# Statistics 5601 (Geyer, Fall 2003) Examples: Wilcoxon Signed Rank Test

## General Instructions

To do each example, just click the "Submit" button. You do not have to type in any R instructions or specify a dataset. That's already done for you.

## The Wilcoxon Signed Rank Test

### Example 3.1 in Hollander and Wolfe.

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### Summary

• Lower-tailed signed rank test
• Test statistic: T+ = 5
• Sample size: n = 9
• P-value: P = 0.01953

• Line one assigns the value of the parameter (population median) assumed under the null hypothesis. Usually zero.
• There is no need to sort the z values in line two. It just makes the data easier to look at.
• The vector `r` defined in line four is the (absolute) ranks.
• Line five prints the differences `z` and the ranks `r` stuffed into a matrix so they line up, each difference over the corresponding rank.
• Line six prints the test statistic (sum of the positive signed ranks).
• For an upper-tailed test the seventh line would be replaced by any of the following, which all do the same thing,
```1 - psignrank(tplus - 1, n)
psignrank(tplus - 1, n, lower.tail=FALSE)
psignrank(n * (n + 1) / 2 - tplus, n)
```
the first line here does exactly the same as line seven in the example, but is less accurate for very small P-values. The second does exactly the same as line seven of the example because of the symmetry of the distribution of the signed rank test statistic.
• For handling zeros and tied ranks, see Hollander and Wolfe, the class discussion, and below.

## The Associated Point Estimate (Median of the Walsh Averages)

The Hodges-Lehmann estimator associated with the signed rank test is the median of the Walsh averages, which are the n (n + 1) / 2 averages of pairs of differences

(Zi + Zj) / 2,     i ≤ j

The following somewhat tricky code computes the Walsh averages and their median.

### Example 3.3 in Hollander and Wolfe.

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### Summary

• Point Estimate (sample median of Walsh averages): -0.46

• There is no need to sort the z values in line one. It just makes the data easier to look at.
• There is no need to sort the Walsh averages in line three. It just makes them easier to look at. Having them sorted is necessary later on when we use them to construct a confidence interval.

## The Associated Confidence Interval

Very similar to the confidence interval associated with the sign test, the confidence interval has the form

(W(k), W(m + 1 - k))

where m = n (n + 1) / 2 is the number of Walsh averages, and, as always, parentheses on subscripts indicates order statistics, in this case, of the Walsh averages Wk. That is, one counts in k from each end in the list of sorted Walsh averages to find the confidence interval.

### Example 3.4 in Hollander and Wolfe.

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### Summary

• Achieved confidence level: 0.9609375
• Confidence interval for the population median: (-0.786, -0.010)

• Some experimentation may be needed to achieve the confidence level you want. The possible confidence levels are shown by
```1 - 2 * psignrank(k - 1, n)
```
for different values of `k`. The vectorwise operation of R functions can give them all at once
```k <- seq(1, 100)
conf <- 1 - 2 * psignrank(k - 1, n)
conf[conf > 1 / 2]
```
If one adds these lines to the form above, one sees that the choice is fairly restricted. There are nine possible achieved levels between 0.99 and 0.80 are
0.9883, 0.9805, 0.9727, 0.9609, 0.9453, 0.9258, 0.9023, 0.8711, 0.8359.
• Alternatively, you can just assign `k` to be any integer between one and `n / 2` just before the second to last line in the form (`cat . . .`). A confidence interval with some achieved confidence level will be produced.
• For a one-tailed confidence interval (called upper and lower bounds by Hollander and Wolfe) just use `alpha` rather than `alpha / 2` in the fifth line of the form. Then make either the lower limit minus infinity or the upper limit plus infinity, as desired.

## The R Functions `wilcox.test` and `wilcox.exact`

All of the above can be done in one shot with the R function `wilcox.test` (on-line help).

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Only one complaint. It does not report the actual achieved confidence level (here 96.1%) but rather the confidence level asked for (here 95%, the default). If you want to know the actual achieved confidence level, you'll have to use the code in the confidence interval section above. But you can use `wilcox.test` as a convenient check (the intervals should agree).

