17.6 – ANCOVA – analysis of covariance

Introduction

Analysis of covariance (ANCOVA) is intended to help with analysis of designs with categorical treatment variables on some response (dependent) variable, but a known confounding variable is also present. Thus, the researcher is also likely to know of additional ratio scale variables that covary with the response variable and, moreover, must be included in the experimental design in some way.

Take for example the well-known relationship between body size and whole-animal metabolic rate as measured by rates of carbon dioxide production or rates of oxygen consumption for aerobic organisms. We may be interested in how addition or blocking of stress hormones affects resting metabolism; we may be interested in comparing men and women for activity metabolism, and so on. We’d like to know if the regressions were the same (eg, metabolic rate covaried with body mass in the same way — that is, the slope of the relationship was the same).

This situation arises frequently in biology. For example, we might want to know if male and female birds have different mean field metabolic rates, in which case we might be tempted to use a one-way ANOVA or t-test (since one factor with two levels). However, if males and females also differ for body size, then any differences we might see in metabolic rate could be due to differences in metabolic rate are confounded by differences in the covariable body size. We already discussed one approach to correction: calculate a ratio. Thus, a logical approach to correcting or normalizing for the covariation would be to divide body mass (units of kilograms) into metabolic rate (eg, volume of oxygen, O₂, consumed), and make comparisons, say, among different species, on mass-specific trait (). However, because the regression between mass and metabolic rate is allometric, i.e., not equal to one, the ratio does not, in fact normalize for body mass.  We made this point in Chapter 6.2, and remarked that analysis of covariance ANCOVA was a solution.

ANCOVA allows you to test for mean differences in traits like metabolic rate between two or more groups, but only after first accounting for covariation due to another variable (eg, body size). However, ANCOVA makes the assumption that relationship between the covariable and the response variable is the same in the two groups. This is the same as saying that the regression slopes are the same. We discussed how to use t-test to test hypothesis of equal slopes between regression models in Chapter 17.5, but a more elegant way is to include this in your model.

Example

We return to our sample of 13 tadpoles (Rana pipiens), hatched in the laboratory. I’ve repeated the data set in this page, scroll or click here.

Our linear model was and a scatterplot of the data set is shown in Figure 1 (a repeat of Figure 3 from Chapter 17.5, but now points identified to developmental group).

Figure 1. Scatterplot of oxygen consumption by R. pipiens tadpoles vs body mass (g) by developmental group (Gosner stages I or II).

The project looked at whether metabolism as measured by oxygen consumption was consistent across two developmental stages. Metamorphosis in frogs and other amphibians represents profound reorganization of the organism as the tadpole moves from water to air. Thus, we would predict some cost as evidenced by change in metabolism associated with later stages of development. Figure 2 shows a box plot of tadpole oxygen consumption by Gosner (1960) developmental stage (Figure 2 is a repeat of Figure 4 Chapter 17.5).

Figure 2. Copy of Figure 4, Chapter 17.5; boxplot of oxygen consumption of R. pipiens tadpoles by Gosner developmental stages.

Looking at Figure 2 we see a trend consistent with our prediction; developmental stage may be associated with increased metabolism. However, older tadpoles also tend to be larger, and the plot in Figure 2 does not account for that. Thus, body mass is a confounding variable in this example. There are several options for analysis here (eg, ANCOVA), but one way to view this is to compare the slopes for the two developmental stages. While this test does not compare the means, it does ask a related question: is there evidence of change in rate of oxygen consumption relative to body size between the two developmental stages? The assumption that the slopes are equal is a necessary step for conducting the ANCOVA.

So, divide the data set (Table 1) into two groups by developmental stage (12 tadpoles could be assigned to one of two developmental stages; one was at a lower Gosner stage than the others and so is dropped from the subset.

Table 1. and of 12 R. pipiens tadpole larvae by Gosner developmental stage.

Gosner stage I

1.76	109.41
1.88	329.06
1.95	82.35
2.13	198
2.26	607.7

Gosner stage II

Body mass
2.28	362.71
2.35	556.6
2.62	612.93
2.77	514.02
2.97	961.01
3.14	892.41
3.79	976.97

The slopes and standard errors we obtained in Chapter 17.5 were

Make a plot (Figure 3).

Figure 3. Scatterplot with best-fit regression lines of by for Gosner State I (closed circle, solid line) and Gosner Stage II (open circle, dashed line) R. pipiens tadpoles.

R code for plot in Figure 3.

#Used Rcmdr scatterplot(), then modified code
scatterplot(VO2~Body.mass | Gosner, regLine=FALSE, smooth=FALSE, 
boxplots=FALSE, xlab="Body mass (g)", ylab="Oxygen consumption (ml/h)", 
main="", cex=1.4, cex.axis=1.5, cex.lab=1.5, pch=c(19,19), by.groups=TRUE,
col=c("blue","red"), grid=FALSE, 
legend=list(coords="bottomright"), data=Tadpoles)
#Get regression equations for groups, subset by Gosner
abline(lm(VO2~Body.mass, data=Stage01), lty=1, lwd=2, col="blue")
abline(lm(VO2~Body.mass, data=Stage02), lty=1, lwd=2, col="red")

Returning to the important question, are the two slopes statistically indistinguishable (Ho: b_I = b_II), where b_I is the slope for the Gosner Stage I subset and b_II is the slope for the Gosner Stage II subset? We look at the plot, and since the lines cross, we tend to see a difference. Of course, we need to consider that our perception of slope differences may simply be chance, especially because the sample size is small. Proceed to test.

R code

The ANCOVA is a new ANOVA model where the factor variables are adjusted or corrected for the effects of the continuous variable.

R code for ANCOVA example, crossed or interaction model.

tadpole.1 <- lm(VO2 ~ Body.mass*Gosner, data=example.Tadpole)
summary(tadpole.1)
Anova(tadpole.1, type="II")

Output

summary(tadpole.1)

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

Residuals:
    Min      1Q   Median      3Q      Max 
-167.80 -117.93    13.81   94.66   214.65

Coefficients:
                       Estimate   Std. Error   t value    Pr(>|t|) 
(Intercept)             -1231.6       783.0     -1.573    0.1544 
Body.mass                 750.0       390.7      1.919    0.0912 .
Gosner[T.II]              790.4       859.2      0.920    0.3845 
Body.mass:Gosner[T.II]   -350.1       409.5     -0.855    0.4174 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 155.8 on 8 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.821, Adjusted R-squared: 0.7539 
F-statistic: 12.23 on 3 and 8 DF, p-value: 0.002336

The coefficients for the first factor (GII) and then the differences in the coefficient for the second factor. You can just add the second coefficient to the first so they’re on the same scale.

Anova Table (Type II tests)

Response: VO2
                  Sum Sq   Df    F value     Pr(>F) 
Body.mass         330046    1    13.6030   0.006146 **
Gosner              5630    1     0.2321   0.642908 
Body.mass:Gosner   17736    1     0.7310   0.417423 
Residuals 194102 8 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Suggests interaction is not significant, i.e., the slopes are not different.

We can then proceed to check to see if the intercepts are different, now that we’ve confirmed no significant difference in slope.

