18.5 – Selecting the best model

Introduction

This is a long entry in our textbook, many topics to cover. We discuss aspects of model fitting, from why model fitting is done to how to do it and what statistics are available to help us decide on the best model. Model selection very much depends on what the intent of the study is. For example, if the purpose of model building is to provide the best description of the data, then in general one should prefer the full (also called the saturated) model. On the other hand, if the purpose of model building is to make a predictive statistical model, then a reduced model may prove to be a better choice. The text here deals mostly with the later context of model selection, finding a justified reduced model.

From Full model to best Subset model

Model building is an essential part of being a scientist. As scientists, we seek models that explain as much of the variability about a phenomenon as possible, but yet remain simple enough to be of practical use.

Having just completed the introduction to multiple regression, we now move to the idea of how to pick best models.

We distinguish between a full model, which includes as many variables (predictors, factors) as the regression function can work with, returning interpretable, if not always statistically significant output, and a saturated model.

The saturated model is the one that includes all possible predictors, factors, interactions in your experiment. In well-behaved data sets, the full model and the saturated model will be the same model. However, they need not be the same model. For example, if two predictor variables are highly collinear, then you may return an error in regression fitting.

For those of you working with meta-analysis problems, you are unlikely to be able to run a saturated model because some level of a key factor are not available in all or at least most of the papers. Thus, in order to get the model to run, you start dropping factors, or you start nesting factors. If you were unable to get more things in the model, then this is your “full” model. Technically we wouldn’t call it saturated because there were other factors, they just didn’t have enough data to work with or they were essentially the same as something else in the model.

Identify the model that does run to completion as your full model and proceed to assess model fit criteria for that model, and all reduced models thereafter.

In R (Rcmdr) you know you have found the full model when the output lacks “NA” strings (missing values) in the output. Use the full model to report the values for each coefficient, i.e., conducting the inferential statistics.

Get the estimates directly from the output from running the regression function. You can tell if the effect is positive (look at the estimate for sample — it is positive) so you can say — more samples, greater likelihood to see more cases of cancer.

Remember, the experimental units are the papers themselves, so studies with larger numbers of subjects are going to find more cases of diabetes. We would worry big time with your project if we did not see statistically significant and positive for sample size.

Example: A multiple linear model

For illustration, here’s an example output following a run with the linear model function on a portion of the Kaggle Cardiovascular disease data set.

The variables (and data types) were

SBP = systolic blood pressure (from ap_hi)

age.years = Independent variable, ratio scale (calculated from age in days)

BMI = Dependent variable, ratio scale (calculated from height and weight variables)

active = Independent variable, physical activity, binomial

alcohol = Independent variable, alcohol use, binomial

cholesterol = Independent variable, cholesterol levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

gender = Independent variable, binomial

glucose = Independent variable, glucose levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

smoke = Independent variable, binomial

R output

Call:
lm(formula = SBP ~ BMI + gender + age.years + active + alcohol + 
cholesterol + gluc + smoke, data = kaggle.train)

Residuals:
Min 1Q Median 3Q Max 
-33.359 -9.701 -1.065 8.703 45.242

Coefficients:
             Estimate Std. Error t value   Pr(>|t|) 
(Intercept)   76.7664    14.9311   5.141  0.00000155 ***
BMI            0.8395     0.2933   2.862  0.00522 ** 
gender        -2.0744     1.6102  -1.288  0.20092 
age.years      0.2742     0.2403   1.141  0.25672 
active         7.1411     3.4934   2.044  0.04383 * 
alcohol       -7.9272     9.1854  -0.863  0.39039 
cholesterol    7.4539     2.3200   3.213  0.00182 ** 
gluc          -1.2701     2.9533  -0.430  0.66817 
smoke         11.4132     6.7015   1.703  0.09197 . 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 14.96 on 91 degrees of freedom
Multiple R-squared: 0.2839, Adjusted R-squared: 0.221 
F-statistic: 4.51 on 8 and 91 DF, p-value: 0.0001214

Question. Write out the equation in symbol form.

We see from the modest R² (adjusted) that the model explains little variation in systolic blood pressure; only BMI, physical activity (yes, no), and serum cholesterol levels, were statistically significant at 5% level. Smoker (yes, no) approached significance (p-value = 0.092), but I wouldn’t go on and on about positive or negative coefficients and effects.

We report the value (I would round to 7.5) for cholesterol on SBP (Table 1), and note that those who had elevated serum cholesterol levels tended to have higher systolic blood pressure (e.g., the sign of the coefficient — and I related the coefficient back to the most important thing about your study — the biological interpretation). Repeat the same for the other significant coefficients.

