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)

18.5 – Selecting the best model

Introduction

This is a long entry in our textbook, many topics to cover. We discuss aspects of model fitting, from why model fitting is done to how to do it and what statistics are available to help us decide on the best model. Model selection very much depends on what the intent of the study is. For example, if the purpose of model building is to provide the best description of the data, then in general one should prefer the full (also called the saturated) model. On the other hand, if the purpose of model building is to make a predictive statistical model, then a reduced model may prove to be a better choice. The text here deals mostly with the later context of model selection, finding a justified reduced model.

From Full model to best Subset model

Model building is an essential part of being a scientist. As scientists, we seek models that explain as much of the variability about a phenomenon as possible, but yet remain simple enough to be of practical use.

Having just completed the introduction to multiple regression, we now move to the idea of how to pick best models.

We distinguish between a full model, which includes as many variables (predictors, factors) as the regression function can work with, returning interpretable, if not always statistically significant output, and a saturated model.

The saturated model is the one that includes all possible predictors, factors, interactions in your experiment. In well-behaved data sets, the full model and the saturated model will be the same model. However, they need not be the same model. For example, if two predictor variables are highly collinear, then you may return an error in regression fitting.

For those of you working with meta-analysis problems, you are unlikely to be able to run a saturated model because some level of a key factor are not available in all or at least most of the papers. Thus, in order to get the model to run, you start dropping factors, or you start nesting factors. If you were unable to get more things in the model, then this is your “full” model. Technically we wouldn’t call it saturated because there were other factors, they just didn’t have enough data to work with or they were essentially the same as something else in the model.

Identify the model that does run to completion as your full model and proceed to assess model fit criteria for that model, and all reduced models thereafter.

In R (Rcmdr) you know you have found the full model when the output lacks “NA” strings (missing values) in the output. Use the full model to report the values for each coefficient, i.e., conducting the inferential statistics.

Get the estimates directly from the output from running the regression function. You can tell if the effect is positive (look at the estimate for sample — it is positive) so you can say — more samples, greater likelihood to see more cases of cancer.

Remember, the experimental units are the papers themselves, so studies with larger numbers of subjects are going to find more cases of diabetes. We would worry big time with your project if we did not see statistically significant and positive for sample size.

Example: A multiple linear model

For illustration, here’s an example output following a run with the linear model function on a portion of the Kaggle Cardiovascular disease data set.

The variables (and data types) were

SBP = systolic blood pressure (from ap_hi)

age.years = Independent variable, ratio scale (calculated from age in days)

BMI = Dependent variable, ratio scale (calculated from height and weight variables)

active = Independent variable, physical activity, binomial

alcohol = Independent variable, alcohol use, binomial

cholesterol = Independent variable, cholesterol levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

gender = Independent variable, binomial

glucose = Independent variable, glucose levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

smoke = Independent variable, binomial

R output

Call:
lm(formula = SBP ~ BMI + gender + age.years + active + alcohol + 
cholesterol + gluc + smoke, data = kaggle.train)

Residuals:
Min 1Q Median 3Q Max 
-33.359 -9.701 -1.065 8.703 45.242

Coefficients:
             Estimate Std. Error t value   Pr(>|t|) 
(Intercept)   76.7664    14.9311   5.141  0.00000155 ***
BMI            0.8395     0.2933   2.862  0.00522 ** 
gender        -2.0744     1.6102  -1.288  0.20092 
age.years      0.2742     0.2403   1.141  0.25672 
active         7.1411     3.4934   2.044  0.04383 * 
alcohol       -7.9272     9.1854  -0.863  0.39039 
cholesterol    7.4539     2.3200   3.213  0.00182 ** 
gluc          -1.2701     2.9533  -0.430  0.66817 
smoke         11.4132     6.7015   1.703  0.09197 . 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 14.96 on 91 degrees of freedom
Multiple R-squared: 0.2839, Adjusted R-squared: 0.221 
F-statistic: 4.51 on 8 and 91 DF, p-value: 0.0001214

Question. Write out the equation in symbol form.

We see from the modest R² (adjusted) that the model explains little variation in systolic blood pressure; only BMI, physical activity (yes, no), and serum cholesterol levels, were statistically significant at 5% level. Smoker (yes, no) approached significance (p-value = 0.092), but I wouldn’t go on and on about positive or negative coefficients and effects.

We report the value (I would round to 7.5) for cholesterol on SBP (Table 1), and note that those who had elevated serum cholesterol levels tended to have higher systolic blood pressure (e.g., the sign of the coefficient — and I related the coefficient back to the most important thing about your study — the biological interpretation). Repeat the same for the other significant coefficients.

Now’s a good time to be clear about HOW you report statistical results. DO NOT SIMPLY COPY AND PASTE EVERYTHING into your report. Now, for the estimates above, you would report everything, but not all of the figures. Here’s how the output should be reported in your Project paper.

Table 1. Coefficients from full model.

term     Coefficient Error  P-value* 
Intercept       76.8  14.93 < 0.00001
BMI              0.8   0.29   0.00522
gender          -2.1   1.61   0.20092 
age.years        0.3   0.24   0.25672 
active           7.1   3.49   0.04383
alcohol         -7.9   9.19   0.39039 
cholesterol      7.5   2.32   0.00182 
gluc            -1.3   2.95   0.66817 
smoke           11.4   6.70   0.09197

* t-test

Looks better, doesn’t it?

Once you have the full model, we should run simple diagnostics on the regression fit to the data (VIF, Cook’s Distance, basic diagnostic plots), then proceed to use this model for the inferential statistics.

> vif(model.1)
BMI gender age.years active alcohol cholesterol gluc smoke 
1.112549 1.091154 1.081641 1.048519 1.096382 1.178635 1.168442 1.131081

No VIF was greater than 2, suggesting little multicollinearity among the predictor variables.

> cooks.distance(model.1)[which.max(cooks.distance(model.1))]
60 
0.355763

This person had high SBP (180), was a smoker, but active and normal BMI.

Use the significance tests of each parameter in the model from the corresponding ANOVA table. Now, where is the ANOVA table? Remember, right after running the linear regression,

Rcmdr: Models → Hypothesis testing → ANOVA tables

Accept the default (partial marginality), and, Boom! … the ANOVA table you should be familiar with.

