Experimental & Behavioral Economics Lecture 6: Nonparametric tests and selection of sample size


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1 Experimental & Behavioral Economics Lecture 6: Nonparametric tests and selection of sample size Based on Siegel, Sidney, and N. J. Castellan (1988) Nonparametric statistics for the behavioral sciences, McGrawHill, New York, and teaching material by John Duffy (University of Pittsburgh) Bernd Rönz (HU Berlin) David Danz Summer term
2 Contents 1. Introduction (recap hypothesis testing) 2. Common (nonparametric) tests in experimental economics 3. Selection of sample size (power analysis) 2
3 Hypothesis testing The research hypothesis is the prediction derived from the theory under test. Null hypothesis (H 0 ) is an hypothesis of no effect (e.g., μ 1 = μ 2 ) usually formulated for the purpose of being rejected If rejected, the alternative hypothesis (H 1 ) is supported (not necessarily true) Alternative hypothesis (H 1 ) is the operational statement of the experimenter's research hypothesis. nature of the research hypothesis determines how H 1 should be stated (e.g., μ 1 μ 2, or μ 1 < μ 2, or μ 1 > μ 2 ) 3
4 Hypothesis testing The region of rejection is a region of the sampling distribution under H0 includes all possible values that a test statistic can take on. consists of a set of possible values which are so extreme that when H0 is true the probability of observing them is very small (α) Distribution of some test statistic under H0 4
5 NONPARAMETRIC TESTS 5
6 Nonparametric tests + If the sample size is very small, there may be no alternative to using a nonparametric statistical test (unless the nature of the population distribution is known exactly) Make usually fewer assumptions about the data Interpretation of nonparametric statistical tests is often more straightforward than the interpretation of parametric tests (easier to learn and to apply than are parametric tests) If assumptions of a parametric statistical model are met in the data, then parametric statistical tests are usually more efficient (lower powerefficiency with nonparametric tests) parametric statistical tests have been systematized: different tests are simply variations on a central theme (nonparametric tests less systematic) 6
7 Nonparametric tests Two independent samples (e.g., betweensubject design: same measure for each subject in two treatments) Fisher s Exact Test / ChiSquare Test of independence Median test WilcoxonMannWhitney Test / Robust Rank Order Test KolmogorovSmirnov Test Two dependent samples (e.g., withinsubject design: two measures or repeated measure for each subject) McNemar test Sign test / Wilcoxon Signed Ranks Tests 7
8 Scales 1. Nominal (or categorical) scale numbers or other symbols are used to classify an object, person, or characteristic (i.e., to identify the groups to which various objects belong) Example: Gender 2. Ordinal (or ranking) scale (1) + objects in one category of a scale stand in some kind of relation >R to objects in other categories ( higher, more preferred, more difficult, etc.) Example: Socioeconomic status, grades 3. Interval Scale (2) + distances or differences between any two numbers on the scale have can be interpreted in a meaningful way Example: Temperature 4. Ratio Scale (3) + has a true zero point as its origin, thus the ratio of any two scale points is independent of the unit of measurement Example: Weight, age 8
9 NONPARAMETRIC TESTS INDEPENDENT SAMPLES 9
10 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test Two independent samples Binary variables 10
11 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test H 0 : No relation between the variables (independence) under H0, the conditional probability of observing success for one variable is independent of the realization of the other variable. i.e., Pr(+ I) = Pr(+ II) = Pr(+) 11
12 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test Hypergeometric distribution describes the probability of k successes in n draws without replacement from a finite population of size N containing exactly K successes. In our contingency table: K = (A+C) N = (A + B + C + D) k = A n = (A+B) 12
13 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test Idea: Regard marginal totals as fixed A finite population of size N has (A+C) elements of group I and (B+D) elements of group II We draw a random sample of size (A+B) without replacement V is a random variable = number of observations sampled from group I In our sample, the realization of V is V = A Under H 0, the probability that V takes on the value A is given by the hypergeometric distribution 13
