How to Conduct the Wilcoxon Sign Test - 12-21-2010 by James Lani - Statistics Solutions - http://www.statisticssolutions.com
How to Conduct the Wilcoxon Sign Test by James Lani http://www.statisticssolutions.com/how-to-conduct-the-wilcox-sign-test/ Click here for to get help with your Thesis or Dissertation. Click here for FREE Thesis and Dissertation resources (templates, samples, calculators).
The Wilcoxon sign test signed rank test is a close sibling of the dependent samples t-test. Because the dependent samples t-tests analyzes if the average difference of two repeated measures is zero; it requires metric (interval or ratio) and normally distributed data; the Wilcoxon sign test uses ranked or ordinal data. Thus it is a common alternative to the dependent samples t-test when its assumptions are not met. The Wilcoxon signed rank test relies on the W-statistics. For large samples with n>10 paired observations the W-statistics approximates a Normal Distribution. The W statistics is a non-parametric test, thus it does not need multivariate normality in the data. A research team wants to test whether a new teaching method increases the literacy of children. Therefore the researchers take measure the literacy of 20 children before and after the teaching method has been applied. The literacy is measured on a scale from 0 to 10, with 10 indicating high literacy. The initial baseline shows an average literacy score of 5.9 and after the method has been used the average increases to 7.6. A dependent samples t-test can not be used, as the distribution does not approximate a normal distribution. Also both measurements are not independent from each other and therefore the MannWhitney U-test can not be used. The first step of the Wilcoxon sign test is to calculate the differences of the repeated measurements and to calculate the absolute differences. The next step of the Wilcoxon sign test is to order the cases by increasing absolute differences. For the Wilcoxon signed rank test we can ignore cases where the difference is zero. For all other cases we assign their relative rank. In case of tied ranks the average rank is calculated. That is if rank 10 and 11 have the same observed differences both are assigned rank 10.5. The next step of the Wilcoxon sign test is to sign each rank. If the original difference < 0 then the rank is multiplied by -1; if the difference is positive the rank stays positive. The W-statistic is simply the sum of the signed ranks. In our example W = -172.0 for n = 20. For large samples with n>10 the W-statistics approximates normal distribution, with
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How to Conduct the Wilcoxon Sign Test - 12-21-2010 by James Lani - Statistics Solutions - http://www.statisticssolutions.com
Since the Wilcoxon signed rank test has the null hypothesis that there is on average no difference between the two measurements, it is assumed that with In our example the research team observed a sample of 20 pupils. Therefore, the z-ratio shown above includes a continuity correction, to account for the fact that we use a continuous probability function (the normal distribution) and our test value is based on discrete (ordinal, ranked) data. If the difference WmW is negative the continuity correction is +0.5 (this is the case when W < mW). If the difference WmW is positive the correction is -0.5 (this is the case when W > mW). The shortcut to the hypothesis testing of the Wilcoxon signed rank-test is knowing the critical z-value for a 95% confidence interval (or a 5% level of significance) which is z = 1.96 for a two-tailed test and directionality. Whenever a test is based the normal distribution the sample z value needs to be 1.96 or higher to reject the null hypothesis. Alternatively, we could compare the Wilcoxon sign test's z-value with the tabulated z values for large samples. In our example z = 3.20 which gives the probability of P(0..z) = .4990. Remember that this table shows the one-tailed area under the curve from 0 to z. To calculate the two-tailed significance p we would need to multiply the number by 2 and deduct from 1 so that we get the two tailed probability of z. Therefore p = 1 – 2 * P(0…z) = 1 – 2 * 0.499 = 0.002. Therefore we can reject our null hypothesis (with p < .002) that the average difference of the two observed measurements is 0. Thus we can say that the treatment caused significantly different results. Statistics Solutions can assist with your quantitative analysis by assisting you to develop your methodology and results chapters. The services that we offer include: Data Analysis Plan Edit your research questions and null/alternative hypotheses Write your data analysis plan; specify specific statistics to address the research questions, the assumptions of the statistics, and justify why they are the appropriate statistics; provide references Justify your sample size/power analysis, provide references Explain your data analysis plan to you so you are comfortable and confident Two hours of additional support with your statistician Quantitative Results Section (Descriptive Statistics, Bivariate and Multivariate Analyses, Structural Equation Modeling, Path analysis, HLM, Cluster Analysis) Clean and code dataset Conduct descriptive statistics (i.e., mean, standard deviation, frequency and percent, as appropriate)
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How to Conduct the Wilcoxon Sign Test - 12-21-2010 by James Lani - Statistics Solutions - http://www.statisticssolutions.com
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