From the ANOVA table you will tell me whether a Factor is significant or not. You report the ANOVA table in your paper. You describe it.

Now, the next step is to decide what is the best model. It then guides you to the next step which is to decide whether a better model (fewer parameters, Occam’s razor) can be found. Identify the parameter from the ANOVA table with the highest P-value and remove it from the model when you run the regression again. Repeat the steps above, return the ANOVA table, checking the estimates and P-values, until you have a model with only statistically significant parameters.

Find the best model

We may be tempted to run the linear model again, but with non-significant predictors excluded. For example, re-run the model on SBP but with just BMI, active, cholesterol and excluding smoke.

We can calculate the AIC, Akaike Information Criterion, the lower value is considered the better model.

AIC(model.1)
[1] 835.4938

AIC(model.2)
[1] 834.0782

The full model had eight predictor variables, excluding the intercept, whereas the reduced model had just three predictor variables, again, not including the intercept. Unsurprisingly, the reduced model is better. But now, a stronger test — what about smoking, where the p-value approached 5%? We ran the linear model (model.3).

> AIC(model.3)
[1] 831.9495

We can use an anova test or the likelihood ratio test to compare the models. The anova test, the models must be nested.

stats::anova(model.2, model.3)
Analysis of Variance Table

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
  Res.Df   RSS Df Sum of Sq      F   Pr(>F) 
1     96 22205 
2     95 21307  1     898.1 4.0043 0.04824 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

and

lmtest::lrtest((model.2,model.3)
Likelihood ratio test

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
 #Df  LogLik Df  Chisq Pr(>Chisq) 
1  5 -412.04 
2  6 -409.97  1 4.1286 0.04216 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Both tests tell us the same thing — our model with four predictors, including smoking, is better.

2 December 2024 — Past this point, edits pending

But first, I want to take up an important point about your models that you may not have had a chance to think about. The order of entry of parameters in your model can effect the significance and value of the estimates themselves. The order of parameter model entry above can be read top to bottom. Age was first, followed in sequence by CalsPDay, CholPDay, and so on. By convention, enter the covariates first (the ratio-scale predictors), that’s what I did above.

Here’s the output from a model in which I used a different order of parameters.

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

Response: BMI
              Sum Sq    Df    F value     Pr(>F)
Sex             1.52     1     0.0478    0.82764
Smoke           8.84     1     0.2782    0.59964
Age             0.62     1     0.0196    0.88918
CalsPDay       10.79     1     0.3394    0.56215
CholPDay       92.37     1     2.9065    0.09286 .
Sex:Smoke       3.35     1     0.1055    0.74631
Residuals    2129.34    67

The output is the same!!! So why did I give you a warning about parameter order? Run the ANOVA table summary command again, but this time select Type III type of test, i.e., ignore marginality

> Anova(LinearModel.2, type="III")
Anova Table (Type III tests)

Response: BMI
               Sum Sq    Df    F value       Pr(>F)
(Intercept)   1544.34     1    48.5929   1.708e-09 ***
Sex              4.69     1     0.1475     0.70214
Smoke           11.82     1     0.3720     0.54400
Age              0.62     1     0.0196     0.88918
CalsPDay        10.79     1     0.3394     0.56215
CholPDay        92.37     1     2.9065     0.09286 .
Sex:Smoke        3.35     1     0.1055     0.74631
Residuals     2129.34    67

The output has changed — and in fact it now reports the significance test of the intercept. This output is the same as the output from the linear model. Try again, this time selecting Type I, sequential

> anova(LinearModel.2)
Analysis of Variance Table

Response: BMI
           Df   Sum Sq    Mean Sq  F value    Pr(>F)
Sex         1     0.68      0.681   0.0214   0.88402
Smoke       1     2.82      2.816   0.0886   0.76690
Age         1     3.44      3.436   0.1081   0.74333
CalsPDay    1     2.27      2.272   0.0715   0.78998
CholPDay    1    96.30     96.299   3.0301   0.08633
Sex:Smoke   1     3.35      3.354   0.1055   0.74631
Residuals  67  2129.34     31.781

Here, we see the effect of order. So, as we are working to learn all of the issues of statistics and in particular mode fitting, I have purposefully restricted you to Type II analyses — obeying marginality correctly handles most issues about order of entry.

Easy there…. Take a deep breath, and guess what? Your best model needs to have significant parameters in it, right? Your best fit model then is Model 5. And that model will be your candidate for best fit as we proceed to complete our model building.

Now we proceed to gain some support evidence for our candidate best model. We are going to use an information criterion approach.

Use a fit criterion for determining model fit

To help us evaluate evidence in favor of one model over another there are a number of statistics one may calculate to provide a single number for each model for comparison purposes. The criteria model evaluators available to us include Mallow’s Cp, adjusted R², Akaike Information Criterion (AIC) or the Bayesian Information Criterion (BIC) to select best model.

We already introduced the coefficient of determination R² as a measure of fit – in general we favor models with larger values of R². However, values of R² will always be larger for models with more parameters. Thus, the other evaluators attempt to adjust for the parameters in the model and how they contribute to increased model fit.  For illustrative purposes we will use Mallow’s Cp. The equation for Mallow’s Cp in linear regression is

where p is the number of parameters in the model. Mallow’s Cp is thus equal to the number of parameters in the model plus an additional amount due to lack of fit of the model (i.e., large residuals). All else being equal we favor the model in which the Cp is close to the number of parameters in the model.

In Rcmdr, select Models → Subset model selection … (Fig. 1)

Figure 1. Rcmdr popup menu, Subset model selection…

From the menu, select the criterion and how many models to return. The function returns a graph that can be used to interpret which model is best given the selection criterion used. Below is an example (although for a different data set!) for Mallow’s Cp (Fig. 2).

