18.3 – Logistic regression

Introduction

We briefly introduced logistic regression in the previous chapter on nonlinear regression. We expand our discussion of logistic regression here.

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 may be 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.

This is an odds ratio. It represents the effect of the predictor variable on the chance that the event will occur.

The logistic regression model then very much resembles the same general linear models 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

    \begin{align*} G \sim Model \end{align*}

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 (see Chapter 20.10 for related growth and related models). 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 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 1).

lifespan mice, Yuan et al 2012 PNAS 109:8224

Figure 1. Lifespan of 1881 mice from 31 inbred strains (Data from Yuan et al [2012] available at https://phenome.jax.org/projects/Yuan2).

Note 1: I labeled Y axis labeled “Survival Probability”; “Inverse Survival Probability” would be more accurate.

We ask — of all the possible models we could draw — which model best fits the data? The curve fitting process is called the logistic regression. The sample data set is listed at end of this page (scroll down of click here). Create data.frame called yuan.

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 calling the Generalized Linear Model (Fig 2).

Screenshot Rcmdr, select Statistics, Fit models, Generalized linear model...

Figure 2. Access Generalized Linear Model via R Commander

R results

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

Rcmdr: Statistics → Fit models → Generalized linear model.

Screenshot Rcmdr GLM

Figure 3. 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

Note 2: The %in% operator is an infix function used to determine if elements of a vector or object on the left-hand side are present within another vector or object on the right-hand side. Quietly, we introduced use of %in% in our code to subset data set in a BONUS problem.

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 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.

Nonlinear regression

Nonlinear regression, nls() function, may be a better choice. The following

attach(yuan)
logisticModel <-nls(cumFreq~DD/(1+exp(-(CC+bb*Months))), start=list(DD=1,CC=0.2,bb=.5),data=yuan,trace=TRUE)
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

Get fit statistics

AIC(logisticModel)
[1] -71.54679

Because AIC for the nonlinear model much smaller (more negative) than AIC for logistic model, we may be tempted to judge fit of the nonlinear regression as best. However, this comparison of models is not valid because the Y variables are different between the two models and the fit families are different. One option, evaluate fit of models by plots of residuals (see 17.7 – Regression model fit).

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 18.3

Logistic regression

Data set this page

MonthsfreqcumFreq
000
30.0106330.010633
60.0170120.027645
90.0451890.072834
120.0643270.137161
150.0648590.20202
180.097820.299841
210.1185540.418394
240.1711860.58958
270.1621480.751728
300.1371610.888889
330.0696440.958533
360.0244550.982988
390.0116960.994684
420.0053161

Chapter 18 contents

18.6 – Compare two linear models

Introduction

Rcmdr (R) provides a very useful tool to compare models. Now, you can compare any two models, but this would be a poor strategy. Use this tool to perform in effect a stepwise test by hand. As one of the models, select for example the saturated model, then for the second model, select one in which you drop one model factor. In the example below, I dropped the two-way interaction from the saturated model (a logistic regression model, actually):

The model was Type.II diabetes = Treatment + Samples + Gender + BMI + Age + Gender:Treatment

where Type.II diabetes is a binomial (Yes,No) dependent variable and Treatment and Gender were categorical factors. The ANOVA table is shown below.

Anova(GLM.1, type="II", test="LR")
Analysis of Deviance Table (Type II tests)
Response: Type.II
                LR Chisq  Df Pr(>Chisq)
Treatment          0.266  7    0.9999
Samples           38.880  1    4.508e-10 ***
Gender             0.671  1    0.4127
BMI                2.259  1    0.1329
Age                2.064  1    0.1508
Treatment:Gender   1.803  1    0.1794

From this output we see that there are a number of terms that are not significant (P < 0.05), but with one exception (Treatment) they seem to contribute to the total variation (P values are between 0.13 and 0.4). So, we conclude that the saturated model is not the best fit model, and proceed to evaluate alternative models in search of the best one.