14 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test With marginal totals being fixed, we can write down all possible contingency tables possible tables will be completely determined by alternative values for A (V) Pvalue is the probability (under H0) of sampling the observed or a more extreme contingency table Let A be the observed frequency in the cell where the row and column containing the smallest and second smallest marginal frequencies intersect. Observed or more extreme contingency tables (twosided): D D observed = A/(A+C) B/(B+D) Reject H0, if Pr( D D observed ) < α 14
15 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test Example II = observed: D =
16 Two independent samples Binary variables (nominal or ordinal) Fisher s Exact Test Example II = observed: D = Pr( D D observed ) = 16
17 Two independent samples Nominal or ordinal scaling Chisquare test of independence Two variables, independent observations Generalization of Fisher s exact test to more than two discrete categories Expected frequencies in each discrete category should not be too small expected frequencies of each cell must exceed 1 at most 20% of the cells with expected frequencies less than 5 17
18 Two independent samples Nominal or ordinal scaling Chisquare test of independence H 0 : The variables are statistically independent = no relation = groups are sampled from the same population 18
19 Two independent samples Nominal or ordinal scaling Chisquare test of independence Idea: Test whether the deviations of observed cell proportions (conditional probabilities) from cell proportions expected under H 0 (independence) exceed what we can expect by chance (random deviations) 19
20 Two independent samples Nominal or ordinal scaling Chisquare test of independence Test statistic: n ij = observed number of cases categorized in the ith row of the jth column E ij = number of cases expected in the ith row of the jth column when H 0 is true 20
21 Two independent samples Nominal or ordinal scaling Chisquare test of independence Asymptotically (as N gets large), X 2 follows a chisquare distribution with df = (r 1)(c 1), where r is the number of rows and c is the number of columns in the contingency table 21
22 Two independent samples Nominal or ordinal scaling Chisquare test of independence Example 22
23 Two independent samples Nominal or ordinal scaling Chisquare test of independence Example 23
24 Two independent samples Nominal or ordinal scaling Chisquare test of independence Example df = (r 1)(c 1) = 2 Reject H 0 since value of X 2 is beyond the critical value with df = 2 and α =
25 Two independent samples Nominal or ordinal scaling Chisquare test of independence Remark for 2x2 tables: if N not too large, use Fisher s exact test If N large (say N > 30), use chisquare test, but employ test statistic with continuity correction (Yates): 25
26 Two independent samples At least ordinal scale The median test Two independent groups At least ordinal scale H 0 : Groups do not differ in central tendency = groups have been drawn from populations with the same median 26
27 Two independent samples At least ordinal scale The median test Idea: first determine the median score for the combined group (i.e., the median for all scores in both samples) if both groups are samples from populations whose medians are the same, we would expect about half of each group's scores to be above the combined median and about half to be below 27
28 Two independent samples At least ordinal scale The median test Under H 0, the sampling distribution of the number of the m cases in group I that fall above the combined median (A) and the number of the n cases in group II that fall above the combined median (B) is the hypergeometric distribution: 28
29 Two independent samples At least ordinal scale The median test Remarks When several scores may fall right at the combined median: i. The groups may be dichotomized as those scores that exceed the median and those that do not. ii. If m + n is large, and if only a few cases fall at the combined median, those few cases may be dropped from the analysis. Better do (i) and see whether it makes a difference when analysis based on greater than or equal to or greater than. There may be no alternative to the median test, even for intervalscale data, e.g., with censored data (some observations may be off the scale and therefore measured as the maximum (or minimum) previously assigned to the observations.) 29
