How to Read a Wilcoxon Signed Rank Test

Inferential Statistics Iii: Nonparametric Hypothesis Testing

Andrew P. Rex , Robert J. Eckersley , in Statistics for Biomedical Engineers and Scientists, 2019

6.3 Wilcoxon Signed-Rank Examination

Although the sign test tin can be used to test both i-sample and ii-sample paired information, the Wilcoxon signed-rank test is more than powerful than the sign exam for these tasks because it makes use of the magnitudes of the differences rather than just their signs.

The Wilcoxon signed rank examination was developed past Frank Wilcoxon ^ane in 1945. Nosotros will illustrate its use using ii-sample paired data. Following our checklist from Section v.2, the basic idea behind the Wilcoxon signed-rank exam is:

■

Class nothing and alternative hypotheses and choose a degree of conviction. The null hypothesis is that the median of the population of differences between the paired data is nix. The alternative hypothesis is that it is not.

■

Compute the test statistic. We practice this past first computing the differences between the paired data samples; then nosotros rank the differences according to magnitude only, that is, without regard to their sign; next, nosotros sum the ranks of the positive and negative differences; finally, nosotros choice the minimum of the sums as our test statistic.

■

Compare the test statistic to a disquisitional value. If the test statistic is less than the critical value, and so we pass up the cipher hypothesis.

Again, we volition illustrate this procedure with an example. We volition revisit the case nosotros introduced in Section 5.vii.1 for testing claret pressure data from patients suffering from hypertension. Two sets of data accept been gathered: before handling with a new drug and later on treatment. The researchers now doubt that their data are normally distributed then wish to try a nonparametric hypothesis test.

Our null hypothesis is that the median of the population of differences between the ii sets of blood pressure level information is nothing. The alternative hypothesis is that it is not zippo. Denoting the before treatment data as the control information and the after treatment information as the exam information, our two samples are:

Control:	175.4	188.iii	147.4	178.6	173.2	156.ix	165.seven	173.4
Test:	152.3	159.7	155.seven	166.2	149.1	162.3	163.5	146.0

To compute our test statistic, we beginning compute the differences between the paired data, that is, test minus control. These are (−23.i −28.half-dozen 8.3 −12.4 −24.ane five.4 −2.2 −27.iv).

Next, we rank these differences by magnitude and assign ranks to each departure value. This process is illustrated in Table half dozen.two. Note that the signs of the deviation values are ignored when ranking them, that is, they are ranked by magnitude only, just we recollect the signs.

Table 6.two. Wilcoxon signed-rank exam: differences between 2 paired samples ranked by magnitude, with corresponding rankings.

Differences	−23.ane	−28.half dozen	8.3	−12.4	−24.one	5.four	−ii.2	−27.4
Ranked differences	−2.2	5.4	8.three	−12.4	−23.one	−24.one	−27.4	−28.6
Ranks	one	2	3	iv	5	half dozen	7	8

Side by side, nosotros sum the rankings for the positive and negative differences (i.east. using the remembered signs). Referring to the ranks in Table half dozen.two, the sum of ranks for the positive differences is

$T_{+}=2+3=\mathrm{five}$ ,

and the sum of ranks for the negative differences is

$T_{-}=1+4+5+6+7+8=31$ .

The minimum of these rank sums is $T_{+}$ , which is v, so this is our test statistic.

Now we look up a critical value in Tabular array A.3 (meet Appendix): for $n=8$ and a significance level of 0.05 in a 2-tailed test, we have a critical value of three. We compare our calculated and critical values: if the calculated value is less than the critical value, then we pass up the zero hypothesis. Since five is not less than 3, we cannot reject the null hypothesis in this case, and so we cannot conclude that at that place is a pregnant difference between the two samples. Therefore, based on this exam, we take not demonstrated that the new drug reduces blood pressure. Note the divergence hither with the parametric test on the aforementioned data (see Section 5.7.1): in that case, we were able to testify statistical significance (on the assumption that the distributions were normal). Generally, significance is more probable to exist shown with parametric tests than with nonparametric ones.

As we pointed out earlier, it is possible to use both the sign test or the Wilcoxon signed-rank test in either the i-sample case or the two-sample paired data instance. To apply the Wilcoxon signed-rank test in the i-sample case, we simply compute the differences by subtracting the expected median value being tested against from the sample. However, although both tests are applicable in both situations, the Wilcoxon signed-rank test is the preferred method equally it makes use of the magnitudes of the differences rather than simply the signs. It is important to think that the Wilcoxon signed-rank exam does make a stronger supposition about the data sample(s) being tested: it requires that the distribution of the differences is symmetric. If this tin exist shown to be not the case or if nosotros have good reason to dubiety that information technology is the case, so the sign test should be used.

The Intuition. Wilcoxon Signed-Rank Test

We have seen how the Wilcoxon signed-rank test tin exist used to exam hypotheses almost data that are not usually distributed, but why does it piece of work like this?

The best manner to understand the process underlying this nonparametric test is to think of it equally converting the problem into one that tin be addressed using a parametric arroyo. As noted before, the Wilcoxon signed-rank test makes the assumption that the distribution of differences between the two samples is symmetric. This assumption is important every bit it allows usa to convert the original nonparametric problem into one that can be addressed past a parametric examination. Note that the assumption is not saying that the distributions of the ii samples are symmetric – we are but talking about the differences computed by subtracting the paired values (or the values from the expected median in the i-sample case).

