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Nonparametric and Distribution- Free Statistics By Lecture Series on Biostatistics No. Bio-Stat_12 Date – 16.09.2008 Dr. Bijaya Bhusan Nanda, M. Sc (Gold Medalist) Ph. D. (Stat.) Topper Orissa Statistics & Economics Services, 1988 bijayabnanda@yahoo.com

CONTENTS: 

CONTENTS Introduction The Sign Test The Wilcoxon Signed- Rank Test for Location The Median Test The Mann-Whitney Test The Kolmogorov- Smirnov Goodness-of-fit Test The Kruskal-Wallis one-way analysis of variance by ranks

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Nonparametric Test: Those procedures that test hypotheses that tests hypotheses that are not statements about population parameters are classified as nonparametric. Distribution free procedure: Those procedures that make no assumption about the sampled population are called distribution free procedures. However, both these terms Nonparametric and distribution free are used interchangeably and under the common heading ‘Nonparametric’. Introduction

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Advantages: Nonparametric statistics allow for the testing of hypothesis that are not statements about population parameter values. It may be used when the form of the sampled population is unknown. Computationally easier. May be applied when the data being analyzed consists merely of rankings or classifications. Disadvantages: Nonparametric procedure with data that can be handled with a parametric procedure results in a waste of data. Application of the Nonparametric tests may be laborious for large samples.

The Sign Test: 

The Sign Test The ‘ t’ test is not strictly valid for testing hypothesis regarding single population mean or mean difference between pairs of observations unless the relevant populations are at least approximately normally distributed. A frequently used nonparametric test that doesn’t depend on the assumptions of ‘t’ test is the sign test. This test focuses on the median rather than the mean as a measure of central tendency or location. The only assumption underlying the test is that the distribution of the variable of the interest is continuous.

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Theoretical background of the sign test: Let x 1 , x 2 , ..., x n be a random sample form a population with unknown median M. H0: M = M 0 , Ha: M > M 0 . The probability that a x-value selected from the population is larger than M is 0.5, i.e., P(x i > M) = 0.5 if the null hypothesis is true. Then we should expect to observe approximately half the sample x-value greater than M= M 0 . Test Statistics : S = { number of values x i that exceed M 0 }. Notice that S depends only on the sign (positive or negative) of the difference xi - M 0 . If S is “too large” the we will reject H 0 in favor of Ha: M > M 0 .

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Rejection region: Let each sample difference x i - M 0 denote the outcome of a single trial in an experiment consisting of ‘n’ identical trials. If we call a positive difference a “Success” and a negative difference a “Failure”, then S is the number of successes in n trials. Under H 0 the probability of observing a success on any one trial is p = P(Success) = P(xi - M0 > 0) = P(xi > M0) = 0.5 Since the trials are independent, the properties of a binomial distribution, are satisfied. Therefore, S has a binomial distribution with parameters n and p = 0.5. We can use this fact to calculate the observed significance level (p-value ) of the sign test.

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SIGN TEST FOR A POPULATION MEDIAN ONE-TAILED TEST Test statistic: S = Number of sample observations greater than M 0 ( or S less than M0 ) Observed significance level: p-value = P(S  Sc) TWO-TAILED TEST Test statistic: S = max ( S 1 , S 2 ), where S 1 = No. of sample observations greater than M 0 , S 2 = No. of sample observations less than M 0 [ Note: By definition S 2 = n – S 1 ] Observed significance level: p-value =2 P(S  S c ) where S c is the computed value of the test statistic and S has a binomial distribution with parameters n and p = 0.5. Rejection region: Reject H 0 if  > p -value.

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Example: Researchers wished to know if instruction in personal care and grooming would improve the appearance of mentally retarded girls. In a school for the mentally retarded, 10 girls selected at random reciveed special instruction in personal care and grooming. Two weeks after completion of the course of instruction the girls were interviewed by a nurse and a social worker who assigned each girl a score based on her general appearance. The investigators believed that the score achieved the level of an ordinal scale. The scores are shown in the table. We wish to know if we can conclude that tha median score of the population from which we assume this sample to have been drawn is different from 5 Table: General appearance Scores of 10 mentally Retarded Girls Girl Score Girl Score 1 4 6 6 2 5 7 10 3 8 8 7 4 8 9 6 5 9 10 6

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Assumption: We assume that that the measurements are taken on a continuous variable. Hypothesis: H 0 : The population median is 5 H: The population median is not 5 Let =0.05 Test Statistic: The test statistic for the sign test is either the observed number of +sign or –sign, which one is less.

