Nonparametric Methods Marc H. Mehlman marcmehlman@yahoo.com - - PowerPoint PPT Presentation

Marc Mehlman

Nonparametric Methods

Marc H. Mehlman

marcmehlman@yahoo.com

University of New Haven

Nonparametric Methods, or Distribution–Free Methods is for testing from a population without knowing anything about the population’s distribution.

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Marc Mehlman

Table of Contents

1

Sign Test for Median of Ordinal Data

2

Wilcoxon Signed–Ranks Test for Matched Pairs

3

Wilcoxon Rank–Sum Test for Two Independent Samples

4

Kruskal–Wallis Test for Multiple Independent Samples

5

Spearman Rank Correlation Test for Bivariate Data

6

Chapter #12 R Assignment

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Advantages and Disadvantages to Nonparametric Methods

Advantages of Nonparametric Methods Less assumptions so nonparametric methods can be used in more situations. Computations are often simple and easier to understand. Less sensitive to outliers that are actually incorrect observations. Disadvantages of Nonparametric Methods Information is wasted when numerical data is converted into rank data so conclusions tend to be weaker. Underestimates the effect of correct outliers.

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Definition Let T1 and T2 be two tests statistics of H0 versus H1. Let n1 and n2 be the minimum sample sizes to achieve a test of power β and size α using T1 or T2 respectively. The Pitman asymptotic relative efficiency, ARE(T1, T2), of test statistic, T1, to test statistic, T2, is the limit, if it exists, of the ratios n2/n1 as n1 ↑ ∞. The ARE of a parametric Test to a nonparametric test is generally less than one because one must give up some efficiency in return for less assumptions.

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Sign Test for Median of Ordinal Data

Sign Test for Median of Ordinal Data

Sign Test for Median of Ordinal Data

A nonparametric version of the one sample t–test, with ARE = 2

π = 0.64.

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Sign Test for Median of Ordinal Data

Sign Test

Situations for using the Sign Test Test of Magnitude between Matched Pairs, (x1, y1), (x2, y2), · · · (xn.yn). For any (xj, yj)’s such that xj = yj, delete (xj, yj) from the sample and readjust n, the sample size. Let H0 : P(X > Y ) = P(Y > X) dj

def

= yj − xj x def = min(# of positive dj’s, # of negative dj’s). Test for Median of a x1, x2, · · · , xn. Fixed M. For andy xj’s such that xj = M, delete xj from the sample and readjust n, the sample size. Let H0 : median = M dj

def

= xj − M x def = min(# of positive dj’s, # of negative dj’s).

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Sign Test for Median of Ordinal Data

Sign Test

Situations for using the Sign Test Test of Distribution of Nominal Data where x1, · · · , xn. Let H0 : P(X = first value) = P(X = second value) dj

def

=

• 1

if xj = first value if xj = second value x def = min(

n

• j=1

dj, 1 −

n

• j=1

dj) = min (# first values, # second values) . Nominal data need not be numbers. The two “Situations for using the Sign Test” on the previous slide are just special cases of the situation above were what was being counted was +’s and −’s.

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Sign Test for Median of Ordinal Data

Sign Test

Theorem (Sign Test for x ≤ 25) The test statistic is x and the critical values are found in Table A–7. Theorem (Sign Test for x > 25) The test statistic is z = (x+0.5)−n/2

√n/2

which is approximately standard normal.

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Sign Test for Median of Ordinal Data Marc Mehlman (University of New Haven) Nonparametric Methods 9 / 44

Marc Mehlman

Sign Test for Median of Ordinal Data

Sign Test

22

Sign Test Example

• You’re a marketing

analyst for Chefs-R-Us. You’ve asked 8 people to rate a new ravioli on a 5-point Likert scale 1 = terrible to 5 = excellent The ratings are: 2 4 1 2 1 1 2 1 At the .05 level, is there evidence that the median rating is at least 3?

H0 : median = 3 versus HA : median < 3 Table A–7 says to reject at 0.05 significance level since x = 1.

