UCLA STAT 110B Why Use Applied Statistics for Engineering Nonparametric Statistics? and the Sciences zParametric tests are based upon assumptions that may include the following: zInstructor: Ivo Dinov, „ The data have the same variance, regardless of the treatments or Asst. Prof. In Statistics and Neurology conditions in the experiment. „ The data are normally distributed for each of the treatments or zTeaching Assistants: Brian Ng, UCLA Statistics conditions in the experiment.

University of California, Los Angeles, Spring 2003 zWhat happens when we are not sure that these http://www.stat.ucla.edu/~dinov/courses_students.html assumptions have been satisfied?

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The Wilcoxon Rank Sum Test How Do Nonparametric Tests Compare with the Usual z, t, and F Tests? z Suppose we wish to test the hypothesis that two distributions have the same center. z Studies have shown that when the usual z We select two independent random samples from each assumptions are satisfied, nonparametric tests are population. Designate each of the observations from about 95% efficient when compared to their population 1 as an “A” and each of the observations parametric equivalents. from population 2 as a “B”.

z When normality and common variance are not z If H0 is true, and the two samples have been drawn satisfied, the nonparametric procedures can be from the same population, when we rank the values in much more efficient than their parametric both samples from small to large, the A’s and B’s equivalents. should be randomly mixed in the rankings.

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What happens when H0 is true? What happens if H0 is not true? •Suppose we had 5 measurements from • If the observations come from two different population 1 and 6 measurements from populations, perhaps with population 1 lying population 2. to the left of population 2, the ranking of the observations might take the following ordering. •If they were drawn from the same population, the rankings might be like this. AAABABABBB ABABBABABBA •In this case if we summed the ranks of the A In this case the sum of the ranks of the B measurements and the ranks of the B observations would be larger than that for the measurements, the sums would be similar. A observations.

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1 The Wilcoxon Rank Sum Test How to Implement Wilcoxon’s Rank Test H : the two population distributions are the same H00: the two population distributions are the same •Rank the combined sample from smallest to HHa:: thethe twotwo populationspopulations areare inin somesome wayway differentdifferent largest. a *. •Let T1 represent the sum of the ranks of the z The test statistic is the smaller of T1 and T1 first sample (A’s). z Reject H if the test statistic is less than the critical •Then, * defined below, is the sum of the 0 T 1 value found in Table 7(a). ranks that the A’s would have had if the observations were ranked from large to small. z Table 7(a) is indexed by letting population 1 be the one associated with the smaller sample size n1, and * T1 = n 1 (n 1+ n 2 +1) − T 1 population 2 as the one associated with n2, the larger sample size.

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Example The Bee Problem The wing stroke frequencies of two Can you conclude that the distributions of wing strokes differ for these two species? α = .05. speciesIfIf severalseveral of measurements measurementsbees were arearerecorded tied,tied, for a sample of n1 each gets the average of the ranks = 4each from gets species the average 1 ofand the nranks= 6 from species 2. SpeciesRejectReject 1 Species HH0 2.. DataDataCalculate providesprovides T sufficientsufficient= 7 + 8 + 9 +10 = 34 they would have gotten, if they2 0 1 they would have gotten, if they 235 evidence(10) 180 (3.5) indicating a difference in Canwere you not tied!conclude (See x = that 180) the distributions of wing evidence indicating a difference* in were not tied! (See x = 180) 225 the(9) distributions169 (1) of wing Tstroke1 = n 1(n1 + n2 +1) −T1 strokes differ for these two species? Use α = .05. the distributions of wing stroke H : the two species are the same 190 frequencies.(8)frequencies.180 (3.5) = 4(4 + 6 +1) − 34 =10 Species 1 Species 2 H00: the two species are the same H : the two species are in some way different 188 (7) 185 (6) 235 180 Ha : the two species are in some way different (10) (3.5) a 178 (2) 1. The test statistic is T = 10. 225 169 1. The sample with the smaller sample (9) (1) 182 (5) size is called sample 1. 190 (8) 180 (3.5) 2. The critical value of T from 188 (7) 185 (6) 2. We rank the 10 observations from Table 7(b) for a two-tailed test 178 (2) smallest to largest, shown in with α/2 = .025 is T = 12; H0 is 182 (5) parentheses in the table. rejected if T ≤ 12. Slide 9 Stat 110B, UCLA, Ivo Dinov Slide 10 Stat 110B, UCLA, Ivo Dinov

