This document provides an overview of non-parametric tests, including their advantages over parametric tests and their equivalents. It also discusses the statistical power of non-parametric tests and provides examples of their application in different scenarios.
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Non-parametric tests Non-parametric tests Student name: Instructor: 1|P a g e
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Non-parametric tests PART A QUESTION 1: Reasons for using non-parametric tests over parametric tests We have parametric and non-parametric tests. The choice of test to be employed in a given test depends on the distribution of a given data. For example, parametric tests are normally very sensitive to distribution of data (normality). Therefore it is not appropriate to use a parametric test in a distribution that is not normal (Dallal, 2012). The simple reason is that a skewed distribution will greatly affect the measure of central tendency (mean). Thus if a parametric test is employed in this case, the results will not be accurate. It is therefore necessary to use a non-parametric test where data is not normally distributed. Secondly, when the requirements of parametric test about sample size are not met, then a non-parametric test becomes an alternative. Small sample sizes make it difficult for one to sometimes determine the distribution. In this case the non-parametric tests come in handy. Parametric tests are only capable of assessing continuous data which are again not affected by outliers. However, non-parametric tests can be employed in ordinal and ranked data but not affected by outliers. QUESTION 2:Statistical Power in Non-Parametric Tests According to central limit theorem, samples that appear to be large tend to be normally distributed as opposed to small sample sizes (Field, 2013). In such cases parametric tests are employed to conduct various tests Armstrong (2010). The large sample sizes translate to more power making parametric tests to be considered more powerful than non-parametric tests. Though calculating the power of tests is not popular with statisticians, (Forman, 2017) asserts that the power of parametric tests can be calculated using a textbook formulae. However, according to (Speer, 2006), some non-parametric tests might have more than parametric tests. 2|P a g e
Non-parametric tests QUESTION 3 The table below shows some parametric tests and their non-parametric equivalents Parametric testNon-parametric test equivalents Dependent t-testWilcoxonranksumtest Independent samples t-testMann-WhitneyUtest Repeated measures ANOVA (one-variable)Friedmantwo-wayANOVA One-way ANOVA (independent)Kruskal-Wallistest Pearson correlationSpearmancorrelation Table1. Source: HealthKnowledge.org.uk (n.d.) and Dallal (2012). Thedependentt-testmeasuresthedifferenceinmeansoftwopairedvariables. Independent t-test on the other hand assesses the difference in means of two independent variables. One way anova tests whether there is difference in the means of three variables. Pearson correlation measures the extent of relationship between two variables while repeated measures ANOVA compares variable groups that have been classified into two factors. PART B Non-Parametric Version of the Dependent T-test Table of dependent t-test Hypothesis H0:Having undergone the creative course does not lead to higher marks in creative writing. VERSUS H1:Having undergone the creative course leads to higher marks in creative writing. 3|P a g e
Non-parametric tests Source: Author Table 2 The dependent t-test made us conclude that there was a major statistical difference in the means of the two variables (scores before creative writing course and scores after creative writing course). The null hypothesis asserts that the writing course does not translate into increased scores for creativity. This means that the mean mark of creative writing after the training course remain almost the same. The alternative hypothesis asserts that the writing course did translate into increased scores for creativity.From table 2 above, it can be observed that mean difference of the two variables is 0.2. The p-values calculated (0.508 and 0.708) are all greater than the level of significance which is 0.05. The decision therefore is that the null hypothesis is not rejected. The conclusion of the test is thathaving undergone the creative course does not lead to higher marks in creative writing. It can therefore be concluded that the non- parametric equivalent of dependent t-test (Wilcoxon rank sum test) suggests that the null hypothesis should be accepted. Thus there are no differences in conclusion. 4|P a g e
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Non-parametric tests Non-Parametric Version of the Independent T-test Hypothesis H0:Creativity score for the two tests are not independent. VERSUS H1:Creativity score for the two tests are independent. Source: Author Table 3 The table above represents the results of the Mann-Whitney test which is the counterpart of the independent t-test. The null hypothesis states that the two variables are dependent while the alternative hypothesis states that the two variables are independent. From the table above, it can be observed that the p-value calculated is 0.1 while the significance level is 0.05. The p- value is less compared to the level of significance. This therefore means that the null hypothesis is rejected. It is concluded therefore that creativity score for the two tests are independent. The independent t-test results are t (78) = -1.595, p=0.115. Thus the conclusion from the two tests is the same. 5|P a g e
Non-parametric tests Non-Parametric Version of the Single Factor ANOVA Hypothesis H0:The mean Systolic BP and Diastolic BP recorded at each location of the population. VERSUS H1:At least one mean is different. Ranks SettingNMean Rank Systolic Blood Pressure Home (control)1014.25 Doctor's office1022.80 Classroom109.45 Total30 Diastolic Blood Pressure Home (control)1015.75 Doctor's office1016.30 Classroom1014.45 Total30 Source: Author Test Statisticsa,b Systolic Blood Pressure Diastolic Blood Pressure Chi-Square11.851.237 df22 Asymp. Sig..003.888 a. Kruskal Wallis Test b. Grouping Variable: Setting Source: Author Table 4 The table above represents the results of the Kruskal Wallis test which is the counterpart of the anova test. The null hypothesis state that the three variables have the same mean while the alternative hypothesis states that at least one variable has a different mean. From the table above, it can be observed that the p-value calculated is 0.003 while the significance level is 0.05. The p- 6|P a g e
Non-parametric tests value is less compared to the level of significance. This therefore means that the null hypothesis is rejected. It is concluded therefore that at least one mean is different. The anova test results are F (2, 27) = 9.964, p=0.001. Thus the conclusion from the two tests is the same. 7|P a g e
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Non-parametric tests References Field,A.(2013).DiscoveringstatisticsusingIBMSPSSstatistics.WashingtonD.C.: SagePublications, Inc. Speer,J.(2016).Comparisonofparametricandnonparametrictestsfordifferencesin distribution.Proceedings of the NCUR 2016 Armstrong, R. (2010). Sample size estimation and statistical power analysis HealthKnowledge.org.uk (n.d.). Parametric and non-parametric tests for comparing two or more groups.Retrievedfromhttps://www.healthknowledge.org.uk/public-health-textbook/research- methods/1b-statistical-methods/parametric-nonparametric-tests Dallal,G.E.(2012,May22).Nonparametricstatistics.Retrievedfrom http://www.jerrydallal.com/lhsp/npar.htm 8|P a g e