Figure 2. Mallow’s Cp plot

Lets break down the plot. First define the axes. The vertical axis is the range of values for the Cp calculated for each model. The horizontal axis is categorical and reads from left to right: Intercept, Inspection[T.Yes], etc., up to Suspended. Looking into the graph itself we see horizontal bars — the extent of shading indicates which model corresponds to the Cp value. For example, the lowest bar which is associated with the Cp value of 8 extends all the way to the right of the graph. This says that the model evaluated included all of the variables and therefore was the saturated or full model. The next bar from the bottom of the graph is missing only one block (lgCarIns), which tells us the Cp value 6 corresponds to a reduced model, and so forth.

Cross-validation

Once you have identified your Best Fit model, then, your proceed to run the diagnostics plots. For the rest of the discussion we return to our first example.

Rcmdr: Models → Graphs → Basic diagnostic plots.

We’ll just concern ourselves with the first row of plots (Fig. 3).

Figure 3. Diagnostic plots

The left one shows the residuals versus the predicted values — if you see a trend here, the assumption of linearity has been violated. The second plot is a test of the assumption of normality of the residuals. Interpret them (residuals OK, Residuals normally distributed? Yes/ No), and you’re done. Here, I would say I see no real trend in the residuals vs fitted plot, so assumption of linear fit is OK. For normality, there is a tailing off at the larger values of residuals, which might be of some concern (and I would start thinking about possible leverage problems), but nothing dramatic. I would conclude that our Model 6 is a good fitting model and one that could be used to make predictions.

Now, if you think a moment, you should identify a logical problem. We used the same data to “check” the model fit as we did to make the model in the first place. In particular if the model is intended to make predictions it would be advisable to check the performance of the model (e.g., does it make reasonable predictions?) by supplying new data, data not used to construct the model, into the model. If new data are not available, one acceptable practice is to divide the full data set into at least two subsets, one used to develop the model (sometimes called the calibration or training dataset) and the other used to test the model. The benefits of cross-validation include testing for influence points, over fitting of model parameters, and a reality check on the predictions generated from the model.

A note on best practices

In the 1980s, researchers often “removed” the effect of a covariate by first fitting a regression and then saving the residuals to use in separate analyses, such as t-tests, ANOVAs, or variance component estimates (eg, Dohm et al 2001). For example, in a study of plant growth, scientists might regress final plant height on initial size to remove the effect of size, then compare the residuals between fertilizer groups. While this approach adjusts for the covariate, it is problematic because residuals are not raw data: they are correlated, have constrained sums, and underestimate uncertainty. This can distort group differences, especially if the covariate distributions differ among groups, and it ignores the variability introduced by the first regression. This approach was common in ecology, evolutionary biology, physiology, and psychology before modern mixed modeling and generalized regression workflows became standard.

Modern best practices instead recommend fitting a single unified model that includes both the covariate and the grouping variable directly. Using the same plant growth example, a linear model such as estimates the effect of fertilizer while properly accounting for initial plant size. This approach provides valid standard errors, allows correct inference about group differences, and can accommodate interactions (eg, if the relationship between and differs by group) or hierarchical data structures using mixed-effects models. Additionally, modern workflows include diagnostics such as residual plots, Cook’s Distance, and variance inflation factors to check model assumptions and influential points. Use mixed models when data are hierarchical or repeated data. Random effects handle variation among groups, individuals, years, etc., in a single coherent model. When covariates need to be “controlled for,” we use partial regression. This happens inside the model, not by subtracting covariate effects first. By analyzing all relevant predictors in a single model, researchers obtain more accurate estimates, properly reflect uncertainty, and avoid the statistical pitfalls of the old residual-based two-stage approach.

Thus, there are several statistical reasons the older approach fails under modern scrutiny. First, Residuals are not independent observations. Residuals are model-dependent quantities, not raw data. Once you fit a regression, the residuals are correlated with each other, constrained to sum to zero, and, likely heteroscedastic if the model was imperfect. Using them in a new test violates standard assumptions of the second test.

Second,  you underestimate uncertainty. If you perform tests on residuals, the standard errors do not include uncertainty from the first regression, which may lead to inflated Type I error and an overly confident “adjusted” group comparisons.

Third, the old approach creates a “two-stage model” that breaks the likelihood framework. A two-stage analysis discards the correct joint likelihood and therefore produces biases p-values, and interferes with valid confidence intervals or predictions.

Under the modern one-stage models include ML/REML, generalized estimating equations, and fully integrated Bayesian models, which keeps the model coherent. Use of mixed models make residualization unnecessary. In studies with repeated measures, blocks, sites, subjects, or years, the old approach cannot partition sources of variation correctly.

In short, the advantages of the one-stage approach is that resulting models have valid uncertainty, correct group comparisons, better handling of interactions and hierarchy, full likelihood-based inference. They fit a single unified regression or mixed-effects model with all relevant predictors, examine diagnostics, and make inferences directly. This is one more benefit of the increased power and availability of more powerful computers and more complete development of mixed methods modeling.

Questions

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

Quiz Chapter 18.5

Selecting the best regression model

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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)

18.2 – Nonlinear regression

Introduction

The linear model is incredibly relevant in so many cases. A quick look “linear model” in PUBMED returns about 22 thousand hits; 3.7 million in Google Scholar; 3 thousand hits in ERIC database. These results compare to search of “statistics” in the same databases: 2.7 million (PUBMED), 7.8 million (Google Scholar), 61.4 thousand (ERIC). But all models are not the same.

Fit of a model to the data can be evaluated by looking at the plots of residuals (Fig 1), where we expect to find random distribution of residuals across the range of predictor variable.

Figure 1. Ideal plot of residuals against values of X, the predictor variable, for a well-supported linear model fit to the data.

However, clearly, there are problems for which assumption of fit to line is not appropriate. We see this, again, in patterns of residuals, eg, Figure 2.

Figure 2. Example of residual plot; pattern suggests nonlinear fit.

Fitting of polynomial linear model

Fit simple linear regression, data set Yuan et al lifespan, listed below.

R code

LinearModel.1 <- lm(cumFreq~Months, data=yuan)

summary(LinearModel.1)

Call:
lm(formula = cumFreq ~ Months, data = yuan)

Residuals:
     Min       1Q   Median      3Q     Max 
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
                 Estimate Std. Error t value Pr(>|t|) 
(Intercept)     -0.132709   0.045757   -2.90  0.0124 * 
Months           0.029605   0.001854   15.97  6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.09308 on 13 degrees of freedom
Multiple R-squared: 0.9515, Adjusted R-squared: 0.9477 
F-statistic: 254.9 on 1 and 13 DF, p-value: 6.374e-10

We see from the R² (95%), a high degree of fit to the data. However, residual plot reveals obvious trend (Fig 3)

Figure 3. Residual plot, residuals against fitted values.