Now’s a good time to be clear about HOW you report statistical results. DO NOT SIMPLY COPY AND PASTE EVERYTHING into your report. Now, for the estimates above, you would report everything, but not all of the figures. Here’s how the output should be reported in your Project paper.

Table 1. Coefficients from full model.

term     Coefficient Error  P-value* 
Intercept       76.8  14.93 < 0.00001
BMI              0.8   0.29   0.00522
gender          -2.1   1.61   0.20092 
age.years        0.3   0.24   0.25672 
active           7.1   3.49   0.04383
alcohol         -7.9   9.19   0.39039 
cholesterol      7.5   2.32   0.00182 
gluc            -1.3   2.95   0.66817 
smoke           11.4   6.70   0.09197

* t-test

Looks better, doesn’t it?

Once you have the full model, we should run simple diagnostics on the regression fit to the data (VIF, Cook’s Distance, basic diagnostic plots), then proceed to use this model for the inferential statistics.

> vif(model.1)
BMI gender age.years active alcohol cholesterol gluc smoke 
1.112549 1.091154 1.081641 1.048519 1.096382 1.178635 1.168442 1.131081

No VIF was greater than 2, suggesting little multicollinearity among the predictor variables.

> cooks.distance(model.1)[which.max(cooks.distance(model.1))]
60 
0.355763

This person had high SBP (180), was a smoker, but active and normal BMI.

Use the significance tests of each parameter in the model from the corresponding ANOVA table. Now, where is the ANOVA table? Remember, right after running the linear regression,

Rcmdr: Models → Hypothesis testing → ANOVA tables

Accept the default (partial marginality), and, Boom! … the ANOVA table you should be familiar with.

From the ANOVA table you will tell me whether a Factor is significant or not. You report the ANOVA table in your paper. You describe it.

Now, the next step is to decide what is the best model. It then guides you to the next step which is to decide whether a better model (fewer parameters, Occam’s razor) can be found. Identify the parameter from the ANOVA table with the highest P-value and remove it from the model when you run the regression again. Repeat the steps above, return the ANOVA table, checking the estimates and P-values, until you have a model with only statistically significant parameters.

Find the best model

We may be tempted to run the linear model again, but with non-significant predictors excluded. For example, re-run the model on SBP but with just BMI, active, cholesterol and excluding smoke.

We can calculate the AIC, Akaike Information Criterion, the lower value is considered the better model.

AIC(model.1)
[1] 835.4938

AIC(model.2)
[1] 834.0782

The full model had eight predictor variables, excluding the intercept, whereas the reduced model had just three predictor variables, again, not including the intercept. Unsurprisingly, the reduced model is better. But now, a stronger test — what about smoking, where the p-value approached 5%? We ran the linear model (model.3).

> AIC(model.3)
[1] 831.9495

We can use an anova test or the likelihood ratio test to compare the models. The anova test, the models must be nested.

stats::anova(model.2, model.3)
Analysis of Variance Table

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
  Res.Df   RSS Df Sum of Sq      F   Pr(>F) 
1     96 22205 
2     95 21307  1     898.1 4.0043 0.04824 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

and

lmtest::lrtest((model.2,model.3)
Likelihood ratio test

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
 #Df  LogLik Df  Chisq Pr(>Chisq) 
1  5 -412.04 
2  6 -409.97  1 4.1286 0.04216 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Both tests tell us the same thing — our model with four predictors, including smoking, is better.

2 December 2024 — Past this point, edits pending

But first, I want to take up an important point about your models that you may not have had a chance to think about. The order of entry of parameters in your model can effect the significance and value of the estimates themselves. The order of parameter model entry above can be read top to bottom. Age was first, followed in sequence by CalsPDay, CholPDay, and so on. By convention, enter the covariates first (the ratio-scale predictors), that’s what I did above.

Here’s the output from a model in which I used a different order of parameters.