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Sign Test for Median of Ordinal Data

Sign Test

Example (example solution using R)

> install.packages("BSDA") > library(BSDA) > dat=c(2,4,1,2,1,1,2,1) > SIGN.test(dat,md=3,alternative="less") One-sample Sign-Test data: dat s = 1, p-value = 0.03516 alternative hypothesis: true median is less than 3 95 percent confidence interval:

• Inf

2 sample estimates: median of x 1.5 Conf.Level L.E.pt U.E.pt Lower Achieved CI 0.8555

• Inf

2 Interpolated CI 0.9500

• Inf

2 Upper Achieved CI 0.9648

• Inf

2

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Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test for Matched Pairs

A nonparametric version of the matched pair one sample t–test, with ARE = 3

π = 0.955.

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Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test

Given (x1, y1), (x2, y2), · · · , (xn, yn), define dj

def

= yj − xy. Discard all bivariate data with dj = 0 (and readjust the sample size n). Next rank the |dj|’s from smallest to largest. For ties, reassign them the average of their would be ranks. (For ties, there is better way than averaging the ranks, but it is more complicated.) Definition The Wilcoxon Signed–Rank Statistics are t+(ω) def = sum of ranks of the positive dj’s, t−(ω) def = sum of ranks of the negative dj’s and t def = min(t−, t+). Frank Wilcoxon (1892–1965), American Since the Wilcoxon Signed-Ranks Test takes into account both the signs and the ranks of the dj’s, it obtains a higher ARE.

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Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test

Example Let (7.4, 5.5), (5.5, 3.2), (5.6, 6.0), (7.9, 4.5), (6.7, 6.7), (6.5, 3.0), (7.8, 4.8), (3.5, 6.3) be a matched pair random sample. Calculate the Wilcoxon Signed–Rank Statistics. Solution: After eliminating d5 = 0 and reducing the sample size to 7, d1 = 1.9, d2 = 2.3, d3 = −0.4, d4 = 3.4, d5 = 3.5, d6 = 3.0, d7 = −2.8 so one has rank 1 2 3 4 5 6 7 magnitude |d3| |d1| |d2| |d7| |d6| |d4| |d5| p/m − + + − + + + . Thus t+ = 2 + 3 + 5 + 6 + 7 = 23 t− = 1 + 4 = 5, t = 5.

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Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test

Theorem (Wilcoxon Signed–Rank Test) Given a random bivariate sample, (x1, y1), (x2, y2), · · · , (xn, yn), such that the distribution of the differences, dj

def

= yj − xj, is approximately symmetric about zero and has a median of M, consider H0 : M = 0 versus HA : not H0. Then a conservative test of H0 versus HA is is to use    t if n ≤ 30 z =

t−n(n+1)/4

√

n(n+1)(2n+1)/24

if n > 30 as the test statistic. Use Table A–8 if n ≤ 30 and assume that Z ∼ N(0, 1) if n > 30.

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Wilcoxon Signed–Ranks Test for Matched Pairs Marc Mehlman (University of New Haven) Nonparametric Methods 16 / 44

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Wilcoxon Signed–Ranks Test for Matched Pairs

Wilcoxon Signed–Ranks Test

Example (continued) Using data from the above Example, find p–value of H0 : population differences have a median of 0 versus HA : not H0. Solution: Here n = 7. The approximate p–value is smallest α such that t = 5 < tα. From A–8 one has the p–value > 0.1. Using R:

> xdat=c(7.4,5.5,5.6,7.9,6.5,7.8,3.5) > ydat=c(5.5,3.2,6.0,4.5,3.0,4.8,6.3) > wilcox.test(xdat,ydat,paired=TRUE) Wilcoxon signed rank test data: xdat and yydat V = 23, p-value = 0.1563 alternative hypothesis: true location shift is not equal to 0

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

A nonparametric version of the two sample t–test, with ARE of = 3

π = 0.955.

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

This test is equivalent to the Mann–Whitney Test. With a high ARE it is very efficient. Assume one has a random sample from two different populations, namely x1, · · · , xn, and an independent population y1, · · · , ym. One wishes to test H0 : the two populations have equal medians versus HA : not H0.