R code for ANCOVA as additive model

tadpole.2 <- lm(VO2 ~ Body.mass + Gosner, data=example.Tadpole)
summary(tadpole.2)
Anova(tadpole.2, type="II")

Output

> summary(tadpole.2)

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

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

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

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

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

Response: VO2
              Sum Sq    Df    F value      Pr(>F) 
Body.mass     330046     1    14.0221   0.004593 **
Gosner          5630     1     0.2392   0.636482 
Residuals     211839     9 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Instead of the interaction, or as added, try a nested model, with body mass nested within stage.

tadpole.3 <- lm(VO2 ~ Body.mass/Gosner, data=example.Tadpole)

summary(tadpole.3)

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

Residuals:
    Min       1Q    Median      3Q      Max 
-168.66  -131.14    -20.28   90.33   225.36

Coefficients:
                 Estimate      Std. Error    t value Pr(>|t|) 
(Intercept)           -575.10      319.51     -1.800   0.1054 
Body.mass              423.65      162.50      2.607   0.0284 *
Body.mass:Gosner[T.II]  21.95       63.73      0.344   0.7384 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 154.4 on 9 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.8021, Adjusted R-squared: 0.7581 
F-statistic: 18.24 on 2 and 9 DF, p-value: 0.0006823

Gets the true coefficient (nested lm() version).

The two test different hypotheses.

lm(VO2 ~ Body.mass * Gosner) tests whether or not the regression has a nonzero slope.

lm(VO2 ~ Body.mass */Gosner) test whether or not the slopes and intercepts from different factors are statistically significant.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
2. An OLS approach was used for the analysis of tadpole oxygen consumption body mass. Consider the RMA approach — would that be a more appropriate regression model? Explain why or why not.
3. Consider an experiment in which you plan to administer a treatment that has a carry-over effect. For example, Compare and contrast “crossed” and “nested” designs.
4. True or False. The nested design option for the ANCOVA assumes the slopes for the two groups of tadpoles for the regression line of by are equal. Explain your choice.
5. Metabolic rates like oxygen consumption over time are well-known examples of allometric relationships. That is, many biological variables (eg, is related as aM^b, where M is body mass, slope b is scaling exponent), and best evaluated on log-log scale. Repeat the analysis above on log₁₀-transformed and for
   • crossed design (eg, tadpole.1 model)
   • added design (eg, tadpole.2 model)
   • nested design (eg, tadpole.3 model)
6. Create the plot and add the fitted lines from crossed design to the plot.

lgVO2 <- log10(VO2)

lm(log10(VO2) ~ log10(Body.mass ), data=example.Tadpole)

Quiz Chapter 17.6

ANCOVA - analysis of covariance

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Data sets

Oxygen consumption, , of Anuran tadpoles, dataset=example.Tadpole

Gosner refers to Gosner (1960), who developed a criteria for judging metamorphosis staging.

Chapter 17 contents

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

18.2 – Nonlinear regression

Introduction

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

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

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

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

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

Fitting of polynomial linear model

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

R code

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

summary(LinearModel.1)

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

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

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

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

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

Figure 3. Residual plot, residuals against fitted values.

We can fit a polynomial regression.

First, a second order polynomial

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

summary(LinearModel.2)

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

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

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

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

Second, try a third order polynomial

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

summary(LinearModel.3)

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

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

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

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

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

R code/results

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

Smaller the AIC, better fit.

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

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

Logistic regression

The Logistic regression is a classic example of nonlinear model.

R code

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

summary(logisticModel)

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

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

Residual standard error: 0.01908 on 12 degrees of freedom

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

> AIC(logisticModel)
[1] -71.54679

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

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

where ln refers to the natural logarithm.

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

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

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

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

We write our generalized linear model

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

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

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

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

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

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

Rcmdr: Statistics → Fit models → Generalized linear model.

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

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

R output

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

> summary(GLM.1)

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

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

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

(Dispersion parameter for gaussian family taken to be 0.008663679)

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

Number of Fisher Scoring iterations: 2

Assessing fit of the logistic regression model

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

The Wald test statistic

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

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

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

Questions

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

Quiz Chapter 18.2

Nonlinear regression

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Data set

Yuan data

Chapter 18 content

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

17.8 – Assumptions and model diagnostics for Simple Linear Regression

Introduction

The assumptions for all linear regression:

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

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

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

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

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

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

Model diagnostics

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

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

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

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

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

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

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

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

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

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

Graph options

Using residual plots to diagnose regression equations

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

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

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

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

Here are examples of POSSIBLE outcomes:

Figure 1. An ideal plot of residuals

Solution: Proceed! Assumptions of linear regression met.

Compare to plots of residuals that differ from the ideal.

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

Solution. Try a transform like the log₁₀-transform.

Figure 3. Problem. Residual plot shows systematic trend.

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

Figure 4. Problem. Residual plot shows nonlinear trend.

Solution. Transform data or use more complex model

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

Diagnostic plot examples

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

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

In brief, we look at plots

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

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

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

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

Statistical test options

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

Results for the model were:

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

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

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

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

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

In R Commander, the diagnostic test options are available via

Rcmdr: Models → Numerical diagnostics

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

R code:

vif(RegModel.2)

R output

Body.mass Gosner 
2.186347 2.186347 

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

Our interpretation

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

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

R code:

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

R output

Breusch-Pagan test

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

Our interpretation

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

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

R code:

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

R output

Durbin-Watson test

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

Our interpretation

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

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

R code:

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

R output

RESET test

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

Our interpretation

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

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

R code:

outlierTest(LinearModel.3)

R output

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

Our interpretation

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

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

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

R code:

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

R output

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

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

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

Our interpretation

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

Questions

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

Quiz Chapter 17.8

Assumptions and model diagnostics for Simple Linear Regression

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Chapter 17 contents

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

17.7 – Regression model fit

Introduction

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

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

summary(RegModel.1)

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

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

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

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

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

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

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

fit the data?

Model fit statistics

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

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

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

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

Standard error of regression

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

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

Coefficient of determination

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

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

Adjusted R-squared

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

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

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

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

Questions

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

Quiz Chapter 17.7

Regression model fit

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Data set

USA population year 2010 and year 2019 with Electoral College counts

Chapter 17 contents

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

17.3 – Estimation of linear regression coefficients

Introduction

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

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

Assess fit of a linear model

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

LinearModel.1 <- lm(Matings ~ Body.Mass, data=bird_matings)
summary(LinearModel.1)
Call: lm(formula = Matings ~ Body.Mass, data = bird_matings)
Residuals:       Min       1Q    Median       3Q      Max
            -2.29237 -1.34322 .-0.03178..1.33792 .2.70763
Coefficients:
             Estimate  Std. Error  t value  Pr(>|t|)
(Intercept)   -8.4746      4.6641   -1.817    0.1026 
Body.Mass      0.3623      0.1355    2.673    0.0255 * 
--- 
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.776 on 9 degrees of freedom 
Multiple R-squared: 0.4425, Adjusted R-squared: 0.3806 
F-statistic: 7.144 on 1 and 9 DF, p-value: 0.02551

Request R to print the ANOVA table.

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

With a little rounding we have the following statistical model

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

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

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

estimate + critical value X standard error of the estimate

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

Option 1.

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

               2.5 %    97.5 %
Body.Mass 0.05565899 0.6689173

Option 2.

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

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

names(LinearModel.1)

R output

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

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

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

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

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

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

which returns

[1] 0.1355472

Let’s use this information to calculate confidence intervals

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

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

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

And for the intercept

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

Results

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

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

Extract additional statistics from R’s linear model function

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

At the R prompt type and submit

summary.lm(LinearModel.1)

returns the following R output

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

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

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

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

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

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

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

summary.lm(LinearModel.1)$coefficients

we get

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

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

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

Questions

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

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

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

Quiz Chapter 17.3

Estimation of linear regression coefficients

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Chapter 17 contents

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

20.10 – Growth equations and dose response calculations

Introduction

In biology, growth may refer to increase in cell number, change in size of an individual across development, or increase of number of individuals in a population over time. Nonlinear, time-series, several models proposed to fit growth data, including the Gompertz, logistic, and the von Bertalanffy. These models fit many S-shaped growth curves. These models are special cases of generalized linear models, also called Richard curves.

