form of generalized least squares. In order to estimate w_i, however, multiple values of Y for each observed X must be available.

Reduced Major Axis

There are many alternative methods available when OLS may not be justified. These approaches, collectively, may be called Model II regression methods. These methods are invariably invoked in situations in which both Y and X variables have random error associated with them. In other words, the OLS assumption that the predictor variables are measured without error is violated. Among the more common methods is one called Reduced Major Axis or RMA.

Smoothing functions

data set: atmospheric carbon dioxide (CO₂) readings Mauna Loa. Source: https://gml.noaa.gov/ccgg/trends/data.html

Fit curves without applying a known formula. This technique is called smoothing and, while there are several versions, the technique involves taking information from groups of observations — weighted averaged — and using these groups to estimate how the response variable changes with values of the independent variable. Smoothing is used to help reveal patterns, to emphasize trends by reducing noise — clearly, caution needs to be employed as smoothing necessarily hides outlier data, which can themselves be important. Smoothing techniques by name include kernel, loess, and spline. Default in the scatter plot command is loess.

CO₂ in parts per million (ppm) plotted by year from 1958 to 2014 the first CO₂ readings were recorded in April 1958 (Fig 1).

A few words of explanation for Figure 1. The green line shows the OLS line, the red line shows the loess smoothing with a smoothing parameter of 0.5 (in Rcmdr the slider value reads “5”).

Figure 1. CO₂ in parts per million (ppm) plotted by year from 1958 to 2014

R command was started with option settings available in Rcmdr context menu for scatterplot, then additional commands were added

scatterplot(CO2~Year, reg.line=lm, grid=FALSE, smooth=TRUE, spread=FALSE, boxplots=FALSE, span=0.05, lwd=2, xlab="Months since 1958", ylab="CO2 ppm", main="CO2 at Mauna Loa Observatory, April 1958 - April 2014", cex=1, cex.axis=1.2, cex.lab=1.2, pch=c(16), data=StkCO2MLO)

The next plot is for ppm CO₂ by month for the year 2013. The plot shows the annual cycle of atmospheric CO₂ in the northern hemisphere.

Figure 2. Plot of ppm CO₂ by month for the year 2013.

Again, the smoothing parameter was set to 0.5 and the loess function is plotted in red (Fig 2).

Loess is an acronym short for local regression. Loess is a weighted least squares approach which is used to fit linear or quadratic functions of the predictors at the centers of neighborhoods. The radius of each neighborhood is chosen so that the neighborhood contains a specified percentage of the data points. This percentage of data points is referred to as the smoothing parameter and this parameter may differ for different neighborhoods of points. The idea of loess, in fact, any smoothing algorithm, is to reveal pattern within a noisy sequence of observations. The smoothing parameter can be set to different values, between 0 and 1 is typical.

To get a sense of what the parameter does, Figure 3 takes the same data as in Figure 2, but with different values of the smoothing parameter (Fig 3).

Figure 3. Plot with different smoothing values (0.5 to 10.0).

plot key:

The R code used to generate Figure 3 plot was

spanList = c(0.5, 0.75, 1, 2, 10)
reg1 = lm(ppm~Month)
png(filename = "RplotCO2mo.png", width = 400, height = 400, units = "px", pointsize = 12, bg = "white")
plot(Month,ppm, cex=1.2, cex.axis=1.2, cex.lab=1.2, pch=c(16), xlab="Months", ylab="CO2 ppm", main="CO2 levels 2013")
abline(reg1,lwd=2,col="green") 
for (i in 1:length(spanList))
{
ppm.loess <- loess(ppm~Month, span=spanList[i], Dataset)  
ppm.predict <- predict(ppm.loess, Month)
lines(Month,ppm.predict,lwd=2,col=i)
}

The CO2 data constitutes a time series. Instead of loess, a simple moving average would be a more natural way to reveal trends. In principle, take a set of nearby points (odd number of points best, keeps the calculation symmetric) and calculate the average. Next, shift the points by a specified time interval (eg, 7 days), and recalculate the average for the new set of points. See Chapter 20.5 for Time series analysis.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. This is bio-statistics, so I gotta ask: What ecological/environmental process explains the shape of the relationship between ppm CO₂ and months of the year as shown in Figure 2? Hint: NOAA Global Monitoring Laboratory responsible for the CO2 data is located at Mauna Loa Observatory, Hawaiʻi (lat: 19.52291, lon: -155.61586).
3. As I wrote this question (January 2022), we were 22 months since WHO declared Covid-19 a pandemic (CDC timeline). Omicron variant is now dominant; Daily case counts State of Hawaiʻi from 1 November 2021 to 15 January 2022 reported in data set table.
   1. Make plot like Figure 2 (days instead of months)
   2. Apply different loess smoothing parameters and re-plot the data. Observe and describe the change to the trend between case reports and days.

Quiz Chapter 17.4

OLS, RMA, and smoothing functions

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Data set

Covid-19 cases reported State of Hawaiʻi from 1 November 2021 to 15 January 2022 (data extracted from Wikipedia)

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)

17.1 – Ordinary least squares regression

Introduction

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

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

We start our LM model as

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

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

Figure 1. R commander menu interface for linear model. 

Our model was

Matings ~ Body.Mass

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

Example

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

Our causal model looks like

Let’s review the case for ANOVA first.

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

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

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

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

Least squares regression explained

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

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

The data set was

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

And a scatterplot of the data (Fig 2)

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

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

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

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

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

= the Y-intercept coefficient and it is defined as

solve for intercept by setting X = 0.

 = the regression coefficient (slope)

ε = the error or “residual”

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

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

the linear regression predicts Y

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

There will be as many residuals as there were observations.

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

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

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

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

Models used to predict new values

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

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

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

This is what people do when they say

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

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

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

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

R Code

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

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

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

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

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

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

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

What are the values for the coefficients? See Estimate .

Table 2. Coefficient estimates.

	Value
, Y-intercept	-8.4746
, Slope	0.3623

Get the sum of squares from the ANOVA table

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

Output from R, the ANOVA table

Analysis of Variance Table

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

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

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

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

And the “statistical significance?” See Pr(>|t|) .There were tests for each coefficient, and :

Table 3. Coefficient estimates.

	P-value
, Y-intercept	0.1026
, Slope	0.0255

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

Make a prediction:

How many matings expected for a 40 g male?

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

R output

1 
6.016949

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

See below for predicting confidence limits.

How to extract information from an R object

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

str(LinearModel.3)

Output from R

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

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

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

RegModel.1$fitted.values

and R returns

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

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

Output from R

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

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

myAOV.full $"Sum Sq"

Output from R

[1] 22.52773 28.38136

Get the residual sum of squares

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

Output from R

[1] 22.52773

Get the regression sum of squares

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

Output from R

[1] 50.90909

Now, get the total sums of squares for the model

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

Calculate the coefficient of determination

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

Output from R

[1] 0.4425091

Which matches what we got before, as it should.

Regression equations may be useful to predict new observations

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

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

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

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

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

How I got this graph

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

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

Regression through the origin

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

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

R code for regression without the Y-intercept:

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

And the R output:

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

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

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

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

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

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

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

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

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

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

Assumptions of OLS, an introduction

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

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

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

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

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

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

Questions

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

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

Anscombe’s quartet

Quiz Chapter 17.1

Ordinary Least Squares

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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)