We can fit a polynomial regression.

First, a second order polynomial

LinearModel.2 <- lm(cumFreq ~ poly( Months, degree=2), data=yuan)

summary(LinearModel.2)

Call:
lm(formula = cumFreq ~ poly(Months, degree = 2), data = yuan)

Residuals:
     Min       1Q   Median      3Q     Max 
-0.13996 -0.06720 -0.02338 0.07153 0.14277

Coefficients:
                           Estimate Std. Error  t value   Pr(>|t|) 
(Intercept)                 0.48900    0.02458   19.891   1.49e-10 ***
poly(Months, degree = 2)1   1.48616    0.09521   15.609   2.46e-09 ***
poly(Months, degree = 2)2   0.06195    0.09521    0.651   0.528 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.09521 on 12 degrees of freedom
Multiple R-squared: 0.9531, Adjusted R-squared: 0.9453 
F-statistic: 122 on 2 and 12 DF, p-value: 0.0000000106

Second, try a third order polynomial

LinearModel.3 <- lm(cumFreq ~ poly(Months, degree = 3), data=yuan)

summary(LinearModel.3)

Call:
lm(formula = cumFreq ~ poly(Months, degree = 3), data = yuan)

Residuals:
      Min        1Q   Median       3Q      Max 
-0.052595 -0.021533 0.001023 0.025166 0.048270

Coefficients:
                           Estimate Std. Error t value   Pr(>|t|) 
(Intercept)                0.488995   0.008982  54.442   9.90e-15 ***
poly(Months, degree = 3)1  1.486157   0.034787  42.722   1.41e-13 ***
poly(Months, degree = 3)2  0.061955   0.034787   1.781   0.103 
poly(Months, degree = 3)3 -0.308996   0.034787  -8.883   2.38e-06 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.03479 on 11 degrees of freedom
Multiple R-squared: 0.9943, Adjusted R-squared: 0.9927 
F-statistic: 635.7 on 3 and 11 DF, p-value: 1.322e-12

Which model is best? We are tempted to compare R-squared among the models, but R² turn out to be untrustworthy here. Instead, we compare using Akaike Information Criterion, AIC

R code/results

AIC(LinearModel.1,LinearModel.2, LinearModel.3)
              df       AIC
LinearModel.1  3 -24.80759
LinearModel.2  4 -23.32771
LinearModel.3  5 -52.83981

Smaller the AIC, better fit.

anova(RegModel.5,LinearModel.3, LinearModel.4)
Analysis of Variance Table

Model 1: cumFreq ~ Months
Model 2: cumFreq ~ poly(Months, degree = 2)
Model 3: cumFreq ~ poly(Months, degree = 3)
  Res.Df   RSS    Df Sum of Sq       F      Pr(>F)
1     13 0.112628
2     12 0.108789  1  0.003838  3.1719 0.1025
3     11 0.013311  1  0.095478 78.9004 0.000002383 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Logistic regression

The Logistic regression is a classic example of nonlinear model.

R code

logisticModel <-nls(yuan$cumFreq~DD/(1+exp(-(CC+bb*yuan$Months))), start=list(DD=1,CC=0.2,bb=.5),data=yuan,trace=TRUE)
5.163059 : 1.0 0.2 0.5
2.293604 : 0.90564552 -0.07274945 0.11721201
1.109135 : 0.96341283 -0.60471162 0.05066694
0.429202 : 1.29060000 -2.09743525 0.06785993
0.3863037 : 1.10392723 -2.14457296 0.08133307
0.2848133 : 0.9785669 -2.4341333 0.1058674
0.1080423 : 0.9646295 -3.1918526 0.1462331
0.005888491 : 1.0297915 -4.3908114 0.1982491
0.004374918 : 1.0386521 -4.6096564 0.2062024
0.004370212 : 1.0384803 -4.6264657 0.2068853
0.004370201 : 1.0385065 -4.6269276 0.2068962
0.004370201 : 1.0385041 -4.6269822 0.2068989

summary(logisticModel)

Formula: yuan$cumFreq ~ DD/(1 + exp(-(CC + bb * yuan$Months)))

Parameters:
      Estimate  Std. Error  t value      Pr(>|t|)
DD    1.038504    0.014471    71.77   < 2e-16 ***
CC   -4.626982    0.175109   -26.42  5.29e-12 ***
bb    0.206899    0.008777    23.57  2.03e-11 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.01908 on 12 degrees of freedom

Number of iterations to convergence: 11
Achieved convergence tolerance: 0.000006909

> AIC(logisticModel)
[1] -71.54679

Logistic regression is a statistical method for modeling the dependence of a categorical (binomial) outcome variable on one or more categorical and continuous predictor variables (Bewick et al 2005).

The logistic function is used to transform a sigmoidal curve to a more or less straight line while also changing the range of the data from binary (0 to 1) to infinity (-∞,+∞). For event with probability of occurring p, the logistic function is written as

where ln refers to the natural logarithm.

The logit link function is used in models to transform a probability, which is restricted between 0 and 1, into the log-odds scale, which ranges from negative infinity to positive infinity. This transformation makes the relationship between predictors and the outcome linear. In other words, this is an odds ratio. It represents the effect of the predictor variable on the chance that the event will occur. Much of our contingency table work in Chapter 7 would be best analyzed with the logistic regression.

The logistic regression model then very much resembles the same as we have seen before.

In R and Rcmdr we use the glm() function to model the logistic function. Logistic regression is used to model a binary outcome variable. What is a binary outcome variable? It is categorical! Examples include: Living or Dead; Diabetes Yes or No; Coronary artery disease Yes or No. Male or Female. One of the categories could be scored 0, the other scored 1. For example, living might be 0 and dead might be scored as 1. (By the way, for a binomial variable, the mean for the variable is simply the number of experimental units with “1” divided by the total sample size.)