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

To obtain test statistics, combine the 2 random samples and order them according to increasing size. One frequently assumes the distributions, FX(·) and FY (·), to be continuous to prevent the possibility of ties when

• rdering by size (if there are ties average their ranks).

Example (Drug Example) Consider

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Example (Five x’s and Four y’s Example)

Let a random sample of X be 1.2 0.8 0.39 0.7 −0.1 and a random sample of Y be −0.2 1.3 0.5 0.6 . Then the combined sample is rank 1 2 3 4 5 6 7 8 9 value −0.2 −0.1 0.39 0.5 0.6 0.7 0.8 1.2 1.3 x/y y x x y y x x x y .

Note that in general, there exists n+m

n

• possible rankings and under H0

they are all equally likely.

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Definition After creating the combined, ranked sample, let W n,m

x def

= sum of the ranks of the x’s and W n,m

y def

= sum of the ranks of the y’s. The Wilcoxon statistics are W n,m

x

and W n,m

y

. Example (Five x’s and Four y’s Example, continued) w5,4

x

= 2 + 3 + 6 + 7 + 8 = 26, w5,4

y

= 1 + 4 + 5 + 9 = 19,

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Theorem One has µW n,m

x

= n(n + m + 1) 2 and σW n,m

x

=

• nm(n + m + 1)

12 . Theorem (Wilcoxon Rank–Sum Test) To test H0 : the two populations have equal medians versus HA : not H0. let Z def = W n,m

x

− µW n,m

x

σW n,m

x

be the test statistic. Then if n, m > 10 then Z is approximately N(0, 1).

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Example (Drug Example, continued) µW 12,12

x

= 12(12 + 12 + 1) 2 = 150 σW 12,12

x

=

• 12 ∗ 12(12 + 12 + 1)

12 = 10 √ 3 Z = 119.5 − 150 10 √ 3 = −1.760918. Using a two–sided test, p–value = 2P(Z ≤ −1.760918) = 0.078.

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Wilcoxon Rank–Sum Test for Two Independent Samples

Wilcoxon Rank–Sum Test for Two Independent Samples

Example (Drug Example continued with R)

> xdat=c(11,15, 9, 4,34,17,18,14,12,13,26,31) > ydat=c(34,31,35,29,28,12,18,30,14,22,10,29) > sum(rank(c(xdat,ydat))[1:12]) [1] 119.5 > wilcox.test(xdat,ydat,correct=FALSE) Wilcoxon rank sum test data: xdat and ydat W = 41.5, p-value = 0.07786 alternative hypothesis: true location shift is not equal to 0 Warning message: In wilcox.test.default(xdat, ydat, correct = FALSE) : cannot compute exact p-value with ties

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test for Multiple Independent Samples

Nonparametric version of the ANOVA F–test, with ARE of = 3

π = 0.955.

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test

This nonparametric test is for a one–factor ANOVA design without any assumptions of normal error. Furthermore, there is also no assumption that the one–way layout is balanced. Definition The Kruskal–Wallis Test (also called the H test) is a nonparametric ANOVA-like test that uses the ranks for sample data from k independent populations (k ≥ 2). One test H0 : the populations have the same median versus HA : not H0. The test is similar to the Wilcoxon Rank–Sum Test in that it takes rankings into consideration. In fact, if k = 2, the Kruskal–Wallis test is equivalent to the Wilcoxon Rank–Sum Test!

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test

Conditions for Kruskal–Wallis Test

1 There are k independent random samples where k ≥ 2. 2 The sample size of the jth sample is nj ≥ 5. 3 The total number of observations is N = k

j=1 nj.

Procedure for Finding Value of Test Statistic H

1 Combine all the samples into one big sample, sort for lowest to

highest, and assign a rank to each sample value. In the cases of ties, assign to each observation the average of the ranks involved.

2 For each sample, let Rj

def

= the sum of ranks for sample j.