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

Response: BMI
              Sum Sq    Df    F value     Pr(>F)
Sex             1.52     1     0.0478    0.82764
Smoke           8.84     1     0.2782    0.59964
Age             0.62     1     0.0196    0.88918
CalsPDay       10.79     1     0.3394    0.56215
CholPDay       92.37     1     2.9065    0.09286 .
Sex:Smoke       3.35     1     0.1055    0.74631
Residuals    2129.34    67

The output is the same!!! So why did I give you a warning about parameter order? Run the ANOVA table summary command again, but this time select Type III type of test, i.e., ignore marginality

> Anova(LinearModel.2, type="III")
Anova Table (Type III tests)

Response: BMI
               Sum Sq    Df    F value       Pr(>F)
(Intercept)   1544.34     1    48.5929   1.708e-09 ***
Sex              4.69     1     0.1475     0.70214
Smoke           11.82     1     0.3720     0.54400
Age              0.62     1     0.0196     0.88918
CalsPDay        10.79     1     0.3394     0.56215
CholPDay        92.37     1     2.9065     0.09286 .
Sex:Smoke        3.35     1     0.1055     0.74631
Residuals     2129.34    67

The output has changed — and in fact it now reports the significance test of the intercept. This output is the same as the output from the linear model. Try again, this time selecting Type I, sequential

> anova(LinearModel.2)
Analysis of Variance Table

Response: BMI
           Df   Sum Sq    Mean Sq  F value    Pr(>F)
Sex         1     0.68      0.681   0.0214   0.88402
Smoke       1     2.82      2.816   0.0886   0.76690
Age         1     3.44      3.436   0.1081   0.74333
CalsPDay    1     2.27      2.272   0.0715   0.78998
CholPDay    1    96.30     96.299   3.0301   0.08633
Sex:Smoke   1     3.35      3.354   0.1055   0.74631
Residuals  67  2129.34     31.781

Here, we see the effect of order. So, as we are working to learn all of the issues of statistics and in particular mode fitting, I have purposefully restricted you to Type II analyses — obeying marginality correctly handles most issues about order of entry.

Easy there…. Take a deep breath, and guess what? Your best model needs to have significant parameters in it, right? Your best fit model then is Model 5. And that model will be your candidate for best fit as we proceed to complete our model building.

Now we proceed to gain some support evidence for our candidate best model. We are going to use an information criterion approach.

Use a fit criterion for determining model fit

To help us evaluate evidence in favor of one model over another there are a number of statistics one may calculate to provide a single number for each model for comparison purposes. The criteria model evaluators available to us include Mallow’s Cp, adjusted R², Akaike Information Criterion (AIC) or the Bayesian Information Criterion (BIC) to select best model.

We already introduced the coefficient of determination R² as a measure of fit – in general we favor models with larger values of R². However, values of R² will always be larger for models with more parameters. Thus, the other evaluators attempt to adjust for the parameters in the model and how they contribute to increased model fit.  For illustrative purposes we will use Mallow’s Cp. The equation for Mallow’s Cp in linear regression is

where p is the number of parameters in the model. Mallow’s Cp is thus equal to the number of parameters in the model plus an additional amount due to lack of fit of the model (i.e., large residuals). All else being equal we favor the model in which the Cp is close to the number of parameters in the model.

In Rcmdr, select Models → Subset model selection … (Fig. 1)

Figure 1. Rcmdr popup menu, Subset model selection…

From the menu, select the criterion and how many models to return. The function returns a graph that can be used to interpret which model is best given the selection criterion used. Below is an example (although for a different data set!) for Mallow’s Cp (Fig. 2).

Figure 2. Mallow’s Cp plot

Lets break down the plot. First define the axes. The vertical axis is the range of values for the Cp calculated for each model. The horizontal axis is categorical and reads from left to right: Intercept, Inspection[T.Yes], etc., up to Suspended. Looking into the graph itself we see horizontal bars — the extent of shading indicates which model corresponds to the Cp value. For example, the lowest bar which is associated with the Cp value of 8 extends all the way to the right of the graph. This says that the model evaluated included all of the variables and therefore was the saturated or full model. The next bar from the bottom of the graph is missing only one block (lgCarIns), which tells us the Cp value 6 corresponds to a reduced model, and so forth.

Cross-validation

Once you have identified your Best Fit model, then, your proceed to run the diagnostics plots. For the rest of the discussion we return to our first example.

Rcmdr: Models → Graphs → Basic diagnostic plots.

We’ll just concern ourselves with the first row of plots (Fig. 3).

Figure 3. Diagnostic plots

The left one shows the residuals versus the predicted values — if you see a trend here, the assumption of linearity has been violated. The second plot is a test of the assumption of normality of the residuals. Interpret them (residuals OK, Residuals normally distributed? Yes/ No), and you’re done. Here, I would say I see no real trend in the residuals vs fitted plot, so assumption of linear fit is OK. For normality, there is a tailing off at the larger values of residuals, which might be of some concern (and I would start thinking about possible leverage problems), but nothing dramatic. I would conclude that our Model 6 is a good fitting model and one that could be used to make predictions.