Growth example

This page describes how to use R to analyze growth curve data sets.

Key to parameter estimates: y0 is the lag, mumax is the growth rate, and K is the asymptotic stationary growth phase. The spline function does not return an estimate for K.

R code

require(growthrates)
#Enter the data. Replace these example values with your own 
#time variable (Hours)
Hours <- c(0.000, 0.274, 0.384, 0.658, 0.986, 1.260, 1.479, 1.699, 1.973, 2.137, 2.466, 2.795, 3.123, 3.671, 4.110, 4.548, 5.151, 5.534, 5.863, 6.521, 7.068, 7.671, 8.164, 8.877, 9.644, 10.466, 11.233, 11.890, 12.493, 13.205, 14.082)
#absorbance or concentration variable (Abs)
Abs <- c(0.002207, 0.010443, 0.033688, 0.063257, 0.111848, 0.249240, 0.416236, 0.515578, 0.572632, 0.589528, 0.619091, 0.608486, 0.621136, 0.616850, 0.614689, 0.614643, 0.612465, 0.606082, 0.603933, 0.595407, 0.589006, 0.578372, 0.567749, 0.559217, 0.546451, 0.537907, 0.537826, 0.529300, 0.516551, 0.505905, 0.491013)

#Make a dataframe and check the data; If error, then check that variables have equal numbers of entries
Yeast <- data.frame(Hours,Abs); Yeast

#Obtain growth parameters from fit of a parametric growth model
#First, try some reasonable starting values
p <- c(y0 = 0.001, mumax = 0.5, K = 0.6)
model.par <- fit_growthmodel(FUN = grow_logistic, p = p, Hours, Abs, method=c("L-BFGS-B"))
summary(model.par)
coef(model.par)

#Obtain growth parameters from fit of a nonparametric smoothing spline
model.npar <- fit_spline(Hours,Abs)
summary(model.npar)
coef(spline.md)

#Make plots
par(mfrow = c(2, 1))
plot(Yeast, ylim=c(0,1), cex=1.5,pch=16, main="Parametric Nonlinear Growth Model", xlab="Hours", ylab="Abs")
lines(model.par, col="blue", lwd=2)
plot(model.npar, ylim=c(0,1), lwd=2, main="Nonparametric Spline Fit", xlab="Hours", ylab="Abs")

Results from example code

Parameters:
      Estimate Std. Error t value Pr(>|t|)    
y0    0.003765   0.001049    3.59  0.00125 ** 
mumax 3.954458   0.238810   16.56 5.37e-16 ***
K     0.579354   0.008092   71.60  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.03833 on 28 degrees of freedom

Parameter correlation:
            y0   mumax        K
y0     1.00000 -0.8755  0.05847
mumax -0.87550  1.0000 -0.14623
K      0.05847 -0.1462  1.00000
         y0       mumax           K 
0.003764518 3.954457995 0.579354378 
Fitted smoothing spline:
Call:
smooth.spline(x = time, y = ylog)

Smoothing Parameter  spar= 0.4409789  lambda= 1.221762e-05 (12 iterations)
Equivalent Degrees of Freedom (Df): 14.70929
Penalized Criterion (RSS): 0.5024203
GCV: 0.05868796

Parameter values of exponential growth curve:
Maximum growth at x= 0.0004255438 , y= 0.003072228 
y0 = 0.003065586 
mumax = 5.085845 

r2 of log transformed data= 0.9906254 
         y0       mumax 
0.003065586 5.085845036

Figure 1. Top: Parametric Nonlinear Growth Model; Bottom: Nonparametric Spline Fit

LD₅₀

In toxicology, the dose of a pathogen, radiation, or toxin required to kill half the members of a tested population of animals or cells is called the lethal dose, 50%, or LD₅₀. This measure is also known as the lethal concentration, LC₅₀, or properly after a specified test duration, the LCt₅₀ indicating the lethal concentration and time of exposure. LD₅₀ figures are frequently used as a general indicator of a substance’s acute toxicity. A lower LD₅₀ is indicative of increased toxicity.

The point at which 50% response of studied organisms to range of doses of a substance (e.g., agonist, antagonist, inhibitor, etc.) to any response, from change in behavior or life history characteristics up to and including death can be described by the methods described in this chapter. The procedures outlined below assume that there is but one inflection point, i.e., an “s-shaped” curve, either up or down; if there are more than one inflection points, then the logistic equations described will not fit the data well and other choices need to be made (see Di Veroli et al 2015). We will use the drc package (Ritz et al 2015).

Example

First we’ll work through use of R. We’ll follow up with how to use Solver in Microsoft Excel.

After starting R, load the drc library.

library(drc)

Consider some hypothetical 24-hour survival data for yeast exposed to salt solutions. Let resp equal the variable for frequency of survival (e.g., calculated from OD₆₆₀ readings) and NaCl equal the millimolar (mm) salt concentrations or doses.

At the R prompt type

resp <- c(1,1,1,.9,.7,.3,.4,.2,0,0,0)
NaCl=seq(0,1000,100) 
#Confirm sequence was correctly created; alternatively, enter the values.
NaCl 
[1] 0 100 200 300 400 500 600 700 800 900 1000 
#Make a plot
plot(NaCl,resp,pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

And here is the plot (Fig. 2).

Figure 2. Hypothetical data set, survival of yeast in different salt concentrations.

Note the sigmoidal “S” shape — we’ll need an logistic equation to describe the relationship between survival of yeast and NaCl doses.

The equation for the four parameter logistic curve, also called the Hill-Slope model, is

where c is the parameter for the lower limit of the response, d is the parameter for the upper limit of the response, e is the relative EC₅₀, the dose fitted half-way between the limits c and d, and b is the relative slope around the EC₅₀. The slope, b, is also known as the Hill slope. Because this experiment included a dose of zero, a three parameter logistic curve would be appropriate. The equation simplifies to

EC₅₀ from 4 parameter model

Let’s first make a data frame

dose <- data.frame(NaCl, resp)

Then call up a function, drm, from the drc library and specify the model as the four parameter logistic equation, specified as LL.4(). We follow with a call to the summary command to retrieve output from the drm function. Note that the four-parameter logistic equation

model.dose1 = drm(dose,fct=LL.4())
summary(model.dose1)

And here is the R output.

Model fitted: Log-logistic (ED50 as parameter) (4 parms) 
Parameter estimates: 
                   Estimate  Std. Error    t-value  p-value 
b:(Intercept)      3.753415    1.074050   3.494636   0.0101 
c:(Intercept)     -0.084487    0.127962  -0.660251   0.5302 
d:(Intercept)      1.017592    0.052460  19.397441   0.0000 
e:(Intercept)    492.645128   47.679765  10.332373   0.0000 

Residual standard error: 
0.0845254 (7 degrees of freedom)

The EC₅₀, technically the LD₅₀ because the data were for survival, is the value of e: 492.65 mM NaCl.

You should always plot the predicted line from your model against the real data and inspect the fit.

At the R prompt type

plot(model.dose1, log="",pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

As long as the plot you made in earlier steps is still available, R will add the line specified in the lines command. Here is the plot with the predicted logistic line displayed (Fig. 3).

Figure 3. Logistic curve added to Figure 1 plot.

