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.

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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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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)

18.1 – Multiple Linear Regression

Introduction

Last time we introduced simple linear regression:

• one independent X variable
• one dependent Y variable.

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

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

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

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

R code

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

Rcmdr: Statistics → Fit model → Linear model

You may, however, access linear regression through R Commander

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

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

Rcmdr: Models → Hypothesis testing → ANOVA tables

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

Our full multiple regression model becomes

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

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

Selecting the best model

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

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

Let’s get the full regression model

The statistical model is

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

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

Rmdr: Statistics → Fit models → Linear model

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

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

The output

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

R code for Figure 3.

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

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

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

Diagnostic plots

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

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

Rcmdr: Models → Graphs → Basic diagnostics plots

Four plots are returned

Figure 5. R’s default regression diagnostic plots.

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

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

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

Interpreting the diagnostic plots for this problem

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

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

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

The multicollinearity problem

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

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

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

Tolerance and Variance Inflation Factor

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

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

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

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

Rcmdr: Models → Numerical diagnostics → Variation-inflation factors

vif(RegModel.2)
      BMI     Dose 
1.010256 1.010256

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

Solutions for multicollinearity

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

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

Questions

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

Quiz Chapter 18.1

Multiple Linear Regression

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Chapter 18 contents

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

17.3 – 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)

17 – Linear Regression

Introduction

Regression is a toolkit for developing models of cause and effect between one ratio scale data type dependent response variables, and one (simple linear regression) or more or more (multiple linear regression) ratio scale data type independent predictor variables. By convention the dependent variable(s) is denoted by Y, the independent variable(s) represented by X₁, X₂, X_n for n independent variables. Like ANOVA, linear regression is simply a special case of the general linear model, first introduced in Chapter 12.7.

Components of a statistical model

Regression statistical methods return model estimates of the intercept and slope coefficients, plus statistics of regression fit (eg, R², aka “R-squared,” the coefficient of determination).

Chapter 17.1 – 17.9 cover the simple linear model

Chapter 18.1 – 18.5 cover the multiple regression linear model

where α or β₀ represent the Y-intercept and β or β₁, β₂, … β_n represent the regression slopes.

Regression and correlation test linear hypotheses

We state that the relationship between two variables is linear (the alternative hypothesis) or it is not (the null hypothesis). The difference? Correlation is a test of linear association (are variables correlated, we ask?), imply possible causation, but are not sufficient evidence for causation: we do not imply that one variable causes another to vary, even if the correlation between the two variables is large and positive, for example. Correlations are used in statistics on data sets not collected from explicit experimental designs incorporated to test specific hypotheses of cause and effect.

Linear regression, however, is to cause and effect as correlation is to association. With regression and ANOVA, we are indeed making a case for a particular understanding of the cause of variation in a response variable: modeling cause and effect is the goal. Regression, ANOVA, and other general linear models are designed to permit the statistician to control for the effects of confounding variables provided the causal variables themselves are uncorrelated.

When to use correlation and when to apply linear regression to data set? For two ratio scale variables, one can always apply either approach. Correlation is always appropriate if both variables are measured in the course of the study whereas a regression modeling approach would be implied where one variable was the result of manipulation by the researcher.

Assumptions of linear regression

The key assumption in linear regression is that a straight line indeed is the best fit of the relationship between dependent and independent variables. The additional assumptions of parametric tests (Chapter 13) also hold. In Chapter 18 we conclude with an extension of regression from one to many predictor variables and the special and important topic of correlated predictor variables or multicollinearity.

Build a statistical model, make predictions

In our exploration of linear regression we begin with simple linear regression, also called ordinary least squares regression, starting with one predictor variable. Practical aspects of model diagnostics are presented. Regression may be used to describe or to provide a predictive statistical framework. In Chapter 18 we conclude with an extension of regression from one to many predictor variables. We conclude with a discussion of model selection. Throughout, use of Rcmdr and R have multiple ways to analyze linear regression models are presented; we will continue to emphasize the general linear model approach, but note that use of linear model in Rcmdr provides a number of default features that are conveniently available.

Questions

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

Suggested homework

Homework 11: Simple linear regression.

Quiz

In addition to the quiz on this page, a total of 81 questions among the several subchapters, a mix of true or false and multiple choice question format.

Quiz Chapter 17

Linear Regression

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References

Linear regression is a huge topic; references I include are among my favorite on the subject, but are only a small and incomplete sampling. For simplicity, I merged references for Chapter 17 and Chapter 18 into one page at References and suggested readings (Ch17 & 18)

Chapter 17 contains

• 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