Now, if you think a moment, you should identify a logical problem. We used the same data to “check” the model fit as we did to make the model in the first place. In particular if the model is intended to make predictions it would be advisable to check the performance of the model (e.g., does it make reasonable predictions?) by supplying new data, data not used to construct the model, into the model. If new data are not available, one acceptable practice is to divide the full data set into at least two subsets, one used to develop the model (sometimes called the calibration or training dataset) and the other used to test the model. The benefits of cross-validation include testing for influence points, over fitting of model parameters, and a reality check on the predictions generated from the model.

A note on best practices

In the 1980s, researchers often “removed” the effect of a covariate by first fitting a regression and then saving the residuals to use in separate analyses, such as t-tests, ANOVAs, or variance component estimates (eg, Dohm et al 2001). For example, in a study of plant growth, scientists might regress final plant height on initial size to remove the effect of size, then compare the residuals between fertilizer groups. While this approach adjusts for the covariate, it is problematic because residuals are not raw data: they are correlated, have constrained sums, and underestimate uncertainty. This can distort group differences, especially if the covariate distributions differ among groups, and it ignores the variability introduced by the first regression. This approach was common in ecology, evolutionary biology, physiology, and psychology before modern mixed modeling and generalized regression workflows became standard.

Modern best practices instead recommend fitting a single unified model that includes both the covariate and the grouping variable directly. Using the same plant growth example, a linear model such as estimates the effect of fertilizer while properly accounting for initial plant size. This approach provides valid standard errors, allows correct inference about group differences, and can accommodate interactions (eg, if the relationship between and differs by group) or hierarchical data structures using mixed-effects models. Additionally, modern workflows include diagnostics such as residual plots, Cook’s Distance, and variance inflation factors to check model assumptions and influential points. Use mixed models when data are hierarchical or repeated data. Random effects handle variation among groups, individuals, years, etc., in a single coherent model. When covariates need to be “controlled for,” we use partial regression. This happens inside the model, not by subtracting covariate effects first. By analyzing all relevant predictors in a single model, researchers obtain more accurate estimates, properly reflect uncertainty, and avoid the statistical pitfalls of the old residual-based two-stage approach.

Thus, there are several statistical reasons the older approach fails under modern scrutiny. First, Residuals are not independent observations. Residuals are model-dependent quantities, not raw data. Once you fit a regression, the residuals are correlated with each other, constrained to sum to zero, and, likely heteroscedastic if the model was imperfect. Using them in a new test violates standard assumptions of the second test.

Second,  you underestimate uncertainty. If you perform tests on residuals, the standard errors do not include uncertainty from the first regression, which may lead to inflated Type I error and an overly confident “adjusted” group comparisons.

Third, the old approach creates a “two-stage model” that breaks the likelihood framework. A two-stage analysis discards the correct joint likelihood and therefore produces biases p-values, and interferes with valid confidence intervals or predictions.

Under the modern one-stage models include ML/REML, generalized estimating equations, and fully integrated Bayesian models, which keeps the model coherent. Use of mixed models make residualization unnecessary. In studies with repeated measures, blocks, sites, subjects, or years, the old approach cannot partition sources of variation correctly.

In short, the advantages of the one-stage approach is that resulting models have valid uncertainty, correct group comparisons, better handling of interactions and hierarchy, full likelihood-based inference. They fit a single unified regression or mixed-effects model with all relevant predictors, examine diagnostics, and make inferences directly. This is one more benefit of the increased power and availability of more powerful computers and more complete development of mixed methods modeling.

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Selecting the best regression model

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Chapter 18 contents

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

9.1 – Chi-square test: Goodness of fit

Introduction

We ask about the “fit” of our data against predictions from theory, or from rules that set our expectations for the frequency of particular outcomes drawn from outside the experiment. Three examples to illustrate goodness of fit, GOF, , A, B, and C follow.

A. For example, for a toss of a coin, we expect heads to show up 50% of the time. Out of 120 tosses of a fair coin, we expect 60 heads, 60 tails. Thus, our null hypothesis would be that heads would appear 50% of the time. If we observe 70 heads in an experiment of coin tossing, is this a significantly large enough discrepancy to reject the null hypothesis?

B. For example, simple Mendelian genetics makes predictions about how often we should expect particular combinations of phenotypes in the offspring when the phenotype is controlled by one gene, with 2 alleles and a particular kind of dominance.

For example, for a one locus, two allele system (one gene, two different copies like R and r) with complete dominance, we expect the phenotypic (what you see) ratio will be 3:1 (or round, wrinkle). Our null hypothesis would be that pea shape will obey Mendelian ratios (3:1). Mendel’s round versus wrinkled peas (RR or Rr genotypes give round peas, only rr results in wrinkled peas).

