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

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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)

18.1 – Multiple Linear Regression

Introduction

Last time we introduced simple linear regression:

• one independent X variable
• one dependent Y variable.

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

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

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

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

R code

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

Rcmdr: Statistics → Fit model → Linear model

You may, however, access linear regression through R Commander

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

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

Rcmdr: Models → Hypothesis testing → ANOVA tables

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

Our full multiple regression model becomes

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

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

Selecting the best model

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

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

Let’s get the full regression model

The statistical model is

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

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

Rmdr: Statistics → Fit models → Linear model

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

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

The output

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

R code for Figure 3.

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

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

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

Diagnostic plots

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

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

Rcmdr: Models → Graphs → Basic diagnostics plots

Four plots are returned

Figure 5. R’s default regression diagnostic plots.

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

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

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

Interpreting the diagnostic plots for this problem

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

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

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

The multicollinearity problem

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

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

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

Tolerance and Variance Inflation Factor

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

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

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

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

Rcmdr: Models → Numerical diagnostics → Variation-inflation factors

vif(RegModel.2)
      BMI     Dose 
1.010256 1.010256

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

Solutions for multicollinearity

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

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

Questions

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

Quiz Chapter 18.1

Multiple Linear Regression

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Data set

Chapter 18 contents

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

17.8 – Assumptions and model diagnostics for Simple Linear Regression

Introduction

The assumptions for all linear regression:

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

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

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

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

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

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

Model diagnostics

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

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

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

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

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

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

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

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

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

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

Graph options

Using residual plots to diagnose regression equations

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

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

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

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

Here are examples of POSSIBLE outcomes:

Figure 1. An ideal plot of residuals

Solution: Proceed! Assumptions of linear regression met.

Compare to plots of residuals that differ from the ideal.

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

Solution. Try a transform like the log₁₀-transform.

Figure 3. Problem. Residual plot shows systematic trend.

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

Figure 4. Problem. Residual plot shows nonlinear trend.

Solution. Transform data or use more complex model

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

Diagnostic plot examples

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

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

In brief, we look at plots

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

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

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

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

Statistical test options

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

Results for the model were:

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

Residuals:
    Min      1Q Median    3Q    Max 
-163.12 -125.53 -20.27 83.71 228.56

Coefficients:
           Estimate Std. Error t value  Pr(>|t|) 
(Intercept) -595.37     239.87  -2.482 0.03487 * 
Body.mass    431.20     115.15   3.745 0.00459 **
Gosner[T.II]  64.96     132.83   0.489 0.63648 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 153.4 on 9 degrees of freedom
  (1 observation deleted due to missingness)
Multiple R-squared: 0.8047, Adjusted R-squared: 0.7613 
F-statistic: 18.54 on 2 and 9 DF, p-value: 0.0006432

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

In R Commander, the diagnostic test options are available via

Rcmdr: Models → Numerical diagnostics

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

R code:

vif(RegModel.2)

R output

Body.mass Gosner 
2.186347 2.186347 

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

Our interpretation

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

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

R code:

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

R output

Breusch-Pagan test

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

Our interpretation

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

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

R code:

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

R output

Durbin-Watson test

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

Our interpretation

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

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

R code:

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

R output

RESET test

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

Our interpretation

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

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

R code:

outlierTest(LinearModel.3)

R output

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

Our interpretation

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

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

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

R code:

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

R output

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

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

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

Our interpretation

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

Questions

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

Quiz Chapter 17.8

Assumptions and model diagnostics for Simple Linear Regression

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Chapter 17 contents

• Introduction
• Simple Linear Regression
• Relationship between the slope and the correlation
• Estimation of linear regression coefficients
• OLS, RMA, and smoothing functions
• Testing regression coefficients
• Regression model fit
• Assumptions and model diagnostics for Simple Linear Regression
• References and suggested readings (Ch17 & 18)

18 – Multiple Linear Regression

Introduction

This is the second part of our discussion about general linear models. In this chapter we extend linear regression from one (Chapter 17) to many predictor variables. We also introduce logistic regression, which uses logistic function to model binary outcome variables. Extensions to address ordinal outcome variables are also presented. We conclude with a discussion of model selection, which applies to models with two or more predictor variables.

For linear regression models with multiple predictors, in addition to our LINE assumptions, we add the assumption of no multicollinearity. That is, we assume our predictor variables are not themselves correlated.