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Distribution of Test statistic: If we assign a +sign to those score that lies above the hypothesized median of 5 and a – sign to those that fall below. Then the table is From the table we constitute a set of n independent random variable from the Bernoulli population with parameter p. If we let S= the test statistic, the sampling distribution of S is the binomial distribution with parameter p=0.5 if the null hypothesis is true. The decision rule is reject H 0 if the ‘p’ value for the computed test statistic is less than or equal to 0.05. Calculation of test statistic: Determine the probability of observing S or fewer –signs when given a sample of size n and parameter p=0.5. Girl 1 2 3 4 5 6 7 8 9 10 Score - 0 + + + + + + + +

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Decision rule: The decision rule depend on the alternative hypothesis. For H A : P(+)  P(-) reject H 0 if (given H 0 is true) the probability of obtaining a value of ‘S’ as extreme as or more extreme than the actually computed is equal to or less than α /2. P(k  x n, p) =  n C k p k q n-k For the example 9 C 0 (0.5) 0 (0.5) 9-0 + 9 C 1 (0.5) 1 (0.5) 9-1 = .00195+.01758 = .0195 Here there are fewer minuses so we focus our attention on minuses rather than pluses. Statistical decision: If the number of minuses is so small that the probability of observing this few of fewer is less than 0.025, we will reject the null hypothesis. The probability .0915 is less than .025 so reject the null hypothesis. Conclusion: The median score is not 5. p value: The p value for this test is 2(.0195) = 0.0390

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Sign Test-Paired Data: Conditions The data to be analyzed consist of observation in matched pairs The measurement scale is weak. Sign test may be employed to test the null hypothesis that the median difference is zero Say (Yi, Xi) are pairs of matched scores, i = 1,2,……..n One of the matched scores, say Yi is subtracted from the other score Xi. If Yi<Xi, the sign of the difference is + and if Yi>Xi the sign of the difference is - Then the null hypothesis is H 0 : P(+) = P(-) = 0.5 or P (Xi > Yi) = P (Xi<Yi) =0.5 H A : P(+) ≠ P(-) or H A : P(+) > P(-) or HA: P(+) < P(-)

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Example: A dental research team wished to know if teaching people how to brush their teeth would be beneficial. 12 pairs of patients seen in a dental clinic were obtained by carefully matching on such factors as age, sex, intelligence and initial oral hygienic scores. One member of each pair received instruction on hoe to brush the teeth and other oral hygienic matters. Six months later all 24 subjects were examined and assigned an oral hygienic score by a dental hygienist on aware of which subjects had received the instruction. A low score indicates a high level of oral hygienic. The results are shown in Table. Pair No 1 2 3 4 5 6 7 8 9 10 11 12 Inst. (Xi) 1.5 2.0 3.5 3.0 3.5 2.5 2.0 1.5 1.5 2.0 3.0 2.0 Not inst. (Yi) 2.0 2.0 4.0 2.5 4.0 3.0 3.5 3.0 2.5 2.5 2.5 2.5

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Solution: Assumption: We assume that the population of difference between two pairs of scores is a continuous variable. Hypothesis: H 0 : The median of the differences is zero [P(+)≥P(-)] H A : The median of the differences is negative [P(+)<P(-1) Let =0.05. Test Statistic: The test statistic is the number of +signs. Distribution of Test Statistic: The sampling distribution of k is the binomial distribution with parameters n and 0.5 if H 0 is true.

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Decision rule: Reject H 0 if P(S  211,5)0.05. Calculation Of Test Statistic: The score difference have been obtained for each pair. Then the result is in Table. The nature of the hypothesis indicates a one sided test so that all of = 0.05 is associated with the rejection region, which consists of all values of S (equal to number of +signs). The experiment yielded 1zero , 2pluses and 9 minuses. When we eliminate 0, the effective sample size is n=11 with 2 pluses and 9minuses. Since a small no of +signs will cause rejection of the null hypothesis, test statistic is S=2. Pair 1 2 3 4 5 6 7 8 9 10 11 12 Score - 0 - + - - - - - - + -