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test

Definition The Kruskal–Wallis Test Statistic is H def = 12 N(N + 1)

k

• j=1

R2

j

nj − 3 (N + 1) . Theorem (Kruskal–Wallis) For testing H0 : the k populations have equal medians versus HA : not H0. use a right hand test on the test statistic H. The test statistic is approximately χ2(k − 1).

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test

From Nonparametric Statistical Methods, by Myles Hollander, Douglas

• A. Wolfe and Eric Chicken, third edition, 2014, John Wiley & Sons.

H′ def = H 1 − g

j=1(t3 j − tj)/(N3 − N)

, where g def = number of tied groups tj

def

= size of tied group j One can even use this formula for ”groups of one”. If there are no ties,

• ne just has H. When there are ties it is best to use H′ instead of H in the

Kruskal–Wallis Test. Our textbook, Biostatistics for Biology and Health Sciences, by Marc

• M. Triola and Mario F. Triola does not define or use H′ when there are

ties, but R does.

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Kruskal–Wallis Test for Multiple Independent Samples

Kruskal–Wallis Test

Example The built–in R dataset “airquality” contains daily air quality measurements in New York, May to September 1973. One wants to test if the medians of the five monthly ozone counts are equal. H0 : the 5 month’s daily ozone readings have equal medians versus HA : not H0. Using R:

> kruskal.test(Ozone~Month, data=airquality) Kruskal-Wallis rank sum test data: Ozone by Month Kruskal-Wallis chi-squared = 29.267, df = 4, p-value = 6.901e-06

With such a low p–value, one rejects H0.

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Spearman Rank Correlation Test for Bivariate Data

Spearman Rank Correlation Test for Bivariate Data

Spearman Rank Correlation Test for Bivariate Data

A nonparametric version of the Pearson Correlation test, with an ARE of (3/π)2 = 0.91.

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Given a paired random sample, (X1, Y1), · · · , (Xn, Yn), from a continuous joint distribution, FXY (·, ·), one desires to test H0 : There is no monotonic association between the two variables versus HA : not H0. A Monotonic association between X and Y means that there is either a positive association between X and Y or there is a positive association between X and Y .

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Suppose the paired random sample, (x1, y1), · · · , (xn, yn) has been transformed into paired random sample of ranks, (u1, v1), · · · , (un, vn), (if there are ties average their ranks) where uj

def

= the rank order of the xj among all the xi’s and vj

def

= the rank order of the yj among all the yi’s. Definition The Spearman’s Rank Correlation Coefficient, rs, of a matched paired random sample, (x1, y1), · · · , (xn, yn) is the Pearson Product–Moment Correlation Coefficient, rp, of the matched paired random sample, (u1, v1), · · · , (un, vn), formed from the ranks of the original matched paired random sample. Charles Spearman (1863–1945), English psychologist

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Advantages of Spearman over Pearson

1 One can use Spearman’s Rank Correlation Coefficient for matched

pair ordinal data.

2 Spearman’s Rank Correlation Coefficient can detect monotonic

non–linear tendencies between two variables. Disadvantages

1 Less efficient than Pearson’s Rank Correlation Coefficient. Marc Mehlman (University of New Haven) Nonparametric Methods 35 / 44

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Theorem (Spearman Rank Correlation Test) Given a paired random sample, (X1, Y1), · · · , (Xn, Yn) let H0 : There is no monotonic association between the two variables versus HA : not H0. Then rs =          1 −

6 n

j=1 d2 j

n(n2−1)

if there are no ties

n n

j=1 xj yj −

n

j=1 xj

n

j=1 yj

• n

n

j=1 x2 j

• −

n

j=1 xj

2 n n

j=1 y2 j

• −

n

j=1 yj

2

if there are ties be the test statistic. For critical values, use Table A-9 if n ≤ 30. If n > 30. the critical values are r∗

s =

±z∗ √n − 1 where ±z∗ are the corresponding critical values from N(0, 1).