30 Two independent samples At least ordinal scale Two independent groups At least ordinal scale Asymptotically equivalent to a ttest. H 0 : X and Y come from the same population, Pr(X>Y)= ½ =Pr(X<Y). the median is the same in both groups (assuming that variances of the distributions in both groups are equal) H 1 (onetail): Wilcoxon MannWhitney Test (a.k.a. Mann Whitney U test, Wilcoxon ranksum test, or Wilcoxon Mann Whitney test) X is stochastically larger than Y, Pr(X>Y) > ½ the bulk of the elements in X are larger than the bulk of the elements in Y H 1 (twotail): Pr(X>Y) ½ 30
31 Two independent samples At least ordinal scale Wilcoxon MannWhitney Test Idea: m = number of observations in the sample from group X n = number of observations in the sample from group Y combine the observations from both groups and rank them in order of increasing size lowest ranks are assigned to the largest negative values (if any) Note that the sum of the first N = (m+n) integers is N = N(N + 1)/2 W x is the sum of the ranks in group X W y is the sum of the ranks in group Y Thus, W x + W y = N(N + 1)/2 31
32 Two independent samples At least ordinal scale Wilcoxon MannWhitney Test Idea: If H0 is true, we would expect the average ranks in each of the two groups to be about equal. If W x is very large (or very small), then we may have reason to suspect that the samples were not drawn from the same population. The sampling distribution of W x (together with m and n) when H0 is true is known Hence, we can determine the probability associated with the occurrence under H0 of any W x as extreme as the observed value. 32
33 Two independent samples At least ordinal scale Wilcoxon MannWhitney Test Example W x = = 15 W Y = = 13 Pr(W x 15, n = 4, m = 3 ) =.20 (Pr(W x 15, n = 4, m = 3 ) =.8857) Do not reject H0 33
34 Two independent samples At least ordinal scale Wilcoxon MannWhitney Test Remarks Use normal approximation for large samples (m > 10 or n > 10) Then, is asymptotically normally distributed with zero mean and unit variance. Wilcoxon test has greater power than the median test The Wilcoxon test considers the rank value of each observation rather than simply its location with respect to the combined median, and, thus, uses more of the information in the data. 34
35 Two independent samples At least ordinal scale Wilcoxon MannWhitney Test Remarks When ties occur each of the tied observations the average of the ranks they would have had if no ties had occurred Correction of test statistic may be necessary (see Siegel & Castellan, 1988) Wilcoxon MannWhitney Test may be regarded as a permutation test applied to the ranks of the observations and, thus, constitutes a good approximation to the permutation test. 35
36 Two independent samples At least ordinal scale Robust Rank Order Test In order to interpret Wilcoxon tests as a test for equality of medians, we have to assume equal variances in both groups The robust Rank Order Test relaxes the assumption of the same variances, i.e., the underlying distributions may be different when testing equality of medians As before: Two independent groups, at least ordinal scale m = number of observations in the sample from group X n = number of observations in the sample from group Y combine the observations from both groups and rank them in order of increasing size, were lowest ranks are assigned to the largest negative values (if any) 36
37 Two independent samples At least ordinal scale Robust Rank Order Test Procedure For each observation in X [Y] we count the number of observations of Y [X] with a lower rank ( placement of Xi [Yj] ) =: U(YXi) [=: U(XYj)] Calculate the mean of the placements in X [and Y]: Calculate the index of variability of U(YXi) and U(XYj): Test statistic with known distribution: 37
38 Two independent samples At least ordinal scale Robust Rank Order Test Example U(YX) = 3 U(XY) =.75 V x = 2 V y = 2.75 Ù = 1.13 Pr(Ù > 1.13) > 0.1 do not reject H 0 (same conclusion with Wilcoxon MannWhitney test) 38
39 More than two independent samples At least ordinal scale KruksalWallis Test Do k > 2 independent samples (ordinal/ordered data) come from the same or different populations? Extension of MannWhitney to three or more samples. Analogue to the Ftest used in analysis of variance, but without the assumption that all populations under comparison are normally distributed. H0: All k samples have the same distribution functions. H1: At least two of the samples have different distribution functions. 39
40 Two independent samples At least interval scale KolmogorovSmirnov Test Prerequisites Here: Two independent samples/groups At least interval scale H0: samples have been drawn from the same population (i.e., from populations with the same distribution) sensitive to any kind of difference in the distributions from which the two samples were drawn differences in location (central tendency), in dispersion, in skewness, etc. 40
