# 10.1.2 - Example: Therapeutic Value of Vitamin C

10.1.2 - Example: Therapeutic Value of Vitamin C

## Example: Vitamin C Study - Revisited

Recall the Vitamin C study, a $$2 \times 2$$ example from Lesson 3. From the earlier analysis, we learned that the treatment and response for this data are not independent:

$$G^2 = 4.87$$, df = 1, p−value = 0.023,

And from the estimated odds ratio of 0.49, we noted that the odds of getting a cold were about twice as high for those who took the placebo, compared with those who took vitamin C. Next, we will consider the same question by fitting the log-linear model of independence.

There are (at least) two ways of fitting these models in SAS and R. First, we will see the specification for two procedures in SAS (PROC CATMOD and PROC GENMOD); see VitaminCLoglin.sas and VitaminCLoglin.R.

#### The PROC CATMOD procedure

##### INPUT:
proc catmod data=ski order=data;
weight count;
model treatment*response=_response_;
loglin treatment response;
run;

The MODEL statement (line) specifies that we are fitting the model on two variables treatment and response specified on the left-hand side of the MODEL statement. The MODEL statement only takes independent variables on its right-hand side. In the case of fitting a log-linear model, we need to use a keyword _response_ on the right-hand side.

The LOGLIN statement specifies that it's a model of independence because we are stating that the model has only main effects, one for the "treatment" variable and the other for the "response" variable, in this case.

##### OUTPUT:
Population Profiles and Response profiles
The responses we're modeling are four cell counts:

Population Profiles
Sample Sample Size
1 279

Response Profiles
Response treatment response
1 placebo cold
2 placebo nocold
3 absorbic cold
4 absorbic nocold
Maximum Likelihood Analysis of Variance
As with ANOVA, we are breaking down the variability across factors. More specifically, the main effect TREATMENT tests if the subjects are evenly distributed over the two levels of this variable. The null hypothesis assumes that they are, and the alternative assumes that they are not. Based on the output below, are the subjects evenly distributed across the placebo and the vitamin C conditions?

Maximum Likelihood Analysis of Variance
Source DF Chi-Square Pr > ChiSq
treatment 1 0.00 0.9523
response 1 98.12 <.0001
Likelihood Ratio 1 4.87 0.0273
The main effect RESPONSE tests if the subjects are evenly distributed over the two levels of this variable. The Wald statistic $$X^2=98.12$$ with df = 1 indicates highly uneven distribution.

The LIKELIHOOD RATIO (or the deviance statistic) tells us about the overall fit of the model. Does this value look familiar? Does the model of independence fit well?
Analysis of Maximum Likelihood Estimates
Gives the estimates of the parameters. We focus on the estimation of odds.

Analysis of Maximum Likelihood Estimates
Parameter   Estimate Standard
Error
Chi-
Square
Pr > ChiSq
treatment placebo 0.00358 0.0599 0.00 0.9523
response cold -0.7856 0.0793 98.12 <.0001
For example,
$$\lambda_1^A=\lambda_{placebo}^{TREATMENT}=0.00358=-\lambda_{ascobic}^{TREATMENT}=-\lambda_2^A$$
What are the odds of getting cold? Recall that PROC CATMOD, by default, uses zero-sum constraints. Based on this output the log odds are
\begin{align}
\text{log}(\mu_{i1}/\mu_{i2}) &=\text{log}(\mu_{i1})-\text{log}(\mu_{i2})\\
&=[\lambda+\lambda_{placebo}^{TREATMENT}+\lambda_{cold}^{RESPONSE}]-[\lambda+\lambda_{placebo}^{TREATMENT}+\lambda_{nocold}^{RESPONSE}]\\
&=\lambda_{cold}^{RESPONSE}-\lambda_{nocold}^{RESPONSE}\\
&=\lambda_{cold}^{RESPONSE}+\lambda_{cold}^{RESPONSE}\\
&=-0.7856-0.7856=-1.5712\\
\end{align}
So the odds are $$\exp(-1.5712) = 0.208$$. Notice that the odds are the same for all rows.

What are the odds of being in the placebo group? Note that this question is not really relevant here, but it's a good exercise to figure out the model parameterization and how we come up with estimates of the odds. Hint: log(odds)=0.00716.