### Warning About Ties and Zeros

Do not use the `wilcox.test` function when there are ties or zeros in the data. See the following section.

There is a function `wilcox.exact` explained below that does do hypothesis tests correctly in the presence of ties and zeros (if you accept the zero fudge).

Neither `wilcox.test` nor `wilcox.exact` calculate point estimates or confidence intervals correctly in the presence of ties or zeros.

Do not use for confidence intervals or point estimates!

## Ties and Zeros

What if the continuity assumption is false and there are tied absolute Z values or zero Z values?

First, neither ties nor zeros should make any difference in calculating point estimators or confidence intervals.

This is a bit of programmer brain damage (PBD) in the implementation of the `wilcox.test` and `wilcox.exact` functions. They change the way they calculate point estimates and confidence intervals when there are ties or zeros. But they shouldn't.

### The Zero Fudge

What we called the "zero fudge" in the context of the sign test (because it is fairly bogus there) makes much more sense in the context of the signed rank test. Zero values in the vector Z = Y - X of paired differences should get the smallest possible absolute rank because zero is smaller in absolute value than any other number. We might as well give them rank zero, starting counting at zero rather than at one. Then they make no contribution to the sum of ranks statistic.

Here's another way to see the same thing. Let's compare three procedures: Student t, signed rank, and sign.

• The t procedures use the actual numbers, both size and sign. Because of this they are highly non-robust (breakdown point zero).
• The sign procedures completely ignore the size of the numbers using only the sign (are the differences positive or negative). Because of this they are highly robust (breakdown point one-half).
• The Wilcoxon signed rank procedures use both size and sign. But they don't use the actual magnitudes of the numbers, only the ranks of their magnitudes. So they pay some attention to size, but not as much attention as the parametric Student t procedures. Because of this they have in between robustness (breakdown point 0.293).

This analysis explains why the Wilcoxon should treat differences of zero size just like the Student t, that is, they don't count at all.

There is another issue about the zero fudge. As explained in the section about the zero fudge for the sign test, the null hypothesis for the sign test with zero fudge is

pr(Zi < μ) = pr(Zi > μ).

which generally is neither true nor interesting for the sign test but which is true under the usual assumptions for the Wilcoxon signed rank test. We are already assuming the distribution of the differences is symmetric, and that implies the probability to either side of the population median μ is the same.

All that having been said, I still feel that there is something wrong with deliberately ignoring data favoring the null hypothesis. It still seems like cheating.

Thus we won't argue (at least not much) with the zero fudge for the signed rank test. We will start our hypothesis test calculations (don't do this for point estimates or confidence intervals) with

```z <- y - z
z <- z - mu
z < z[z != 0]
```

### Tied Ranks

The preceding section takes care of one kind of violation of the continuity assumption. But there's a second kind of violation that causes another problem.

If there are ties among the magnitudes of the Z values, then

• there is no way to unambiguously assign ranks,
• the theorem about the equivalence of the T+ and W+ statistics is no longer true, and
• the distribution of T+ tabulated in Table A.4 in Hollander and Wolfe and computed by the `psignrank` and `qsignrank` functions in R is no longer correct in the presence of ties.

So there are really two issues to be resolved.

1. What test statistic?
2. What is the sampling distribution of the test statistic under the null hypothesis.

The standard solution to the first problem is to use so-called "tied ranks" in which each of a set of tied magnitudes is assigned the average of the ranks they otherwise would have gotten if they had been slightly different (and untied). The R `rank` function automatically does this. So the ranks are done the same as before. And the test statistic is calculated from the ranks the same as before.

```r <- rank(abs(z))
tplus <- sum(r[z > 0])
```

### The Wrong Thing

The Wrong Thing (with a capital "W" and a capital "T") is to just ignore the fact that tied ranks change the sampling distribution and just use tables in books or computer functions that are based on the assumption of no ties.

This is not quite as bad as it sounds, because tied ranks were thought up in the first place with the idea of not changing the sampling distribution much. So this does give wrong answers, but not horribly wrong.

