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)

13.1 – ANOVA Assumptions

Introduction

Like all parametric tests, assumptions are made about the data in order to justify and trust estimates and inferences drawn from ANOVA. While acknowledging the “experimental design” message suggested by the xkcd panel in Figure 1 — the presence of a large effect size between groups mitigates the technical necessities of inferential statistics — of course, not all of our projects will have such obvious results. Understanding the impact of assumptions on inference is a key concept in (bio)statistics: when the p-value is close to the Type I error rate, deviations from assumptions may reduce our confidence in the “statistical significance” of the result.

Figure 1. xkcd.com “Statistics,” https://xkcd.com/2400/.

ANOVA assumptions include:

1. Data come from normal distributed population. View with a histogram or Q-Q plot. Test with Shapiro-Wilks or other appropriate goodness of fit test (see Question 1 below). Normality tests are the subject of Chapter 13.3.
2. Sample size equal among groups.
   • This is an example of a potentially confounding factor — If sample sizes differ, then any difference in means could be simply because of differences in sample size! This gets us into weighted versus unweighted means.
   • You shouldn’t be surprised that modern implementations of ANOVA in software easily handle (adjust for) these known confounding factors. Depending on the program, you’ll see “Adjusted means,” “least squares means,” “Marginal means” etc. This just implies that the group means are compared after accounting for confounding factors.
   • Importantly, as long as sample sizes among the groups are roughly equivalent, normality assumption is not a big deal (low impact on risk of Type I error).
3. Independence of errors. One consequence of this assumption is that you would not view 100 repeated observations of a trait on the same subject as 100 independent data points. We’ll return to this concept more in the next two lectures. Some examples:

• • Colorimetric assay where the signal changes over time, and you measure in order (eg, samples from group 1 first, samples from group 2 second, etc.) — this confounds group with time.
    • The consequence is that you are far more likely to reject the null hypothesis, committing a Type I error.
  • Let’s say you are observing running speeds of ten mongoose. However, it turns out that five of your subjects are actually from the same family, identical quintuplets! Do you really have ten subjects?
  • Compare brain-body mass ratio among different species; this is a classic comparative method problem (Fig 3). Since 1985 (Felsenstein 1985), it was recognized that the hierarchical evolutionary relationships among the species must be accounted for to control for lack of independence among the taxa tested. See Phylogenetically independent contrasts, Chapter 20.12.

4. Equal variances (homogeneity) among groups. See Chapter 13.4  for how to test this for multiple groups.

Impact of assumptions

R — and pretty much all statistics packages — will calculate the ANOVA table and the p-value, but it is up to you to recognize that the P-value is accurate only if the assumptions are met. Violation of the assumption of normality can lead to Type I errors occurring more often than the 5% level. What to do if the assumptions are violated?

If the violation is due to only a handful of the data, you might proceed anyway. But following a significant test for normality, we could avoid the ANOVA in favor of nonparametric alternatives (Chapter 15), or, we might try to transform the data.

Consider a histogram of brains weight measures in grams for a variety of mammals (Fig 2).

Figure 2. Histogram of body mass (g) for 24 mammals (data from Boddy et al 2012).

We will introduce a variety of statistical tests of the assumption of normality in Chapter 13.3, but looking at a histogram as part of our data exploration, we clearly see the data are right-skewed (Fig 2). Is this an example of normal distributed sample of observations? Clearly not. If we proceed with statistical tests on the raw data set, then we are more likely to commit a Type I error (ie, we will reject the null hypothesis more often than we should).

A comment on normality and biology. It is very possible that data may not be normally distributed or have equal variances on the original scale, but a simple mathematical manipulation may take care of that. In fact, in many case in biology that involve growth, many types of variables are expected to not be normal on the original scale. For example, while the relationship between body mass, , and metabolic rate, , in many groups of organisms is allometric and increases positively, the relationship

is not directly proportional (linear) on the original scale. By taking the logarithm of both body mass and metabolic rate, however, the relationship is linear

In fact, taking the logarithm (base 10, base 2, or base e) is often a common solution to both non-normal data (Fig 2) and unequal variances.