#### The PROC GENMOD procedure

Fits a log-linear model as a Generalized Linear Model (GLM) and by default uses indicator variable coding. We will mostly use PROC GENMOD in this course.

##### INPUT:
proc genmod data=ski order=data;
class treatment response;
model count = response*treatment /link=log dist=poisson lrci type3 obstats;
run;

The CLASS statement (line) specifies that we are dealing with two categorical variables: treatment and response.

The MODEL statement (line) specifies that our responses are the cell counts, and link=log specifies that we are modeling the log of counts. The random distribution is Poisson, and the remaining specifications ask for different parts of the output to be printed.

##### OUTPUT:
Model Information
Gives assumptions about the model. In our case, tells us that we are modeling the log, i.e., the LINK FUNCTION, of the responses, i.e. DEPENDENT VARIABLE, that is counts which have a Poisson distribution.

Model Information
Data Set WORK.SKI
Distribution Poisson
Dependent Variable count
 Number of Observations Read 4 4
Class Level Information
Information on the categorical variables in our model

Class Level Information
Class Levels Values
treatment 2 placebo absorbic
response 2 cold nocold
Criteria For Assessing Goodness of Fit
Gives the information on the convergence of the algorithm for MLE for the parameters, and how well the model fits. What is the value of the deviance statistic $$G^2$$? What about the Pearson $$X^2$$? Note that they are the same as what we got when we did the chi-square test of independence; we will discuss "scaled" options later with the concept of overdispersion. Note also the value of the log-likelihood for the fitted model, which in this case is independence.

Criteria For Assessing Goodness Of Fit
Criterion DF Value Value/DF
Deviance 1 4.8717 4.8717
Scaled Deviance 1 4.8717 4.8717
Pearson Chi-Square 1 4.8114 4.8114
Scaled Pearson X2 1 4.8114 4.8114
Log Likelihood   970.6299
Analysis of Parameter Estimates
Gives individual parameter estimates. Recall that PROC GENMOD does not use zero-sum constraints but fixes typically the last level of each variable to be equal to zero:
$$\hat{\lambda}=4.75,\ \hat{\lambda}^A_1=0.0072,\ \hat{\lambda}^A_2=0,\ \hat{\lambda}^B_1=-1.5712,\ \hat{\lambda}^B_2=0$$

Analysis Of Maximum Likelihood Parameter Estimates
Parameter   DF Estimate Standard
Error
Likelihood Ratio 95% Confidence Limits Wald Chi-Square Pr > ChiSq
Intercept   1 4.7457 0.0891 4.5663 4.9158 2836.81 <.0001
treatment placebo 1 0.0072 0.1197 -0.2277 0.2422 0.00 0.9523
treatment absorbic 0 0.0000 0.0000 0.0000 0.0000 . .
response cold 1 -1.5712 0.1586 -1.8934 -1.2702 98.11 <.0001
response nocold 0 0.0000 0.0000 0.0000 0.0000 . .
Scale   0 1.0000 0.0000 1.0000 1.0000

Note:The scale parameter was held fixed.

So, we can compute the estimated cell counts for having cold given vitamin C based on this model:

$$\mu_{21}=\exp(\lambda+\lambda_2^A+\lambda_1^B)=\exp(4.745+0-1.5712)=23.91$$

What about the other cells: $$\mu_{11}$$, $$\mu_{12}$$, and $$\mu_{22}$$?

What are the estimated odds of getting a cold?
$$\exp(\lambda_1^B-\lambda_2^B)=\exp(-1.571-0)=0.208$$

LR Statistics for Type 3 Analysis
Testing the main effects is similar to the Maximum Likelihood Analysis of Variance in the CATMOD, except this is the likelihood ratio (LR) statistic and not the Wald statistic. When the sample size is moderate or small, the likelihood ratio statistic is the better choice. From the CATMOD output, the chi-square Wald statistic was 98.12, with df=1, and we reject the null hypothesis. Here, we reach the same conclusion, except, that LR statistic is 130.59 with df=1.

LR Statistics For Type 3 Analysis
Source DF Chi-Square Pr > ChiSq
treatment 1 0.00 0.9523
response 1 130.59 <.0001
Observation Statistics
Gives information on expected cell counts and five different residuals for further assessment of the model fit (i.e., "Resraw"=observed count - predicted (expected) count, "Reschi"=pearson residual, "Resdev"= deviance residuals, "StReschi"=standardized pearson residuals, etc.)

