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 Ho: 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 Ho: 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 Ho = 0; slope Ho = 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:

A good residual plot, no obvious trends.

Figure 1. An ideal plot of residuals

Solution: Proceed! Assumptions of linear regression met.

Compare to plots of residuals that differ from the ideal.

We have a problem. Residual plot shows unequal variance (heteroskedasticity).

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

Solution. Try a transform like the log10-transform.

Residual plot shows systematic trend.

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.

Residual plot shows systematic trend.

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

Basic diagnostic plots from R

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.

A note of caution about over-thinking with these plots. R provides a red line to track the points. However, these lines are guides, not judges. We humans are generally good at detecting patterns, but with data visualization, there is the risk of seeing patterns where none exits. In particular, recognizing randomness is not easy. If anything, we may tend to see patterns where none exist, termed apophenia. So yes, by all means look at the graphs, but do so with a plan: red line more or less horizontal? Then no pattern and the regression model is a good fit to the data.

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 \dot{V} O_{2} \sim \text{Body.mass}+\text{Gosner} 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 \chi^2 \approx nR^2. 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

Chapter 17 contents

10.2 – Digging deeper into t-test Plus the Welch test

Introduction

We need to spend some more time with the independent sample t-test; by tearing it apart, we can learn about how parametric tests work in general.

Our assumptions for the independent sample t-test are like the one-sample t-test the data must be continuous and normally distributed (one of our standard assumptions for parametric tests, see Chapter 13). The formula is very similar to the one-sample t-test, except that now we have two sample means and the formula for the standard error (SE) has also changed.

    \begin{align*} t = \frac{\bar{X}_{1}-\bar{X}_{2}}{s_{\bar{X}_{1}-\bar{X}_{2}}} \end{align*}

We see that the test statistic t is large if the numerator is large compared to the denominator. Large values of t will be evidence in favor of rejecting the null hypothesis.

The numerator is straightforward: we subtract one sample mean from the other — if there is no difference between the samples, then this difference will be close to zero.

The denominator requires additional discussion. What is it called? It is the pooled standard error of the mean (pooled SEM). Provided the assumption of equal variances holds — an additional standard assumptions for parametric tests, see Chapter 13 — then sample variances estimate the population variance and we can use this information to our advantage to best test the hypothesis about the sample means. In other words, we don’t have to lose a degree of freedom to account for differences in variability for the two groups (see Welch test). More degrees of freedom means more statistical power to test the null hypothesis and at the same time, more confidence that the test is performing to its best.

Lets break down the pooled standard error of the mean in order to see how the assumption of equal variances affects the t-test. We assume that s_{1}^2 = s_{2}^2

Recall that

    \begin{align*} H_{0}: \bar{X}_{1} = \bar{X}_{2} \end{align*}

Now we want a “pooled SE” that is the pooled standard error for both samples. The variance of the difference between the means can be written as

    \begin{align*} \sigma^2_{\bar{X}_{1}-\bar{X}_{2}} = \frac{\sigma_{1}^2}{n_{1}} + \frac{\sigma_{2}^2}{n_{2}} \end{align*}

First we need to calculate the pooled variance. where v1 and v2 are degrees of freedom for sample one and sample two, respectively.

Note 1This is just simply a combined formula of the sample variance.

    \begin{align*} s^2_{p} = \frac{SS_{1} + SS_{2}}{v_{1} + v_{2}} \end{align*}

What is SS? The sum of squares , where SS1 and SS2 refer to sum of squares for each of the two groups.

    \begin{align*} SS = \sum_{i=1}^{n}\left ( X_{i} - \bar{X}\right )^2 \end{align*}

You should recognize SS from your definition of the sample variance, Chapter 3.3.

And the standard error for the difference between two means can now be written as So the 2-sample t-test can be written to reflect the pooled sample standard error of the difference between two sample means. We can see how unequal sample size is accommodated by the t-test.

Conducting the test by hand follows the same form as the one-sample t-test. Find the degrees of freedom (DF), but now for each sample.

Finally we evaluate with the critical value in Table (Appendix, Table of Student’s t distribution) and compare the t test statistic against the critical value with the appropriate degrees of freedom.