Why do we demand to make this assumption? An case of a distribution of differences is illustrated in Fig. 6.iiiA. If this is symmetric and the null hypothesis is true (i.e. the median of this distribution is nix), then at that place volition be the aforementioned number of difference values to a higher place and beneath zilch (this is the definition of a median). Therefore, the sums of the positive and negative ranks of these differences volition be the same, that is, $T_{+}=T_{-}$ . What would be the expected value of these rank sums?

Figure half dozen.iii. The intuition behind the Wilcoxon signed-rank test: (A) shows the (symmetric) distribution of differences between paired values from the ii samples – the exam statistics T ₊ or T ₋ will be equal if the null hypothesis is truthful; (B) shows the distribution of rank sums, which is normal according to the primal limit theorem – the computed examination statistic is not beyond the critical value for the 0.05 significance level, so nosotros cannot reject the null hypothesis.

We can compute this using knowledge of the properties of an arithmetics serial. The ranks for a sample of n numbers are simply the arithmetics series $\mathrm{i}+2+\dots +n$ . The sum of this series is $\frac{n(\mathrm{north}+\mathrm{one})}{2}$ , so we would expect the sums of either the positive or negative ranks to be $\frac{n(n+1)}{4}$ (i.east. half of the overall sum) if the aught hypothesis were truthful. This value $\frac{\mathrm{northward}(\mathrm{northward}+\mathrm{i})}{4}$ is the mean of the distribution of rank sums. Importantly, if the sample size n is large enough, then this distribution will be normal according to the fundamental limit theorem. The central limit theorem as nosotros described information technology in Department iv.v referred to the sample hateful, but the same principle applies to the sum (which is simply the mean multiplied by the sample size). Again, using noesis of the properties of arithmetic series and some basic algebra, the standard error of the (normal) distribution of rank sums can be calculated every bit $\sqrt{\frac{\mathrm{due\; north}(\mathrm{north}+\mathrm{i})(2n+\mathrm{ane})}{24}}$ . This is illustrated in Fig. 6.3B.

Now we have converted our nonparametric problem into a parametric one, and we can go on in a similar manner to our procedure for the Student's t-exam (run across Section 5.5). We compute our actual sum of ranks (i.e. from our data – for our blood pressure example, $T_{+}$ was v) and compare this to the (now known) population distribution of rank sums. For our example, nosotros take a population distribution of rank sums that has the following hateful and standard error:

$\begin{array}{ccc}\hfill \mu & \hfill =\hfill & \frac{n(\mathrm{northward}+1)}{\mathrm{iv}}=\frac{8(8+1)}{4}=18,\hfill \\ \hfill \text{s.e.}& \hfill =\hfill & \sqrt{\frac{\mathrm{north}(\mathrm{northward}+1)(2\mathrm{northward}+1)}{24}}=\sqrt{\frac{\mathrm{eight}(8+1)(\mathrm{two}\times \mathrm{eight}+\mathrm{i})}{24}}=\mathrm{vii.14}.\hfill \end{array}$

Following the procedure outlined in Section 5.5, nosotros compute our examination statistic using Eq. (5.1):

$t=\frac{\overline{x}-\mu }{s/\sqrt{n}}=\frac{5-18}{\mathrm{vii.xiv}}=-\mathrm{one.82}.$

This tells us that the rank sum we got from our data is 1.82 standard errors away from the expected rank sum. How likely is it that we would go this figure (or a more farthermost ane) if the cypher hypothesis were true?

For a one-sample Pupil's t-test with a sample size of 8 (i.east. 7 degrees of freedom), the disquisitional value from Table A.i for 0.05 significance is ii.365 (this is also shown in Fig. half-dozen.3B). The magnitude of our test statistic (1.82) is not greater than the critical value (2.365), and then we cannot turn down the null hypothesis. This is the same result as we got using the simplified process outlined in a higher place. Notation that the critical values in Table A.three are but precomputed values of the rank sums for the critical values of t-tests for different significance levels and sample sizes.

Activity 6.2

A new drug has been developed that is intended to lower the cholesterol level of patients who take high cholesterol. Information have been gathered from 9 patients who had high cholesterol. The information consist of their cholesterol levels (in mmol/L) before and after taking the drug. The data are shown in the table.

O6.B

Cholesterol levels (mmol/L)
Earlier	6.four	4.2	three.8	3.vi	4.1	3.7	5.0	4.four	four.vii
Afterwards	3.4	2.6	3.1	iii.iii	3.3	3.three	2.half dozen	4.3	4.ix

There is good reason to suspect that the data are non commonly distributed. Perform a Wilcoxon signed-rank test to determine if at that place was a difference in cholesterol level every bit a result of taking the drug. Clearly country what your hypotheses are, testify all working, and use a 95% degree of confidence in your examination.

Read full affiliate

URL:

https://www.sciencedirect.com/science/commodity/pii/B9780081029398000153

Nonparametric Methods

Rudolf J. Freund , ... Donna L. Mohr , in Statistical Methods (3rd Edition), 2010

fourteen.ii One Sample

In Department 4.5 we considered an alternative approach to analyzing some income data that had a single extreme observation. This approach was based on the fact that the median is not affected by extreme observations. Call back that in this example we converted the income values into either a "success" (if above the specified median) or a "failure" (if below), and the then-called sign examination was based on the proportion of successes. In other words, the test was performed on a set of information that were converted from the ratio to the nominal scale.