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Statistical Decision: The probability of obtaining no more than 2 pluses out of eleven tries when the null hypothesis is true is P(S  2 11, 5) =  11 C S (.5) S (.5) 11-S . This probability is to be .0327. since .0327 is less than .05, we must reject H 0 . Conclusion: Then we conclude that the median difference is negative and the instruction was beneficial. p value: The p value is 0.0327

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Sample size When the sample size is large and p is close to 0.5 the binomial distribution may be approximated by normal distribution. According to the rule of thumb when both np and nq are greater than 5, normal approximation is appropriate. Under H 0 p = 0.5, a sample size of 12 would satisfy normal approximation. In this situation normal z test can be applied for sign test. The test statistic here z = ( k ± 0.5 )- 0. 5n 0 . 5 √n

The Wilcoxon Signed- Rank Test for Location: 

The Wilcoxon Signed- Rank Test for Location To test a null hypothesis about a population mean , but for some reason neither z nor t is an appropriate test statistic. Wilcoxon Signed- Rank Test is a appropriate test which makes use of the magnitudes of the differences between measurements and a hypothesized location parameter rather than just the signs of the differences. Assumptions: The sample is random The variable is continuous. The population is symmetrically distributed about its mean . The measurement scale is at least interval.

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Hypothesis: The hypothesis are : (a) H 0 : = 0 (b) H 0 :  0 (c ) H 0 :  0 H A :  0 H A : < 0 H A : > 0 Calculation: 1. Subtract the hypothesized mean 0 from each observation xi, to obtain di = xi- 0. 2. Rank the usable di from the smallest to the largest without regard to the sign of di. If two or more  di  are equal, assign each tied value the mean of the rank positions the tied values occupy. 3. Assign each rank the sign of the di that yields that rank. 4. Find T+, the sum of the ranks with +ve sign and T-, the sum of the ranks with –ve sign.

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The Test Statistic: The Wilcoxon test statistic is either T+ or T-, depending on the nature of the alternative hypothesis. When the alternative hypothesis is two-sided ( μ ≠ μ 0) either a sufficiently small value of T+ or a sufficiently small value of T- will cause us to reject H 0 : = 0. The test statistic then is T+ or T-, whichever is smaller. When H 0 :  0 is true we expect our sample to yield a large value of T+ . Therefore, when the one-sided alternative hypothesis state that the true population mean is less than the hypothesized mean ( < 0 ), a sufficiently small value of T+ will cause rejection of H 0 . and T+ is t he test statistic. When H 0 :  0 is true we expect our sample to yield a large value of T- . Therefore, for the one-sided alternative H A : > 0 , a sufficiently small value of T- will cause rejection of H 0 and T- is t he test statistic.

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Critical Value: The following are the decision rules for the three possible alternative hypothesis: (a) H A :  0 . Reject H 0 at the α level of significance if the calculated T is smaller that or equal to the tabulated T for n and preselected α /2. Alternatively, we may enter Table K with n and our calculated T is less than or equal to our Stated level of significance. If so, we may reject H 0 . (b) H A : < 0. Reject H 0 at the α level of significance if the calculated T+ is less than or equal to the tabulated T for n and preselected α . (c) H A : > 0. Reject H 0 at the α level of significance if the calculated T- is less than or equal to the tabulated T for n and preselected α .

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Example: Cardiac output(lit/min) was measured by thermodilution in a simple random sample of 15 postcardiac surgical patients in the left lateral position. The results were We wish to know if we can conclude on the basis of these data that the population mean is different from 5.05. Solution: Assumption: We assume that the requirements for the application of the Wilcoxon signed-ranks test are met. Hypothesis: H 0 :  = 5.05 H A :   5.05 and Let  = 0.05 Test Statistic: The test statistic will be T+ or T-, whichever is smaller. 4.91 4.10 6.74 7.27 7.42 7.50 6.56 4.64 5.98 3.14 3.23 5.80 6.17 5.39 5.77

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Distribution of Test Statistic: Critical values of the test statistics are given in Table K. Decision Rule: The decision rule is reject H 0 if the computed value of T is less than or equal to 25, the critical value for n=15, and /2 = .0240, the closest value to .0250 in Table K. Calculation of test statistic: The calculation of the test statistic is in the table-