Minitab Output Large Sample Approximation: Wilcoxon Rank Sum Test * Recall T1 = 34;T1 = 10. When n1 and n2 are large (greater than 10 is large

Mann-Whitney Test and CI: Species1, Species2 enough), a normal approximation can be used to

Species1 N = 4 Median = 207.50 approximate the critical values in Table 7. Species2 N = 6 Median = 180.00 * * Point estimate for ETA1-ETA2 is 30.50 1. Calculate T1 and T1 . Let T = min(T1,T1 ). 95.7 Percent CI for ETA1-ETA2 is (5.99,56.01) * W = 34.0 T1 = n 1 (n 1+ n 2 +1) − T 1 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant T − µT at 0.0142 2. The statistic z = has an approximate The test is significant at 0.0139 (adjusted for ties) σT T1 = sum of the ranks MinitabMinitab calls calls the the procedure procedure the the Mann-Whitney Mann-Whitney U U z distribution with of sample 1 (A’s). Test,Test, equivalent equivalent to to the the Wilcoxon Wilcoxon Rank Rank Sum Sum Test. Test. n1(n1 + n2 +1) 2 n1n2 (n1 + n2 +1) The test statistic is W = T = 34 and has p-value µT = and σT = The test statistic is W = T11 = 34 and has p-value = .0142. Do not reject H for α = .05. 2 12 = .0142. Do not reject H00 for α = .05. Slide 11 Stat 110B, UCLA, Ivo Dinov Slide 12 Stat 110B, UCLA, Ivo Dinov

2 Some Notes The Sign Test

•When should you use the Wilcoxon Rank Sum zThe sign test is a fairly simple test instead of the two-sample t test for procedure that can be used to compare two independent samples? populations when the samples consist of 9when the responses can only be ranked and paired observations. not quantified (e.g., ordinal qualitative data) zIt can be used 9when the F test or the Rule of Thumb shows a 9when the assumptions required for the paired- problem with equality of variances difference test are not valid or 9when a normality plot shows a violation of 9 when the responses can only be ranked as “one the normality assumption better than the other”, but cannot be quantified.

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The Sign Test The Sign Test

99ForFor eacheach pair,pair, measuremeasure whetherwhether thethe firstfirst H : the two populations are identical versus response—say,response—say, A—exceedsA—exceeds thethe secondsecond H00: the two populations are identical versus H : one or two-tailed alternative response—say,response—say, B.B. Haa: one or two-tailed alternative is equivalent to 99TheThe testtest statisticstatistic isis xx,, thethe numbernumber ofof timestimes thatthat is equivalent to A exceeds B in the n pairs of observations. H : p = P(A exceeds B) = .5 versus A exceeds B in the n pairs of observations. H00: p = P(A exceeds B) = .5 versus 9Only pairs without ties are included in the test. H : p (≠, <, or >) .5 9Only pairs without ties are included in the test. Haa: p (≠, <, or >) .5 99CriticalCritical valuesvalues forfor thethe rejectionrejection regionregion oror exactexact TestTest statistic:statistic: xx== numbernumber ofof plusplus signssigns pp-values-values cancan bebe foundfound usingusing thethe cumulativecumulative RejectionRejection regionregion,, pp-values-valuesfrom from Bin(n=size,Bin(n=size, p).p). binomialbinomial distributiondistribution (SOCR(SOCR resourceresource online).online).