Of form the conversion of the variable from a ratio to a nominal scale with only two values implies a loss of information; hence, the resulting test is likely to accept less power. However, converting a nominal variable to ranks preserves more of the information and thus a test based on ranks should provide more power. One such test is known as the Wilcoxon signed rank test.

The Wilcoxon signed rank test is used to test that a distribution is symmetric near some hypothesized value, which is equivalent to the test for location. We illustrate with a test of a hypothesized median, which is performed every bit follows:

ane.

Rank the magnitudes (absolute values) of the deviations of the observed values from the hypothesized median, adjusting for ties if they exist.

2.

Assign to each rank the sign ( $+$ or $-$ ) of the deviation (thus, the name "signed rank").

3.

Compute the sum of positive ranks, $T\left(+\right)$ , or negative ranks, $T\left(-\right)$ , the choice depending on which is easier to calculate. The sum of $T\left(+\right)$ and $T\left(-\right)$ is $n\left(\mathrm{north}+\mathrm{one}\right)∕2$ , then either can exist calculated from the other.

4.

Choose the smaller of $T\left(+\right)$ and $T\left(-\right)$ , and call this $T$ .

5.

Since the exam statistic is the minimum of $T\left(+\right)$ and $T\left(-\right)$ , the disquisitional region consists of the left tail of the distribution, containing a probability of at nigh $\alpha ∕2$ . For small samples $\left(\mathrm{north}\le 50\right)$ , critical values are institute in Appendix Table A.9. If $n$ is large, the sampling distribution of $T$ is approximately normal with

$\begin{array}{c}\mu =n(\mathrm{north}+1)∕4,\text{and}\\ {\sigma }^2=n(n+1)(\mathrm{two}n+\mathrm{ane})∕24,\end{array}$

which can be used to compute a $z$ statistic for the hypothesis test.

Example 14.2

Case iv.7, particularly the data in Table iv.5, concerned a test for the mean family unit income of a neighborhood whose results were unduly influenced by an extreme outlier. A exam for the median was used to overcome the influence of that observation. Nosotros now apply that example to illustrate the Wilcoxon signed rank test. The hypothesis of interest is

$H_0:\text{the\hspace{0.17em}distribution\hspace{0.17em}on\hspace{0.17em}incomes\hspace{0.17em}is\hspace{0.17em}symmetric\hspace{0.17em}about}13.0,$

with a ii-tailed alternative,

$H_{\mathrm{one}}:\text{the\hspace{0.17em}distribution\hspace{0.17em}is\hspace{0.17em}symmetric\hspace{0.17em}most\hspace{0.17em}some\hspace{0.17em}other\hspace{0.17em}value}.$

Solution

The deviations of the observed values from 13.0 (the specified $H_0$ value) are given in Table 14.2 in the column labeled "Diff," followed past the signed ranks respective to the differences. Note that several ties are given average ranks, and that zero is arbitrarily given a positive sign. A quick inspection shows that there are fewer negative signed ranks and then we showtime compute $T\left(-\right)$ :

$\begin{array}{lll}\hfill T\left(-\right)=4.\mathrm{v}+\mathrm{ix}+\mathrm{vii}+2.\mathrm{v}=23.& & \hfill \\ \hfill \end{array}$

The total sum of $n$ ranks is $\left(\mathrm{north}\right)\left(n+\mathrm{ane}\right)∕\mathrm{ii}$ ; hence, it follows that $T\left(+\right)+T\left(-\right)=\left(20\right)\left(\mathrm{ii}\mathrm{ane}\right)∕2=2\mathrm{i}0$ . Thus $T\left(+\right)=210-2\mathrm{three}=\mathrm{i}87$ . The examination statistic is the smaller of the two, $T=T\left(-\right)=23$ . From Appendix Table A.9, using $n=20$ and $\alpha =0.01$ , we encounter that the critical value is 37. We decline $H_0$ if the calculated value is less than 37; hence, we reject the hypothesis and conclude that the population is not symmetric about 13.0.

Table fourteen.2. Deviations from the Median and Signed Ranks

Obs	Diff	Rank Signed	Obs	Unequal	Rank Signed
1	four.i	18	11	ii.7	14
2	$-\text{ 0.iii}$	$-\text{ four.v}$	12	80.4	20
three	three.5	16	13	1.9	13
iv	one.0	eleven	14	0.0	one
5	one.2	12	15	0.eight	10
6	$-\text{ 0.vii}$	$-\text{ 9}$	16	three.2	15
7	0.2	2.5	17	0.6	8
8	0.3	4.5	eighteen	$-\text{ 0.2}$	$-\text{ 2.5}$
9	4.9	19	nineteen	0.4	6
10	$-\text{ 0.5}$	$-\text{ vii}$	20	three.vi	17

Alternately, we can utilise the large sample normal approximation. Nether the null hypothesis, $T$ is approximately normally distributed with

$\begin{array}{ll}\hfill \mu & =\left(\mathrm{ii}0\right)\left(21\right)∕4=10\mathrm{v},\text{and}\\ \hfill {\sigma }^2& =\left(20\right)\left(\mathrm{ii}1\right)\left(\mathrm{four}\mathrm{i}\right)∕24=7\mathrm{one}7.\mathrm{v};\end{array}$

hence $\sigma =26.7\mathrm{nine}$ . These values are used to compute the examination statistic

$\begin{array}{l}\hfill z=\frac{23-105}{\mathrm{two}6.\mathrm{vii}9}=-3.0\mathrm{six}.\end{array}$

Using Appendix Table A.one, we find a (two-tailed) $p$ value of approximately 0.002; hence, the goose egg hypothesis is readily rejected. Even so, the sample is rather small; hence, the $p$ value calculated from the large sample approximation should not exist taken besides literally.