With the addition of a binary response variable, we are now really close to the Generalized Linear Model. Now we can handle statistical models in which our predictor variables are either categorical or ratio scale. All of the rules of crossed, balanced, nested, blocked designs still apply because our model is still of a linear form.

We write our generalized linear model

just to distinguish it from a general linear model with the ratio-scale Y as the response variable.

Think of the logistic regression as modeling a threshold of change between the 0 and the 1 value. In another way, think of all of the processes in nature in which there is a slow increase, followed by a rapid increase once a transition point is met, only to see the rate of change slow down again. Growth is like that. We start small, stay relatively small until birth, then as we reach our early teen years, a rapid change in growth (height, weight) is typically seed (well, not in my case … at least for the height). The fitted curve I described is a logistic one (other models exist too). Where the linear regression function was used to minimize the squared residuals as the definition of the best fitting line, now we use the logistic as one possible way to describe or best fit this type of a curved relationship between an outcome and one or more predictor variables. We then set out to describe a model which captures when an event is unlikely to occur (the probability of dying is close to zero) AND to also describe when the event is highly likely to occur (the probability is close to one).

A simple way to view this is to think of time being the predictor (X) variable and risk of dying. If we’re talking about the lifetime of a mouse (lifespan typically about 18-36 months), then the risk of dying at one months is very low, and remains low through adulthood until the mouse begins the aging process. Here’s what the plot might look like, with the probability of dying at age X on the Y axis (probability = 0 to 1) (Fig 4).

Figure 4. Lifespan of 1881 mice from 31 inbred strains (Data from Yuan et al (2012) available at https://phenome.jax.org/projects/Yuan2).

We ask — of all the possible models we could draw, which best fits the data? The curve fitting process is called the logistic regression.

With some minor, but important differences, running the logistic regression is the same as what you have been doing so far for ANOVA and for linear regression. In Rcmdr, access the logistic regression function by invoking the Generalized Linear Model (Fig 5).

Rcmdr: Statistics → Fit models → Generalized linear model.

Figure 5. Screenshot Rcmdr GLM menu. For logistic on ratio-scale dependent variable, select gaussian family and identity link function.

Select the model as before. The box to the left accepts your binomial dependent variable; the box at right accepts your factors, your interactions, and your covariates. It permits you to inform R how to handle the factors: Crossed? Just enter the factors and follow each with a plus. If fully crossed, then the interactions may be specified with “:” to explicitly call for a two-way interaction between two (A:B) or a three-way interaction between three (A:B:C) variables. In the later case, if all of the two way interactions are of interest, simply typing A*B*C would have done it. If nested, then use %in% to specify the nesting factor.

R output

> GLM.1 <- glm(cumFreq ~ Months, family=gaussian(identity), data=yuan)

> summary(GLM.1)

Call:
glm(formula = cumFreq ~ Months, family = gaussian(identity),
data = yuan)

Deviance Residuals:
Min 1Q Median 3Q Max
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.132709 0.045757 -2.90 0.0124 *
Months 0.029605 0.001854 15.97 6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 0.008663679)

Null deviance: 2.32129 on 14 degrees of freedom
Residual deviance: 0.11263 on 13 degrees of freedom
AIC: -24.808

Number of Fisher Scoring iterations: 2

Assessing fit of the logistic regression model

Some of the differences you will see with the logistic regression is the term deviance. Deviance in statistics simply means compare one model to another and calculate some test statistic we’ll call “the deviance.” We then evaluate the size of the deviance like a chi-square goodness of fit. If the model fits the data poorly (residuals large relative to the predicted curve), then the deviance will be small and the probability will also be high — the model explains little of the data variation. On the other hand, if the deviance is large, then the probability will be small — the model explains the data, and the probability associated with the deviance will be small (significantly so? You guessed it! P < 0.05).

The Wald test statistic

where n and β refers to any of the n coefficient from the logistic regression equation and SE refers to the standard error if the coefficient. The Wald test is used to test the statistical significance of the coefficients. It is distributed approximately as a chi-squared probability distribution with one degree of freedom. The Wald test is reasonable, but has been found to give values that are not possible for the parameter (eg, negative probability).

Likelihood ratio tests are generally preferred over the Wald test. For a coefficient, the likelihood test is written as

where L₀ is the likelihood of the data when the coefficient is removed from the model (ie, set to zero value), whereas L₁ is the likelihood of the data when the coefficient is the estimated value of the coefficient. It is also distributed approximately as a chi-squared probability distribution with one degree of freedom.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. What is the primary difference between linear regression and logistic regression?
3. What type of dependent variable is used in logistic regression?
4. What is the purpose of the sigmoid (or logistic) function in this model?
5. What is the role of a significance test (eg, a Wald test) in logistic regression?
6. Why is the LRT generally considered more reliable than the Wald test for logistic regression, especially with smaller sample sizes?

Quiz Chapter 18.2

Nonlinear regression

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Data set

Yuan data

Chapter 18 content

• 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)

17.7 – Regression model fit

Introduction

In Chapter 17.5 and 17.6 we introduced the example of tadpoles body size and oxygen consumption. We ran a simple linear regression, with the following output from R

RegModel.1 <- lm(VO2~Body.mass, data=example.Tadpole)

summary(RegModel.1)

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

Residuals:
    Min      1Q    Median       3Q       Max 
-202.26 -126.35     30.20    94.01    222.55

Coefficients:
                Estimate     Std. Error    t value    Pr(>|t|) 
(Intercept)      -583.05         163.97     -3.556     0.00451 ** 
Body.mass         444.95          65.89      6.753   0.0000314 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 145.3 on 11 degrees of freedom
Multiple R-squared: 0.8057, Adjusted R-squared: 0.788 
F-statistic: 45.61 on 1 and 11 DF, p-value: 0.00003144

You should be able to pick out the estimates of slope and intercept from the table (intercept was -583 and slope was 445). Additionally, as part of your interpretation of the model, you should be able to report how much variation in VO2 was explained by tadpole body mass (coefficient of determination, , was 0.81, which means about 81% of variation in oxygen consumption by tadpoles is explained by knowing the body mass of the tadpole.