Because this is the t-test, again, we are assuming that the variances are the same between the two populations (homoscedasticity) and this allows us to pool the variance. As it turns out, the T-test is not overly sensitive to other deviations from the assumptions, but if the variances are in fact different, then the standard formula may yield incorrect Type I error rates compared to stated probability level (α).

However, it would be poor statistical choice to use a test where there are alternatives. This is why in part that R (Rcmdr) sets as the default for the t-test that variances are unequal! In fact, R does not do a t-test unless you change the default to assume equal variances, which, as we now know, is the t-test.

Welch test

What to do when assumptions to t-test are not met? Many options have been proposed, and Welch’s approximate t is a good alternative to the two-sample t-test — it would be appropriate if the normal assumption still held.

    \begin{align*} t = \frac{\left ( \bar{X}_{1}-\bar{X}_{2} \right )}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} \end{align*}

The degrees of freedom for the Welch’s test are now

    \begin{align*} df \approx \frac{\left ( \frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}} \right )^2}{\frac{s_{1}^4}{n_{1}^2 \cdot v_{1}} + \frac{s_{2}^4}{n_{2}^2\cdot v_{2}}} \end{align*}

Note 2: The Welch test removes the pooled estimate of the standard error, replaces with both variance estimates directly.

As a default option, R and Rcmdr uses a variation of Welch’s test when you select to do the t-test without making the assumption of equal variance (Fig 1).

Screenshot Rcmdr t-test, Welch test option

Figure 1. Screenshot Rcmdr t-test options. Default is “No” for Assume equal variances, ie, the Welch test.

The Welch test is not a nonparametric test, it is a different formulation of the t-test.

Justification for beginning with t-test

It’s unlikely that you will need the t-test in today’s research climate. Data sets are large, experiments are complex with multiple variables and samples. Why do we have to consider the t-test, and then a separate test in the case for unequal sample size? I view it as a teaching moment. It makes the general point that ALL statistical tests make assumptions about how the calculations are done and as to the nature of the data set.

This is our first experience with what to do if there is a violation of an assumption of parametric tests (see Chapter 13). Here, the assumption is that the two groups have equal sample size. When they do not, the standard t-test tends to biased estimates. On the other hand, if the assumptions are met, the standard test is the best test because it has more power to do what we intended — that is, it is best at the actual test of the null hypothesis! Take heart — this point is not always appreciated even by scientists (cf. Fagerland 2012).

Note 3: Expert advice will differ here, but you can always calculate the test statistic, even if the assumptions do not hold. The problem, the concern is with the interpretation.

Let’s us approach the problem of violation of assumptions in a couple of ways as an introduction to how, in general, to approach choice of statistical tests.

  1. Power of a test. Much current statistical research focuses on learning about how a particular statistical test works when assumptions are violated. Thus, in addition to learning what tests are designed to do, we need to consider the effects of violations of assumptions on the performance of the test (namely, is the Type I error at the stated alpha level?). This is a matter of statistical power; power of a test reflects how able the test is to get you the correct result even if assumptions are violated. Often tests perform well if sample sizes are large despite violation of assumptions. We mention without proving that the 2-sample T-test is robust to violations of normality assumption, to requirement for equal sample size, and even to unequal variances. But good experimental design attempts to meet the assumptions because the test does better!
  2. Alternate forms of some tests are available to handle some aspects of test violations. For example, the simple 2-sample T-test can be modified to accommodate different variances (Welch’s formula). Or you must find a different test (eg, nonparametric tests).

In conclusion, all tests begin with consideration of the assumptions. In some cases we can test our assumptions. For example, we learned about testing the assumption of normal distributions of sample data. We can also test the assumption of equal variances.

Questions

  1. Consider a clinical trial in which resting blood pressure is recorded on hypertensive subjects at the start of the trial, then 6 weeks after subjects have received daily supplements of flaxseed. Thus, for each subject there are two measures of blood pressure, BEFORE and AFTER.
    • Write the null hypothesis
    • Write the alternative hypothesis
    • Justify why or why not the hypothesis should be two tailed. Explain why or why not an independent sample t-test may be used to compare groups.

Quiz Chapter 10.2

Digging deeper into t-test Plus the Welch test

Chapter 10 contents