Observation Statistics
Observation count treatment response Predicted Value Linear Predictor Standard Error of the Linear Predictor HessWgt Lower Upper Raw Residual Pearson Residual Deviance Residual Std Deviance Residual Std Pearson Residual Likelihood Residual Leverage CookD DFBETA_Intercept DFBETA_treatmentplacebo DFBETA_responsecold DFBETAS_Intercept DFBETAS_treatmentplacebo DFBETAS_responsecold
1 31 placebo cold 24.086031 3.1816321 0.1561792 24.086031 17.734757 32.711861 6.9139686 1.4087852 1.3483626 2.0994112 2.1934897 2.1551805 0.5875053 2.2842489 -0.060077 0.1197239 0.3491947 -0.674251 0.9998856 2.2013658
2 109 placebo nocold 115.91398 4.7528483 0.0888123 115.91398 97.395437 137.95359 -6.913978 -0.642185 -0.648733 -2.215859 -2.193493 -2.195419 0.9142868 17.107471 -0.060077 -0.576172 0.3491952 -0.674252 -4.811954 2.2013689
3 17 absorbic cold 23.913989 3.1744636 0.1563437 23.913989 17.602407 32.488674 -6.913989 -1.413848 -1.491801 -2.314435 -2.193496 -2.244533 0.5845377 2.2564902 -0.060077 0.1197243 -0.346701 -0.674253 0.9998885 -2.185648
4 122 absorbic nocold 115.08602 4.7456799 0.0891012 115.08602 96.645029 137.04577 6.913978 0.6444908 0.638194 2.172062 2.1934927 2.1916509 0.9136701 16.973819 0.6358195 -0.576172 -0.346701 7.1359271 -4.811954 -2.185645

#### The LOGLIN and GLM functions in R

To do the same analysis in R, the loglin() function corresponds roughly to PROC CATMOD, and glm() function corresponds roughly to PROC GENMOD. See VitaminCLoglin.R for the commands and to generate the output.

Here is a brief comparison to loglin() and glm() output with respect to model estimates.

The following command fits the log-linear model of independence, and from the part of the output that gives parameter estimates, we see that they correspond to the output from CATMOD. For example, the odds of getting a cold are estimated to be

$$\exp(\lambda_{cold}^{response} - \lambda_{no cold}^{response})=\exp(-0.7856-0.7856)=\exp(-1.5712)=0.208$$

loglin(ski, list(1, 2), fit=TRUE, param=TRUE)
$param$"(Intercept)"
[1] 3.963656

$param$Treatment
Placebo     VitaminC
0.003584245 -0.003584245

$param$Cold
Cold     NoCold
-0.7856083  0.7856083 

The following command fits the log-linear model of independence using glm(), and from the part of the output that gives parameter estimates, we see that they correspond to the output from GENMOD but with the signs reversed, i.e., odds of $$nocold=\exp(\lambda_{nocold}^{response})=\exp(1.5712)$$. Thus, the odds of cold are $$\exp(-1.5712)$$.

glm(ski.data$Freq~ski.data$Treatment+ski.data$Cold, family=poisson()) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.181632 0.156179 20.372 <2e-16 *** ski.data$TreatmentVitaminC -0.007168   0.119738  -0.060    0.952
ski.data$ColdNoCold 1.571217 0.158626 9.905 <2e-16 *** #### Assessing Goodness of Fit The ANOVA function gives information on how well the model fits. The value of the deviance statistic is $$G^2=4.872$$ with 1 degree of freedom, which is not significant evidence to reject this model. Note that this is the same as what we got when we did the chi-square test of independence; we will discuss options for dealing with overdispersion later as well. > ski.ind Call: glm(formula = ski.data$Freq ~ ski.data$Treatment + ski.data$Cold,      family = poisson())

Coefficients:
(Intercept)  ski.data$TreatmentVitaminC ski.data$ColdNoCold
3.181632                   -0.007168                    1.571217

Degrees of Freedom: 3 Total (i.e. Null);  1 Residual
Null Deviance:	    135.5
Residual Deviance: 4.872 	AIC: 34 

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