Why would I use ANOVA instead of a Rank-Sum test?

A colleague of mine with little statistics experience is trying to perform an experimental evaluation of a computer program. He created a between subjects design and solicited test subjects.

11 people were given his "new and improved" computer program to use. 10 others got the "old and boring" computer program to use, for the same task.

He asked me, and several other people around the lab how to analyze his data.

I told him he should examine the data for normality. If it was normally distributed, he should use a t-test. If it was not, he should use a Wilcoxon rank sum test.

One of my colleagues told him he should use ANOVA, even though he only has 2 groups. Apparently using ANOVA on non-normal data in R produces a new degree of freedom measure which can be put into a t-test.

I've never heard of such a thing. Is this true? Is it statistically valid? Why would anyone use it instead of just doing a rank-sum test?

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What is his outcome variable? There are a variety of tests that are suited for particular types of outcomes. For example, time to task completion, or number of mistakes, etc all can benefit from specialized methods. –  Aniko Oct 8 '11 at 2:36
His outcome is a score on a multiple choice written test. –  John Doucette Oct 8 '11 at 21:02
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2 Answers

Using ANOVA in R does not produce anything different from using ANOVA in another program, and with two groups the results will be equivalent to an equal variance t-test. The t-test is known to be robust to deviations from normality, though with unequal variances a Welch's t-test is probably preferrable.

In the special case of a score based on the number of correct answers on a multiple choice test, the distribution of the score is probably an overdispersed binomial. In that case the "correct" analysis might be a GLM model with a quasi-binomial distribution. Of course, the results might be quite similar to the unequal variance t-test.

Here is a simple simulation based example with 20 questions, and unequal variances. Welch's t-test gives a result much closer to the overdispersed binomial regression.

set.seed(3413)
#generate first sample
p1 <- 1/(1+exp(-1+rnorm(10,sd=1)))
x1 <- rbinom(10, size=20, p=p1)
#generate second sample
p2 <- 1/(1+exp(-3+rnorm(10, sd=1)))
x2 <- rbinom(10, size=20, p=p2)
#combine two sets
x <- c(x1,x2)
g <- gl(2,10)

#summaries:
tapply(x, g, mean)
1    2
12.6 19.2
tapply(x, g, sd)
1        2
3.921451 1.032796

#t-test:
t.test(x ~ g, var.equal=TRUE)

Two Sample t-test

data:  x by g
t = -5.1468, df = 18, p-value = 6.765e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-9.294136 -3.905864
sample estimates:
mean in group 1 mean in group 2
12.6            19.2

#without equal variances:
t.test(x ~ g, var.equal=FALSE)

Welch Two Sample t-test

data:  x by g
t = -5.1468, df = 10.243, p-value = 0.0004016
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-9.448128 -3.751872
sample estimates:
mean in group 1 mean in group 2
12.6            19.2

#overdispersed binomial regression:
summary(glm(cbind(x, 20-x) ~ g, family="quasibinomial") )

Call:
glm(formula = cbind(x, 20 - x) ~ g, family = "quasibinomial")

Deviance Residuals:
Min       1Q   Median       3Q      Max
-2.5340  -0.8386  -0.2199   1.2778   2.7581

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   0.5322     0.2242   2.374 0.028946 *
g2            2.6458     0.5962   4.438 0.000318 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasibinomial family taken to be 2.343713)

Null deviance: 120.242  on 19  degrees of freedom
Residual deviance:  45.197  on 18  degrees of freedom
AIC: NA

Number of Fisher Scoring iterations: 5

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Thanks. Welch's test and the GLM were also things we recommended to him (though he said they were too complicated), so it's good to see that we were on the right track. The ANOVA suggestion seems stranger and stranger. –  John Doucette Oct 10 '11 at 15:36
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First, the data should not be checked for "normality". This is a useless test. The t test is quite robust to departures from this assumption anyway.

Second, the t test is just a special case of ANOVA for two groups.

If you believe the data to likely be close to a normal distribution (based on experience of the variable in question, visualization of a histogram, but NOT using a normality test), just use the t test. If not, use the rank sum test.

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This doesn't really answer my question. You've suggested exactly what I told him: If you're data looks normal, use a t-test. If it doesn't, use a rank sum test. His data is hetroskedastic, and has many outliers. It doesn't look normal at all. In spite of this, a colleague told him to use ANOVA (in R?), stating that this would somehow provide him with a correction factor. What I want to know is: Is this true, and why would you want to use it? –  John Doucette Oct 7 '11 at 21:02
Then why don't you ask him to send you the reference for the putative "correction"? –  pmgjones Oct 7 '11 at 21:56
Two reasons: First, his "reference" is "That's how we do things in my subfield." This appears to be true. Second, it seemed to me that if I thought this procedure odd, then perhaps I was not alone. Posting it on Cross Validated thus served two purposes: If the procedure is widely used in certain places, then having a question for it on CV would help others find out why. If it is not widely used, then this CV question is likely to go unanswered, which says something in its own right. –  John Doucette Oct 7 '11 at 23:03
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