Rcmdr and R have multiple ways to analyze linear regression models; we will continue to emphasize the general linear model approach, which allow us to handle continuous and categorical predictor variables.

Practical aspects of model diagnostics were presented in Chapter 17; these rules apply for multiple predictor variable models. Regression and correlation (Chapter 16) both test linear hypotheses: we state that the relationship between two variables is linear (the alternative hypothesis) or it is not (the null hypothesis). The difference? Correlation is a test of association (are variables correlated, we ask?), but are not tests of causation: we do not imply that one variable causes another to vary, even if the correlation between the two variables is large and positive, for example. Correlations are used in statistics on data sets not collected from explicit experimental designs incorporated to test specific hypotheses of cause and effect. Regression is to cause and effect as correlation is to association. 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.

Models

Chapter 17 covered the simple linear model

Chapter 18 covers multiple regression linear model

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

And the logistic regression model

where L refers to the upper or maximum value of the curve, refers to the rate of change at the steepest part of the curve, and refers to the inflection point of the curve. Logistic functions are S-shaped, and typical use involves looking at population growth rates (eg, Fig 1), or in the case of logistic regression, how a treatment effects the rate of growth.

Figure 1. Growth of bacteria over time (optical density at 600 nm UV spectrophotometer), fit by logistic function (dashed line).

Suggested homework

Homework 12: Multiple linear regression.

Quizzes in this chapter

A total of 59 questions among the several subchapters, a mix of true or false and multiple choice question format.

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 – 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

14.6 – Some other ANOVA designs

Introduction

There are several additional ANOVA models in common use. The crossed, balanced design is but one example of the two-way ANOVA. And, from a consideration of two factors it logically follows that there can be more than two factors as part of the design of an experiment. As the number of factors increase, the number of two-way, three-way, and even higher-order interactions are possible and at least in principle may be estimated.

Our purpose here is to highlight several, but certainly not all possible experimental designs from the perspective of ANOVA. Examples are provided. Keep in mind that the general linear model approach unifies these designs.

Some of the classical experimental ANOVA designs one sees include

• Two-way randomized complete block design
• Two-way factorial with no replication design
• Repeat-measures ANOVA with one factor
• Nested ANOVA
• Three-way ANOVA
• Split-plot ANOVA
• Latin squares ANOVA

Put simply, these designs differ in how the groups are arranged and how members of the groups are included.

Two-way randomized complete block design

This design refers to the “textbook” design. For each, factor A and factor B, there are multiple levels, in this example three levels of Factor A and three levels of Factor B,  and subjects (sampling units) are randomly assigned to each level. However, one of the factors is, perhaps, of less interest, yet certainly accounts for variation in the response variable.

		Factor B
		1	2	3
Factor A	1
	2
	3

where n_1,1, n_2,1, etc. represents the number of subjects in each cell. Thus, in this design there are nine groups. Typically, minimum replication would be three subjects per group.

Two-way factorial with no replication design

While it may seem obvious that a good experiment should have replication, there are situations in which replication is impossible. While this seems rather odd, this scenario very much describes a typical microarray, gene expression project.

		Factor B
		1	2	3
Factor A	1
	2
	3

where, again, n1,1, n2,1, etc., represents the number of subjects in each cell and there are nine groups in this study. With no replication, then no more than one subject per group.

Repeated-measures ANOVA

When subjects in the study are measured multiple times for the dependent variable, this is called a repeated-measures design. We introduced the design for the simple case of before and after measures on the same individuals in Chapter 12.3. It’s straight-forward to extend the design concept to more than two measures on the subjects. The blocking effect is the individual (see Chapter 14.4), and, therefore, a random effect (see Chapters 12.3 and 14.3) in this type of experimental design.

Although straightforward in concept, repeated measure designs have many complications in practice. For example, long-term studies can expect for subjects to drop out of the study, resulting in censored data. Another complication, the assumption is that there is no carry over effect — it doesn’t matter the order different treatments are applied to the subjects. Think of this assumption as akin to the equal variances assumption in ANOVA; just like unequal variances effects Type I error rates in ANOVA, deviations from sphericity inflate Type I error rates in repeated-measures designs.

Sphericity assumption is described in two ways:

Assumption of sphericity — the ranking of individuals remains the same across treatment levels — no interaction between individual and treatment. Sphericity assumption is always met if there are just two levels of the repeated measure, eg, before and after.