The $p$ value obtained for the sign test in Section 4.v was 0.012. Thus, for $\alpha =0.0\mathrm{one}$ the Wilcoxon signed rank test rejected the cypher hypothesis while the sign test did not. ³

Some texts recommend discarding nada differences, such every bit the one arbitrarily assigned a positive value in Table 14.2. This discard is done before the ranking, and the exam statistic computed using the remaining observations with the correspondingly smaller sample size. Run across Higgins (2004) for a discussion.

A popular awarding of the signed rank test is for comparison means from paired samples. In this application the differences between the pairs are computed as is done for the paired $t$ test (Section 5.4). The hypothesis to be tested is that the distribution of differences is symmetric most 0.

Example 14.3

To determine the effect of a special diet on activity in pocket-size children, 10 children were rated on a scale of 1 to 20 for degree of activity during luncheon hr past a school psychologist. After vi weeks on the special nutrition, the children were rated once again. The results are give in Table 14.3. We examination the hypothesis that the distribution of differences is symmetric about 0 against the alternative that it is non.

Table 14.3. Event of Diet on Activity

Child	Earlier Rating	After Rating	$\mathfrak{|}\text{ Difference}\mathfrak{|}$	Signed Rank
1	xix	11	viii	$-\text{ 10}$
two	14	15	1	$+\text{ i}$
3	xx	17	iii	$-\text{ 3.5}$
iv	6	12	half dozen	$+\text{ 8}$
5	12	8	4	$-\text{ 5}$
6	4	9	v	$+\text{ half-dozen.5}$
7	10	7	three	$-\text{ 3.5}$
eight	thirteen	6	seven	$-\text{ 9}$
9	15	10	5	$-\text{ vi.5}$
x	9	11	2	$+\text{ 2}$

Solution

The sum of the positive ranks is $T\left(+\right)=17.5$ ; hence $T\left(-\right)=\mathrm{v}5-17.5=\mathrm{three}7.5$ . Using $\alpha =0.05$ , the rejection region is for the smaller of $T\left(+\right)$ and $T\left(-\right)$ to be less than eight (from Appendix Tabular array A.nine). Using $T\left(+\right)$ as our examination statistic, we cannot pass up the null hypothesis, so we conclude that at that place is insufficient evidence to conclude that the diet afflicted the level of activity.

The Randomization Arroyo for Example xiv.3

Considering this data contains ties and is a small sample, we might request an exact $p$ value computed using a randomization test. How should the randomization be done? That the values are paired by kid is an inherent characteristic of this data, and nosotros must maintain it. When nosotros randomize, the only possibility is that the before-and-later on values within each child might switch places. This would cause the signs on the rank to switch, though it would not disturb the magnitude of the rank. Hence, nosotros would need to list all the ${\mathrm{ii}}^{10}=102\mathrm{iv}$ possible sets where the signed ranks in Table xiv.3 are free to opposite their signs. For each of these hypothetical (or pseudo) information sets, we compute the pseudo-value of T. In 33.two% of the sets, the pseudo-T is at or beneath our observed value of 17.five. Hence, our $p$ value is 0.3320, which agrees with the value from the SAS Organization's Proc UNIVARIATE.

Case Study fourteen.ane

Gumm et al. (2009) studied the preferences of female zebrafish for males with several possible fin characteristics. Each zebrafish tin can be long fin, brusk fin, or wildtype. Practice females have a preference for a item fin type? In each trial, a female zebrafish (the focal private) was placed in the central part of an aquarium. At one finish, behind a divider, was a male of ane of the fin types. At the other finish, behind a divider, was a male of a contrasting fin type. The males are referred to as the stimulus fish. The researchers recorded the corporeality of time each female spent in the vicinity of each stimulus fish, yielding two measurements for each trial.

We would prefer to utilize a paired t test to compare the preference of females for one type of fin versus the other. However, the authors state:

The information were non normally distributed after all attempts at transformations and thus nonparametric statistics were used for inside treatment analysis. Total fourth dimension spent with each stimulus was compared within treatments with a Wilcoxon-Signed Rank test.

The results of their analysis are summarized every bit follows:

Treatment	Wilcoxon Signed Rank Examination
wildtype female: wildtype vs. long fin male	$n=19,z=-1.81,p=0.8\mathrm{six}$
wildtype female: brusque fin vs. wildtype male	$n=20,z=-0.\mathrm{i}3\mathrm{ane},p=0.90$
long fin female person: wildtype vs. long fin male person	$n=20,z=-2.427,p=0.02$
short fin female person: curt fin vs. wildtype male	$n=20,z=-0.08,p=0.45$

(Notation the inconsistency in the $p$ value for brusk fin females.) The authors conclude:

The preference for males with longer fins was observed only in females that also take long fins. This unique preference for longer fins by long fin females may suggest that the mutation controlling the expression of the long fin trait is also playing a role in controlling female association preferences.