Thus, out of 100 individuals, we would expect 75 round and 25 wrinkled. If we observe 84 round and 16 wrinkled, is this a significantly large enough discrepancy to reject the null hypothesis?

C. For yet another example, in population genetics, we can ask whether genotypic frequencies (how often a particular copy of a gene appears in a population) follow expectations from Hardy-Weinberg model (the null hypothesis would be that they do).

This is a common test one might perform on DNA or protein data from electrophoresis analysis. Hardy-Weinberg is a simple quadratic expansion:

If p = allele frequency of the first copy, and q = allele frequency of the second copy, then p + q = 1,

Given the allele frequencies, then genotypic frequencies would be given by 1 = p² + 2pq + q².

Deviations from Hardy-Weinberg expectations may indicate a number of possible causes of allele change (including natural selection, genetic drift, migration).

Thus, if a gene has two alleles, and , with the frequency for , and for ,  (equivalently q = 1 – p) in the population, then we would expect 36 , 16 , and 48 individuals. (Nothing changes if we represent the alleles as A and a, or some other system, eg, dominance/recessive.)

Question. If we observe the following genotypes: 45  individuals, 34 individuals, and 21 individuals, is this a significantly large enough discrepancy to reject the null hypothesis?

Table 1. Summary of our Hardy Weinberg question

Recall from your genetics class that we can get the allele frequency values from the genotype values, eg,

We call these chi-square tests, tests of goodness of fit. Because we have some theory, in this case Mendelian genetics, or guidance, separate from the study itself, to help us calculate expected values in a chi-square test.

goodness of fit

For k groups, the equation for the chi-square test may be written as

where f_i is the frequency (count) observed (in class i) and f_i<hat> is the frequency (count) expected if the null hypothesis is true, sum over all k groups. Alternatively, here is a format for the same equation that may be more familiar to you… ?

where O_i is the frequency (count) observed (in class i) and E_i is the frequency (count) expected if the null hypothesis is true.

The degrees of freedom, df, for the GOF are simply the number of categories minus one, k – 1.

Explaining GOF

Why am I using the phrase “goodness of fit?” This concept has broad use in statistics, but in general it applies when we ask how well a statistical model fits the observed data.

At least for the chi-square test it is simple to see how the test statistic increases from zero as the agreement between observed data and expected data depart, where zero would be the case in which all observed values for the categories exactly match the expected values.

The goodness of fit test is designed to evaluate whether or not your data agree with a theoretical expectation (there are additional ways to think about this test, but this is a good place to start). Let’s take our time here and work with an example. The other type of chi-square problem or experiment is one for the many types of experiments in which the response variable is discrete, just like in the GOF case, but we have no theory to guide us in deciding how to obtain the expected values. We can use the data themselves to calculate expected values, and we say that the test is “contingent” upon the data, hence these types of chi-square tests are called contingency tables.

You may be a little concerned at this point that there are two kinds of chi-square problems, goodness of fit and contingency tables. We’ll deal directly with contingency tables in the next section, but for now, I wanted to make a few generalizations.

1. Both goodness of fit and contingency tables use the same chi-square equation and analysis. They differ in how the degrees of freedom are calculated.
2. Thus, what all chi-square problems have in common, whether goodness of fit or contingency table problems
   1. You must identify what types of data are appropriate for this statistical procedure? Categorical (nominal data type).
   2. As always, a clear description of the hypotheses being examined.

For goodness of fit chi-square test, the most important type of hypothesis is called a Null Hypothesis: In most cases the Null Hypothesis (H_O) is “no difference” “no effect”…. If H_O is concluded to be false (rejected), then an alternative hypothesis (H_A) will be assumed to be true. Both are specified before tests are conducted. All possible outcomes are accounted for by the two hypotheses.

From above, we have

A. H_O: Fifty out of 100 tosses will result in heads.

H_A: Heads will not appear 50 times out of 100.

B. H_O: Pea shape will equal Mendelian ratios (3:1).

H_A: Pea shape will not equal Mendelian ratios (3:1).

C. H_O: Genotypic frequencies will equal Hardy-Weinberg expectations.

H_A: Genotypic frequencies will not equal Hardy-Weinberg expectations

Assumptions: In order to use the chi-square, there must be two or more categories. Each observation must be in one and only one category. If some of the observations are truly halfway between two categories then you must make a new category (eg, low, middle, high) or use another statistical procedure. Additionally, your expected values are required to be integers, not ratio. The number of observed and the number of expected must sum to the same total.