Compound symmetry assumption — the variances and covariances are equal across the study: the changes experienced by the subjects are the same across the study regardless of the order of treatments.

Tests for sphericity include:

Mauchly test: mauchly.test(object)

If results of tests for violations of sphericity warrant, corrections are available. One recommended correction is called Greenhouse-Giesser correction, which adjusts the degrees of freedom and so results in a better p-value estimate. A second correction is called Huyhn-Feldt correction; this correction, too, adjusts the degrees of freedom to improve the p-value estimate.

Three-way ANOVA

It is relatively straightforward to imagine an experiment that involves three or more factors. The analysis and interpretation of such designs, while feasible, becomes somewhat complicated especially for the mixed models (Model III).

Consider just the case of a fixed-effects 3-way ANOVA. How many tests of null hypotheses are there?

1. Three tests for main effects.
2. Three tests of two-way interactions.
3. A test for a three-way interaction.

Thus, there are seven separate null hypotheses from a three-way ANOVA with fixed effects! As you can imagine, large sample sizes are needed for such designs, and the “higher-order” interactions (eg, three-way interaction) can be difficult to interpret and may lack biological significance.

ANOVA designs without random assignment to treatment levels

Latin square design

We have introduced you to several ANOVA experimental designs that employed randomization for assignment of subjects to treatment groups. The purpose of randomization is even out differences due to confounding variables. However, if we know in advance something about the direction of the influence of these confounding variables, strictly random assignment is not in fact the best design. For example, the Latin square design is common in agriculture research and is very useful for situations in which two gradients are present (eg, soil moisture levels, soil nutrient levels).

Split-Plot Design

Another design from agriculture research is especially useful to ecotoxicology research. We mentioned the repeated measures design in which individuals are measured more than once and each individual receives all levels of the treatment in a random order (cross-over design). However, this design assumes that there are no carry-over effects (see Hills and Armitage 1979;  For ecology/evolution definition see O’Connor et al 2014). While this assumption may hold for many experiments, we can also imagine many more situations in which this is undoubtedly false. For example, if we wish to measure the effects of ozone and relative humidity on frog behavior, we might consider using the individual as its own control. But we also wish to compare frog behavior following ozone exposure against behavior exhibited in clean air. But we are likely to violate the carry-over assumption. If a frog receives ozone then air, the effects of ozone may inhibit activity for several days after the initial exposure, which would then influence subsequent measures. The solution to this dilemma is to use what’s called a split plot design. The design combines elements of nesting.

Consider our frog experiment. There would be three factors:

Factor 1 = Exposure (air or ozone),

Factor 2 = Saturation (dry, intermediate, wet),

Factor 3 = Individual (each frog is measured 3 times).

The design table would look like

Thus, the design is crossed for one factor (saturation), but nested for another factor (individuals are nested within Exposure factor).

Questions

1. Which of the study designs mentioned so far are sensitive to carry-over effects?
2. With respect to how levels of Factors are assigned, distinguish the split-plot design from the Latin square design.

Quiz Chapter 14.6

Some other ANOVA designs

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Chapter 14 contents

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

10.1 – Compare two independent sample means

Introduction

We introduced the concept of comparing a sample statistic (mean) against a population parameter (Chapter 6.7, Normal deviate) or one-sample t-test against a specified mean (eg, from published data or from theory, Chapter 8.5).

Consider now a basic experimental design, the randomized control trial, or RCT (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Control Trial.

Subjects randomly selected from population of interest, then again — random assignment — once recruited into one of two treatment groups. Importantly, subjects belong to one treatment arm only: no subject simultaneously receives the treatment and the control. This is in contrast to the paired design, in which subjects receive both treatments (see Chapter 10.3).

In inferential statistics about an experiment, we are more likely trying to test if sample means are different. For example

• two species grown in common garden, do they differ in growth rate?
• human subjects given a new treatment have better outcomes compare to those receiving a control treatment (eg, placebo).

The equivalent null hypothesis is that two samples are pulled from the same population. We write the null hypothesis as

and the corresponding alternative hypothesis, H_A, then must be

Question: Is this a one tailed or two-tailed hypothesis?