Read full affiliate

URL:

https://www.sciencedirect.com/science/article/pii/B9780123749703000147

Statistics, Nonparametric

Joseph W. McKean , Simon J. Sheather , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

I.B.5 Sample Size Determination

Consider the signed-rank Wilcoxon test for the one-sided hypothesis (replace α past α/2 for a ii-sided alternative hypothesis). Suppose the level, α, and the power, γ, for a particular alternative Δ _A are specified. Let north denote the number of pairs or blocks to be selected. Under these weather, the recommended number of pairs is given by

(25) $\text{n}={\left(\frac{{\text{z}}_{\alpha }-{\text{z}}_{\gamma }}{{\Delta }_{\text{A}}}\right)}^2{\tau }^2.$

Notation that information technology does depend on τ which, in applications, would have to be guessed or estimated in a pilot study. As in the two-sample location trouble, if the underlying density of the errors is assumed to be normal with standard deviation σ then $\tau =\sqrt{\pi /3}\sigma$ . For LS, the formula for n would be the same, except τ would exist replaced by σ.

Read full chapter

URL:

https://www.sciencedirect.com/science/commodity/pii/B0122274105007328

Nonparametric Statistics

Kandethody M. Ramachandran , Chris P. Tsokos , in Mathematical Statistics with Applications in R (Tertiary Edition), 2021

12A Comparison of Wilcoxon tests with normal approximation

(i)

For the Wilcoxon signed rank test, compare the results from the Wilcoxon signed rank test table with the normal approximation using several sets of data of various sample sizes. Also, if the sample size is very minor, compare the results from the Wilcoxon signed rank test with a small sample t-test.

(ii)

For the Wilcoxon rank sum examination, compare the results from the Wilcoxon rank sum exam table with the normal approximation using several sets of data (from pairs of samples) of various sample sizes. Also, if the sample sizes are very modest, compare the results from the Wilcoxon rank sum test with small sample t-test for two samples.

Read full chapter

URL:

https://world wide web.sciencedirect.com/science/article/pii/B9780128178157000129

RANK-BASED AND NONPARAMETRIC METHODS

Rand R. Wilcox , in Applying Contemporary Statistical Techniques, 2003

15.half-dozen.3 Wilcoxon Signed Rank Test

The sign test provides an interesting, useful, and reasonable perspective on how two groups differ. Still, a common criticism is that its ability can be low relative to other techniques that might be used. Ane alternative approach is the Wilcoxon signed rank exam, which tests

$H_0:F_1\left(x\right)=F_2\left(\mathrm{10}\right),$

the hypothesis that two dependent groups have identical distributions. To apply information technology, first class departure scores equally was washed in conjunction with the paired T-exam in Chapter xi and discard whatever divergence scores that are equal to cipher. It is assumed that there are n departure scores not equal to zero. That is, for the ith pair of observations, compute

$D_i={\mathrm{Ten}}_{i1}-X_{i2},$

i = 1, …, north and each D_i value is either less than or greater than nada. Adjacent, rank the |D_i| values and permit U_i denote the result for |D_i|. So, for example, if the Di values are 6, − 2, 12, 23, −eight, so U1 = two, because later taking accented values, 6 has a rank of ii. Similarly, U₂ = 1, because subsequently taking accented values, the second value, −2, has a rank of 1. Next prepare

$R_i=U_i,$

if Di > 0; otherwise

$R_i=-U_i,$

Positive numbers are said to have a sign of one and negative numbers a sign of −1, then Ri is the value of the rank corresponding to |Di| multiplied by the sign of D_i.

If the sample size (north) is less than or equal to xl and there are no ties among the |Di| values, the test statistic is West, the sum of the positive Ri values. For example, if R₁ =four,R₂ = −3, R_three = 5,R₄ = 2, and R₅ = −1, and then

$W=4+5+\mathrm{ii}-\mathrm{eleven.}$

A lower critical value, cL, is read from Table 12 in Appendix B. So for α = .05 and n = 5, the critical value corresponds to α/ii = .025 and is 0, so reject if W ≤ 0. The upper disquisitional value is

${}^{c}U=\frac{n(n+\mathrm{one})}{2}-{}^c\mathrm{Fifty}.$

In the illustration, because cL, = 0,

${}^{c}U=\frac{5\left(6\right)}{2}-0=15,$

pregnant that you refuse if W ≥ xv. Because W = 11 is between i and 15, fail to reject.

If at that place are ties among the |Di| values or the sample size exceeds 40, the test statistic is

$W=\frac{\sum R_i}{\sqrt{\sum R_i^{\mathrm{two}}}}.$

If there are no ties, this last equation simplifies to

$W=\frac{\sqrt{\mathrm{vi}}\sum R_i}{\sqrt{n(n+\mathrm{one})(2n+1)}}.$

For a two-sided test, turn down if |Due west|equals or exceeds Z_1-α/2 the 1 − α/2 quantile of a standard normal distribution.

Rejecting with the signed rank test indicates that two dependent groups have dissimilar distributions. Although the signed rank exam can have more power than the sign examination, a criticism is that it does not provide sure details about how the groups differ. For instance, in the cork boring instance, rejecting indicates that the distribution of weights differs for the north versus due east side of a tree, but how might nosotros elaborate on what this difference is? One possibility is to gauge p, the probability that the weight from the north side is less than the weight from the east side. So despite lower power, one might argue that the sign test provides a useful perspective on how groups compare.