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Table: Calculation of the test statistic Cardiac output di =xi-5.05 Rank of  di  Signed rank of  di  4.91 -.14 1 -1 4.10 -.95 7 -7 6.74 +1.69 10 +10 7.27 +2.22 13 +13 7.42 +2.37 14 +14 7.50 +2.45 15 +15 6.56 +1.52 9 +9 4.64 _.41 3 -3 5.98 +.93 6 +6 3.14 -1.91 12 -12 3.23 -1.82 11 -11 5.80 +.75 5 +5 6.17 +1.12 8 +8 5.39 +.34 2 +2 5.77 +.72 4 +4

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T+ = 86 , T- =34, T = 34. Statistical Decision: Since 34 is greater than 25, we are unable to reject H 0 . Conclusion: We conclude that the population mean may be 5.05 p value: From Table K we see that the p value is p = 2(.0757) = 0.1514.

The Median Test : 

The Median Test A nonparametric procedure that may be used to test the null hypothesis that two independent samples have been drawn from populations with equal medians is the median test. Example: Do urban and rural male junior high school students differ with respect to their level of mental health? Urban Rural Urban Rural 35 29 26 50 27 43 21 22 27 42 38 47 23 42 25 32 25 50 27 37 45 34 46 31 33 26 46 41

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Solution: Assumption: The assumptions are the samples are selected independently ,at random from their respective population and the populations are of the same form, differing only in location and the variable of interest is continuous . Hypothesis: H 0 :M U =M R and H A : M U  M R. M U is the median score of the sampled population of urban students and M R is the median score of the sampled population of rural students. Let  = 0.05. Test Statistic: The test statistic is X 2 as computed. Distribution of test statistic: When H 0 is true and the assumptions are met, X2 with 1 degree of freedom. Decision Rule: The decision rule is reject H 0 if the computed value of X 2 is 3.841(since =0.05)

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Urban Rural Total No. of scores above median 6 8 14 No. of scores below median 10 4 14 Total 16 12 28 Calculation of test statistic: The first step in calculating the test statistic is to compute the common median i.e. (33+34)/2 = 33.5. The result according to for each group the number of observations falling above and below the common median is The test statistic is Statistical decision: Since 2.33< 3.841, the critical value of X 2 with =.05 and 1 degree of freedom, we are unable to reject the null hypothesis on the basis of these data. Conclusion: The two samples may have been drawn from populations with equal median. p value: The p value is p>0.10 since 2.33 < 2.706. χ 2 = 28[(6)(4) – (8)(10)]2 (16)(12)(14)(14) = 2.33

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Handling Values Equal to the Median: When one or more observed values will be exactly equal to the common median and, hence, will fall neither above nor below it and if n 1 +n 2 is odd, at least one value will always be exactly equal to the median one solution is drop them from the analysis if n 1 +n 2 is large and there are only a few values that fall at the combined median or we may dichotomize the scores into those that exceed the median and those that do not, in which case the observations that equal to the median will be counted in the second category. Median Test extension: The median test extends logically to the case where it is desired to test the null hypothesis that k ≥3 samples are from populations with equal medians. For this test a 2×k contingency table may be constructed by using the frequencies that fall above and below the median computed from combined samples. If conditions as to sample size and expected frequencies are met, χ 2 may be computed and compared with the critical χ 2 with k-1 degrees of freedom.

The Mann-Whitney Test: 

The Mann-Whitney Test If for testing the desired hypothesis, there is available a procedure that makes use of more of the information inherent in the data , that procedure should be used if possible Such a nonparametric procedure that can often be used instead of the median test is called the Mann-Whitney Test. Assumption: The two sample of size n and m, respectively available for analysis have been independently and randomly drawn from their respective populations. measurement scale is at least ordinal. variable of interest is continuous. If the population differ at all, they differ only with respect to their medians.

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Hypothesis: When these assumptions are met we may test the null hypothesis that the two populations have equal medians against either of the three possible alternatives: (1) the populations don’t have equal medians(two sided test) (2) the median of population 1 is larger than the median of population 2(one sided test) (3) the median of population 1 is smaller than the median of population 2(one-sided test). Example: A researcher designed an experiment to asses the effects of prolonged inhalation of cadmium oxide. 15 laboratory animals served as experimental subjects, while 10 similar animals served as controls. The variable of interest was hemoglobin level following the experiment. If we can conclude that prolonged inhalation of cadmium oxide reduces hemoglobin level.