41 Two independent samples At least interval scale KolmogorovSmirnov Test Idea If the two samples have been drawn from the same population distribution, then the cumulative distribution functions (CDF) of both samples are expected to be close to each other If the two sample CDFs are "too far apart" at any point, this suggests that the samples come from different populations large deviations between the two sample CDFs is evidence against H0 41
42 Two independent samples At least interval scale KolmogorovSmirnov Test Procedure determine the empirical CDF for each sample by using the same intervals for both distributions for each interval we subtract one step function from the other test focuses on the largest of these observed deviations S m (X) := empirical CDF for sample A (of size m), i.e., S m (X) =K/m, where K is the number of observations equal to or less than X S n (X) := empirical CDF for sample B (of size n) KolmogorovSmirnov twosample test statistic onesided: D m,n = max[s m (X) S n (X)] twosided: D m,n = max[ S m (X) S n (X) ] Reject H0 if D m,n is too large (sampling distributions of D m,n are known, depend on nature of H1) 42
43 Two independent samples At least interval scale KolmogorovSmirnov Test Example 43
44 Two independent samples At least interval scale KolmogorovSmirnov Test Example D m,n = 0.70, m = 9, n = 10 Value of test statistic greater than critical value > reject H0. 44
45 Two independent samples At least interval scale KolmogorovSmirnov Test Remarks Can also be used to test an empirical distribution from one sample against some theoretical distribution (as the corresponding chisquare test); then the theoretical distribution must be continuous 45
46 NONPARAMETRIC TESTS DEPENDENT SAMPLES 46
47 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Two related (dependent) samples Binary variable Test for the significance of changes in some binary response (e.g., by treatment manipulation) Often used in the context of before and after designs 47
48 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Idea B, C: # individuals who responded the same on each treatment (+ and, respectively) A, D: # individuals whose responses changed between treatments (from + to, and from to +, respectively) 48
49 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Idea Thus, (A + D) is the total number of people whose responses changed. Focus on cells in which changes may occur: Without any treatment effect, the number of changes in each direction would be equally likely. H0: Expected number of observations in each cell is (A + D)/2 49
50 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Remember: Test statistic for the Chisquare test of independence O i = number of cases observed in category i E i = number of cases expected in category i (under H0) Applied to cells counting changes, we yield McNemar s statistic: which (approximately) follows a chisquare distribution with df = 1 50
51 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Remarks Correction for continuity (Yates) gives better approximation (correction is necessary because a continuous distribution (chisquare) is used to approximate a discrete distribution): If the total number of changes (A+D) is less than 10, use the binomial test rather than the McNemar test. 51
52 Two dependent samples Binary variable (nominal or ordinal scale) McNemar test Example Under H 0, Pr(X 2 > 1.25) > 0.05 Do not reject H 0 52
53 Two dependent samples At least ordinal scale Sign test Two related samples Variable under consideration has a continuous distribution Xi: score of subject i in treatment X Yi: score of subject i in treatment Y H0: Pr(Xi > Yi) = Pr(Xi < Yi) = ½ = median difference between X and Y is zero 53
54 Two dependent samples At least ordinal scale Sign test Idea focus on the direction of the difference between every Xi and Yi, noting whether the sign of the difference is positive or negative When H0 is true, we would expect the number of pairs which have (Xi > Yi) to be equal to the number of pairs which have (Xi < Yi). H0 is rejected if too few differences of one sign occur. 54
55 Two dependent samples At least ordinal scale Sign test The probability associated with the occurrence of a particular number of positive (and negative) differences can be determined by the binomial distribution with p = 1/2, N = the number of pairs. If a matched pair shows no difference (i.e., the difference is zero and has no sign), it is dropped from the analysis and N is reduced accordingly. 55
