Download The Wilcoxon Signed Ranks Test with Instructions for Vassarstats

HANDOUT -­‐ The Wilcoxon Signed Ranks Test (Wilcoxon T-­‐test) The Wilcoxon Signed Ranks Test is named for Frank Wilcoxon (1892–1965) who, in a single paper,
proposed both it and the rank-sum test for two independent samples (Wilcoxon, 1945).[2] The test is
sometimes referred to as the Wilcoxon T-test, and the test statistic is reported as a value of T. The
Wilcoxon signed-rank test is NOT the same as the Wilcoxon rank-sum test, although both are
nonparametric and involve the summation of ranks.
When carrying out the internal assessment, nothing seems to worry students (and teachers) more than the application of inferential statistics. And yet -­‐ it is actually not as difficult as it may seem. The Wilcoxon Signed Ranks test is used when we have a repeated measures design. It also requires that the data be at least Ordinal in nature. Even though I may use this test for interval or ratio data, the test will convert my data to ordinal in order to carry out the test. This means that part of the test (which we will not have to do manually!) determines the rank of each piece of data within the sample. We do this because we tend to have small sample sizes for IB Psychology IA's and because often we do not have a standard distribution of data -­‐ but rather, the data tends to contain outliers. When we calculate the Wilcoxon signed ranks test -­‐ we get a "W" value. We then have to determine whether the W value is "significant" or not. The concept of significance is simply -­‐ what is the probability that my results are only due to chance? When we discuss probability, psychologists look for a maximum probability of .05 that the results are due only to chance. In other words, I want to be able to say with a 95% level of confidence that my results are not due to chance. The typical "p values" are p < 0.05, 0.25 and .01. The W value is always compared to a "critical value." The idea of critical values is demonstrated in the graph below. This diagram demonstrates what happens with a one-­‐tailed hypothesis. Let's say that my hypothesis is that "Noise will increase the number of words that an individual will be able to memorize from a list of 30 words." If the value for W that I get is in the "white" area of the graph under the curve, then we can say that my results were simply due to chance. If that is the case, then I have to "retain the null hypothesis" -­‐ it would appear that noise does not have any effect on the number of words that an individual will be able to memorize from a list of 30 words. However, if my W value falls into the striped area under the curve, then I know that it has met or exceeded the critical value (t) and thus I can reject my null hypothesis at p < 0.05 -­‐ or whichever value t represents. One simple program that helps us to easily calculate U values is Vasserstats.net 1. See below for more information 2. Watch the screencast found on Haiku to demonstrate how to use this simple online program. Vassarstats: Wilcoxon T Test (Signed Ranks) – Worked Example 1. First, go to the Vassarstats site. 2. To calculate the Wilcoxon Signed Ranks Test, click on "Ordinal Data" in the left-­‐hand margin. 3. The Wilcoxon Signed Ranks Test is used for repeated measures designs. Click on the test. The program will ask for the size (n) of the sample. In order to practice using the test, enter a size of 9. 4. Enter the following data in the boxes for Xa: 15, 12, 13, 15, 17, 18, 14, 19, 20. 5. Then enter the following data for Xb: 25, 27, 29, 30, 30, 22, 28, 29, 30. 6. After you have entered the data, click on the "calculate" button. 7. Now for the final step. The IA's require that students predict the direction of the hypothesis -­‐ that is, their hypothesis argues that x increases the effect on y or decreases the effect on y. Therefore, you are going to use the values under "directional test." 8. To interpret the results, you need to compare the W score -­‐ in this case -­‐45 to the values in the table below. You will see that for a directional test at a significance level of p<0.05 for a sample of 9 a W of 29 or higher is required. You should always use the absolute value of the W score -­‐ that is, ignore the fact that in this case it is negative. Since our value is greater than 29, we can say that our data is significant at p< 0.05. If we then move across the chart, we see that our data is greater than 35, 39, and 43. Therefore, we can conclude that our data is significant at p< 0.005. That is, there is less than a .05 percent chance that our results are due only to chance. 9. If, on the other hand, our W value had been 22, then we would not have exceeded the value for p<0.05. Then we would have concluded that our data is not significant, and that our null hypothesis is retained.