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Solution: Assumption: We presume that the assumptions of the Mann- Whitney test are met. Table: Hemoglobin determination (grams) for 25 laboratory animals Exposed animals (X) Unexposed animals (Y) 14.4 14.2 13.4 16.5 14.1 16.6 15.9 15.6 14.1 15.3 15.7 16.7 13.7 15.3 14.0 17.4 16.2 17.1 17.5 15.0 16.0 16.9 15.0 16.3 16.8

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The null and alternative hypothesis are as follows: H 0 : M X M Y H A :M X < M Y where M X is the median of a population of animals exposed to cadmium oxide and M Y is the median of a population of animals not exposed to the substance. Let  = 0.05. Test statistic : 1. Two samples are combined 2. All observations are ranked from smallest to largest keeping track of the sample to which each observations belongs. 3. Tied observations are assigned a rank equal to the mean of the rank position for which they are tied. The test statistic: Where n is the size of sample X and S is corresponding sum of ranks. T = S - n (n+1) 2

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Distribution Of test: Critical values from the distribution of the test statistic are given in Table L for various levels of α . Decision Rule: In general, for one-sided tests of the type illustrated here the decision rule is : Reject H 0 : M X M Y if the computed t is less than w α is the critical value of T obtained by entering Table L with n, the number of X observations, m, the number of Y observations, and α , the chosen level of significance. Calculation of test statistic: S=145, so that T = 145 - 15 (15+1) 2 = 25

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Statistical decision: when we enter Table L with n=15, m= 10, and α = 0.05, we find the critical value of w α to be 45. since 25<45, we reject H 0 . Conclusion: We conclude that M X is smaller than M Y . This leads to the conclusion that prolonged inhalation of cadmium oxide does reduce the hemoglobin level. p value: since 22<25<30, we have for this test 0.005>p>0.001.

The Kolmogorov- Smirnov Goodness-of-fit Test: 

The Kolmogorov - Smirnov Goodness-of-fit Test The Kolmogorov’s work concerned with the one-sample case as discussed and Smirnov’s work deals with the case involving two samples in which interest centers on testing the hypothesis that the distributions of the two parent populations are identical . Test Statistic: Kolmogorov- Smirnov goodness-of-fit test a comparison between some theoritical cumulative distribution function, F T (x), and a sample cumulative distribution function F S (x). The difference between the theoretical cumulative distribution function, F T (x), and the sample cumulative distribution function F S (x), is measured by the statistic D which is the greatest vertical distance between F S (x) and F T (x)

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When a two sided test is appropriate, that is when the hypothesis are H 0 : F(x) =F T (x) for all x from -  to  H A : F(x)  F T (x) for at least one x the statistic is D = sup  F s (x) - F t (x)  x which is read, “ d equals the supremum (greatest) over all x, of the absolute value of the difference F s (x) - F t (x).” The null hypothesis is rejected at the  level of significance if the computed value of D exceed the for 1-(two sided) and the sample size n. Assumption: The assumptions are 1.Sample is a random sample 2. Hypothesized distribution F T (x) is continuous .

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When values of D are based on a discrete theoretical distribution, the test is conservative. When the test is used with discrete data, then, the true probability of committing a type-I error is at most equal to , the stated level of significance. Example: Fasting, blood glucose determinations made on 36 nonobese, apparently healthy, adult males in table. We have to test if we may conclude that these data are not from a normally distributed population with a mean of 80 and a standard deviation of 6. Solution: Assumption: We assume that the sample available is a simple random sample from a continuous population distribution. Hypothesis: H 0 : F(x) =F T (x) for all x from -  to  H A : F(x)  F T (x) for at least one x and Let =0.05.