Read total chapter

URL:

https://www.sciencedirect.com/science/commodity/pii/B9780127515410500365

Nonparametric Methods

Donna Fifty. Mohr , ... Rudolf J. Freund , in Statistical Methods (Fourth Edition), 2022

Exercises

1.

In eleven test runs a brand of harvesting machine operated for 10.1, 12.two, 12.4, 12.4, 9.4, eleven.2, 14.8, 12.6, 10.1, 9.2, and xi.0   h on a tank of gasoline.

(a)

Use the Wilcoxon signed rank test to make up one's mind whether the machine lives up to the manufacturer's claim of an average of 12.5   h on a tank of gasoline. (Use $\alpha =0.05$ .)

(b)

For the sake of comparison, utilize the i-sample $t$ test and compare results. Annotate on which method is more appropriate.

2.

Twelve adult males were put on a liquid nutrition in a weight-reducing plan. Weights were recorded before and afterwards the diet. The data are shown in Tabular array fourteen.10. Apply the Wilcoxon signed rank exam to define whether the program was successful. Do you lot retrieve the use of this test is appropriate for this set of data? Comment.

Table xiv.x. Data for Practise two.

	SUBJECT
	i	two	3	4	5	half-dozen	seven	8	9	10	11	12
Before	186	171	177	168	191	172	177	191	170	171	188	187
After	188	177	176	169	196	172	165	190	165	180	181	172

3.

The test scores shown in Table 14.11 were recorded past two different professors for two sections of the same form. Using the Mann–Whitney test and $\alpha =0.05$ , determine whether the locations of the two distributions are equal. Why might the median be a better measure of location than the mean for these data?

Tabular array fourteen.11. Data for Do 3.

PROFESSOR
A	B
74	75
78	80
68	87
72	81
76	72
69	73
71	80
74	76
77	68
71	78

4.

Inspection of the data for Exercise xi in Chapter 5 suggests that the information may non be normally distributed. Redo the trouble using the Isle of man–Whitney examination. Compare the results with those obtained by the pooled $t$ examination.

5.

8 human tooth teeth were sliced in half. For each tooth, one randomly chosen half was treated with a compound designed to wearisome loss of minerals; the other one-half served as a control. All tooth halves were and so exposed to a demineralizing solution. The response is per centum of mineral content remaining in the molar enamel. The data are given in Table 14.12.

(a)

Perform the Wilcoxon signed rank test to decide whether the handling maintained a higher mineral content in the enamel.

(b)

Compute the paired $t$ statistic and compare the results. Comment on the differences in the results.

Table 14.12. Data for Practise v.

	Mineral Content
Control	66.1	79.iii	55.3	68.8	57.viii	71.8	81.3	54.0
Treated	59.1	58.9	55.0	65.9	54.i	69.0	60.2	55.5

half dozen.

Three teaching methods were tested on a group of 18 students with homogeneous backgrounds in statistics and comparable aptitudes. Each student was randomly assigned to a method and at the end of a half dozen-week program was given a standardized exam. Because of classroom space, the students were non as allocated to each method. The results are shown in Table 14.13.

(a)

Test for a difference in distributions of test scores for the different instruction methods using the Kruskal–Wallis test.

(b)

If in that location are differences, explain the differences using a multiple comparison exam.

Table fourteen.13. Data for Do half-dozen.

METHOD
i	2	3
94	82	89
87	85	68
90	79	72
74	84	76
86	61	69
97	72
	80

7.

Hail damage to cotton, in pounds per planted acre, was recorded for 4 counties for three years. The data are shown in Table xiv.xiv. Using years as blocks use the Friedman test to determine whether there was a divergence in hail harm among the iv counties. If a departure exists, determine the nature of this difference with a multiple comparison examination. Also discuss why this examination was recommended.

Table 14.14. Data for Exercise vii.

Canton	Year
	1	2	three
P	49	141	82
B	13	64	viii
C	175	thirty	7
R	179	9	7

8.

To be as off-white as possible, nigh county fairs employ more than one guess for each type of event. For example, a pie-tasting competition may have two judges testing each entered pie and ranking information technology according to preference. The Spearman rank correlation coefficient may be used to make up one's mind the consistency between the judges (the interjudge reliability). In ane such competition at that place were 10 pies to be judged. The results are given in Table 14.15.

(a)

Summate the Spearman correlation coefficient between the 2 judges' rankings.

(b)

Test the correlation for significance at the 0.05 level.

Table 14.xv. Ranking of pies by judges.

Pie	Judge A	Approximate B
1	four	five
2	7	vi
iii	v	4
4	8	9
5	10	8
6	one	1
7	ii	three
viii	9	10
ix	iii	2
10	vi	seven

ix.

An agriculture experiment was conducted to compare 4 varieties of sweet potatoes. The experiment was conducted in a completely randomized design with varieties equally the treatment. The response variable was yield in tons per acre. The data are given in Table xiv.16. Exam for a divergence in distributions of yields using the Kruskal–Wallis examination.

Table 14.16. Yield of sweet potatoes.

Diversity A	Variety B	Variety C	Diverseness D
eight.3	9.1	10.1	7.8
ix.4	9.0	x.0	8.2
nine.1	8.1	nine.vi	8.i
9.1	8.2	9.3	7.9
9.0	8.viii	9.eight	7.7
8.9	8.4	nine.5	eight.0
8.ix	8.iii	9.four	eight.1

x.