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Example The Gourmet Chefs

Meal 1 2 3 4 5 6 7 8 Two gourmet chefs each tasted and rated Chef A 6 4 7 8 2 4 9 7 eight different meals from 1 to 10. Does it appear Chef B 8 5 4 7 3 7 9 8 that one of the chefs tends to give higher ratings Sign - - + + pp-value-value- - =.454 =.4540 is -is too too large large to to rejectreject H H0. .There There is is insufficient insufficient than the other? Use α = .01. H : p = .5 0 H00: p = .5 evidenceevidence to to indicate indicate that that one one H : p ≠ .5 with n = 7 (omit the tied pair) Meal 1 2 3 4 5 6 7 8 Haa: p ≠ .5 with n = 7 (omitchef chefthe tiedtends tends pair) to to rate rate one one meal meal Chef A 6 4 7 8 2 4 9 7 TestTest Statistic: Statistic: x x== number number of of higherplus higherplus signs signs than than = =the the2 2 other. other. Chef B 8 5 4 7 3 7 9 8 UseUse Table Table 1 1 with with n n == 7 7 and and p p = = .5. .5. Sign - - + + - - 0 - pp-value-value = = P(observe P(observe x x = = 2 2 or or something something equally equally as as unlikely) unlikely) == P( P(xx ≤ ≤2)2) + + P( P(xx≥≥5)5) = = 2(.227) 2(.227) = = .454 .454 HH0:: the the rating rating distributions distributions are are the the same same ( (pp== .5) .5) 0 k 0 1 2 3 4 5 6 7 HHa:: the the ratings ratings are are different different ( (pp≠≠.5).5) a P(x ≤ k) .008 .062 .227 .500 .773 .938 .992 1.000

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3 Y~Bin(n, p) Î Large Sample Approximation: E(Y) = np Example The Sign Test Var(Y) = np(1-p) You record the number of accidents per day at a large When n ≥ 25, a normal approximation can be manufacturing plant for both the day and evening shifts for n = used to approximate the critical values of 100 days. You find that the number of accidents per day for the

evening shift xE exceeded the Forcorresponding a two tailed number test, we of reject accidents H Binomial distribution. For a two tailed test, we reject H00 in the day shift xD on 63 of theifif 100| z|z| |> > days. 1.96 1.96 (5% Do(5% these level). level). results provide sufficient evidence to indicateH thatis morerejected. accidents There istend evidence to occur 1. Calculate x = number of plus signs. H00 is rejected. There is evidence on one shift than on the other?ofof a a difference difference between between the the day day x − .5n and night shifts. 2. The statistic z = has an approximate H : the distributions (# of accidents)and night are shifts. the same (p = .5) H00: the distributions (# of accidents) are the same (p = .5) .5 n H : the distributions are different (p ≠ .5) Haa: the distributions are different (p ≠ .5) z distribution. x − .5n 63 − .5(100) Test statistic: z = = = 2.60 .5 n .5 100

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Which test should you use? Which test should you use? z We compare statistical tests using z If all parametric assumptions have been met, the Definition: Power = 1 − β parametric test will be the most powerful.

= P(reject H0 when Ha is true) z If not, a nonparametric test may be more powerful. z The power of the test is the probability of rejecting the null z If you can reject H0 with a less powerful nonparametric hypothesis when it is false and some specified alternative is true. test, you will not have to worry about parametric z The power is the probability that the test will do what it was assumptions. designed to do—that is, detect a departure from the null z If not, you might try hypothesis when a departure exists. „ more powerful nonparametric test or „ increasing the sample size to gain more power

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The Wilcoxon Signed-Rank Test The Wilcoxon Signed-Rank Test – different form Wilcoxon Rank Sum Test – different form Wilcoxon Rank Sum Test