As a matter of practice I first drop the interaction term. Here’s the ANOVA table for the second model now without the interaction

Anova(GLM.1, type="II", test="LR")
Analysis of Deviance Table (Type II tests)
Response: Type.II
              LR Chisq Df Pr(>Chisq)
Treatment        0.266  7    0.9999
Samples         37.086  1    1.13e-09 ***
Gender           0.671  1    0.4127
BMI              2.017  1    0.1556
Age              1.794  1    0.1804

Both models look about the same. Which one is best? We now wish to know if dropping the interaction harms the model in any way. We will use the AIC (Akaike Information Criterion) to evaluate the models. AIC provides a way to assess which among a set of nested models is better. The preferred model is the one with the lowest AIC value.

To access the AIC calculation, just enter the script AIC(model name), where model name refers to one of the models you wish to evaluate (e.g., GLM.1), then submit the code

AIC(GLM.1)
50.65518

AIC(GLM.2)
50.45793

Thus, we prefer the second model (GLM.2) because the AIC is lower.

AIC does not provide a statistical test of model fit. To access the model comparison tool, simply select

Models → Hypothesis tests → Compare two models…

and the following screen will appear (Fig 1).

Figure 1. Screenshot Rcmdr compare models menu.

Select the two models to compare (in this case, GLM1 and GLM2), then press OK button. R output

anova(GLM.1, GLM.2, test="Chisq")
Analysis of Deviance Table
Model 1: Type.II ~ Treatment + Samples + Gender + BMI + Age + Gender:Treatment
Model 2: Type.II ~ Treatment + Samples + Gender + BMI + Age
    Resid. Df  Resid. Dev  Df  Deviance P(>|Chi|)
1          49      24.655
2          50      26.458  -1   -1.8027   0.1794

We see that P >0.05 (= 0.1794), which means the fit of the model is fine if we lose the one term.

Deviance

Those of you working with logistic regressions will see this new term, “deviance.” Deviance is a statistical term relevant to model fitting. Think of it like a chi-square test statistic. The idea is that you compare your fitted model against the data in which the only thing estimate is the intercept. Do the additional components of the model add significantly to the prediction of the original data? If they do, dropping the term will have a significant effect on the model fit and the P-value would be less than 0.05. In this example, we see that dropping the interaction term had little effect on the deviance score and in agreement, the P value is larger than 0.05. It means we can drop the term and the new model lacking the term is in some sense better: fewer predictors, a simpler model.

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 18.6

Compare two linear models

Chapter 18 contents

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.

A good residual plot, no obvious trends.

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.

Residual plot shows systematic trend.

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)

Residual plot of nonlinear data on simple linear model

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

    \begin{align*} logit(p) = ln\left( \frac{p}{1-p} \right) \end{align*}

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.

    \begin{align*} logit(p) = logit(p) = \beta_{0}+\beta_{1}X_{1}+\beta_{2}X_{2}+\cdots + \beta_{n}X_{n} \end{align*}

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

    \begin{align*} G \sim Model \end{align*}

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).

lifespan mice, Yuan et al 2012 PNAS 109:8224

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.

Screenshot Rcmdr GLM menu.

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

Data set

Yuan data

MonthsfreqcumFreq
000
30.0106330.010633
60.0170120.027645
90.0451890.072834
120.0643270.137161
150.0648590.20202
180.097820.299841
210.1185540.418394
240.1711860.58958
270.1621480.751728
300.1371610.888889
330.0696440.958533
360.0244550.982988
390.0116960.994684
420.0053161

Chapter 18 content

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…

Menus to seelct linear model function, lm()

Figure 1. Linear model menu in Rcmdr, version 2.7.0

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

    \begin{align*} Y\sim model \end{align*}

Response Y 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 \%in\% 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

    \begin{align*} Y\sim A \end{align*}

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

 

    \begin{align*} Y\sim A + B + A:B \end{align*}

or equivalently

    \begin{align*} Y\sim A*B \end{align*}

 

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

Snake Source flick
1 dH2O 3
2 dH2O 0
3 dH2O 0
4 dH2O 5
5 dH2O 1
6 dH2O 2
1 fish 6
2 fish 22
3 fish 12
4 fish 24
5 fish 16
6 fish 16
1 worm 6
2 worm 22
3 worm 12
4 worm 24
5 worm 16
6 worm 16

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.

repeat measure linear model

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.

Note 1: Type I sums of squares tests are affected by the order of the predictors in the model, not by type of contrast coding used. R Commander defaults to Type II tests, which are generally unaffected by coding contrasts used. In contrast, Type III tests are affected by the type of coding contrasts used — Type III tests require sum-to-zero (effects) contrasts.