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Spearman Rank Correlation Test for Bivariate Data Marc Mehlman (University of New Haven) Nonparametric Methods 37 / 44

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Example (Job Salary and Stress Rankings) Job Salary Rank Stress Rank dj d2

j

Stockbroker 2 2 Zoologist 6 7 1 1 Electrical Engineer 3 6 3 9 School Principal 5 4

• 1

1 Hotel Manager 7 5

• 2

4 Bank Officer 10 8

• 2

4 Occupational Safety Inspector 9 9 Home Economist 8 10 2 4 Psychologist 4 3

• 1

1 Airline Pilot 1 1 TOTAL 24 rs = 1 − 6 n

j=1 d2 j

n(n2 − 1) = 1 − 6(24) 10(102 − 1) = 0.855. Referring to Table A–9 for n = 10, one sees that ±0.648 corresponds to a significance level of 0.05. Since 0.855 > 0.648, one rejects H0 (no monotonic association between stress and salary) with a significance level of 0.05.

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Example (Job Salary and Stress Rankings, using R)

> xdat=c(2,6,3,5,7,10,9,8,4,1) > ydat=c(2,7,6,4,5,8,9,10,3,1) > cor.test(xdat,ydat,method="spearman") Spearman’s rank correlation rho data: xdat and ydat S = 24, p-value = 0.003505 alternative hypothesis: true rho is not equal to 0 sample estimates: rho 0.8545455

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Spearman Rank Correlation Test for Bivariate Data

Rank Correlation

Example (Air Quality)

> cor(airquality$Wind,airquality$Temp) [1] -0.4579879 > cor(airquality$Wind,airquality$Temp,method="spearman") [1] -0.4465408 > cor.test(airquality$Wind,airquality$Temp,method="spearman") Spearman’s rank correlation rho data: airquality$Wind and airquality$Temp S = 863450, p-value = 7.229e-09 alternative hypothesis: true rho is not equal to 0 sample estimates: rho

• 0.4465408

Warning message: In cor.test.default(airquality$Wind, airquality$Temp, method = "spearman") : Cannot compute exact p-value with ties

With such a small p–value, one rejects the hypothesis that there is no correlation between temperature and wind.

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Chapter #12 R Assignment

Chapter #12 R Assignment

Chapter #12 R Assignment

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Chapter #12 R Assignment 1 Compute the p-value for a two sided sign–test for the null hypothesis

that the population median for ’x’ is 6.5. The alternative hypothesis is that the median is not 6.5.

x = c(7.8, 6.6, 6.5, 7.4, 7.3, 7., 6.4, 7.1, 6.7, 7.6, 6.8) 2 In order to investigate whether adults report verbally presented

material more accurately from the right than from their left ear, a dichotic listening task was carried out.

LEar=c(15,29,10,31,27,24,26,29,30,32,20,5) REar=c(32,30,8,32,20,32,27,30,32,32,30,32)

Use the Wilcoxon Signed–Rank Test for Matched Pairs to find the p–value of H0 : the left and right ear data have the same median, versus HA : not H0.

3 Consider the following sets of data on the latent heat of the fusion of

ice (cal/gm). Enter the following three lines into R:

AA=c(79.98, 80.04, 80.02, 80.04, 80.03, 80.03, 80.04, 79.97, 80.05, 80.03, 80.02, 80.00, 80.02) BB=c(80.02, 79.94, 79.98, 79.97, 79.97, 80.03, 79.95, 79.97)

Using the Wilcoxon Rank–Sum Test, test to see if the populations AA and BB were sampled from have equal medians. Give the p–value of the test.

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Chapter #12 R Assignment 4 Using the R built in dataset, iris, do a Krushal–Wallis Test of

Sepal.Width according to Species.

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Chapter #12 R Assignment 5 Enter in R the data from the following frigatebird example (paste the

following block into R):

Input = (" Volume Pitch 1760 529 2040 566 2440 473 2550 461 2730 465 2740 532 3010 484 3080 527 3370 488 3740 485 4910 478 5090 434 5090 468 5380 449 5850 425 6730 389 6990 421 7960 416 ") Data = read.table(textConnection(Input),header=TRUE)

Using the Spearman Rank Correlation test, find the p–value of H0 there isn’t a monotonic association between Volume and Pitch versus HA : not H0.

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