In a study of student beliefs, a school psychologist randomly sampled four students from each of five classes. He then gave each student one of 4 different tasks to perform and recorded the time, in seconds, necessary to complete the assigned chore. The information from the study are listed in Table 14.17. Using classes as blocks use the Friedman exam to make up one's mind whether at that place is a difference in tasks. Utilise a level of significance of 0.10. Explain your results.

Table 14.17. Time to perform assigned task.

Course	Task
	1	2	3	4
1	43.2	45.8	45.4	44.7
2	48.3	48.vii	46.9	48.8
3	56.6	56.one	55.3	54.6
four	72.0	74.1	89.v	82.vii
5	88.0	88.half dozen	91.5	88.2

11.

Table fourteen.18 shows the total number of birds of all species observed by birdwatchers for routes in three different cities observed at Christmas for each of the 25 years from 1965 through 1989. Our interest centers on a possible alter over the years within cities, that is, cities are blocks.

Tabular array xiv.18. Bird counts for Twenty-Five years.

Year	Route			Year	Road
	A	B	C		A	B	C
65	138	815	259	78	201	1146	674
66	331	1143	202	79	267	661	494
67	177	607	102	fourscore	357	729	454
68	446	571	214	81	599	845	270
69	279	631	211	82	563	1166	238
70	317	495	330	83	481	1854	98
71	279	1210	516	84	1576	835	268
72	443	987	178	85	1170	968	449
73	1391	956	833	86	1217	907	562
74	567	859	265	87	377	604	380
75	477	1179	348	88	431	1304	392
76	294	772	236	89	459	559	425
77	292	1224	570

An inspection of the data indicates that the counts are not normally distributed. Since the responses are frequencies, a possible alternative is to employ the square root transformation, but another alternative is to use a nonparametric method. Perform the analysis using the Friedman exam. Compare results with those obtained in Practice ten.10. Which method appears to provide the nearly useful results?

12.

The ratings by respondents on the visual impact of wind farms (Tabular array 12.24 for Exercise 12.sixteen) are on an ordinal scale that makes rankings possible. Apply a nonparametric examination to compare the ratings from residents of Gigha to those of Kintyre. How does the interpretation of these results compare to the interpretation of the analysis in Do 12.16?

13.

The data in Table 5.one give the price-to-earnings ratio (PE) for samples of stocks on the NYSE and NASDAQ exchanges.

Is there bear witness that the typical PE values differ on the two exchanges? (Rather than depending on a logarithmic transformation to achieve normality, as in Example v.1, use a technique that does non require normality.)

fourteen.

Compare the variability in the test scores for the iii educational activity methods given in Table 14.v. To practice this, implement a nonparametric version of Levene's examination by first computing the absolute differences of each value from its grouping median. Compare the typical magnitudes of the accented differences using a nonparametric test from this chapter. What do you conclude?

Read total chapter

URL:

https://world wide web.sciencedirect.com/science/commodity/pii/B978012823043500014X

Estimating Measures of Location and Calibration

Rand Wilcox , in Introduction to Robust Interpretation and Hypothesis Testing (3rd Edition), 2012

3.nine The Hodges–Lehmann Estimator

Chapter 2 mentioned some practical concerns about R-measures of location in general and the Hodges and Lehmann (1963) estimator in particular. But the Hodges–Lehmann estimator plays a central role when applying standard rank-based methods (in particular, the Wilcoxon signed-rank test), so for completeness the details of this computer are given here.

The Walsh averages of n observations refers to all pairwise averages: (X _i + X _j )/2, for all i ≤ j. The Hodges–Lehmann computer is the median of all Walsh averages, namely,

${\widehat{\theta }}_{\text{HL}}={\text{med}}_{i\le j}\frac{{\mathrm{10}}_i+{\mathrm{Ten}}_j}{2}.$

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780123869838000032

Estimating Measures of Location and Scale

Rand R. Wilcox , in Introduction to Robust Interpretation and Hypothesis Testing (Fifth Edition), 2022

iii.9 The Hodges–Lehmann Computer

Affiliate 2 mentioned some practical concerns nearly R-measures of location in full general and the Hodges–Lehmann (1963) figurer in particular. But the Hodges–Lehmann estimator plays a cardinal function when applying standard rank-based methods (in particular, the Wilcoxon signed rank examination), so for completeness, the details of this calculator are given here.

The Walsh averages of n observations refers to all pairwise averages, $(X_i+X_j)/2$ , for all $i\le j$ . The Hodges–Lehmann reckoner is the median of all Walsh averages, namely,

${\hat{\theta }}_{\mathrm{HL}}={\mathrm{med}}_{i\le j}\frac{{\mathrm{Ten}}_i+{\mathrm{Ten}}_j}{2}.$

As noted in Department 2.2, in that location are conditions where this measure of location, also as R-estimators in general, has good backdrop. Merely there are general weather condition under which they perform poorly (due east.thousand., Bickel and Lehmann, 1975; Huber, 1981, p. 65; Morgenthaler and Tukey, 1991, p. 15).

Read total chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128200988000099

Nonparametric Tests

Ronald N. Forthofer , ... Mike Hernandez , in Biostatistics (Second Edition), 2007

9.3 The Wilcoxon Signed Rank Test

Another much more recently developed examination that can exist used to examine whether or not there is reversion toward the hateful in the data in Instance 9.1 is the Wilcoxon Signed Rank (WSR) test. An American statistician, Frank Wilcoxon, who worked in the chemical manufacture, developed this test in 1945. Unlike the sign test which can exist used with nonnumeric information, the WSR test requires that the differences in the paired data come from a continuous distribution.