For the ANOVA table, we call up the command via

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

Rcmdr menu: Model --> Hypothesis testing --> ANOVA table

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

Drug Source Chol
1 1 202.6
1 1 207.8
1 1 190.2
1 1 211.7
1 1 201.5
1 2 189.3
1 2 198.5
1 2 208.4
1 2 205.3
1 2 210
2 3 212.3
2 3 204.4
2 3 221.6
2 3 209.2
2 3 222.1
2 4 203.6
2 4 209.8
2 4 204.1
2 4 201.8
2 4 202.6
3 5 189.1
3 5 219.9
3 5 196
3 5 205.3
3 5 204
3 6 194.7
3 6 192.8
3 6 226.5
3 6 200.9
3 6 219.7

Note 2: If you were working with a CROSSED model, then you would enter the two factors and indicate the interaction by typing Drug*Source (if these are the two factors involved in the interaction).

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

Data sets

Table 1. The snake data set

Snake Source flick
1 dH2O 3
2 dH2O 0
3 dH2O 0
4 dH2O 5
5 dH2O 1
6 dH2O 2
1 fish 6
2 fish 22
3 fish 12
4 fish 24
5 fish 16
6 fish 16
1 worm 6
2 worm 22
3 worm 12
4 worm 24
5 worm 16
6 worm 16

 


Chapter 14 contents

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

    \begin{align*} \mathbf{\hat{Y}}=\mathbf{Xb} \end{align*}

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

    \begin{align*} \mathbf{Y}=\mathbf{X}\beta +e \end{align*}

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).

ohia image from Wikipedia

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, P = G + E

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).

Rcmdr data view

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).

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.

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).

Rcmdr menu, select linear model

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.

Rcmdr, Input linear model formula

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.

Note 1: Brief description of linear model output; these will be discussed fuller in Chapter 17 and Chapter 18. The residual standard error is a measure of how well a model fits the data. The Adjusted R-squared is calculated by dividing the residual mean square error by the total mean square error. The result is then subtracted from 1.

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).

Rcmdr: To retrieve an ANOVA table, select Models, Hypothesis tests, then ANOVA table...

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).

ptions for types of tests.

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.

Note 2: Marginal or partial effects are slopes (or first derivatives): they quantify the change in one variable given change in one or more independent variables. Type I tests are sequential: sums of squares are calculated in the order the predictor variables are entered into the model. Type II tests the sums of squares are calculated after adjusting for some of the variables in the model. Type III, every sum of square calculation is adjusted for all other variables in the model. Sandwich estimator refers to algorithms for calculating the structure of errors or residuals remaining after the predictor variables are fitted to the data. The assumption for ordinary least square estimation (See Chapter 17) is that errors across the predictors are equal, ie, equal variances assumption. HC refers to “heteroscedasticity consistent” (Hayes and Chai 2007), which infers fitting of a model does not assume equal error variance across all observations.

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

Data set and R code used in this page

Corn and Hiesey (1973)

SiteHeightWidth
M-112.556719.1264
M-113.201913.1547
M-18.069916.032
M-16.095222.8586
M-111.387911.0105
M-112.224221.8102
M-116.014711.0488
M-119.740325.9756
M-136.482425.2867
M-113.123320.0487
M-121.772524.8511
M-114.201343.7679
M-137.762937.3438
M-12.86522.5549
M-10.645622.8013
M-129.62320.0194
M-110.581229.0328
M-118.304622.2867
M-119.052824.684
M-12.569335.74
M-245.016214.3878
M-240.840418.8396
M-227.103221.0547
M-229.803616.9327
M-263.831630.7037
M-242.1073.2491
M-230.032247.4412
M-234.051642.239
M-215.766432.8354
M-235.126250.9698
M-243.698819.3897
M-226.758513.8168
M-236.78950.5817
M-230.945853.7757
M-226.846515.4137
M-240.38839.2161
M-230.655556.8456
M-219.973644.9411
M-227.67636.8543
M-244.08424.3396
M-315.264611.4999
M-319.67459.7757
M-323.27512.7825
M-316.11612.4065
M-316.83931.1253
M-323.1073.7349
M-321.53226.9725
M-313.419112.2867
M-314.727311.4841
M-318.424511.9078

Chapter 12 contents