While there are additional steps we can take to decide is the fit of the logistic curve was good to the data, visual inspection suggests that indeed the curve fits the data reasonably well.

More work to do

Because the EC₅₀ calculations are estimates, we should also obtain confidence intervals. The drc library provides a function called ED which will accomplish this. We can also ask what the survival was at 10% and 90% in addition to 50%, along with the confidence intervals for each.

At the R prompt type

ED(model.dose1,c(10,50,90), interval="delta")

And the output is shown below.

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower   Upper
1:10  274.348     38.291 183.803  364.89
1:50  492.645     47.680 379.900  605.39
1:90  884.642    208.171 392.395 1376.89

Thus, the 95% confidence interval for the EC₅₀ calculated from the four parameter logistic curve was between the lower limit of 379.9 and an upper limit of 605.39 mm NaCl.

EC₅₀ from 3 parameter model

Looking at the summary output from the four parameter logistic function we see that the value for c was -0.085 and p-value was 0.53, which suggests that the lower limit was not statistically different from zero. We would expect this given that the experiment had included a control of zero mm added salt. Thus, we can explore by how much the EC₅₀ estimate changes when the additional parameter c is no longer estimated by calculating a three parameter model with LL.3().

model.dose2 = drm(dose,fct=LL.3())
summary(model.dose2)

R output follows.

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
               Estimate Std. Error   t-value p-value
b:(Intercept)   4.46194    0.76880   5.80378   4e-04
d:(Intercept)   1.00982    0.04866  20.75272   0e+00
e:(Intercept) 467.87842   25.24633  18.53253   0e+00
Residual standard error:
 0.08267671 (8 degrees of freedom)

The EC₅₀ is the value of e: 467.88 mM NaCl.

How do the four and three parameter models compare? We can rephrase this as as statistical test of fit; which model fits the data better, a three parameter or a four parameter model?

At the R prompt type

anova(model.dose1, model.dose2)

The R output follows.

1st mode
fct:      LL.3()
2nd model
fct:      LL.4()
ANOVA table
          ModelDf      RSS Df F value p value
2nd model       8 0.054684                   
1st model       7 0.050012  1  0.6539  0.4453

Because the p-value is much greater than 5% we may conclude that the fit of the four parameter model was not significantly better than the fit of the three parameter model. Thus, based on our criteria we established in discussions of model fit in Chapter  16 – 18, we would conclude that the three parameter model is the preferred model.

The plot below now includes the fit of the four parameter model (red line) and the three parameter model (green line) to the data (Fig. 4).

Figure 4. Four parameter (red) and three parameter (green) logistic models fitted to data.

The R command to make this addition to our active plot was

lines(dose,predict(model.dose2, data.frame(x=dose)),col="green")

We continue with our analysis of the three parameter model and produce the confidence intervals for the EC₅₀ (modify the ED() statement above for model.dose2 in place of model.dose1).

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower  Upper
1:10  285.937     33.154 209.483 362.39
1:50  467.878     25.246 409.660 526.10
1:90  765.589     63.026 620.251 910.93

Thus, the 95% confidence interval for the EC₅₀ calculated from the three parameter logistic curve was between the lower limit of 409.7 and an upper limit of 526.1 mm NaCl. The difference between upper and lower limits was 116.4 mm NaCl, a smaller difference than the interval calculated for the 95% confidence intervals from the four parameter model (225.5 mm NaCl). This demonstrates the estimation trade-off: more parameters to estimate reduces the confidence in any one parameter estimate.

Additional notes of EC₅₀ calculations

Care must be taken that the model fits the data well. What if we did not have observations throughout the range of the sigmoidal shape? We can explore this by taking a subset of the data.

dd = dose[1:6,]

Here, all values after dose 500 were dropped

dd
  resp dose
1  1.0    0
2  1.0  100
3  1.0  200
4  0.9  300
5  0.7  400
6  0.3  500

and the plot does not show an obvious sigmoidal shape (Fig. 5).

Figure 5. Plot of reduced data set.

We run the three parameter model again, this time on the subset of the data.

model.dosedd = drm(dd,fct=LL.3())
summary(model.dosedd)

Output from the results are

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
                Estimate Std. Error    t-value p-value
b:(Intercept)   6.989842   0.760112   9.195801  0.0027
d:(Intercept)   0.993391   0.014793  67.153883  0.0000
e:(Intercept) 446.882542   5.905728  75.669344  0.0000
Residual standard error:
 0.02574154 (3 degrees of freedom)

Conclusion? The estimate is different, but only just so, 447 vs. 468 mm NaCl.

Thus, within reason, the drc function performs well for the calculation of EC₅₀. Not all tools available to the student will do as well.

NLopt and nloptr

draft

Free open source library for nonlinear optimization.

Steven G. Johnson, The NLopt nonlinear-optimization package, http://github.com/stevengj/nlopt

https://cran.r-project.org/web/packages/nloptr/vignettes/nloptr.html

Alternatives to R

What about online tools? There are several online sites that will allow students to perform these kinds of calculations. Students familiar with MatLab know that it can be used to solve nonlinear equation problems. An open source alternative to MatLab is called GNU Octave, which can be installed on a personal computer or run online at http://octave-online.net. Students also may be aware of other sites, e.g., mycurvefit.com and IC50.tk.

Both of these free services performed well on the full dataset (results not shown), but fared poorly on the reduced subset: mycurvefit returned a value of 747.64 and IC50.tk returned an EC₅₀ estimate of 103.6 (Michael Dohm, pers. obs.).

Simple inspection of the plotted values shows that these values are unreasonable.

EC₅₀ calculations with Microsoft Excel

Most versions of Microsoft Excel include an add-in called Solver, which will permit mathematical modeling. The add-in is not installed as part of the default installation, but can be installed via the Options tab in the File menu for a local installation or via Insert for the Microsoft Office online version of Excel (you need a free Microsoft account). The following instructions are for a local installation of Microsoft Office 365 and were modified from information provided by sharpstatistics.co.uk.

After opening Excel, set up worksheet as follows. Note that in order to use my formulas your spreadsheet values need to be set up exactly as I describe.

• Dose values in column A, beginning with row 2
• Response values in column B, beginning with row 2
• In column F type b in row 2, c in row 3, d in row 4, and e in row 5.
• In cells G2 – G5, enter the starting values. For this example, I used b = 1, c = 0, d = 1, e = 400.
  • For your own data you will have to explore use of different starting values.
• Enter headers in row 1.
  • Column A Dose
  • Column B Response
  • Column C Predicted
  • Column D Squared difference
  • Column F Constants
• In cell C14 enter sum squares
• In cell D14 enter =sum(D2:D12)

Here is an image of the worksheet (Fig. 6), with equations entered and values updated, but before Solver run.

Figure 6. Screenshot Microsoft Excel worksheet containing our data set (col A & B), with formulas added and calculated. Starting values for constants in column G, rows 2 – 4.

Next, enter the functions.

• In cell C2 enter the four parameter logistic formula — type everything between the double quotes: “=$G$3+(($G$4-$G$3)/(1+(A2/$G$5)^$G$2))“. Next, copy or drag the formula to cell C12 to complete the predictions.
  • Note: for a three parameter model, replace the above formula with “=(($G$4)/(1+(A2/$G$5)^$G$2))“.
• In cell D2 type everything between the double quotes: “=(B2-C2)^2“.
• Next, copy or drag the formula to cell D12 to complete the predictions.
• Now you are ready to run solver to estimate the values for b, c, d, and e.
  • Reminder: starting values must be in cells G2:G5
• Select cell D14 and start solver by clicking on Data and looking to the right in the ribbon.