##### Example 3.2 in Hollander and Wolfe.

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##### Summary
• Upper-tailed signed rank test
• Test statistic: T+ = 62.5
• Sample size: n = 12
• P-value (wrong): P = 0.03857
• Note that when the variable names aren't "x" and "y", you have to use the actual names, here `private` and `government`.
• The lines
```mu <- 0 # hypothesized value of median
```
and
```z <- z - mu
z <- z[z != 0]
```
serve no purpose in this problem where μ = 0 and there are no zero differences. They are only there for the general case.
• The `floor` function in the last line rounds `tplus` down to the nearest integer (in this case from 62.5 to 62) because only integer arguments make sense to `psignrank` and rounding down is conservative for upper tails (makes them bigger).
• For a lower-tailed test replace the last line with
```psignrank(ceiling(tplus), n)
```

### The Right Thing

As long as you have a computer, why not use it? There are only 2n points in the sample space of the sampling distribution of the test statistic under the null hypothesis corresponding to all possible choices of signs to the ranks. In this case, 212 = 4096, not a particularly big number for a computer (although out of the question for hand calculation). So just do it.

There is an R function `wilcox.exact` (on-line help) that does the job.

##### Example 3.2 in Hollander and Wolfe.

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##### Summary
• Upper-tailed signed rank test
• Test statistic: T+ = 62.5
• Sample size: n = 12
• P-value (exact): P = 0.03345
• Note that the "wrong thing" isn't all that wrong. It doesn't even have one correct significant figure, though.  Wrong P = 0.04 Right P = 0.03

• You may be wondering why R has a `wilcox.test` function to do the wrong thing and a `wilcox.exact` function to do the right thing.

I wonder too. Here's my guess. The `wilcox.test` function has been there for more than 10 years and uses 60-year-old ideas. It's well understood. The `wilcox.exact` uses ideas that are very recent research. It may eventually replace the other function, but has to wait a while.

## The Large Sample Approximation

We have been ignoring up to now, large sample (also called "asymptotic") approximations to sampling distributions of test statistics. Why use an approximation when the computer can do it exactly?

However, the computer can't do it exactly for really large n. The functions `psignrank` and `qsignrank` crash when n is larger than about 50. The function `wilcox.exact` may also crash for large n but I haven't experimented with it.

Thus the need for large sample approximation.

It is a fact, which Hollander and Wolfe derive on pp. 44-45, that the mean and variance of the sampling distribution of T+ are

E(T+) = n (n + 1) / 4
var(T+) = n (n + 1) (2 n + 1) / 24

under the assumption of continuity (hence no ties).

When there are ties, the mean stays the same but the variance is reduced by a quantity that, because it has a summation sign in it, doesn't look good in web pages. Just see equation (3.13) in Hollander and Wolfe.

Here's how R uses this approximation.

##### Example 3.2 in Hollander and Wolfe.

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##### Summary
• Upper-tailed signed rank test
• Test statistic: T+ = 62.5
• Sample size: n = 12
• P-value (approximate with correction for ties): P = 0.03258
• P-value (approximate with correction for ties and correction for continuity): P = 0.03403
• P-value (approximate with correction for ties and another correction for continuity): P = 0.03554
• The function that calculates the c. d. f. of the normal distribution is `pnorm`.
• The last two lines calculate without and with a "correction for continuity" which is 1 / 4 here rather than 1 / 2 because the discreteness in the distribution is in steps of 1 / 2 rather than integer steps. This is because tied ranks may have half-integer values (and do in this case).

Arrghh!!! The tied ranks may also have only integer values (when the number tied in each tied group is odd). In that case the correction for continuity should be 1 / 2.

• For a lower-tailed test, we need to reverse the sign of the correction for continuity
```pnorm(tplus, et, sqrt(vt))
pnorm(tplus + 1 / 4, et, sqrt(vt))
pnorm(tplus + 3 / 8, et, sqrt(vt))
pnorm(tplus + 1 / 2, et, sqrt(vt))
```
whichever you think is better.

The R function `wilcox.test` always uses 1 / 2 for the correction for continuity even though we can see that 1 / 4 works better in this case.

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