Chapter 14 - Part II

Chapter 14 - Part II - Notes : When should we use...

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Notes : When should we use non-parametric tests?? When the is too small. When the response distribution is not This can happen if the data has . In this case statistical methods which are based on the normality assumption breaks down and we have to use non-parametric tests. Important point : When the population distribution is highly , a better summary of the population is the rather than the . So, the softwares generally tests for and form confidence intervals of the difference of of two groups. Advantage of medians over means : Means are highly influenced by . But medians always remain unaffected by outliers. This is why non- parametric tests are unaffected by outliers if they are based on the medians. Moreover the process of ranking itself is independent of outliers. This is because no matter how small or
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large an observation is with respect to the others, it will still get the same rank. This is because the rank of an observation is dependent only on its with respect to the other observations, NOT on its absolute This is another reason why non-parametric tests (like the Wilcoxon’s test) are unaffected by What if two observations have the same value??? If this happens it is said that the subjects (or observations) are . In this case, we the ranks (what they would have got if they were not tied) in assigning them to the tied subjects. Eg : Suppose we want to compare the grades of two students based on their scores given below: Jack Jill 70 68 72 72 70 68 63 69 65 Here the response variable is which is
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in nature. The two groups are the two sets of exams Jack and Jill took. It is also assumed that the exams are a from all the exams each of them took. As can be seen, there are some similar scores. So, we have ties. Here’s how to proceed : Arrange the observations (scores) from smallest to largest. Rank the observations such that the gets a rank and the largest gets the maximum rank . If there are , assign each observation the average rank they would have received if there were no ties. In the above example, the ranking goes like this : Scores Raw rank Final rank 63 1 1 65 2 2 68 3 68 4 69 5 5 70 6 70 7 72 8 72 9
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**Blue correspond to Jack’s scores. Sum of ranks for Jack : Sum of ranks for Jill : 1 + 2 + 3.5 + 5 = 11.5 Hypotheses : H 0 : Jack and Jill did equally well in the exams i.e the ( of) the score distributions of Jack and Jill are the same. H a : Jack did better than Jill i.e the ( of) the score distribution of is greater than that of P – value : Software to find it for us. Minitab output : Mann-Whitney Test and CI: Jack, Jill                       N       Median  Jack 5         70.000  Jill 4         66.500  Point estimate for ETA1-ETA2 is 4.000  96.3 Percent CI for ETA1-ETA2 is  (-0.001, 8.999)  W = 33.5  Test of ETA1 = ETA2 vs ETA1 > ETA2 is significant at  0.0250   The test is significant at 0.0236 (adjusted for ties) 
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* Wilcoxon
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This note was uploaded on 04/18/2008 for the course EXP 3116 taught by Professor Fasig during the Spring '08 term at University of Florida.

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Chapter 14 - Part II - Notes : When should we use...

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