Solver is not normally installed as part of the default installation of Microsoft Excel 365. If the add-in has been installed you will see Solver in the Analyze box (Fig. 7).

Figure 7. Screenshot Microsoft Excel, Solver add-in available.

If Solver is not installed, go to File → Options → Add-ins (Fig. 8).

Figure 8. Screenshot Microsoft Excel, Solver add-in available and ready for use.

Go to Microsoft support for assistance with add-ins.

With the spreadsheet completed and Solver available, click on Solver in the ribbon (Fig. 7) to begin. The screen shot from the first screen of Solver is shown below (Fig. 9).

Figure 9. Screenshot Microsoft Excel solver menu.

Setup Solver

• Set Objective, enter the absolute cell reference to the sum squares value, \$D$14
• Set To: Min.
• For By Changing Variable Cells, enter the range of cells for the four parameters, $G$2:$G$5
• Uncheck the box by “Make Unconstrained Variables Non-Negative.”
  • Where constraints refers to any system of equalities or inequalities equations imposed on the algorithm.
• Select a Solving method, choose “GRG Nonlinear,” the nonlinear programming solver option (not shown in Fig. 7, select by clicking on the down arrow).
• Click Solve button to proceed.

If all goes well, this next screen (Fig. 10) will appear which shows the message, “Solver has converged to the current solution. All constraints are satisfied.”

Figure 10. Screenshot solver completed run.

Click OK and note that the values of the parameters have been updated (Table 1).

Table 1. Four parameter logistic model, results from Solver 

Note the values obtained by Solver are virtually identical to the values obtained in the drc R package. The differences are probably because of the solver algorithm.

More interestingly, how did Solver do on the subset data set? Here are the results from a three parameter logistic model (Table 2).

Table 2. Three parameter logistic model, results from Solver

The results are again very close to results from the drc R package.

Thus, we would conclude that Solver and Microsoft Excel would be a reasonable choice for EC₅₀ calculations, and much better than IC50.tk and mycurvefit.com. The advantages of R over Microsoft Excel is that the model building is more straightforward than entering formulas in the cell reference format required by Excel.

Questions

[pending]

References and suggested readings

Beck B, Chen YF, Dere W, et al. Assay Operations for SAR Support. 2012 May 1 [Updated 2012 Oct 1]. In: Sittampalam GS, Coussens NP, Nelson H, et al., editors. Assay Guidance Manual [Internet]. Bethesda (MD): Eli Lilly & Company and the National Center for Advancing Translational Sciences; 2004-. Available from: www.ncbi.nlm.nih.gov/books/NBK91994/

Di Veroli G. Y., Fornari C., Goldlust I., Mills G., Koh S. B., Bramhall J. L., Richards, F. M., Jodrell D. I. (2015) An automated fitting procedure and software for dose-response curves with multiphasic features. Scientific Reports 5: 14701.

Ritz, C., Baty, F., Streibig, J. C., Gerhard, D. (2015) Dose-Response Analysis Using R PLOS ONE, 10(12), e0146021

Tjørve, K. M., & Tjørve, E. (2017). The use of Gompertz models in growth analyses, and new Gompertz-model approach: An addition to the Unified-Richards family. PloS one, 12(6), e0178691.

Chapter 20 contents

• Additional topics
• Area under the curve
• Peak detection
• Baseline correction
• Surveys
• Time series
• Dimensional analysis
• Estimating population size
• Diversity indexes
• Survival analysis
• Growth equations and dose response calculations
• Plot a Newick tree
• Phylogenetically independent contrasts
• How to get the distances from a distance tree
• Binary classification
• Meta-analysis

/MD

14.7 – Rcmdr Multiway ANOVA

Introduction

We have been talking about the two-way randomized, balanced, replicated design.  Here, we take you step by step through use of R to conduct the multiway ANOVA.

R code: Multiway ANOVA

Rcmdr: Statistics → Means → Multiway ANOVA… we will review this as Option 1

or

Rcmdr: Statistics → Fit Models → Linear model… we will review this as Option 2

In either case, as a reminder, your data set must be stacked worksheet, like the data in this worksheet

Table 1. Data set, example.14.7†

Option 1

Your first option is to use the ANOVA menus via “Means.” This is a perfectly good way to handle a standard two-way, fully-crossed, fixed effects model.  However, other designs will not run with this command and R will return a report of errors for ANOVA models that do not conform to the replicated, balanced, crossed design.

Rcmdr: Statistics → Means → Multiway Analysis of variance …

Factors: highlight “Diet” AND “Population”

Response variable: pick one (in this window, all we see is “Response”)

Figure 1. Screenshot Rcmdr multi-way ANOVA.

Interpret the output

AnovaModel.2 <- (lm(Response ~ Diet*Population, data=example.14.7))

Anova(AnovaModel.2)
 
Anova Table (Type II tests)
Response: Response
                  Sum Sq    Df    F value      Pr(>F) 
Diet              36.750     1    12.2500    0.008079 **
Population        10.083     1     3.3611    0.104104 
Diet:Population   52.083     1    17.3611    0.003136 **
Residuals         24.000     8
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
tapply(example.14.7 Response, list(Diet=example.14.7 Diet, 
+ Population=example.14.7 Population), mean, na.rm=TRUE) # means
Population
Diet 1 2
A 5.00000 7.333333
B 12.66667 6.666667

tapply(example.14.7 Response, list(Diet=example.14.7 Diet, 
+ Population=example.14.7 Population), sd, na.rm=TRUE) # std. deviations
Population
Diet 1 2
A 1.000000 2.081666
B 2.081666 1.527525

tapply(example.14.7 Response, list(Diet=example.14.71 Diet, 
+ Population=example.14.7 Population), function(x) sum(!is.na(x))) # counts
Population
Diet 1 2
A 3 3
B 3 3

End R output.

Summary of multi-way ANOVA command

The multi-way ANOVA command returns our ANOVA table plus the adjusted means, along with standard deviations and number of observations (counts).  The adjusted means would then be good to put into a chart to present group comparisons following adjustments from the effects of levels within groups.

Rcmdr: Models → Graphs → Predictor effect plots …

Here’s the chart (hint: )

Figure 2. Predictor effect plots, Diet and Population on Response variable.

Option 2

A more general approach is to use the General linear model. This approach can handle the standard 2-way fixed effects ANOVA (above), but any other model as well. The model is Response ~ Diet*Population.

Rcmdr: Statistics → Fit Models → Linear model… 

Figure 3. Screenshot Rcmdr linear model menu.

Interpret the output

LinearModel.1 <- lm(Response ~ Diet * Population, data=example.14.7)

summary(LinearModel.1)

Call:
lm(formula = Response ~ Diet * Population, data = example.14.7)

Residuals:
Min 1Q Median 3Q Max 
-2.3333 -1.1667 0.1667 1.0833 2.3333 

Coefficients:
                        Estimate    Std. Error ..t value .  Pr(>|t|) 
(Intercept) .              5.000         1.000 .   5.000    .0.00105 ** 
Diet[T.B] .                7.667       ..1.414 .   5.421    .0.00063 ***
Population[T.2]           .2.333       ..1.414 .   1.650    .0.13757 
Diet[T.B]:Population[T.2] -8.333       ..2.000    -4.167    .0.00314 ** 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 1.732 on 8 degrees of freedom
Multiple R-squared: 0.8047, Adjusted R-squared: 0.7315 
F-statistic: 10.99 on 3 and 8 DF, p-value: 0.003285 

End R output

Lots to sort through, so let’s begin with what is in common between the two approaches, the Multi-way ANOVA command versus the linear model command.