Answer: Two-tailed (review Chapter 8.4)

Worked example

We introduce the two-sample t-test, or better, the independent sample t-test.

where the numerator is the difference between the two sample means and the denominator is the standard error of the differences between the two groups standard errors. The formula for

The choice of independent sample over two-sample is best because it emphasizes that the two groups (the two samples), must be comprise of independent sampling units. This is a pretty straight-forward requirement; you have randomly assigned twenty individuals to two groups, a control group (n = 10) and a treatment group (n = 10). Individuals are either in the control group or they are in the treatment group — they cannot simultaneously appear in both groups.

We will work our way through this test by example. For starters, let’s use the same lizard dataset (see Example data set, below), four body mass recordings (grams) each for house geckos (Hemidactylus frenatus, Fig 2) and the Carolina anole (Anolis carolensis, Fig 3), two of many lizard species introduced to Hawaiʻi.

Figure 2. Male Hemidactylus frenatus, central Oahu, M. Dohm.

Figure 3. Male Anolis carolinensis, ʻAkaka Falls, Big Island of Hawaiʻi, M. Dohm.

Example data set

Geckos: 3.186, 2.427, 4.031, 1.995

Anoles: 5.515, 5.659, 6.739, 3.184

Question: How would you go about creating a data frame with the values in long form (stacked worksheet), including a label variable and the body mass?

Answer: At the R prompt, type

Geckos <- c(3.186, 2.427, 4.031, 1.995); Anoles = c(5.515, 5.659, 6.739, 3.184) #create two vectors
bmass <- c(Geckos, Anoles)    #combine the two vectors into a single vector holding all of the body mass records
species <- c("gecko", "gecko", "gecko", "gecko", "anole", "anole", "anole", "anole")   #create your label variable
lizards <- data.frame(species, bmass)  #create your data frame
lizards   #print your data frame  species bmass
1   gecko 3.186
2   gecko 2.427
3   gecko 4.031
4   gecko 1.995
5   anole 5.515
6   anole 5.659
7   anole 6.739
8   anole 3.184

Note also that you can enter data into the Data editor by creating the data frame first then adding values. To edit the data frame “lizards” type fix(lizards) at the R prompt, then close the data frame when you have added or changed values as needed.

As always, begin with an exploration of the data, including a graph (Fig 4).

Figure 4. Box plot of lizard body mass.

We can see already that there’s greater spread of data for the Anoles compared to the Geckos, but the median values differ. Small sample sizes can be a problem for analyses as we can only have reduced confidence in our conclusions. However, we press on for the sake of demonstration.

Let’s test the null hypothesis, H₀, ie, the two species of lizards have the same mean body mass.

Rcmdr: Statistics → Means → Independent-samples t-test…

In this next image I posted the Rcmdr menu popup for the Independent Samples t-test (Fig 5). Later versions of Rcmdr split the settings for this command into two tabs; the first tab allows for the selection of the variables and setting the hypotheses whereas the second tab, labeled Options, permits additional choices. The default selections need your attention: to actually conduct the t-test you need to answer “No” to the question, “Assume equal variance?” (Fig 6).

Figure 5. Rcmdr menu for Independent sample t-test.

Select the Options tab (Fig 6) to select null hypothesis and to select the t-test and not the Welch-test (which is the default, ie, No to the prompt “Assume equal variances?”).

Figure 6. Rcmdr Options menu for Independent sample t-test.

Let’s look at the results and break down the parts of the test.

t.test(Body.mass~Lizard, alternative='two.sided', conf.level=.95, 
+ var.equal=TRUE, data=lizards)

Two Sample t-test

data: Body.mass by Lizard 
t = 2.7117, df = 6, p-value = 0.03503
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
0.2308685 4.4981315 
sample estimates:
mean in group Anolis mean in group Gecko 
5.27425 2.90975

Consider the R session output above and answer the following questions.

Questions for the worked example

1. Which lizard group had the greater mean value, Anolis or Gecko?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

Try another example

DNA damage, changes in the chemical structure of nucleotide bases or breakage of the DNA chains, occurs in cells under many circumstances. The comet assay, or single-cell gel electrophoresis, is one method for visualizing and measuring DNA strand breaks in cells. Exposed cells are mixed with a low-melting temperature agarose and placed onto a microscope slide. The cells are then lysed with an alkaline detergent and high salts. When current is applied across the slide, undamaged DNA remains in the nucleus, whereas damaged DNA extends towards the anode to form a comet-like tail, with imaging assisted by including a fluorescent dye like Sybr-Green. Examples of comets are shown below (Fig 7).