56 Two dependent samples At least ordinal scale Sign test Example 56
57 Two dependent samples At least ordinal scale Sign test Example Probability of observing k of n ranks being negative: k pdf cdf With n = 12, p = ½, the probability of observing 2 or less negative (or positive) signs = 2*Pr(X 2) = 2* = Reject H0 57
58 Two dependent samples At least ordinal scale Sign test Remark For large samples (say, N > 35), normal approximation to the binomial distribution is used 58
59 Two dependent samples At least ordinal scale Wilcoxon SignedRank Test (a.k.a. Wilcoxon T test) Sign test uses only information about the direction of the differences within pairs Wilcoxon signedrank test uses also the relative magnitude gives more weight to a pair which shows a large difference between the two conditions than to a pair which shows a small difference. 59
60 Two dependent samples At least ordinal scale Wilcoxon SignedRank Test Idea Calculate the difference di = Xi Yi for each matched pair of observations Rank di's without respect to sign Assign to each rank the sign (+ or ) of the di which it represents. If H0 is true, the sum of ranks having plus signs and summed those ranks having minus signs, are expected to be equal Reject H0 if the sum of the positive ranks is too different from the sum of the negative ranks, (suggesting that treatment X differs from treatment Y) 60
61 Two dependent samples At least ordinal scale Wilcoxon SignedRank Test N = number of nonzero di s. T + = sum of the ranks which have a positive sign T = sum of the ranks which have a negative sign Note: the sum of all of the ranks is N(N + 1)/2 = T + + T Distribution of T + under H0 is known (Wilcoxon SignedRank Test corresponds to permutation test (for paired observations) based on ranks rather than scores di) 61
62 Two dependent samples At least ordinal scale Wilcoxon SignedRank Test Example T + = = 73 2*Pr(T + 73, N=12) = Reject H0 (as with sign test, but note lower pvalue here) 62
63 Two dependent samples At least ordinal scale Wilcoxon SignedRank Test Remarks Ties pairs with di = 0 are dropped from the analysis and the sample size is reduced accordingly. When two or more d's have the same magnitude, their rank is the average of the ranks which would have been assigned if the d's had differed slightly Large Samples T + is approximately normally distributed with 63
64 SELECTION OF SAMPLE SIZE  POWER ANALYSIS 64
65 Power analysis True state of the world (population) H 0 is true H 1 is true Test result (based on sample) Do not reject H 0 Reject H 0 Correct (1 α) Type I Error α Type II Error β Correct (1 β) power 65
66 Power analysis Type I error: rejecting H0 when it is, in fact, true. Pr(Type I error) =: α In experimental economics, common values of α are.05 and.01 Type II error: failing to reject H0 when, in fact, it is false. Pr(Type II error) =: β Distribution of test statistic under H0 True distribution of the test statistic: 66
67 Power analysis Type I error: rejecting H0 when it is, in fact, true. Pr(Type I error) =: α In experimental economics, common values of α are.05 and.01 Type II error: failing to reject H0 when, in fact, it is false. Pr(Type II error) =: β Power of a test: Probability of correctly concluding a significant effect when it really exist in the population = 1  Pr(Type II error) = 1  β Usually desired to be
68 Power analysis Distribution of test statistic under H0: True distribution of the test statistic: 68
69 Power analysis Power depends on Level of significance α (+) True effect size in the population (+) Sample size N (+) Variance in the data ( ) The kind of test (e.g., Sign test versus Wilcoxon signed rank test) The nature of H1 (onesided > twosided) and other variables, depending upon the test being done. 69
70 Power analysis Given The kind of test (and nature of H1) Probability of TypeI error Power (1 β) Presumed size of effect/parameter Variance in the data (and further assumptions, depending on the test) we can determine the lowest sample size we need in order to detect the presumed effect (with probability (1 β)) 70
71 Power analysis Ways to determine power If the (approximate) distribution of the test statistic is known, we may calculate the power for given parameters directly If the distribution of the test statistic is not known or if an analytical solution is to tedious (e.g. some parameter of a structural model), we may determine power by simulation Example: Twosample test of proportions See script 71
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