Figure 3. Histogram of log₁₀-transformed body mass observations from Figure 2.

Other common transformations include taking the square root or the inverse of the square-root for skewed or kurtotic sample distributions, and the arcsine for frequencies (since frequencies can only be from 0 to 1 — need to “stretch the data” to make frequencies fit procedures like ANOVA). There are many issues about data transformation, but keep in mind three points. After completing the transformation, you should check the assumptions (normality, equal variances) again.

You may need to recode the data before applying a transform. For example, you cannot take the square root or logarithm of negative numbers. If you do not recode the data, then you will lose these observations in your subsequent analyses. In many cases, this problem is easily solved by adding 1 to all data points before making the transform. I prefer to make the minimum value 1 and go from there. The justification for data transformation is basically to realize that there is no necessity to use the common arithmetic or linear scale: many relationships are multiplicative or nonadditive (eg, rates in biology and medicine).

Statistical outlier

Another topic we should at least mention here is the concept of outliers. While most observations tend to cluster around the mean or middle of the sample distribution, occasionally one or more observations may differ substantially from the others. Such values are called outliers, and we note that there are two possible explanations for an outlier

1. the value could be an error.
2. it is a true value.

We encountered a clear outlier in the BMI homework. If the reason is (1), then we go back and either fix the error or delete it from the worksheet. If (2), however, then we have no objective reason to exclude the point from our analyses.

We worry if the outlier influences our conclusions — so it is a good idea to run your analyses with and without the outlier. If your conclusions remain the same, then no worries. If your conclusions change based on one observation, then this is problematic. For the most part you are then obligated to include the outlier and the more conservative interpretation of your statistical tests.

ANOVA is robust to modest deviations from assumptions

A comment about ANOVA assumptions ANOVA turns out to be robust to violations of item (1) or (2). That means unless the data are really skewed or the group sizes are very different, ANOVA will perform well (Type I error rate stays close to the specified 5% level). We worry about this however when p-value is very close to alpha!!

The third assumption is more important in ANOVA.

Like the t-test, ANOVA makes the assumption of equal variances among the groups, so it will be helpful to review why this assumption is important to both the t-test and ANOVA. In the two-sample independent t-test, the pooled sample variance, , is taken as an estimate of the population variance, . If you recall,

where refers to the sum of squares for the first group and refers to the second group sum of squares (see our discussion on measures of dispersion) and refers to the degrees of freedom for the first group and refers to the second group degrees of freedom. We make a similar assumption in ANOVA. We assume that the variances for each sample are the same and therefore that they all estimate the population variance . To say it in another way, we are assuming that all of our samples have identical variability

Once we make this assumption, we may pool (or combine) all of the SSs and DFs for all groups as our best estimate of the population variance, . The trick to understanding ANOVA is to realize that there can be two types of variability: there is variability due to being part of a group (eg, even though ten human subjects receive the same calorie-restricted diet, not all ten will loose the same amount of weight) and there is variability among or between groups (eg, on average, all subjects who received the calorie-restricted diet lost more weight than did those subjects who were on the non-restricted diet).

Example

The encephalization index (or encephalization quotient) is defined as the ratio of size the brain compared to body size. While there is a well-recognized increase in brain size given increased body size, encephalization describes a shift of function to cortex (frontal, occipital, parietal, temporal) from noncortical parts of the brain (cerebellum, brainstem). Increased cortex is associated with increased complexity of brain function; for some researchers, the index is taken as a crude estimate of intelligence. Figure 4A shows plot of brain mass in grams versus body size (grams) for 24 mammal species (data sampled from Boddy et al 2012); Figure 4B shows the same data, but following log₁₀-transform of both variables.

Figure 4. Plot of brain and body weights (A) and log₁₀-log₁₀ transform (B) for a variety of species (data from Boddy et al 2012). The ratio is called encephalization index.

