Limitaciones en el acceso al financiamiento

The R markdown document for this section is availablehere⁶².

We learned how the sample mean and SD are susceptible to outliers. The t-test is based on these measures and is susceptible as well. The Wilcoxon rank test (equivalent to the Mann-Whitney test) provides an alternative. In the code below, we perform a t-test on data for which the null is true.

However, we change one sum observation by mistake in each sample and the values incorrectly entered are different. Here we see that the t-test results in a small p-value, while the Wilcoxon test does not:

set.seed(779) ##779 picked for illustration purposes N=25

x<- rnorm(N,0,1) y<- rnorm(N,0,1) Create outliers:

⁶²https://github.com/genomicsclass/labs/tree/master/robust/ranktest.Rmd

x[1] <- 5 x[2] <- 7

cat("t-test pval:",t.test(x,y)$p.value)

## t-test pval: 0.04439948

cat("Wilcox test pval:",wilcox.test(x,y)$p.value)

## Wilcox test pval: 0.1310212

The basic idea is to 1) combine all the data, 2) turn the values into ranks, 3) separate them back into their groups, and 4) compute the sum or average rank and perform a test.

library(rafalib) mypar(1,2)

stripchart(list(x,y),vertical=TRUE,ylim=c(-7,7),ylab="Observations",pch=21,bg=1) abline(h=0)

xrank<-rank(c(x,y))[seq(along=x)]

yrank<-rank(c(x,y))[-seq(along=y)]

stripchart(list(xrank,yrank),vertical=TRUE,ylab="Ranks",pch=21,bg=1,cex=1.25) ws <- sapply(x,function(z) rank(c(z,y))[1]-1)

text( rep(1.05,length(ws)), xrank, ws, cex=0.8)

Data from two populations with two outliers. The left plot shows the original data and the right plot shows their ranks. The numbers are the w values

W <-sum(ws)

Wis the sum of the ranks for the first group relative to the second group. We can compute an exact p-value for W based on combinatorics. We can also use the CLT since statistical theory tells us that this^Wis approximated by the normal distribution. We can construct a z-score as follows:

n1<-length(x);n2<-length(y)

Z <- (mean(ws)-n2/2)/ sqrt(n2*(n1+n2+1)/12/n1) print(Z)

## [1] 1.523124

Here the^Zis not large enough to give us a p-value less than 0.05. These are part of the calculations performed by the R functionwilcox.test.

Exercises

We are going to explore the properties of robust statistics. We will use one of the datasets included in R, which contains weight of chicks in grams as they grow from day 0 to day 21. This dataset also splits up the chicks by different protein diets, which are coded from 1 to 4. We use this dataset to also show an important operation in R (not related to robust summaries):^reshape.

This dataset is built into R and can be loaded with:

data(ChickWeight)

To begin, take a look at the weights of all observations over time and color the points to represent the Diet:

head(ChickWeight)

plot( ChickWeight$Time, ChickWeight$weight, col=ChickWeight$Diet)

First, notice that the rows here represent time points rather than individuals. To facilitate the comparison of weights at different time points and across the different chicks, we will reshape the data so that each row is a chick. In R we can do this with the^reshapefunction:

chick = reshape(ChickWeight, idvar=c("Chick","Diet"), timevar="Time", direction="wide")

The meaning of this line is: reshape the data from long to wide, where the columns Chick and Diet are the ID’s and the column Time indicates different observations for each ID. Now examine the head of this dataset:

head(chick)

We also want to remove any chicks that have missing observations at any time points (NA for “not available”). The following line of code identifies these rows and then removes them:

chick = na.omit(chick)

1. Focus on the chick weights on day 4 (check the column names of ‘chick’ and note the numbers). How much does the average of chick weights at day 4 increase if we add an outlier measurement of 3000 grams? Specifically, what is the average weight of the day 4 chicks, including the outlier chick, divided by the average of the weight of the day 4 chicks without the outlier. Hint: use^cto add a number to a vector.