What’s left to do? We need to evaluate how well our model fits the data, i.e., we evaluate regression model fit. By fit we mean, how well does our model agree to the raw data? A poor fit and the model predictions are far from the raw data; a good fit, the model accurately predicts the raw data. In other words, if we draw a line through the raw data, a good fit line crosses through most of the points — a poor fit the line rarely passes through the cloud of points.

How to judge fit? This we can do by evaluating the error components relative to the portion of the model that explains the data. Additionally, we can perform a number of diagnostics of the model relative to the assumptions we made to perform linear regression. These diagnostics form the subject of Chapter 17.8. Here, we ask how well does the model,

fit the data?

Model fit statistics

The second part of fitting a model is to report how well the model fits the data. The next sections apply to this aspect of model fitting. The first area to focus on is the magnitude of the residuals: the greater the spread of residuals, the less well a fitted line explains the data.

In addition to the output from lm() function, which focuses on the coefficients, we typically generate the ANOVA table also.

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

Response: VO2
                Sum Sq   Df    F value       Pr(>F) 
Body.mass       962870    1     45.605   0.00003144 ***
Residuals       232245   11 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Standard error of regression

S, the Residual Standard Error (aka Standard error of regression), is an overall measure to indicate the accuracy of the fitted line: it tells us how good the regression is in predicting the dependence of response variable on the independent variable. A large value for S indicates a poor fit. One equation for S is given by

In the above example, S = 145.3 (underlined, bold in regression output above). We can see how if  is large, S will be large indicating poor fit of the linear model to the data. However, by itself S is not of much value as a diagnostic as it is difficult to know what to make of 145.3, for example. Is this a large value for S? Is it small? We don’t have any context to judge S, so additional diagnostics have been developed.

Coefficient of determination

, the coefficient of determination, is also used to describe model fit. can take on values between 0 and 1 (0% to 100%). A good model fit has a high  value. In our example above, or 80.57%. One equation for is given by

A value of close to 1 means that the regression “explains” nearly all of the variation in the response variable, and would indicate the model is a good fit to the data. As we mentioned in Chapter 17.2, the coefficient of determination, , is the squared value of r, the product moment correlation.

Adjusted R-squared

Before moving on we need to remark on the difference between and adjusted . For Simple Linear Regression there is but one predictor variable, X; for multiple regression there can be many additional predictor variables. Without some correction,   will increase with each additional predictor variables. This doesn’t mean the model is more useful, however, and in particular, one cannot compare  between models with different numbers of predictors. Therefore, an adjustment is used so that the coefficient of determination remains a useful way to assess how reliable a model is and to permit comparisons of models. Thus, we have the Adjusted , which is calculated as

In our example above, Adjusted R ² = 0.3806 or 38.06%.

Which should you report? Adjusted ,because it is independent of the number of parameters in the model.

Both  and S are useful for regression diagnostics, a topic which we will discuss next (Chapter 17.8).

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. As you may know, the US census is used to reapportion congress and also to determine the number of electoral college votes. In honor of the election for US President that’s just days away, in the next series of questions in this Chapter and subsequent sections of Chapter 17 and 18, I’ll ask you to conduct a regression analysis on the electoral college. For starters, make the regression of Electoral votes on 2010 census population. (Ignore for now the other columns, just focus on POP_2019 and Electoral.) Report the
   • regression coefficients (slope, intercept)
   • percent of the variation in electoral college votes explained by the regression (  ).
3. Make a scatterplot and add the regression line to the plot

Quiz Chapter 17.7

Regression model fit

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Data set

USA population year 2010 and year 2019 with Electoral College counts

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)

17.3 – Estimation of linear regression coefficients

Introduction

In discussing correlations we made the distinction between inference, testing the statistical significance of the estimate, and the process of getting the estimate of the parameter itself. Estimating parameters is possible for any data set; whether or not the particular model is a good and useful model is another matter. Statisticians speak about the fit of a model… that a model with one or more independent predictor variables explains a substantial amount of the variation in the dependent variable, that it describes the relationship between the predictors and the dependent variable without bias.

A number of tools have been developed to assess model fit. For starters, I’ll list just two ways you can approach whether or not a linear model fits your data ore requires some intervention on your part.

Assess fit of a linear model

Recall our R output from the regression of Number of Matings on Body Mass from the bird data set. We used the linear model function.

LinearModel.1 <- lm(Matings ~ Body.Mass, data=bird_matings)
summary(LinearModel.1)
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

Request R to print the ANOVA table.

Anova(RegModel.1, type="II")
Anova Table (Type II tests)
Response: Matings 
            Sum Sq   Df   F value    Pr(>F) 
Body.Mass   22.528    1    7.1438   0.02551 * 
Residuals   28.381    9

With a little rounding we have the following statistical model

and in English, Number of matings equals Body.mass multiplied by 0.36 then subtract 8.5; the intercept was -8.5, the slope was 0.36.

From the R output we see that the estimate of the slope was +0.36, statistically different from zero (p = 0.025). The intercept was -8.5, but not statistically significant (p = 0.103), which means the intercept may be zero.

As a general rule, if you make an estimate of a parameter or coefficient, then you should provide a confidence interval of the form.

estimate + critical value X standard error of the estimate

For confidence interval of regression slope we have a couple of options in R

Option 1.

confint(LinearModel.1, 'Body.Mass', level=0.95)

               2.5 %    97.5 %
Body.Mass 0.05565899 0.6689173

Option 2.

Goal: Extract the coefficients from the output of the linear model and calculate the approximate SE with nine degrees of freedom. This is the big advantage of saving output from functions as objects. Typically, much more information is about the results are available, and, additionally, can be retrieved for additional use. Extracting coefficients from the objects is the best option, but does come with a learning curve.  Let’s get started.