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Table: Fasting blood glucose values(mg/100ml) for 36 Nonobese, apparently healthy, adult males Distribution of test statistic: Critical value of the test statistic for selected values of α are given in Table M. Decision rule: Reject H 0 if the computed value of D exceeds 0.221, the critical value of D for n=36 and =0.05. Calculation of test statistic: First step is to compute values of F S (x). Each value of F S (x) is obtained by dividing the corresponding cumulative frequency by the sample size like first value of F S (x) = 2/36= 0.0556 Value of F T (x) can be obtained by converting each observed value of x to a value of the standard normal variable, z. Then find the area between -  and z. 75 92 80 80 84 72 84 77 81 77 75 81 80 92 72 77 78 76 77 86 77 92 80 78 68 78 92 68 80 81 87 76 80 87 77 86

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X F C.F F S (x) 68 72 75 76 77 78 80 81 84 86 87 92 2 2 2 2 6 3 6 3 2 2 2 4 2 4 6 8 14 17 23 26 28 30 32 36 .0556 .1111 .1667 .2222 .3889 .4722 .6389 .7222 .7778 .8333 .8889 1.000 Table: Value of FS(x) Table: Values of F T (x) Statistical decision: Reference to Table M reveals that a computed D of 0.1547 is not significant at any reasonable level. So we are not willing to reject H 0 . Conclusion: The sample may have come from the specified distribution. p value: Since we have a two-sided test and since 0.1547<0.174 we have p>0.20 X Z=(x-80)/6 F T (x) 68 72 75 76 77 78 80 81 84 86 87 92 -2.00 -1.33 -.83 -.67 -.50 -.33 .00 .17 .67 1.00 1.17 2.00 .0228 .0918 .2033 .2514 .3085 .3707 .5000 .5675 .7486 .8413 .8790 .9772

The Kruskal-Wallis one-way analysis of variance by ranks: 

The Kruskal -Wallis one-way analysis of variance by ranks When the populations from which the samples are drawn are not normally distributed with equal variances and when the data for analysis consists only of ranks, a nonparametric alternative to the one way analysis of variance may be used to test the hypothesis of equal location parameter. The best known of these procedures is the Kruskal-Wallis one way analysis of variance by ranks. The application of the test involves the steps The n 1 , n 2 ,…..,n k observations from the k samples are combined in to a single series of size n and arranged in order of magnitude from smaller to largest. Then the observations are then replaced by ranks from 1 to n from smallest to the largest value.

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The ranks assigned to observations in each of the k groups are added separately to give k rank sums. The test statistic Is computed. k = the number of samples n j = the number of observation in the jth sample n= the number of observations in all samples combined Rj= the sum of the ranks in the jth sample. When there are 3 samples and five or fewer observations in each sample, the significance of the computed H is determined by consulting Table. When there are more than five observations in one or more of the samples, H is compared with tabulated values of  2 with k-1 degrees of freedom. H= 12 n(n+1)  R j 2 n j -3(n+1)

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Sample I II III 17 20 40 31 35 8 7 9 8 2 5 4 3 Example: The effects of two drugs on reaction time to a certain stimulus were studied in three samples of experimental animals. Sample-III served as a control while the animals in sample-I were treated with drug A and those in sample II were treated with drug B prior to the application of the stimulus. Table shows the reaction times in seconds of the animals. Can we conclude that the three populations represented by the three samples differ with respect to reaction time? We can so conclude if we can reject the null hypothesis that the three populations do not differ in their reaction times. Table:1 Reaction time in seconds of 13 experimental anima l

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Solution: Assumption: The samples are independent random samples from their respective populations. The measurement scale employed is at least ordinal. The distribution of the values in the sampled population are identical except for the possibility that one or more of the populations are composed of values that tend to be larger than the of the other populations. Hypothesis: H 0 : The population centers are equal H A : At least one of the populations tends to exhibit larger values than at least one of the other populations. Let α =0.01 Distribution of test statistic : critical values of H for various sample sizes and α levels are given in Table N. Decision rule: The null hypothesis will be rejected if the computed value of H is very large that the probability of obtaining a value that large or larger when H0 is true or equal or less than the chosen significance level, α .

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Calculation of test statistic: When the three samples are combined in to a single series and ranked , the Table of ranks shown in Table 2. From the Table data and equation we obtain Sample I II III 9 10 13 11 12 R 1 =55 6.5 5 8 6.5 R 2 =26 1 4 3 2 R 3 =10 Table:2 The data of Table 1 replaced by Ranks H= 12 13(13+1) [ (55)2 5 (26)2 4 (10)2 4 + + ] - 3(3+1) = 10.68

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Statistical decision: Table N shows that when the n j are 5, 4 and 4, the probability of obtaining a value of H ≥ 10.68 is less than 0.009. The null hypothesis can be rejected at the .01 level of significance. Conclusion: We conclude that there is a difference in the average reaction time among the three populations. p value: p < 0.009.