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Generalized Linear Model Workflow

This example shows how to fit a generalized linear model and analyze the results. A typical workflow involves these steps: import data, fit a generalized linear model, test its quality, modify the model to improve its quality, and make predictions based on the model. In this example, you use the Fisher iris data to compute the probability that a flower is in one of two classes.

Load Data

Load the Fisher iris data.

load fisheriris

Extract rows 51 to 150, which have the classification versicolor or virginica.

X = meas(51:end,:);

Create logical response variables that are true for versicolor and false for virginica.

y = strcmp('versicolor',species(51:end));

Fit Generalized Linear Model

Fit a binomial generalized linear model to the data.

mdl = fitglm(X,y,'linear','Distribution','binomial')
mdl = 
Generalized linear regression model:
    logit(y) ~ 1 + x1 + x2 + x3 + x4
    Distribution = Binomial

Estimated Coefficients:
                   Estimate      SE       tStat      pValue 
                   ________    ______    _______    ________

    (Intercept)     42.638     25.708     1.6586    0.097204
    x1              2.4652     2.3943     1.0296     0.30319
    x2              6.6809     4.4796     1.4914     0.13585
    x3             -9.4294     4.7372    -1.9905    0.046537
    x4             -18.286     9.7426    -1.8769    0.060529


100 observations, 95 error degrees of freedom
Dispersion: 1
Chi^2-statistic vs. constant model: 127, p-value = 1.95e-26

According to the model display, some p-values in the pValue column are not small, which implies that you can simplify the model.

Examine and Improve Model

Determine if 95% confidence intervals for the coefficients include 0. If so, you can remove the model terms with those intervals.

confint = coefCI(mdl)
confint = 5×2

   -8.3984   93.6740
   -2.2881    7.2185
   -2.2122   15.5739
  -18.8339   -0.0248
  -37.6277    1.0554

Only the fourth predictor x3 has a coefficient whose confidence interval does not include 0.

The coefficients of x1 and x2 have large p-values and their 95% confidence intervals include 0. Test whether both coefficients can be zero. Specify a hypothesis matrix to select the coefficients of x1 and x2.

M = [0 1 0 0 0     
     0 0 1 0 0];   
p = coefTest(mdl,M)
p = 0.1442

The p-value is approximately 0.14, which is not small. Remove x1 and x2 from the model.

mdl1 = removeTerms(mdl,'x1 + x2')
mdl1 = 
Generalized linear regression model:
    logit(y) ~ 1 + x3 + x4
    Distribution = Binomial

Estimated Coefficients:
                   Estimate      SE       tStat       pValue  
                   ________    ______    _______    __________

    (Intercept)     45.272     13.612      3.326    0.00088103
    x3             -5.7545     2.3059    -2.4956      0.012576
    x4             -10.447     3.7557    -2.7816     0.0054092


100 observations, 97 error degrees of freedom
Dispersion: 1
Chi^2-statistic vs. constant model: 118, p-value = 2.3e-26

Alternatively, you can identify important predictors using stepwiseglm.

mdl2 = stepwiseglm(X,y,'constant','Distribution','binomial','Upper','linear')
1. Adding x4, Deviance = 33.4208, Chi2Stat = 105.2086, PValue = 1.099298e-24
2. Adding x3, Deviance = 20.5635, Chi2Stat = 12.8573, PValue = 0.000336166
3. Adding x2, Deviance = 13.2658, Chi2Stat = 7.29767, PValue = 0.00690441
mdl2 = 
Generalized linear regression model:
    logit(y) ~ 1 + x2 + x3 + x4
    Distribution = Binomial

Estimated Coefficients:
                   Estimate      SE       tStat      pValue 
                   ________    ______    _______    ________

    (Intercept)     50.527     23.995     2.1057    0.035227
    x2              8.3761     4.7612     1.7592    0.078536
    x3             -7.8745     3.8407    -2.0503    0.040334
    x4              -21.43     10.707    -2.0014     0.04535


100 observations, 96 error degrees of freedom
Dispersion: 1
Chi^2-statistic vs. constant model: 125, p-value = 5.4e-27

The p-value (pValue) for x2 in the coefficient table is greater than 0.05, but stepwiseglm includes x2 in the model because the p-value (PValue) for adding x2 is smaller than 0.05. The stepwiseglm function computes PValue using the fits with and without x2, whereas the function computes pValue based on an approximate standard error computed only from the final model. Therefore, PValue is more reliable than pValue.

Identify Outliers

Examine a leverage plot to look for influential outliers.

plotDiagnostics(mdl2,'leverage')

Figure contains an axes object. The axes object with title Case order plot of leverage contains 2 objects of type line. These objects represent Leverage, Reference Line.

An observation can be considered an outlier if its leverage substantially exceeds p/n, where p is the number of coefficients and n is the number of observations. The dotted reference line is a recommended threshold, computed by 2*p/n, which corresponds to 0.08 in this plot. Some observations have leverage values larger than 10*p/n (that is, 0.40). Identify these observation points.

idxOutliers = find(mdl2.Diagnostics.Leverage > 10*mdl2.NumCoefficients/mdl2.NumObservations)
idxOutliers = 4×1

    19
    21
    57
    85

See if the model coefficients change when you fit a model excluding these points.

oldCoeffs = mdl2.Coefficients.Estimate;
mdl3 = fitglm(X,y,'linear','Distribution','binomial', ...
    'PredictorVars',2:4,'Exclude',idxOutliers);
newCoeffs = mdl3.Coefficients.Estimate;
disp([oldCoeffs newCoeffs])
   50.5268   44.0085
    8.3761    5.6361
   -7.8745   -6.1145
  -21.4296  -18.1236

The model coefficients in mdl3 are different from those in mdl2. This result implies that the responses at the high-leverage points are not consistent with the predicted values from the reduced model.

Predict Probability of Being Versicolor

Use mdl3 to predict the probability that a flower with average measurements is versicolor. Generate confidence intervals for the prediction.

[newf,newc] = predict(mdl3,mean(X))
newf = 0.4558
newc = 1×2

    0.1234    0.8329

The model gives almost a 46% probability that the average flower is versicolor, with a wide confidence interval.

See Also

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