Looking at the two figures the linear relationship between the two variables is obvious in Figure 4B, less so for Figure 4A. Thus, one biological justification for transformation of the raw data is exemplified with the brain-body mass dataset: the association is allometric not additive. The other reason to apply a transform is statistical; the log₁₀-transform improves the normality of the variables. Take a look at the Q-Q plot for the raw data (Fig 5) and for the log10-transformed data (Fig 6).

Figure 5. Q-Q plot, body mass, raw data. Compare to Figure 2.

Figure 5 is a good test of your observation skills — what are you supposed to look for in a Q-Q plot? Follow the dots — the data don’t fall on the straight line. A few even fall outside of the confidence interval (the curved dashed lines), which suggests the data were not normally distributed (see histogram, Fig 2). And for the transformed data, the Q-Q plot is shown in Figure 6.

Figure 6. Q-Q plot same data, log₁₀-transformed, compare to Figure 2.

Compared to the raw data, the transformed data now fall on the line and none are outside of the confidence interval. We would conclude that the transformed data are more normal, thus, better meeting the assumptions of our parametric tests.

Lack of independence among data

Species comparisons are common in evolutionary biology and related fields. As noted earlier, comparative data should not be treated as independent data points (discussed in 20.12 – Phylogenetically independent contrasts). For our 24 species, I plotted the estimate of the phylogeny (timetree.org), represented in Figure 7. Some species are closely related (eg the two elephant shrews, Rhynchocyon genus). Thus, because of the hierarchical, nested nature of evolution — first articulated by Charles Darwin — instead of 24 independent data points, we have eight clusters of points (see Question 4).

Figure 7. Phylogenetic tree of 24 species used in this report.

The conclusion? We don’t have 24 data points, more like 8 points. Because the species are more or less related, there are fewer than 24 independent data points.  Statistically, this would mean that the errors are correlated. Various approaches to account for this lack of independence have been developed; perhaps the most common approach is to apply phylogenetically independent contrasts, a topic discussed in Chapter 20.12. (Boddy et al 2012 used this approach.)

Questions

1. Make a histogram for the brain mass data set.
2. Shapiro-Wilks is one test of normality.
   1. Can you recall the name of the other normality tests we named?
   2. Run tests of normality on both body mass and brain mass data. Interpret the results, frame your response with keywords: “central tendency,” “outliers,” “shape,” and “symmetry.”
3. Calculate moment statistics for the body mass and brain mass data set, both raw and log₁₀-transformed data. Interpret your results, taking care to frame your answer with the following keywords: “shape,” “symmetry,” and “peakedness” (aka “tailedness”).
4. We introduced the lack of independence for comparative data attributed to the evolutionary process. Statistically, the impact is on degrees of freedom; for example, if we ignore the lack of independence in the data set, df = 22 for the slope coefficient in a linear regression. However, given the degrees of freedom would be closer to six! What are the effects on Type I error rate of lack of independence?

Quiz Chapter 13.1

ANOVA assumptions

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Data set and R code used in this page

species, order, body, brain
'Herpestes ichneumon', Carnivora, 1764, 24.1
'Potos flavus', Carnivora, 2620, 31.05
'Vulpes vulpes', Carnivora, 3080, 49.5
'Madoqua kirkii', Cetartiodactyla, 4570, 37
'Sus scrofa', Cetartiodactyla, 1000, 47.7
'Tragulus napu', Cetartiodactyla, 2510, 18.5
'Casinycteris argynnis', Chiroptera, 40.5, 0.92
'Cynopterus brachyotis', Chiroptera, 29, 0.88
'Potorous platyops', Chiroptera, 718, 6.5
'Cercartetus nanus', Diprotodontia, 12, 0.44548
'Dactylopsila palpator', Diprotodontia, 474, 7.15876
'Vombatus ursinus', Diprotodontia, 1902, 11.396
'Crocidura fuscomurina', Eulipotyphla, 5.6, 0.13
'Scalopus aquaticus', Eulipotyphla, 39.6, 1.48
'Sorex hoyi', Eulipotyphla, 2.6, 0.107
'Elephantulus fuscipes', Macroscelidea, 57, 1.33
'Rhynchocyon cirnei', Macroscelidea, 490, 6.1
'Rhynchocyon petersi', Macroscelidea, 717.3, 4.46
'Aotus trivirgatus', Primates, 480, 15.5
'Cebus capucinus', Primates, 590, 53.28
'Leontopithecus rosalia', Primates, 502.5, 11.7
'Dipodomys agilis', Rodentia, 61.4, 1.34
'Peromyscus melanocarpus', Rodentia, 58.8, 1.03
'Sciurus carolinensis', Rodentia, 367, 6.49