9For each pair, calculate the difference d = x -x . z The Wilcoxon Signed-Rank Test is a more powerful 9For each pair, calculate the difference d = x11-x22. nonparametric procedure that can be used to compare EliminateEliminate zerozero differences.differences. two populations when the samples consist of paired 99RankRank thethe absoluteabsolute valuesvalues ofof thethe differencesdifferences fromfrom 11 toto nn.. observations. TiedTied observationsobservations areare assignedassigned averageaverage ofof thethe ranksranks theythey wouldwould havehave gottengotten ifif notnot tied.tied. z It uses the ranks of the differences, d = x -x that we + 1 2 ƒƒTT+== rankrank sumsum forfor positivepositive differencesdifferences used in the paired-difference test. - ƒƒTT-== rankrank sumsum forfor negativenegative differencesdifferences + - 99IfIf thethe twotwo populationspopulations areare thethe same,same, TT+andand TT-shouldshould bebe nearlynearly equal.equal. IfIf eithereither TT++oror TT--isis unusuallyunusually large,large, thisthis providesprovides evidenceevidence againstagainst thethe nullnull hypothesis.hypothesis.

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4 The Wilcoxon Signed-Rank Test Example To compare the densities of cakes using mixes A and B, six pairs of pans (A and B) were H : the two populations are identical versus baked side-by-side in six different oven H00: the two populations are identical versus locations. Is there evidence of a difference in HH :: oneone oror two-tailedtwo-tailed alternativealternative aa density for the two cake mixes? ++ -- TestTest statistic:statistic: TT== minmin (( TT andandTT )) Location 1 2 3 4 5 6 Cake Mix A .135 .102 .098 .141 .131 .144 CriticalCritical valuesvaluesforfor aa oneone oror two-tailedtwo-tailed Cake Mix B .129 .120 .112 .152 .135 .163 d = x -x .006 -.018 -.014 -.011 -.004 -.019 rejectionrejection regionregion cancan bebe foundfound usingusing A B WilcoxonWilcoxon Signed-RankSigned-Rank TestTest Table.Table. H : the density distributions are the same H00: the density distributions are the same H : the density distributions are different Haa: the density distributions are different

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Cake Densities Large Sample Approximation: The Signed-Rank Test Location 1 2 3 Do4 not reject5 H6 .There is Do not reject H00.There is Cake Mix A .135 .102 .098 insufficientinsufficient.141 .131 evidence evidence.144 to to indicate indicate Cake Mix B .129 .120 .112 thatthat.152 there there.135 is is a a difference difference.163 in in When n ≥ 25, a normal approximation can be densities for the two cake mixes. d = xA-xB .006 -.018 -.014 densities-.011 -.004 for the-.019 two cake mixes. used to approximate the critical values in Table 8. Rank 2 5 4 3 1 6 1. Calculate T + and T −. Let T = min(T + ,T − ). ++ -- T − µ RankCalculate:RankCalculate: the the 66 T T == 2and 2andTT == 5+4+3+1+6 5+4+3+1+6 = = 19. 19. 2. The statistic z = T has an approximate differences, ++ -- Thedifferences,Thetesttest statistic statistic isis T T== min min ( (TT ,, TT )) = = 2. 2. σT without regard RejectionwithoutRejection regard region: region: UseUse Table Table 8. 8. For For a a two-tailed two-tailed test test with with to sign. z distribution with αtoα =sign.= .05, .05, reject reject H H0 ifif TT≤≤1.1. 0 n(n +1) n(n +1)(2n +1) µ = and σ 2 = T 4 T 24

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The Kruskal-Wallis – H Test The Kruskal-Wallis – H Test

99RankRank thethe totaltotal measurementsmeasurements inin allall kksamplessamples z The Kruskal-Wallis H Test is a nonparametric fromfrom 11 toto nn.. TiedTied observationsobservations areare assignedassigned averageaverage ofof thethe procedure that can be used to compare more than ranksranks theythey wouldwould havehave gottengotten ifif notnot tied.tied. two populations in a completely randomized 99CalculateCalculate design. th ƒƒTTi == rankrank sumsum forfor thethe iithsamplesample ii== 1,1, 2,…,2,…,kk zNon-parametric equivalent to ANOVA F-test! i ƒƒnn == nn1+n+n2+…+n+…+nk zAll n = n1+n2+…+nk measurements are jointly 1 2 k ranked. 99AndAnd thethe testtest statisticstatistic HH isis (analog(analog to:to: FF == MSST/MSSEMSST/MSSE)) zWe use the sums of the ranks of the k samples to  k 2  12  Ti  compare the distributions. H =   − 3(n +1) n(n +1) ∑ ni  i=1  Slide 29 Stat 110B, UCLA, Ivo Dinov Slide 30 Stat 110B, UCLA, Ivo Dinov