Compare R output from ANOVA and linear model

As a direct output, the linear model option does not provide an ANOVA summary table. Instead of our ANOVA table the linear model returns estimates of coefficients along with t-test results for each coefficient of the model from the lm() command output

Recall that we can get ANOVA tables through the following R commands via Rcmdr.

Rcmdr: Models → Hypothesis tests → ANOVA Table.

Let’s do so for this linear model (accept the default for type of tests = “Type II”).

And the output from lm() stored in the object LinearModel.1 is

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

Response: Response
                Sum Sq Df F value   Pr(>F)   
Diet            36.750  1 12.2500 0.008079 **
Population      10.083  1  3.3611 0.104104   
Diet:Population 52.083  1 17.3611 0.003136 **
Residuals       24.000  8                    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End R output

Now, we’re in business, and, using the lm() function, we have the estimates for each model coefficient plus our ANOVA table.

Same answer! Of course. Which to choose, Option 1 or Option 2? Use the lm() options; more flexible, covers more designs than the multiway ANOVA which is strictly for the crossed fully replicated design.

Questions

1. Write out the two-way model described for the data in Table 1.
2. Write the null hypotheses and provide a summary of the statistical significance of the model.

Quiz Chapter 14.7

Rcmdr Multiway ANOVA

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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

12.4 – ANOVA from “sufficient statistics”

Introduction

By now you should be able to run a one-way ANOVA using R (and R Commander) with ease. As a reminder, You should also be aware that, if you need to, you could use spreadsheet software like Microsoft Excel or LibreOffice Calc to run one way ANOVA on a small data set. Still, there are times when you may need to run a one-way ANOVA on a small data set, and doing so by hand calculator may be just as convenient. What are your available options?

Following the formulas I have given would be one way to calculate ANOVA by hand, but it would be tedious and subject to error. Instead of working with the standard formulas, calculator shortcuts can be derived with a little algebra and this is where I want to draw your attention now. This technique will come in handy in lab classes or other scenarios where you collect some data among a set number of groups and calculate means and standard deviations. The purpose of this posting is to show you how to obtain the necessary statistics to calculate a one-way ANOVA from the available descriptive statistics: means, standard deviations, and sample sizes. In other words, these are the sufficient statistics for one-way ANOVA.

As you recall, a one-way ANOVA yields a single F test of the null hypothesis that all group means are equal. To calculate the F test, you need

• Mean Square Between Groups,
• Mean Squares Within Groups or Error,

F is then calculated as

with degrees of freedom k – 1 for the numerator and N – 1 for the denominator. can also appear as

We can calculate  as

where k is the number of i ^th groups, n_i is the sample size of the ith group,  refers to the overall mean for all of the  sample means.

Next, for the Error Mean Square, , all we need is the average of the sample variances (the square of the sample standard deviation, s).

ANOVA from sufficient statistics

Consider an example data set (Table 1) for which only summary statistics are available (mean and standard deviation, sd). The data set is for life span(days) for several inbred strains of laboratory mice from a study (. Sample size for each group was seven mice.

Table 1. Descriptive statistics female life span (days) for ten different inbred strains of mice (Mus domesticus).

Data from Yuan et al (2021; https://phenome.jax.org).

Spreadsheet calculations.

You have several options at this point, ranging from using your calculator and the formulas above (don’t forget to square the standard deviation to get the variances!), or you could use Microsoft Excel or LibreOffice Calc and enter the necessary formulas by hand (Table 2). You’ll also find many online calculators for one-way ANOVA by sufficient statistics (eg, https://www.danielsoper.com/statcalc/calculator.aspx?id=43).

Table 2. Spreadsheet with formulas for calculating one-way ANOVA from means and standard deviations from statistics presented in Table 1.

For this example, you should get the following

MSB = 219899.5
MSE = 31634.1
F = 6.951
P-value = 4.45E-09

But, R can do it better.

Here’s how. Install the HH package (or RcmdrPlugin.HH for use in Rcmdr) and call the ?aovSufficient function.

Step 1. Install the HH package from a CRAN mirror, eg, cloud.r-project.org, in the usual way.

chooseCRANmirror()
install.packages("HH") 
library(HH)

Step 2. Enter the data. Do this in the usual way (eg, from a text file), or enter directly using the read.table command as follows.

MouseData <- read.table(header=TRUE, sep = ",", text= "
Strain, n.size, average, stdev 
129S1, 32, 787.4, 159.16
A.J, 32, 630.7, 130.20
BALB.c, 32, 734.4, 154.43
BUB.BnJ, 24, 611.3, 218.34
C3H.HeJ, 29, 724.1, 131.48
C57BL.6J, 29, 855.7, 185.34
CBA.J, 30, 622.9, 181.95
FVB.NJ, 26, 750.3, 230.11
P.J, 32, 676.0, 178.82
SWR.J, 31, 831.9, 181.31")
#Check import
head(MouseData)

End of R input

I know, a little hard to read, but from the MouseData to the end bracket ") before the comment line #Check import, that’s all one command.

Of course, you could copy the data and import the data from your computer’s clipboard in Rcmdr: Data → Import data → from text file, clipboard, or URL… 

Once the data set is loaded, proceed to Step 3.

Step 3. In our example, sample size was included for each group. Skip to step 4. If, however, the table lacked the sample size information, you can always add a new variable. For example, if we needed to add sample size to the data frame, we would use the repeat element function, rep().

MouseData$n <- rep(7, 10)

If you check the View data set button in Rcmdr, you will see that the command in Step 3 has added a new variable “n” for each of the eleven rows. The function rep() stands for “replicate elements in vectors” and what it did here was enter a value of 7 for each of the ten rows in the data set. Again, this step is not necessary for this example because sample size is already part of the data frame. Proceed to step 4.

Step 4. Run the one way ANOVA using the sufficient statistics and the HH function aovSufficient

MouseData.aov <- aovSufficient(average ~ Strain, data=MouseData)

Step 5. Get the ANOVA table

summary(MouseData.aov)

Here’s the R output

           Df  Sum Sq Mean Sq F value   Pr(>F) 
Strain      9 1977732  219748   7.123 2.54e-09 ***
Residuals 287 8853743   30849 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End R output

To explore other features of the package, type ?aovSufficient at the R prompt (like all R functions, extensive help is generally available for each function in a package).

Meta-analysis and sufficient statistics.

Meta-analysis (see Chapter 20.15 — Meta-analysis), can be performed using sufficient statistics from different tests. Combining sufficient statistics allows for a more precise overall estimate by effectively merging the information from multiple studies, which can be especially useful when individual studies have small sample sizes or lack a unified outcome measure. Common methods include creating summary effect sizes like the standardized mean difference or using p-value combination methods, while ensuring the underlying statistical assumptions are met through goodness-of-fit tests.

Fisher’s method combines the p-values from multiple independent studies into a single overall p-value to test a joint null hypothesis. The method’s test statistic is calculated as the sum of the negative log-transformed p-values. P-values from number of tests are combined; each p-value, , contributes to a variable, , with degrees of freedom. The chi-square distribution is used to evaluate the significance of the Fisher’s method test statistic.

Fisher’s method is available in several R packages (eg,poolr package); here, we’ll just create a simple R code.

Limitations: Methods that only use p-values or test statistics may not adequately address the effects of heterogeneity and potential publication bias among studies. Because many older studies do not report effect sizes and confidence intervals, using test statistics or p-values is often the only available method to combine their results.

Example.