First, what information is available in the linear model output beyond the default information? To find out, use the names() function

names(LinearModel.1)

R output

 [1] "coefficients"  "residuals"     "effects"       "rank"         
 [5] "fitted.values" "assign"        "qr"            "df.residual"  
 [9] "xlevels"       "call"          "terms"         "model"

Another way is to use the summary() function call.

summary(LinearModel.1)$coefficients
              Estimate Std. Error   t value   Pr(>|t|)
(Intercept) -8.4745763  4.6640856 -1.816986 0.10259256
Body.Mass    0.3622881  0.1355472  2.672781 0.02550595

How can we get just the standard error for the slope? Note that the estimates are reported in a 2X4 matrix like so

Therefore, to get the standard error for the slope we identify that it is stored in cell 2,2 of the matrix and we write

summary(LinearModel.1)$coefficients[2,2]

which returns

[1] 0.1355472

Let’s use this information to calculate confidence intervals

slp=summary(LinearModel.1)$coefficients[2,1]
slpErrs=summary(LinearModel.1)$coefficients[2,2] 
slp + c(-1,1)*slpErrs*qt(0.975, 9)

where qt() is the quantile function for the t distribution and “9” is the degrees of freedom from the regression. Results follow

coef + c(-1,1)*errs*qt(0.975, 9)
[1] 0.05565899 0.66891728

And for the intercept

int=summary(LinearModel.1)$coefficients[1,1]
intErrs=summary(LinearModel.1)$coefficients[1,2] 
int + c(-1,1)*intErrs*qt(0.975, 9)

Results

int + c(-1,1)*intErrs*qt(0.975, 9)
[1] -19.025471   2.076318

In conclusion, part of fitting a model includes reporting the estimates of the coefficients (model parameters). And, in general, when estimation is performed, reporting of suitable confidence intervals are expected.

Extract additional statistics from R’s linear model function

The summary() function is used to report the general results from ANOVA and linear model function output in R software, but additional functions can be used to extract the rest of the output, e.g., coefficient of determination. To complete our example of extracting information from the summary() function, we next turn to summary.lm() function to see what is available.

At the R prompt type and submit

summary.lm(LinearModel.1)

returns the following R output

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

Looks exactly like the output from summary(). Let’s look at what is available in the summary.lm() function

names(summary.lm(LinearModel.1))
 [1] "call"          "terms"         "residuals"     "coefficients" 
 [5] "aliased"       "sigma"         "df"            "r.squared"    
 [9] "adj.r.squared" "fstatistic"    "cov.unscaled"

We see some information we got from summary(), e.g., “coefficients”. If we interrogate the name coefficients like so

summary.lm(LinearModel.1)$coefficients

we get

summary.lm(LinearModel.1)$coefficients
              Estimate Std. Error   t value   Pr(>|t|)
(Intercept) -8.4745763  4.6640856 -1.816986 0.10259256
Body.Mass    0.3622881  0.1355472  2.672781 0.02550595

which, again, is a 2X4 matrix (see above)

So to get the standard error for the slope we identify that it is stored in cell 2,2 of the matrix and call it LinearModel.1$coefficients[2,2].

Questions

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

2. What does a small standard error of the slope for a SLR model tell us?

3. For slope, intercept, for just about any statistical estimate, why are we obligated to provide a confidence interval?

Quiz Chapter 17.3

Estimation of linear regression coefficients

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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)

14.8 – More on the linear model in Rcmdr

Introduction

During the last lectures, we could have used the Two-Way ANOVA command in R or Rcmdr

R code: How to analyze multifactorial ANOVA problems

Rcmdr: Statistics → Means → Multiway ANOVA

To analyze our two factor data sets. As long as the design meets the following conditions, by all means use this command because it is simple and precisely correct.

• Both factors are fixed, not random.
• Each level of first factor is crossed with each level of the second factor.
• No missing data (the design is fully replicated and balanced).

If any of the three points do not fit your two-way design, then you’ll need a different, more general and powerful ANOVA procedure in R and Rcmdr to analyze these types of designs. You’ll need the lm() function (Fig 1).

Rcmdr: Statistics → Fit models → Linear model…

Figure 1. Linear model menu in Rcmdr, version 2.7.0

Some R basics with the lm() function, the general linear model.

Response is specified by linear predictor(s), either factors or covariates (ratio-scale predictor variables). We communicate to R what the model is by using operators. The four more commonly used operators are

the basic way to include the model terms, ie, the mode predictors

which is interpreted as the interaction of all the variables and the factors in the term

is interpreted as factor crossing

Not shown, but indicates the term on the left is nested within the term on the right.

A few examples: we’ll have three factors, A, B, and C. For our one-way ANOVA, the model specification would be

For our crossed, balanced two-way ANOVA, the model specification would be

or equivalently

And for our block ANOVA problem?

Click here to get entire List of model commands in R.

How would we analyze our snake experiment?

Table 1. The snake data set

We have two factors, but one factor is a Block (repeated measures on individuals). We need to tell R and Rcmdr what our model is. We’ll return to talk about models next time, a very important topic!! For now, think of a model as adding the independent variables together to predict the response variable.

In our Snake example, it’s a two-way ANOVA, but one factor is individual snake, the other is a treatment, and we have repeat measures, so there cannot be an interaction.

We tell R and Rcmdr which columns contain the Response, and under Model, we enter the columns with the two factors.

Figure 2. Menu of linear model with repeat measures model, Rcmdr, version 2.7.0.

You must also tell R and Rcmdr which factors in the model (if any) are random to get the correct F statistics. Almost without exception, blocking factors are always treated as Random

The output looks like this (see below). More complicated, true (which means more information!), but things marked in red we’ve seen before.