The Newick code for the tree in Figure 6.

((Vombatus_ursinus:48.94499077,((Potorous_platyops:47.59556667,Cercartetus_nanus:47.59556667)'14':0.66887333,Dactylopsila_palpator:48.26444000)'13':0.68055077)'11':109.65259681,(((((Crocidura_fuscomurina:33.74066667,Sorex_hoyi:33.74066667)'10':33.03022424,Scalopus_aquaticus:66.77089091)'19':22.55292749,(((Herpestes_brachyurus:54.32144118,(Vulpes_vulpes:45.52834967,Potos_flavus:45.52834967)'9':8.79309151)'22':23.43351523,((Tragulus_napu:43.96862857,Madoqua_kirkii:43.96862857)'8':17.99735995,Sus_scrofa:61.96598852)'6':15.78896789)'30':0.77378567,(Casinycteris_argynnis:35.20000000,Cynopterus_brachyotis:35.20000000)'29':43.32874208)'27':10.79507632)'35':7.13857077,(((Peromyscus_melanocarpus:69.89837667,Dipodomys_agilis:69.89837667)'43':0.64655123,Sciurus_carolinensis:70.54492789)'42':19.27825952,((Leontopithecus_rosalia:18.38385647,Aotus_trivirgatus:18.38385647)'40':1.29720005,Cebus_capucinus:19.68105652)'48':70.14213090)'51':6.63920175)'47':8.99739162,(Elephantulus_fuscipes:39.23366667,(Rhynchocyon_cirnei:15.34500000,Rhynchocyon_petersi:15.34500000)'39':23.88866667)'56':66.22611412)'55':53.13780679);

Chapter 13 contents

• Introduction
• ANOVA Assumptions
• Why tests of assumption are important
• Test assumption of normality
• Tests for Equal Variances
• References and suggested readings

4.7 – Q-Q plot

Introduction

Use of graphs by a data analyst may serve different purposes: communication of results or as diagnostics. The Q-Q plot is one example of a graph used as a diagnostic.

The quantile-quantile, or Q–Q plot is a probability plot used to compare graphically two probability distributions. In brief, a set of intervals for the quantiles is chosen for each sample. A point on the plot represents one of the quantiles from the second distribution (y value) against the same quantile from the first distribution (x value).

A common use of Q-Q plot would be to compare data from a sample against a normal distribution. If the sample distribution is similar to a normal distribution, the points in the Q–Q plot will approximately lie on the line y = x.

R code

In R, the Q-Q plot can be obtained directly in Rcmdr.

Figure 1. A Q-Q plot, the default command in Rcmdr

Rcmdr: Graphics → Quantile-comparison plot…

After choosing the variable (in this case, Sales), click on Options tab and make additional selections before making the graph. Here, selected normal distribution.

Figure 2. Screenshot of R Commander menu for Q-Q plot

Another version is available in the KMggplot2 package.

Questions

1. What is a Q-Q plot used for in statistics?
2. Looking at the plot in Figure 1, explain why the confidence lines get further and further away from the straight line.

Quiz Chapter 4.7

Q-Q plots

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Chapter 4 contents

• Graphs and tables
• Bar (column) charts
• Histograms
• Box plot
• Mosaic plots
• Scatter plots
• Adding a second Y axis
• Q-Q plot
• Ternary plots
• Heat maps
• Graph software
• References and suggested readings