One hypothesis about longevity is that the greater the genetic diversity can lead to longer life. The mice in Table 1 were all from inbred mouse strains. By definition, genetic diversity is restricted in inbred mice, compared to outbred strains. Jackson Labs produced a “Diverse Outbred” line of mice, and a few individuals have lived nearly five years, well past the typical 1 – 2 year lifespan of mice (Cohen 2017, Mullis et al 2025). Thus, a natural comparison is to ask: is there a difference in average lifespan between the DO and inbred mouse strains? We’ll run a series of one-sample t-tests against the mean lifespan for the diverse outbred stock of mice, save the p-values, and apply Fisher’s method. While not strictly a meta-analysis, which would draw from multiple studies, this example outlines the basic steps to take given a series of hypothesis of the same kind.

do_avg= 794.5
# dataset MouseData
tsum.test(average[1], stdev[1], n.size[1], mu=do_avg)

t = -0.25235, df = 31, p-value = 0.8024

p_values <- numeric(length(average))

for (i in 1:length(average)) {
     suppressWarnings({
      test_result <- tsum.test(
      mean.x = average[i],
      s.x = stdev[i],
      n.x = n.size[i],
      mu = do_avg
      )
      p_values[i] <- test_result$p.value
      })
     }

print(p_values)

The p-values from all ten tests are sorted by strain in Table 2.

Table 2. Results of one sample t-tests inbred strains (Table 1) compared to average lifespan of the J:DO diversity outbred mouse strain from Jackson labs

R-code for Fisher’s method

my_pValues <- c(0.08623, 1.601e-05, 0.3368, 0.0007307, 0.2598)
X_squared <- -2 * sum(log(my_pValues),base = exp(1))
myDf <- 2 * length(my_pValues)
combinedP <- pchisq(X_squared, myDf, lower.tail = FALSE)
result_table <- as.data.frame(rbind(X_squared, myDf, combinedP))
rownames(result_table) <- c("chi_sqr", "df", "P-value")
colnames(result_table) <- c("Value")
print(result_table)

and the R output

                Value
chi_sqr 44.7017431931
     df 16.0000000000
P-value 0.0001541831

With the small p-value, we would reject the null hypothesis of no difference in lifespan among the inbred strains compared to the genetically diverse DO strain of mice.

Limitations of ANOVA from sufficient statistics.

This was pretty easy, so it is worth asking — Why go through the bother of analyzing the raw data, why not just go to the summary statistics and run the calculator formula? First, the chief reason against the calculator formula and use of only sufficient statistics loses information about the individual values and therefore you have no access to the residuals. The residual of an observation is the difference between the original observation and the model prediction. The residuals are important for determining whether the model fits the data well and are, therefore, part of the toolkit that statisticians need to do proper data analysis. We will spend considerable time looking at residual patterns and it is an important aspect of doing statistics correctly.

Secondly, while it is possible to extend this approach to more complicated ANOVA problems like the two-way ANOVA (Cohen 2002), the statistical significance of the interaction term(s) calculated in this way are only approximate (the main effects are OK to interpret). Thus, ANOVA from sufficient statistics has its place when all you have is access to descriptive statistics, but its use is limited and not at all the preferred option for data analysis when the original, raw observations are in hand.

Questions

1. Calculate the one-way ANOVA for body weight of 47 female (F) and 97 male (M) cats (kilograms, dataset cats in MASS R package) from the following summary statistics.

2. Bonus: Load the cats data set (package MASS, loaded with Rcmdr) and run a one-way ANOVA using the aov() function via Rcmdr. Are the ANOVA from sufficient statistics the same as results from the direct ANOVA calculation? If not, why not.

Quiz Chapter 12.4

ANOVA from “sufficient statistics”

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Chapter 12 contents

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

12.2 – One way ANOVA

Introduction

In ANalysis Of VAriance, or ANOVA for short, we likewise have tools to test the null hypothesis of no difference between between categorical independent variables — often called factors when there’s just a few levels to keep track of — and a single, dependent response variable. But now, the response variable is quantitative, not qualitative like the tests.

Analysis of variance, ANOVA, is such an important statistical test in biology that we will take the time to “build it” from scratch. We begin by reminding you of where you’ve been with the material.

We already saw an example of this null hypothesis. When there’s only one factor (but with two or more levels), we call the analysis of means and “one-way ANOVA.” In the independent sample t-test, we tested whether two groups had the same mean. We made the connection between the confidence interval of the difference between the sample means and whether or not it includes zero (ie, no difference between the means). In ANOVA, we extend this idea to a test of whether two or more groups have the same mean. In fact, if you perform an ANOVA on only two groups, you will get the exact same answer as the independent two-sample t-test, although they use different distributions of critical values (t for the t-test, F for the ANOVA — a nice little fact for you square the t-test statistic, you’ll get the F-test statistic: ).

Let’s say we have an experiment where we’ve looked at the effect of different three different calorie levels on weight change in middle-aged men.

I’ve created a simulated dataset which we will use in our ANOVA discussions. The data set is available at the end of this page (scroll down or click here).

We might graph the mean weight changes ( SEM); Below are two possible outcomes of our experiment (Fig 1).

Figure 1. Hypothetical results of an experiment, box plots. Left, no difference among groups; Right, large differences among groups.

As statisticians and scientists, we would first calculate an overall or grand mean for the entire sample of observations; we know this as the sample mean whose symbol is . But this overall mean is made up of the average of the sample means. If the null hypothesis is true, then the all of the sample means all estimate the overall mean. Put another way, being a member of a treatment group doesn’t matter, ie, there is no systematic effect, all differences among subjects are due to random chance.

The hypotheses among are three groups or treatment levels then are:

Null that there are no differences among the group means. And the alternative hypotheses include any (or all) of the following possibilities

or maybe

or… have we covered all possible alternate outcomes among three groups?

In either case, we could use one-way ANOVA to test for “statistically significant differences.”

Three important terms you’ll need for one-way ANOVA

FACTOR: We have one factor of interest. For example, a factor of interest might be

• Diet fed to hypertensive subjects (men and women)
• Distribution of coral reef sea cucumber species in archipelagos
• Antibiotic drug therapy for adolescents with Acne vulgaris (click see Webster 2002 for review).

Gender <- c("Gender non-conforming", "Men", "Non-binary", "Transgender", "Women", "Declined"); Gender
[1] "Gender non-conforming" "Men" "Non-binary" "Transgender" 
[5] "Women" "Declined"

is.factor(Gender)
[1] FALSE

LEVELS: We can have multiple levels (2 or more) within the single factor. Some examples of levels for the Factors listed:

• Three diets (DASH), diet rich in fruits & vegetables, control diet)
• Five archipelagos (Hawaiian Islands, Line Islands, Marshal Islands, Bonin Islands, and Ryukyu Islands)
• Five antibiotics (ciprofloxacin, cotrimoxazole, erythromycin, doxycycline, minocycline).

RESPONSE: There is one outcome or measurement variable. This variable must be quantitative (ie, on the ratio or interval scale). Continuing our examples then

• Reduction in systolic pressure
• Numbers of individual sea cucumbers in a plot
• number of microcomedo.

The response variable can be just about anything we can measure, but because ANOVA is a parametric test, the response variable must be normally distributed!

Note on experimental design

As we discuss ANOVA, keep in mind we are talking about analyzing results from an experiment. Understanding statistics, and in particular ANOVA, informs how to plan an experiment. The basic experimental design is called a completely randomized experimental design, or CDR, where treatments are assigned to experimental units at random.

