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I am having some trouble running an Anova on categorical variables in R and matching SPSS output. What I need to do is run an anova on the dataset below (its a made up data set). But, I need to know if the mean of each category is significantly from the total mean of all races.

Satisfaction    Race
3   Asian
4   Cacasion
5   African American
2   Other 
5   African American
3   African American
4   African American
5   African American
2   Asian
3   African American
1   Cacasion
1   Cacasion
1   Cacasion
5   Other 
5   Other 
5   Other 
5   African American
5   Asian
4   Asian
5   Other 
5   Other 
5   Other 
1   Cacasion
4   Cacasion

For example, the mean of all races is 3.5 :

> mean(test$Satisfaction)
[1] 3.5 

What I would like to know is if the mean score for each race is significantly different from the total mean of 3.5 and the p-value.

I ran an Anova in R with the following model, but R will set one catagory as the refernce and test is against the others :

> lm.test <- lm(test$Satisfaction ~ test$Race)
> summary(lm.test)

Call:
lm(formula = test$Satisfaction ~ test$Race)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.5714 -1.0000  0.4286  0.8482  2.0000 

Coefficients:
                  Estimate Std. Error t value Pr(>|t|)    
(Intercept)         3.8571     0.5023   7.679 2.18e-07 ***
test$RaceAsian     -0.6071     0.8330  -0.729   0.4745    
    test$RaceCacasion  -1.8571     0.7394  -2.512   0.0207 *  
test$RaceOther      0.7143     0.7103   1.006   0.3266    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 1.329 on 20 degrees of freedom
Multiple R-squared: 0.391,  Adjusted R-squared: 0.2997 
F-statistic:  4.28 on 3 and 20 DF,  p-value: 0.01732 

The output is telling me that the mean for African American is 3.8571 and is significantly different from the mean of the caucasian group. It is not different from the mean of group Asian and Other.

Is there a way for me to set the intercept to 3.5 in R and get significant compared to the mean and not the reference group. Or should I be using another tests altogether? My stats isn't that great so if its another tests a brief explain on which test and how to run it in R would be great.

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2 Answers 2

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I think that the easiest is to center your dependent variable around the grand mean. Given your example:

test$Satisfaction <- scale(test$Satisfaction, center=TRUE)

This way, the grand mean is now 0, and the mean for each ethnic group is the deviation from the grand mean. Then you run your regression as usual, but the four tests that you get are whether each ethnic group's mean differs from the grand mean, because those are tests of whether the mean differs from 0, which is exactly the grand mean after you've centered your dependent variable.

If you retain the intercept in the model (as you did in your example), then the significance test of the intercept is whether the mean of the reference group is significantly different from the grand mean. If you suppress the intercept by using:

lm(test$Satisfaction ~ 0 + test$Race)

then you get exactly the same results (barring some difference on the adjusted R²), but instead of having an intercept, you get the label for your 4th ethnic group, the one that used to be the reference category. (See here for more information on R² calculations when the intercept is removed from the model.)

Mean-centering your DV and re-running your regression is probably your best option. Alternatively, you could compute separate 1-sample t-tests for each ethnic group, comparing the ethnic groups means to the grand mean, e.g.:

t.test(subset(test, Race=="Asian")$Satisfaction, mu=mean(test$Satisfaction))

However, this is a less powerful approach, since both the degrees of freedom and the standard error will be computed based on only one group instead of your whole sample. Therefore, your best bet is to re-run your regression, but with your dependent variable mean-centered.

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  • $\begingroup$ lm(Satisfaction ~ Race -1, data=test) The "-1" removes the intercept. That might be the same thing as the +0 not sure. $\endgroup$
    – rbatt
    May 11, 2013 at 18:37
  • $\begingroup$ Yes it is the same thing $\endgroup$ May 11, 2013 at 18:58
  • $\begingroup$ Thanks for the quick reply. I think the grand mean method worked. My second question is does it make sense to center the anova around the grand mean? I am having an argument with a stats professor and need someone to weight in. Is there a benefit to setting the center of the equation around the grand mean or is it better to run lm(test$Satisfaction ~ test$Race) and an TukeyHSD() on the anova to compare the means of each group. I was under the impression that lm() and TukeyHSD is the way to go. Am I wrong? $\endgroup$ May 12, 2013 at 14:30
  • $\begingroup$ Well, the two analyses don't answer the same question.. The two will yield the same R² and same significance for the R² (with the same p-value as you would get in an ANOVA). If you're interested in group means compared to the overall mean, then centering the DV does the trick. However, much more typical is what you mentioned--performing pairwise comparisons if you find a significant R². Tukey's HSD controls the Type I error rate while being the most powerful way to test all pairwise comparisons. I agree with you that it seems to be a more appropriate (or at least, typical) way to proceed... $\endgroup$ May 13, 2013 at 2:24
  • $\begingroup$ By the way, doing a regression with the centered DV and looking at the significance of the coefficient for each group does NOT appropriately control for the nominal Type I error rate, while following up the test of the overall R² with pairwise comparisons with a Tukey adjustment does. I really have to side with you here (and not with your stats prof!) $\endgroup$ May 13, 2013 at 2:32
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If your data are balanced, the sum contrasts (see contr.sum under ?contrasts) are explicitly the differences you're asking about, so you may want to look at testing those.

More generally, one problem is that "the mean of all races" includes the present race you're comparing it to, so you lose independence, which can make things more complex.

However, note that if the present race has the same mean as the average of all the races apart from itself it has the same mean as all the races including itself. One way to achieve that would be to look at the last helmert contrast with the race you want to compare to the others set as the last category (see contr.helmert); that would require reordering the factor each time you wanted to test a race though.

You might also find the discussion of deviation coding here relevant.

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  • $\begingroup$ Deviation coding is the same as effects coding, right? I thought of effects coding too, but in order for the significance tests to be what the poster wanted, the data would need to be balanced, which they're not in his example.. here the tests would be evaluating the difference between group means and the unweighted grand mean, instead of the (weighted) grand mean (see here) $\endgroup$ May 11, 2013 at 17:41
  • $\begingroup$ The issue of needing it to be balanced to correspond to the hypothesis was mentioned in my first sentence. There's a lot of different names for things around. I can only tell if we're all talking about the same thing if it's defined; in R the deviation contrasts are 'sum contrasts', which is where I started. $\endgroup$
    – Glen_b
    May 11, 2013 at 22:56

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