LinearModel.2 <- lm(flick ~ Snake +Source, data=L16SnakeTaste)

summary(LinearModel.2)

Call:
lm(formula = flick ~ Snake + Source, data = L16SnakeTaste)

Residuals:
Min 1Q Median 3Q Max 
-5.2222 -0.7222 0.0278 1.5694 7.4444

Coefficients:
             Estimate Std. Error t value Pr(>|t|) 
(Intercept)    -4.444      2.523  -1.762 0.10862 
Snake[T.2]      9.667      3.090   3.128 0.01072 * 
Snake[T.3]      3.000      3.090   0.971 0.35451 
Snake[T.4]     12.667      3.090   4.099 0.00215 ** 
Snake[T.5]      6.000      3.090   1.942 0.08085 . 
Snake[T.6]      6.333      3.090   2.050 0.06755 . 
Source[T.fish] 14.167      2.185   6.484 0.0000704 ***
Source[T.worm] 14.167      2.185   6.484 0.0000704 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.784 on 10 degrees of freedom
Multiple R-squared: 0.8858, Adjusted R-squared: 0.8058 
F-statistic: 11.08 on 7 and 10 DF, p-value: 0.0005337

Contrast coding

When a factor (categorical variable) is used in a linear model (lm() or aov()), R must convert the factor into one or more numeric predictor variables. Contrast coding tells us how to represent the factor in the model, how to interpret the coefficients associated with the factor, how the hypothesis tests of the coefficients are handled, and whether or not Type III sums of squares tests are valid. R can handle contrast coding in a couple of ways, either as a dummy contrast (each group if the factor is contrasted to a baseline) or effects coding, where group means are compared to the grand mean. The command summary(lm()) reports contrast coding that was used in the linear model. Every kind of contrast leads to the same fitted values and same overall F-test, but coefficient estimates and t-tests change.

For the ANOVA table, we call up the command via

Rcmdr: Models → Hypothesis tests → ANOVA table… (Fig 3).

Figure 3. Rcmdr: Models → Hypothesis tests → ANOVA table… Rcmdr, version 2.7.0

Confirm that the model object is active (in this case, the object was LinearModel.2), accept the defaults about types of tests and marginality, and submit OK. The output is

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

Response: flick
          Sum Sq Df F value Pr(>F) 
Snake     307.61  5  4.2956 0.02396 * 
Source    802.78  2 28.0256 0.00007954 ***
Residuals 143.22 10 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End R output

You do not need to know all of the output; all of that output is there for a reason, of course, but for now, here’s what R and Rcmdr has to say (from the help menu):

The sequential sums of squares is the added sums of squares given that prior terms are in the model. These values depend upon the model order. The adjusted sums of squares are the sums of squares given that all other terms are in the model. These values do not depend upon the model order.

You should try our snake example again, but this time, remove the tongue flick response to the dH2O for the first snake — a missing value. (just type into the cell NA).

How would you use GLM to analyze a simple two-way ANOVA, a fully crossed, fully replicated (“balanced”) design? We could use the Two-way ANOVA command in R and Rcmdr, or we could use lm(). Try it with the data set from random vs nonrandom lecture.

Another example

Below, you see how the model is entered. To let R know that I wish R and Rcmdr to test the interaction, I need to add a Model term for that source of variation. I accomplish this by typing “Diet*Drug” (without the quotes).

Figure 4. Crossed, balanced design. Linear model menu, Rcmdr, version 1.9.2

After clicking OK, the following output from the lm function is returned. How does this compare to output from the two-way ANOVA command in R and Rcmdr?

You should try both and compare!

Rcmdr: Models → Hypothesis tests → ANOVA table

Anova(LinearModel.17, type="II")
Anova Table (Type II tests)
Response: chol_randomized
             Sum Sq  Df F value   Pr(>F) 
diet         141.08   2  1.3061 0.28745 
drug         351.88   2  3.2577 0.05403 .
diet:drug    235.50   4  1.0901 0.38124 
Residuals   1458.19  27

End R output

Nested ANOVA

The nested ANOVA may be analyzed in multiple ways in R and Rcmdr, but I prefer the lm() function because it is the most general. For Nested ANOVA, we can also use lm(). Here’s where it gets a little tricky. Put in the Response variable (Chol), then click in the box for model: Select both factors, then type in / after the factor that’s nesting factor. For our nested model example (14.5 – Nested designs), Manufacturer Source was nested within Drug.

Table 3. Nested design example data set from Chapter 14 – Nested design

Figure 5. Nested design, linear model menu, Rcmdr, version 1.9.2

Fortunately, R and Rcmdr’s help system is quite extensive here, so when in doubt, check the help box…

Output from the linear model for the Nested Example looks like the one below.

The General Linear Model function in R and therefore Rcmdr returns information about our design plus Sums of Squares, Mean squares, and P-values. R and Rcmdr default’s to use of sequential evaluation of effects. Adjusted evaluation is useful for when you have a covariate (like body size or another confounding variable) that should be evaluated first before the factors are evaluated. We will use the sequential analysis.

Rcmdr: Models → Hypothesis tests → ANOVA table

Anova(LinearModel.15, type="II")
Anova Table (Type II tests)
Response: Chol
             Sum Sq Df F value Pr(>F)
Drug         225.14  2  1.1743 0.3262
Drug:Source  269.27  3  0.9363 0.4385
Residuals   2300.61 24

End R output

Repeatability and ANOVA

We need to tell Rcmdr how to structure the error term; you need the data frame to be arranged so

aovRes <- aov(dH2O ~ Source + Error(Source/Subject), data=SnakeTaste)
#Print the results
aovRes
Anova Table (Type II tests)
Response: dH2O
            Sum Sq Df F value   Pr(>F)    
Subject     307.61  5  4.2956 0.02396 *  
Source      802.78  2 28.0256 0.00007954 ***
Residuals   143.22 10                       
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#What components are available in the aovRes object?
names(aovRes)
[1] "Sum Sq"  "Df"      "F value" "Pr(>F)"

How do I extract the “F value” for Subjects?

str(aovRes)

Questions

1. What are three learning objectives from this page?
2. Be able to distinguish use of lm() and aov() and the summary output with respect to testing of linear models.

Quiz Chapter 14.8

More on the linear model in Rcmdr

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Data sets

Table 1. The snake data set

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

12.7 – Many tests one model

Introduction

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

Brief overview of linear models

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

The general Linear Model, LM

In matrix form, the LM can be written as

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

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

R code

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

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

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

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

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

If you recall from your genetics course,

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

Managing data for linear models

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

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

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

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

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

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

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

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

R code to make Figure 4 plots

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

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

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

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

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

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

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

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

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

Step 3. Click OK to carry out the command.

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

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

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

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

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

End R output

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

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

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

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

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

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

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

Figure 7. Options for types of tests.

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

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

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

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

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

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

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

Why use the linear model approach?

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

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

Output from R

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

Questions

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

Quiz Chapter 12.7

Many tests one model

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Data set and R code used in this page

Corn and Hiesey (1973)

Chapter 12 contents

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