Figure 7. Comet examples. A Intact cell, no DNA damage, B Cell with some DNA damage, a slight tail to the right is evident, C  Cell with significant DNA damage, a large tail is evident. M. Dohm.

In an experiment, immortalized lung epithelial cells were exposed to dilute copper solutions for 30 minutes then washed with PBS. The comet assay was applied to these cells and for comparison, to cells without copper exposure but otherwise treated the same way (controls). The data are available at the bottom of this page (scroll down or click here).

Again, you should begin all analyses with an exploration of the data, including a graph (Fig 8).

Figure 8. Boxplot of comet tail lengths for cells with and without (control) exposure to copper in the cell medium for 30 minutes.

Let’s look at the R output for the t-test analysis.

    Two Sample t-test

data:  CometTail by Treatment
t = -5.8502, df = 38, p-value = 9.139e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -22.39865 -10.88213
sample estimates:
mean in group Control  mean in group Copper 
             11.14533              27.78571

Consider the R session output above and answer the following questions.

Questions for Comet assay data set

1. Which cell group had the greater mean value, Copper-exposed or Control-exposed cells?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

T test from summary statistics

In some cases you may only have to summary statistics for data, eg, the means and the standard deviations. We can use the equations of the t test to write a simple formula, where the user provides the known means, standard deviations, and sample size. For example, create a simple function with readline for user input.

myTtest <- function() {
   mnx <- as.numeric(readline(prompt="Enter mean of x: "))
   stdevx <- as.numeric(readline(prompt="Enter sd of x: "))
   nx <- as.numeric(readline(prompt="Enter n of x: "))
   mny <- as.numeric(readline(prompt="Enter mean of y: "))
   stdevy <- as.numeric(readline("Enter sd of y: "))
   ny <- as.numeric(readline(prompt="Enter n of y: "))
   myTvalue <- abs(((mnx-mny)-0)/sqrt(((stdevx^2)/nx)+(stdevy^2)/ny))
   myDF <- as.integer(nx+ny-2)
   myPvalue <- pt(myTvalue,myDF,lower.tail=FALSE)*2
   myResults <- c(myTvalue, myDF, myPvalue)
   report <- c("T-test: ", "df: ", "two-tailed p-value: ")
   cat(sprintf("%s %3.3f, ", report, myResults))
}

then run the function by typing myTest() at the R prompt and entering the means, standard deviations, and sample size when prompted.

myTtest()
Enter mean of x: 2.91
Enter sd of x: .895
Enter n of x: 4
Enter mean of y: 5.27
Enter sd of y: 1.497
Enter n of y: 4
T-test: 2.706, df: 6.000, two-tailed p-value: 0.035

P-values from confidence intervals

While we expect certain reporting criteria for published, it is not uncommon that one or more elements are missing. For example, while means plus or minus standard deviations for two groups may be reported, the confidence interval of the difference may not be reported, or, an inexact p-value is reported like “< 0.05,” but we “need” an exact p-value for our meta-analysis.  A little math, and these missing statistics can be calculated.

We can use the lizard study as an example; while all of the expected elements were reported, including the p-value (0.03503), let’s say the p value was reported as “statistically significant at p < 0.05.”

We need

lower and upper confidence intervals: 0.2309 and 4.498, respectively.

critical value from t distribution, df = 6 (2 tailed, therefore 0.05/2). A little R code we have:

the standard error, which can be calculated as

and then the p-value,

which is pretty close to the result from R (p = 0.03503). The difference is likely due to the small sample size.

Questions

1. Don’t forget to work through the Questions for the Comet tail data set (scroll up or click here).
2. Microsoft Excel, LibreOffice Calc, and Google sheets spreadsheet software all include t-test functions and return the p-value. Consider two variables big (100, 110, 120, 100, 110, 210, 200) and small (0,1,1,2,0,1,0). (Note — these two groups are obviously very different, calculating a t-test on their difference is silly, just for this question.)  If formatting is set to the default two decimal places for Number cell category, the p-value will return as “0.00.” How should you report the p-value in this case?

Quiz Chapter 10.1

Compare two independent sample means

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Comet assay data set

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading