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CHAPTER 14. Nonparametric Methods. LEARNING OBJECTIVES. 1. Distinguish between: a. parametric and nonparametric methods b. rank-sum tests and signed-rank tests c. Pearson and Spearman correlation coefficients 2. List the advantages and disadvantages of nonparametric methods
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CHAPTER 14 Nonparametric Methods
LEARNING OBJECTIVES • 1.Distinguish between: • a.parametric and nonparametric methods • b.rank-sum tests and signed-rank tests • c.Pearson and Spearman correlation coefficients • 2.List the advantages and disadvantages of nonparametric methods • 3.Give the equation for the sum of the first n integers • 4.List the assumptions necessary to perform hypothesis tests by nonparametric methods • 5.Be able to apply the sign test to paired data • 6.Know when and how to use Fisher’s exact test
RATIONALE FOR NONPARAMETRIC METHODS • A.Parametric methods – statistical techniques enabling us to determine if there is a significant difference between to sample means with underlying assumptions of normality, homogeneity of variances, and linearity • B.Nonparametric methods • 1.developed for conditions in which assumptions necessary for using parametric methods cannot be made • 2.sometimes called distribution-free method because it is not necessary to assume that the observations are normally distributed • 3.appropriate for dealing with data that are measured on a nominal or ordinal scale and whose distribution is unknown
ADVANTAGES AND DISADVANTAGES OF NONPARAMETRIC METHODS A.Nonparametric advantages • 1.They do not have restrictive assumptions such as normality of the observations. In practice, data are often nonormal or the sample size is not large enough to gain the benefit of the central limit theorem. At most, the distribution should be somewhat symmetrical. • 2.Computations can be performed speedily and easily – a prime advantage when quick preliminary indication of results is needed • 3.They are well suited to experiments of surveys that yield outcomes that are difficult to quantify. In such cases, the parametric methods, although statistically more powerful, may yield less reliable results than the nonparametric methods, which tend to be less sensitive to the errors inherent in ordinal measures
ADVANTAGES AND DISADVANTAGES OF NONPARAMETRIC METHODS • B.Nonparametric disadvantages • 1.They are less efficient (i.e. they require a larger sample size to reject a false hypothesis) than comparable parametric tests. • 2.Hypotheses tested with nonparametric methods are less specific than those tested comparably with parametric methods • 3.They do not take advantage of all the special characteristics of a distribution. Consequently, these methods do not fully utilize the information known about the distribution • C.Should be viewed as complementary statistical methods rather than attractive alternatives. An inherent characteristic is that they deal with ranks rather than values of observations.
Parametric Test Nonparametric Test One Sample One Sample t-test One sample sign test Two independent Two-sample independent Wilcoxson rank-sum test Samples test Mann-Whitney U test Two dependent Two-paired t-test Wilcoxson signed-rank test Samples Sign test Correlation Pearson r Spearman (rho) rank-order correlation Multiple Groups One-way ANOVA Kruskal-Wallis one-way One Factor ANOVA
WILCOXIN RANK-SUM TEST AND MANN-WHITNEY U TEST • A.Both are mathematically equivalent procedures • B.Both are used to test the null hypothesis that there is no difference in the two population distributions • C.Based on ranks from two independent samples, they correspond to the t test for two independent samples, except that no assumptions are necessary regarding normality ore equality or variances • D.They are excellent alternatives to the t test if your data are significantly skewed
WILCOXIN RANK-SUM TEST AND MANN-WHITNEY U TEST • E.Procedure • 1.Combine the observations from both samples and arrange them in an array from the smallest to the largest • 2.Assign ranks to each of the observations • 3.List the ranks from one sample separately from those of the other • 4.Separately sum the ranks for the first and second samples
WILCOXIN RANK-SUM TEST AND MANN-WHITNEY U TEST • F.Given the hypothesis that the average of the ranks is approximately equal for both samples, the test statistic (the sum of the ranks of the first sample) should not differ significantly from (the expected sum of the ranks). • Expected sum of ranks • Standard error, , for s obtained from repeated samples of lists is
Wilcoxon Rank Sum Test For Two Independent Samples: • Mothers Bearing Low-Birth-Weight Babies • No. X R(Rank • 3 5.5* • 0 1.5* • 4 7.5* • 0 1.5* • 1 3 • 2 4 • 3 5.5* • Mothers Bearing Normal-Birth-Weight Babies • No. X R(Rank) • 4 7.5* • 5 9 • 6 10 • 11 15 • 7 11 • 8 12 • 10 14 • 9 13 W2 = 91.5 R2 = 11.4 W1 = 28.5 R1 = 4.1
WILCOXIN RANK-SUM TEST AND MANN-WHITNEY U TEST • G.Regardless of the shape of the population distribution, the sampling distribution for the sum of a subset of ranks is approximately normal. • Test of significance regarding the equality of the distribution • H.Obtaining the expected rank to compute the Z score
WILCOXIN RANK-SUM TEST AND MANN-WHITNEY U TEST • I.Determining significant difference between observed sum and expected value
WILCOXON SIGNED RANK TEST • A.Counterpart to the paired t test for matched observations, we assume that we have a series of pairs of dependent observations • B.We wish tot test the hypothesis that the median of the first sample equals the median of the second – that is, that there is no tendency for the differences between the outcomes before and after some condition to favor either the before or the after condition • C.Procedure is to obtain the differences between individual pairs of observations. Pairs yielding a difference of zero are eliminated from the computation; the sample size is reduced accordingly
WILCOXON SIGNED RANK TEST • A.Performing the test • 1.rank the absolute differences by assigning ranks of 1 for the smallest to n for the largest • 2.ties are eliminated from the analysis • 3.signs of original differences are restored to each rank • 4.sum of positive ranks, , are obtained and serve as the test statistic • 5.if the null hypothesis is true, we would expect the sum of positive ranks to equal that of the negative ranks
Wilcoxon Signed Rank-Test Number of Cigarettes Smoked per Day Subject Xb: Before Xa: after d = xa – xb d rd 1 8 5 -3 3 3(-) 2 13 15 +2 2 2(+) 3 24 11 -13 13 9(-) 4 15 19 +4 4 4(+) 5 7 0 -7 7 7(-) 6 11 12 +1 1 1(+) 7 20 15 -5 5 5(-) 8 22 0 -22 22 10(-) 9 6 0 -6 6 6(-) 10 15 6 -9 9 8(-) 11 20 20 0 - - n(n + 1) 10(11) 55 = = rd = rd(+) = W1 = 7 2 2 rd 55 rd(-) = W2 = 48 = = 27.5 We = 2 2
WILCOXON SIGNED RANK TEST • E.Sum of all ranks is n(n + 1)/2 • F.Under the null hypothesis, we assume that the sum of ranks of the positive d’s is equal to the sum of the ranks of the negative d’s; that is, each will be half of the total sum of ranks
WILCOXON SIGNED RANK TEST(Data from Table 14.2) • G.Z test for the difference between sums of matched ranks • H.The test has a power efficiency of 92% as compared with paired t tests, which satisfies the assumption of normality • I.Assumption of normality requires at least eight pairs. For smaller sample size, an exact test is required
KRUSKAL-WALLIS ONE-WAY ANOVA BY RANKS • A.A nonparametric equivalent of the one-way ANOVA used for making comparisons of more than two groups • B.The groups are independent • C.The populations from which the samples are selected are not normally distributed or the samples do not have equal variances • D.Can also be used when ordered outcomes exist – ordinal data rather than interval or ratio data necessary to use an ANOVA
KRUSKAL-WALLIS ONE-WAY ANOVA BY RANKS • E.Procedure • 1.combine the observations of the various groups • 2.arrange them in order of magnitude from lowest to highest • 3.assign ranks to each of the observations and replace them in each of the groups • 4.original ratio data has therefore been converted into ordinal or ranked data • 5.ranks are summed in each group and the test statistic, H is computed • 6.ranks assigned to observations in each of the k groups are added separately to give k rank sums
KRUSKAL-WALLIS ONE-WAY ANOVA BY RANKS • F.Test statistic equation • In this equation the number of groups = the number of observations in the jth group N = the number of observations in all the groups combined = the sum of the ranks in the jth group
KRUSKAL-WALLIS ONE-WAY ANOVA BY RANKS A B C 4 2 7 9 8 13 3 10 14 1 11 12 5 6 -- R1=22 R2=37 R3=46 • G.Calculation for statistic H: A B C 96 68 115 128 124 149 83 132 166 61 135 147 101 109 --
TIED OBSERVATIONS • A.When two scores are tied, each score is given the mean of ranks for which it is tied. • B.Because H is somewhat influenced by ties, you may wish to correct for ties in computing H where T is the number of tied observations in a tied group of scores • C.The result of correcting for ties is to increase the value of H and thus make the results more significant than it would be if H remained uncorrected
THE SIGN TEST • A.One of the simplest of statistical test, it focuses on the median rather than the mean as a measure of central tendency • B.Only assumption made in performing the test is that the variables come from a continuous distribution • C.It is called the sign test because we use pluses and minuses as the new data in performing the calculations • D.We illustrate its use with a single sample and a paired sample • E.It is useful when we are not able to use the t test because the assumption of normality has been violated
SINGLE SAMPLE • A.We wish to test the that the sample mean is equal to the population median m • B.We assign + to observations that fall above the population median and – to those that fall below • C.A tie is given a 0 and is not counted • D.If the is true – the medians are the same – we expect an equal number: 50% pluses and 50% minuses • E.We can use the binomial distribution to determine if the number of positive signs deviates significantly from some expected number Page 267 Example 1>>>>>>>>>
PAIRED SAMPLES • A.The sign test is also suitable for experiments with paired data such as before and after, or treatment and control • B.Only one assumption must be satisfied – the different pairs must be independent; that is, only the direction of change in each pair is recorded as a plus or minus sign • C.An equal number of pluses and minuses if there is no treatment effect • D.The tested by the paired samples sign test is that the median of the observations listed first is the same as that of the observations listed second in each pair Page 268 Example 2>>>>>>>>
Pascal’s Triangle n Binomial Coefficient Denominator of p
Pascal’s Triangle n Binomial Coefficient Denominator of p
Sign test • Matched Pairs (Before and After) • Does not require that the underlying population be normally distributed. • Based on Median Difference of zero. • A binomial distribution.
Partial Binomial Distributionp = 0.50 • nLeft SpRight S • 0 .0000 20 • 1 .0000 19 • 2 .0002 18 • 3 .0013 17 • 4 .0059 16 • 5 .0207 15 • 6 .0577 14 • 7 .1316 13 • 8 .2517 12 • 9 .4119 11 • 10 .5881 10
SPEARMAN RANK-ORDER CORRELATION COEFFICIENT • A.We obtain perfect correlation if the ranks for variables x and y are equal for each individual • B.Conversely, lack of association is measured by examining the differences in the ordered ranks, • C.The Spearman rank-order correlation coefficient, rs, can be derived from the Pearson correlation coefficient, r • D.Like the Pearson correlation coefficient, the Spearman rank-order correlation coefficient may take on values of –1 to +1 • E.Values close to + or –1 indicate high correlation • F.Values close to zero indicate a lack of association • G.+ and – signs indicate whether the correlation coefficient is positive or negative
SPEARMAN RANK-ORDER CORRELATION COEFFICIENT • H.Ties in rank are handled by averaging the ranks • I.Test statistic with n – 2 df
SPEARMAN RANK-ORDER CORRELATION COEFFICIENT • J.Use it whenever you are unable to meet the assumptions for Pearson r • K.It is preferable to the Spearman rs because the power of the latter is not as great as that of r • L.The Spearman rs is most appropriate when you have either ordinal data or data that are sufficiently skewed that the Pearson r assumptions are not met
Patients Ranked by Smoking and Severity of Illness No. of Cigs Smoked/day Severity of Illness Difference In Ranks 24
RHO rrho = 1 – [(df D2) / n(n2 – 1)] rrho = 1 – [6 (24)] / [8(64 – 1)] = .71 t = rho n - 2 / 1-rho2 t = .71 6 / 1-.712 = 1.74 / .7 = 2.49
FISHER’S EXACT TEST • A.For use with a small sample size • B.Computes directly the probability of observing a particular set of frequencies in a 2 x 2 table where a, b, c, and d are the frequencies of a 2 X 2 table and N is the sample size • C.There has been some controversy as to whether it is appropriate to use Fisher’s exact test in the health sciences because the model required that the marginal totals in the 2 x 2 table be fixed – and they seldom are in actual health science settings • D.Some statisticians use this test anyway because the test results tend to give conservative values of P; that is, the true P value is actually less than the computed one
CONCLUSION These are nonparametric methods that correspond to parametric methods such as the t test, paired t test, and correlation coefficient. The primary advantage of these methods is that they do not involve restrictive assumptions such as normality and homogeneity of variance. Their major disadvantage is that they are less efficient than the corresponding parametric methods of five methods described here – the Wilcoxin signed-rank test, Kruskal-Wallis test, the Mann-Whitney U test, the sign test, the Spearman rank-order correlation coefficient, and Fisher’s exact test. These are the nonparametric methods used most frequently in the health sciences.