The chi-square test is a good example of such tests, and we will encounter other examples too. Another common goodness of fit is the coefficient of determination, which will be introduced in linear regression sections (see Chapter 17.7 – Regression model fit). Still other examples are the likelihood ratio test (LRT), Akaike Information Criterion (AIC), and Bayesian Information Criterion (BIC), which are all used to assess fit of models to data. (See Graffelman and Weir [2018] for how to use AIC in the context of testing for Hardy Weinberg equilibrium.)

The likelihood ratio test is used to compare the goodness of fit of two hierarchically nested models to determine if a more complex model provides a significantly better fit to the data. It does this by forming a ratio of the likelihood of the data under the two models (the simpler “null” model and the more complex “alternative” model). A small ratio suggests the simpler model is a poor fit, while a value close to one indicates the added complexity is not statistically significant. AIC (and BIC) is a measure of relative fit — it balances goodness of fit and model complexity: A lower AIC suggests a better model by finding the optimal trade-off between fitting the data well and keeping the model simple. A model that is too complex will have a higher AIC than a simpler model with a nearly as good fit.

We use LRT, AIC and BIC when we talk about selecting best regression models (see Chapter 17.7 – Regression model fit). However, as we discuss the concept of comparing models, we need to distinguish between the concept of model choice — methods used to select a best model from competing options and estimating the “goodness of fit” of a single model.

How well does data fit the prediction?

Frequentist approach interprets the test as, how well does the data fit the null hypothesis, ? When you compare data against a theoretical distribution (eg, Mendel’s hypothesis predicts the distribution of progeny phenotypes for a particular genetic system), you test the fit of the data against the model’s predictions (expectations). Recall that the Bayesian approach asks how well does the model fit the data?

Table 2. A. 120 tosses of a coin, we count heads 70/120 tosses.

Table 3. B. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

Table 4. C. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

For completeness, instead of a goodness of fit test we can treat this problem as a test of independence, a contingency table problem. We’ll discuss contingency tables more in the next section, but or now, we can rearrange our table of observed genotypes for problem C, as a 2X2 table

Table 5. Problem C reported in 2X2 table format.

The contingency table is calculated the same way as the GOF version, but the degrees of freedom are calculated differently: df = number of rows – 1 multiplied by the number of columns – 1.

Thus, for a 2X2 table the df are always equal to 1.

Note that the chi-square value itself says nothing about how any discrepancy between expectation and observed genotype frequencies come about. Therefore, one can rearrange the  equation to make clear where deviance from equilibrium, D, occur for the heterozygote (het). We have

where D² is equal to

φ coefficient

The chi-square test statistic and its inference tells you about the significance of the association, but not the strength or effect size of the association. Not surprisingly, Pearson (1904) came up with a statistic to quantify the strength of association between two binary variables, now called the φ (phi) coefficient. Like the Pearson product moment correlation, the φ (phi) coefficient takes values from -1 to +1.

The formula for the absolute value of φ coefficient is

Thus, for A, B, and C examples, φ coefficient was 0.167, 0.190, and 0.771. Thus, only weak associations in examples A and B, but strong association in C. We’ll provide a formula to directly calculate φ coefficient from the cells of the table in 9.2 – Chi-square contingency tables.

Carry out the test and interpret results

What was just calculated? The chi-square, , test statistic.

Just like t-tests, we now want to compare our test statistic against a critical value — calculate degrees of freedom (df = k – 1, k equals the numbers of categories), and set a rejection level, Type I error rate. We typically set the Type I error rate at 5%. A table of critical values for the chi-square test is available in Appendix Table Chi-square critical values.

Obtaining Probability Values for the goodness-of-fit test of the null hypothesis:

As you can see from the equation of the chi-square, a perfect fit between the observed and the expected would be a chi-square of zero. Thus, asking about statistical significance in the chi-square test is the same as asking if your test statistic is significantly greater than zero.

The chi-square distribution is used and the critical values depend on the degrees of freedom. Fortunately for  and other statistical procedures we have Tables that will tell us what the probability is of obtaining our results when the null hypothesis is true (in the population).

Here is a portion of the chi-square critical values for probability that your chi-square test statistic is less than the critical value (Fig 1).

Figure 1. A portion of critical values of the chi-square at alpha 5% for degrees of freedom between 1 and 10. A more inclusive table is provided in the Appendix, Table of Chi-square critical values.