5 TheThe Kruskal-Wallis Kruskal-Wallis H Test H Test Example

HH :: thethe kkdistributionsdistributions areare identicalidentical versusversus Four groups of students were randomly 00 assigned to be taught with four different H : at least one distribution is different Haa: at least one distribution is different techniques, and their achievement test scores TestTest statistic:statistic: Kruskal-WallisKruskal-Wallis HH were recorded. Are the distributions of test When H is true, the test statistic H has an scores the same, or do they differ in location? When H00 is true, the test statistic H has an approximate χ22distribution with df = k-1. 1 2 3 4 approximate χ distribution with df = k-1. 65 75 59 94 UseUse aa right-tailedright-tailed rerejectionjection regionregion oror pp-value-value 87 69 78 89 based on the Chi-square distribution. 73 83 67 80 based on the Chi-square distribution. 79 81 62 88

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Teaching Methods Teaching Methods

1 2 3 4 H : the distributions of scores are the same H00: the distributions of scores are the same H : the distributions differ in location 65 (3) 75 (7) 59 (1) 94 (16) Haa: the distributions differ in location 87 (13) 69 (5) 78 (8) 89 (15) 12 T 2 73 (6) 83 (12) 67 (4) 80 (10) Test statistic: H = ∑ i − 3(n +1) 79 (9) 81 (11) 62 (2) 88 (14) n(n +1) ni 2 2 2 2 Ti 31 35 15 55 12  31 + 35 +15 + 55  =   − 3(17) = 8.96 16(17)  4  Rank the 16 H : the distributions of scores are the same   Rank the 16 H00: the distributions of scores are the same measurementsmeasurements HHa:: the the distributions distributions differ differ in in location location Reject H . There is sufficient a Rejection region: Use Table 5. Reject H00. There is sufficient fromfrom 1 1 to to 16, 16, Rejection region: Use Table 5. evidence to indicate that there 2 For a right-tailed chi-square test evidence to indicate that there and calculate 12 Ti For a right-tailed chi-square test and calculateTest statistic : H = ∑ − 3(n +1) isis a a difference difference in in test test scores scores for for the four rank n(n +1) n withwith αα== .05 .05 and and df df== 4-1 4-1 =3, =3, the four rank i thethe four four teaching teaching techniques. techniques. 2 2 2 2 rejectreject HH0 ifif H H ≥≥7.81.7.81. sums.sums. 12  31 + 35 +15 + 55  0 =   − 3(17) = 8.96 16(17)  4  Slide 33 Stat 110B, UCLA, Ivo Dinov Slide 34 Stat 110B, UCLA, Ivo Dinov

Key Concepts Key Concepts 2. Use T1 to test for population 1 to the left of population 2 I. Nonparametric Methods Use T * to test for population to the right of population 2. 1. These methods can be used when the data cannot be 1 * measured on a quantitative scale, or when Use the smaller of T1 and T 1 to test for a difference in the locations of the two populations. 2. The numerical scale of measurement is arbitrarily set by the researcher, or when 3. Table 7 of Appendix I has critical values for the rejection of H0. 3. The parametric assumptions such as normality or constant 4. When the sample sizes are large, use the normal variance are seriously violated. approximation: T − µ z = T II. Wilcoxon Rank Sum Test: Independent Random σ T Samples n (n + n +1) n n (n + n +1) µ = 1 1 2 and σ 2 = 1 2 1 2 1. Jointly rank the two samples: Designate the smaller T 2 T 12 sample as sample 1. Then * T1 = Rank of sample1 ⇒ T1 = n1(n1 + n2 +1) − T1 Slide 47 Stat 110B, UCLA, Ivo Dinov Slide 48 Stat 110B, UCLA, Ivo Dinov