2. In exercise 1, we saw how sensitive the mean is to outliers. Now let’s see what happens when we use the median instead of the mean. Compute the same ratio, but now using median instead of mean. Specifically, what is the median weight of the day 4 chicks, including the outlier chick, divided by the median of the weight of the day 4 chicks without the outlier.

3. Now try the same thing with the sample standard deviation (the^sd function in R). Add a chick with weight 3000 grams to the chick weights from day 4. How much does the standard deviation change? What’s the standard deviation with the outlier chick divided by the standard deviation without the outlier chick?

the^madfunction. Note that the mad is unaffected by the addition of a single outlier. The^mad function in R includes the scaling factor 1.4826, such that^madand^sdare very similar for a sample from a normal distribution. What’s the MAD with the outlier chick divided by the MAD without the outlier chick?

5. Our last question relates to how the Pearson correlation is affected by an outlier as compared to the Spearman correlation. The Pearson correlation between x and y is given in R by

cor(x,y). The Spearman correlation is given bycor(x,y,method="spearman").

Plot the weights of chicks from day 4 and day 21. We can see that there is some general trend, with the lower weight chicks on day 4 having low weight again on day 21, and likewise for the high weight chicks.

Calculate the Pearson correlation of the weights of chicks from day 4 and day 21. Now calculate how much the Pearson correlation changes if we add a chick that weighs 3000 on day4 and 3000 on day 21. Again, divide the Pearson correlation with the outlier chick over the Pearson correlation computed without the outliers.

6. Save the weights of the chicks on day 4 from diet 1 as a vector^x. Save the weights of the chicks on day 4 from diet 4 as a vector ^y. Perform a t-test comparing ^xand ^y(in R the functiont.test(x,y)will perform the test). Then perform a Wilcoxon test of ^xand^y(in R the functionwilcox.test(x,y)will perform the test). A warning will appear that an exact p-value cannot be calculated with ties, so an approximation is used, which is fine for our purposes.

Perform a t-test of x and y, after adding a single chick of weight 200 grams to^x(the diet 1 chicks). What is the p-value from this test? The p-value of a test is available with the following code:t.test(x,y)$p.value

7. Do the same for the Wilcoxon test. The Wilcoxon test is robust to the outlier. In addition, it has less assumptions that the t-test on the distribution of the underlying data.

8. We will now investigate a possible downside to the Wilcoxon-Mann-Whitney test statistic.

Using the following code to make three boxplots, showing the true Diet 1 vs 4 weights, and then two altered versions: one with an additional difference of 10 grams and one with an additional difference of 100 grams. Use the x and y as defined above, NOT the ones with the added outlier.

What is the difference in t-test statistic (obtained by t.test(x,y)$statistic) between adding 10 and adding 100 to all the values in the group ‘y’? Take the the t-test statistic with x and y+10 and subtract the t-test statistic with x and y+100. The value should be positive.

9. Examine the Wilcoxon test statistic for x and y+10 and for x and y+100. Because the Wilcoxon works on ranks, once the two groups show complete separation, that is all points from group ‘y’ are above all points from group ‘x’, the statistic will not change, regardless of how large the difference grows. Likewise, the p-value has a minimum value, regardless of how far apart the groups are. This means that the Wilcoxon test can be considered less powerful than the t-test in certain contexts. In fact, for small sample sizes, the p-value can’t be very small, even when the difference is very large. What is the p-value if we compare

c(1,2,3)to^c(4,5,6)using a Wilcoxon test?

10. What is the p-value if we compare^c(1,2,3)toc(400,500,600)using a Wilcoxon test?

We will describe three examples from the life sciences: one from physics, one related to genetics, and one from a mouse experiment. They are very different, yet we end up using the same statistical technique: fitting linear models. Linear models are typically taught and described in the language of matrix algebra.