In this experimental design, subjects (experimental units) must be randomly assigned to each of these levels of the factor. That is, each individual should have had the same probability of being found in any one of the levels of the factor. The design is complete because randomization is conducted for all levels, all factors.

Thinking about how you would describe an experiment with three levels of some treatment, we would have

Table 1. Summary statistics of three levels for some ratio scale response variable.

ANOVA sources of variation

ANOVA works by partitioning total variability about the means (the grand mean, the group means). We will discuss the multiple samples and how the ANOVA works in terms of the sources of variation. There are two “sources” of variation that can occur:

• Within Group Variation
• Among Groups Variation

So let’s look first at the variability within groups, also called the Within Group Variation.

Consider and experiment to see if DASH diet reduces systolic blood pressure in USA middle-aged men and women with hypertension (Moore et al 2001). After eight weeks we have

We get the corrected sum of squares, SS, for within group

and the degrees of freedom, DF, for within groups

where i is the identity of the groups,  is the individual observations within group i, is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

Importantly, this value is also referred to as “error sums of squares” in ANOVA. It’s importance is as follows — In our example, the within group variability would be zero if and only if all subjects within the same diet lost exactly the same amount of weight. This is hardly ever the case of course in a real experiment. Because there are almost always some other unknown factors or measurement error that affect the response variable, there will be some unknown variation among individuals who received the same treatment (within the same group). Thus, the error variance will generally be larger than zero.

The first point to consider: your ANOVA will never result in statistical differences among the groups if the error variance is greater than the second type of variability, the variability between groups.

The second type of variability in ANOVA is that due to the groups or treatments. Individuals given a calorie-restricted diet will lose some weight; individuals allowed to eat a calorie-rich diet likely will gain weight, therefore there will be variability (a difference) due to the treatment. So we can calculate the variability among groups. We get the corrected sum of squares for among groups

and the degrees of freedom for among groups

where i is the identity of the groups, is the grand mean as defined in Measures of Central Tendency (Chapter 3.1), is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

The sums of squares here is simply subtracting the mean of each population from the overall mean.

• If the Factor is not important in explaining the variation among individuals then all the population means will be similar and the sums of squares among populations would be small.
• If the Factor is important in explaining some of the variation among the individuals then all the population means will NOT be the same and the sums of squares among populations would be large.

Finally, we can identify the total variation in the entire experiment. We have the total sum of squares.

Thus, the insight of ANOVA is that variation in the dataset may be attributed to a portion explained by differences among the groups and differences among individual observations within each group. The inference comes from recognizing that if the among group effect is greater than the within group effect, then there will be a difference due to the treatment levels.

Mean squares

To decide whether the variation associated with the among group differences are greater than the within group variation, we calculate ratios of the sums of squares. These are called Mean Squares or MS for short. The ratio of the Mean Squares is called F, the test statistic for ANOVA.

For the one-way ANOVA we will have two Mean Squares  and one F, tested with degrees of freedom for both the numerator , and denominator ).

The Mean Square for (among) groups is

The Mean Square for error is

And finally, the value for F, the test statistic for ANOVA, is

Worked example with R

A factor with three levels, A, B, and C

group <- c("A", "A", "A", "B", "B", "B", "C", "C", "C")

and their responses, simulated

response <- c(10.8, 11.8, 12.8, 6.5, 7, 8, 3.8, 2.8, 3)

We create a data frame

all <- data.frame(group, response)

Of course, you could place the data into a worksheet and then import the worksheet into R. Regardless, we now have our dataset.

Now, call the ANOVA function, aov, and assign the results to an object (eg, Model.1)

Model.1 <- aov(response ~ group, data=all)

Now, visualize the ANOVA table

summary(Model.1)

and the output from R, the ANOVA table, is shown below

Table 2. Output from aov() command, the ANOVA table, for the “Difference” outcome variable.

           Df   Sum Sq    Mean Sq   F value       Pr(>F) 
group       2   111.16      55.58     89.49    0.0000341 ***
Residuals   6     3.73       0.62
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Let’s take the ANOVA table one row at a time

• The first row has subject headers defining the columns.
• The second row of the table “groups” contains statistics due to group, and provides the comparisons among groups.
• The third row, “Residuals” is the error, or the differences within groups.

Moving across the columns, then, for each row, we have in turn,

• the degrees of freedom (there were 3 groups, therefore 2 DF for group),
• the Sums of Squares, the Mean Squares,
• the value of F, and finally,
• the P-value.

R provides a helpful guide on the last line of the ANOVA summary table, the “Signif[icance] codes,” which highlights the magnitude of the P-value.

What to report? ANOVA problems can be much more complicated than the simple one-way ANOVA introduced here. For complex ANOVA problems, report the ANOVA table itself! But for the one-way ANOVA it would be sufficient to report the test statistic, the degrees of freedom, and the p-value, as we have in previous chapters (eg, t-test, chi-square, etc.). Thus, we would report:

F = 89.49, df = 2 and 6, p = 0.0000341

where F = 89.49 is the test statistic, df = 2 (degrees of freedom for the among group mean square) and 6 (degrees of freedom for the within group mean square), and p = 0.0000341 is the p-value.

In Rcmdr, the appropriate command for the one-way ANOVA is simply

Rcmdr: Statistics → Means

Figure 2. Screenshot Rcmdr select one-way ANOVA.

which brings up a simple dialog. R Commander anticipates factor (Groups) and Response variable. Optional, choose Pairwise comparisons of means for post-hoc test (Tukey’s) and, if you do not want to assume equal variances (see Chapter 13), select Welch F-test.

Figure 3. Screenshot Rcmdr select one-way ANOVA options.

Questions

1. Review the example ANOVA Table and confirm the following

• How many levels of the treatment were there?
• How many sampling units were there?
• Confirm the calculation of and  using the formulas contained in the text.
• Confirm the calculation of F using the formula contained in the text.
• The degrees of freedom for the F statistic in this example were 2 and 6 (F_2,6). Assuming a two-tailed test with Type I error rate of 5%, What is the critical value of the F distribution (see Appendix 20.4)

2. Repeat the one-way ANOVA using the simulated data, but this time, calculate the ANOVA problem for the “No.difference” response variable.

3. Leaf lengths from three strains of Arabidopsis thaliana plants grown in common garden are shown in Fig 2. Data are provided for you in the following R script.

arabid <- c("wt","wt","wt","AS1","AS1","AS1","AS2","AS2","AS2")
leaf <- c(4.909,5.736,5.108,6.956,5.809,6.888,4.768,4.209,4.065)
leaves <- data.frame(arabid,leaf)

• Write out a null and alternative hypotheses
• Conduct a test of the null hypothesis by one-way ANOVA
• Report the value of the test statistic, the degrees of freedom, and the P-value
• Do you accept or reject the null hypothesis? Explain your choice.

Figure 4. Box plot of lengths of leaves of a 10-day old plant from on of three strains of Arabidopsis thaliana.

4. Return to your answer to question 7 from Chapter 12.1 and review your answer and modify as appropriate to correct your language to that presented here about factors and levels.

5. Conduct the one-way ANOVA test on the Comet assay data presented question 7 from Chapter 12.1. Data are copied for your convenience to end of this page. Obtain the ANOVA table and report the value of the test statistics, degrees of freedom, and p-value.
a. Based on the ANOVA results, do you accept or reject the null hypothesis? Explain your choice.

Quiz Chapter 12.2

One-way ANOVA

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Data used in this page

Difference, no difference

Table 12.2 Difference or No Difference

Comet assay, antioxidant properties of tea

Data presented in Chapter 12.1

Chapter 12 contents

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