6 Key Concepts Key Concepts IV. Wilcoxon Signed-Rank Test: Paired Experiment III. Sign Test for a Paired Experiment 1. Calculate the differences in the paired observations. Rank the absolute values of the differences. Calculate the rank 1. Find x, the number of times that observation A exceeds sums T − and T + for the positive and negative differences, observation B for a given pair. respectively. The test statistic T is the smaller of the two rank sums. 2. To test for a difference in two populations, test H0 : p = 0.5 versus a one- or two-tailed alternative. 2. Table 8 of Appendix I has critical values for the rejection of 3. Use Table 1 of Appendix I to calculate the p-value for the test. for both one- and two-tailed tests. 4. When the sample sizes are large, use the normal 3. When the sampling sizes are large, use the normal approximation: approximation: T − [n(n +1) 4] z = x − .5n n(n +1)(2n +1) 24 z = [ ] .5 n

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Key Concepts Key Concepts

V. Kruskal-Wallis H Test: Completely Randomized Design

1. Jointly rank the n observations in the k samples. Calculate the VI. The Friedman Fr Test: Randomized Block Design rank sums, Ti = rank sum of sample i, and the test statistic 12 T 2 1. Rank the responses within each block from 1 to k. Calculate the H = ∑ i − 3(n +1) rank sums T1, T2, …, Tk, and the test statistic n(n +1) ni 12 2 2. If the null hypothesis of equality of distributions is false, H Fr = ∑Ti − 3b(k +1) will be unusually large, resulting in a one-tailed test. bk(k +1) 2. If the null hypothesis of equality of treatment distributions is 3. For sample sizes of five or greater, the rejection region for H is false, Fr will be unusually large, resulting in a one-tailed test. based on the chi-square distribution with (k − 1) degrees of 3. For block sizes of five or greater, the rejection region for F is freedom. r based on the chi-square distribution with (k − 1) degrees of freedom.

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Key Concepts Sensitivity vs. Specificity

VII. Spearman's Rank Correlation Coefficient zSensitivity is a measure of the fraction of gold standard 1. Rank the responses for the two variables from smallest to largest. known examples that are correctly classified/identified. 2. Calculate the correlation coefficient for the ranked z Sensitivity= TP/(TP+FN) observations: zSpecificity is a measure of the fraction of negative S 2 xy 6∑ di examples that are correctly classified: rs = or rs =1− 2 if there are no ties S S n(n −1) Ho: no effects (µ=0) xx yy z Specificity = TN/(TN+FP) 3. Table 9 in Appendix I gives critical values for rank zTP = True Positives True Reality

s

correlations significantly different from 0. t l Ho true Ho false

u

zFN = False Negatives s 4. The rank correlation coefficient detects not only significant e Can’t reject TN FN

R linear correlation but also any other monotonic relationship zTN = True Negatives t

s between the two variables. Reject Ho FP TP zFP = False Positives Te

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7 The ROC Curve Receiver-Operating Characteristic curve zReceiver Operating Characteristic (ROC) curve zROC curve demonstrates several things: 1.0 ƒIt shows the tradeoff between sensitivity and specificity Better (any increase in sensitivity will be accompanied by a Test Worse decrease in specificity). Test ƒThe closer the curve follows the left border and then the top border of the ROC space, the more accurate the test.

Bigger area Î higher test accuracy and better ƒThe closer the curve comes to the 45-degree diagonal of Sensitivity discrimination i.e. the ability of the test to the ROC space, the less accurate the test.

Fraction True Positives True Fraction correctly classify those with and without the disease. ƒThe slope of the tangent line at a cut-point gives the 1.0 likelihood ratio (LR) for that value of the test. You can Fraction False Positives check this out on the graph above. Specificity ƒThe area under the curve is a measure of test accuracy. Stat 110B, UCLA, Ivo Dinov Slide 55 Stat 110B, UCLA, Ivo Dinov Slide 56

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