For the first example (A), we have df = 2 – 1 = 1 and we look up the critical value corresponding to the probability in which Type I = 5% are likely to be smaller iff (“if and only if”) the null hypothesis is true. That value is 3.841; our test statistic was 3.330, and therefore smaller than the critical value: so we do not reject the null hypothesis.

Interpolating p-values

How likely is our test statistic value of 3.333 and the null hypothesis was true? (Remember, “true” in this case is a shorthand for our data was sampled from a population in which the HW expectations hold). When I check the table of critical values of the chi-square test for the “exact” p-value, I find that our test statistic value falls between a p-value 0.10 and 0.05 (represented in the table below). We can interpolate

Table 6. Interpolated p-value for critical value not reported in chi-square table.

If we assume the change in probability between 2.706 and 3.841 for the chi-square distribution is linear (it’s not, but it’s close), then we can do so simple interpolation.

We set up what we know on the right hand side equal to what we don’t know on the left hand side of the equation,

and solve for x. Then, x is equal to 0.0724

R function pchisq() gives a value of p = 0.0679. Close, but not the same. Of course, you should go with the result from R over interpolation; we mention how to get the approximate p-value by interpolation for completeness, and, in some rare instances, you might need to make the calculation. Interpolating is also a skill used to provide estimates where the researcher needs to estimate (impute) a missing value.

Interpreting p-values

This is a pretty important topic, so much so that we devote an entire section to this very problem — see 8.2 – The controversy over proper hypothesis testing. If you skipped the chapter, but find yourself unsure how to interpret the p-value, then please go back to Ch 8.2. OK, that commercial message, what does it mean to “reject the chi-square null hypothesis?” These types of tests are called goodness of fit in the following sense — if your data agree with the theoretical distribution, then the difference between observed and expected should be very close to zero. If it is exactly zero, then you have a perfect fit. In our coin toss case, if we say that the ratio of heads:tails do not differ significantly from the 50:50 expectation, then we accept the null hypothesis.

You should try the other examples yourself! A hint, the degrees of freedom are one (1) for example B and two (2) for example C.

R code

Printed tables of the critical values from the chi-square distribution, or for any statistical test for that matter are fine, but with your statistical package R and Rcmdr, you have access to the critical value and the p-value of your test statistic simply by asking. Here’s how to get both.

First, let’s get the critical value.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared quantiles (Fig 2).

Figure 2. R Commander menu for Chi-squared quantiles.

I entered “0.05” for the probability because that’s my Type I error rate α. Enter “1” for Degrees of freedom, then click “upper tail” because we are interested in obtaining the critical value for α. Here’s R’s response when I clicked “OK.”

qchisq(c(0.05), df=1, lower.tail=FALSE)
[1] 3.841459

Next, let’s get the exact P-value of our test statistic. We had three from three different tests:  for the coin-tossing example, for the pea example, and for the Hardy-Weinberg example.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared probabilities… (Fig 3).

Figure 3. R Commander menu for Chi-squared probabilities.

I entered “3.333” because that is one of the test statistics I want to calculate for probability and “1” for Degrees of freedom because I had k – 1 = 1 df for this problem. Here’s R’s response when I clicked “OK.”

pchisq(c(3.333), df=1, lower.tail=FALSE)
[1] 0.06790291

I repeated this exercise for  I got ; for I got .

Nice, right? Saves you from having to interpolate probability values from the chi-square table.

3. How to get the goodness of fit  in Rcmdr.

R provides the goodness of fit  (the command is chisq.text()), but Rcmdr thus far does not provide a menu option to link to the function. Instead, R Commander provides a menu for contingency tables, which also is a chi-square test, but is used where no theory is available to calculate the expected values (see Chapter 9.2). Thus, for the goodness of fit chi-square, we will need to bypass Rcmdr in favor of the script window. Honestly, other options are as quick or quicker: calculate by hand, use a different software (eg, Microsoft Excel), or many online sites provide JavaScript tools.

So how to get the goodness of fit chi-square while in R? Here’s one way. At the command line, type

chisq.test (c(O1, O2, ... On), p = c(E1, E2, ... En))

where O1, O2, … On are observed counts for category 1, category 2, up to category n, and E1, E2, … En are the expected proportions for each category. For example, consider our Heads/Tails example above (problem A).

In R, we write and submit

chisq.test(c(70,30),p=c(1/2,1/2))

R returns

Chi-squared test for given probabilities.
data: c(70, 30) 
X-squared = 16, df = 1, p-value = 0.00006334

Easy enough. But not much detail — details are available with some additions to the R script. I’ll just link you to a nice website that shows how to add to the output so that it looks like the one below.

mike.chi <- chisq.test(c(70,30),p=c(1/2,1/2))

Let’s explore one at time the contents of the results from the chi square function.

names(mike.chi)   #The names function 
[1] "statistic" "parameter" "p.value" "method" "data.name" "observed"
[7] "expected" "residuals" "stdres"

Now, call each name in turn.

mike.chi$residuals
[1] 2.828427 -2.828427
mike.chi$obs
[1] 70 30
mike.chi$exp
[1] 50 50

And finally, let’s get the summary output of our statistical test.

mike.chi
Chi-squared test for given probabilities.
data: c(70, 30)
X-squared = 16, df = 1, p-value = 6.334e-05

GOF and spreadsheet apps

Easy enough with R, but it may even easier with other tools. I’ll show you how to do this with spreadsheet apps and with and online at graphpad.com.

Let’s take the pea example above. We had 16 wrinkled, 84 round. We expect 25% wrinkled, 75% round.

Now, with R, we would enter

chisq.test(c(16,80),p=c(1/4,3/4))

and the R output

Chi-squared test for given probabilities
data: c(16, 80)
X-squared = 3.5556, df = 1, p-value = 0.05935

Microsoft Excel and the other spreadsheet programs (Apple Numbers, Google Sheets, LibreOffice Calc) can calculate the goodness of fit directly; they return a P-value only. If the observed data are in cells A1 and A2, and the expected values are in B1 and B2, then use the procedure =CHITEST(A1:A2,B1:B2).

Table 7. A spreadsheet with formula visible.

The P-value (but not the Chi-square test statistic) is returned. Here’s the output from Calc.

Table 8. Spreadsheet example from Table 7 with calculated P-value.

You can get the critical value from MS Excel (=CHIINV(alpha, df), returns the critical value), and the exact probability for the test statistic =CHIDIST(x,df), where x is your test statistic. Putting it all together, a general spreadsheet template for goodness of fit calculations calculations of test statistic and p-value might look like

Table 9. Spreadsheet template with formula to calculate chi-square goodness of fit on two groups.

	A	B	C	D	E
1	f1	0.75
2	f2	0.25
3	N	=SUM(A5,A6)
4	Obs	Exp	Chi.value	Chi.sqr
5	80	=B1*3	=((A5-B5)^2)/B5	=SUM(C5,C6)
6	16	=B2*3	=((A6-B6)^2)/B6		=CHIDIST(D5, COUNT(A5:A6-1)
7
8

Microsoft Excel can be improved by writing macros, or by including available add-in programs, such as the free PopTools, which is available for Microsoft Windows 32-bit operating systems only.

Another option is to take advantage of the internet — again, many folks have provided java or JavaScript-based statistical routines for educational purposes. Here’s an easy one to use www.graphpad.com.

In most cases, I find the chi-square goodness-of-fit is so simple to calculate by hand that the computer is redundant.

Questions

1. A variety of p-values were reported on this page with no attempt to reflect significant figures or numbers of digits (see Chapter 8.2). Provide proper significant figures and numbers of digits as if these p-values were reported in a science journal.

1. 0.0724
2. 0.0679
3. 0.03766692
4. 0.004955794
5. 0.00006334
6. 6.334e-05
7. 0.05935
8. 0.058714340077662

2. For a mini bag of M&M candies, you count 4 blue, 2 brown, 1 green, 3 orange, 4 red, and 2 yellow candies.

1. What are the expected values for each color?
2. Calculate using your favorite spreadsheet app (eg, Numbers, Excel, Google Sheets, LibreOffice Calc)
3. Calculate using R (note R will reply with a warning message that the “Chi-squared approximation may be incorrect”; see 9.2 Yates continuity correction)
4. Calculate using Quickcalcs at graphpad.com
5. Construct a table and compare p-values obtained from the different applications

3. CYP1A2 enzyme involved with metabolism of caffeine. Folks with C at SNP rs762551 have higher enzyme activity than folks with A. Populations differ for the frequency of C. Using R or your favorite spreadsheet application, compare the following populations against global frequency of C is 33% (frequency of A is 67%).

1. 286 persons from Northern Sweden: f(C) = 26%, f(A) = 73%
2. 4532 Native Hawaiian persons: f(C) = 22%, f(A) = 78%
3. 1260 Native American persons: f(C) = 30%, f(A) = 70%
4. 8316 Native American persons: f(C) = 36%, f(A) = 64%
5. Construct a table and compare p-values obtained for the four populations.

Quiz Chapter 9.1

Chi-square test: Goodness of fit

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• References and suggested readings