To utilize the WSR exam to examine whether or not there is reversion toward the hateful, we prepare the information as follows: The data for the fourteen boys with an farthermost day 1 value are shown in Tabular array 9.2. In this table, the differences between solar day ane and mean solar day two values are shown as either a change in the direction of the mean (+) or away from the mean (−). If the 24-hour interval 1 and day 2 values for a male child are the same, then nosotros cannot assign a sign, and such a pair would be excluded from the analysis. The absolute differences are ranked from smallest to largest, and the ranks are summed separately for those changes in the direction of mean and for those changes away from the hateful. We use R _WSRto stand for the signed rank sum statistic for the positive differences — in this case, those changes toward the mean.

Table 9.ii. Days 1 and two caloric intakes for the 14 boys with the more farthermost caloric intakes on day 1.

					Rank
ID	Twenty-four hours 1	Twenty-four hours 2	Alter (+) Toward the Mean	Change (−) Away from the Hateful	+	−
xiii	ane,053	2,484	i,431		13
14	4,322	2,926	i,396		12
16	i,753	ane,054		699		10
27	3,532	3,289	243		3
thirty	2,842	2,849		seven		i
33	1,505	1,925	420		4
41	3,076	two,431	645		9
50	1,292	810		482		7
51	iii,049	two,573	476		6
101	ane,277	2,185	ane,092		eleven
118	ane,781	1,844	63		ii
130	2,773	three,236		463		5
149	1,645	2,269	624		viii
150	i,723	iii,163	i,440		14
				Sum of Ranks	82	23

We now consider the logic behind the testing of R _WSR. When there are n observations or pairs of data, the sum of the ranks is the sum of the integers from 1 to n and that sum is northward(n + ane)/two. The average rank for an observation is therefore (n + 1)/ii.

The aught hypothesis is that the differences have a median of zero and the alternative hypothesis that the median is non equal to zippo for a two-sided examination or greater (or smaller) than zero for a 1-sided test. If the null hypothesis is truthful, the distribution of the differences will exist symmetric, and there should be n/2 positive differences and n/2 negative differences. Therefore, if the naught hypothesis is true, the sum of the ranks for positive (or negative) differences, R_WSR, should exist (n/two) times the boilerplate rank: (n/2)(n + 1)/2 = n(due north + 1)/four.

The examination statistic is the sum of the ranks of positive (or negative) differences, R _WSR. For a small-scale sample, Table B9 (north < 30) provides boundaries for the critical region for the sum of the ranks of the positive (or negative) differences. To give an thought how these boundaries were determined, let us consider five pairs of observations. The boundaries result from the enumeration of possible outcomes as shown in Table 9.three.

Table nine.3. Positive ranks for a sample of size 5 for 0, 1, and two positive ranks.

Number of Positive Ranks	Possible Ranks	Sum of Positive Ranks	Sum of Negative Ranks
0		0	15
1	1	1	14
	ii	2	13
	3	3	12
	iv	4	11
	five	five	10
2	one, 2	iii	12
	one, three	four	eleven
	1, iv	5	x
	1, v	half dozen	9
	2, 3	v	ten
	2, 4	6	9
	2, v	7	8
	3, four	7	viii
	3, 5	8	seven
	4, v	9	6

In Table nine.3, there is no need to show the sum of ranks for iii, 4, and 5 positive ranks because their values are already shown under the sum of the negative rank column. For example, when there are 0 positive ranks, in that location are five negative ranks with a sum of 15. But the sum of 5 positive ranks must also exist 15. When at that place is 1 positive rank, there are four negative ranks with the indicated sums. Simply these are also the sum for the possibilities with 4 positive ranks. The same reasoning applies for 2 and 3 positive ranks.

Based on Table 9.3, nosotros can class Table 9.4, which shows all the possible values of the sum and their relative frequency of occurrence. Using Table 9.four, we meet that the smallest rejection region for a ii-sided examination is 0 or 15, and this gives the probability of a Type I fault of 0.062. Thus, in Table B9, there is no rejection region shown for a sample size of 5 and a significance level of 0.05. If the test of interest were a 1-sided test, then information technology would be possible to have a Type I error probability less than 0.05.

Table ix.4. All possible sums and their relative frequency.

Sum	Frequency	Relative Frequency
0 or 15	1	0.031
i or fourteen	i	0.031
2 or 13	1	0.031
3 or 12	2	0.063
4 or eleven	2	0.063
5 or 10	3	0.094
6 or nine	iii	0.094
7 or 8	3	0.094

Case 9.three

Let united states of america return to the data prepared for the 14 pairs in Table 9.2. We shall perform the exam at the 0.05 significance level, the same level used in the sign test. Since this is a ane-sided test, we read the boundary to a higher place α ≤ 0.05 nether ane-sided comparisons shown at the bottom of the table, which is equivalent to α ≤ 0.x under two-sided comparisons. Using the row north = 14, the critical values are (25, 80). Since our examination statistic is 82, greater than lxxx, we reject the null hypothesis of no regression toward the hateful in favor of the culling that there is regression toward the hateful.

This result is inconsistent with the issue of the sign test in Example 9.one and reflects the greater power of the WSR examination. This greater ability is due to the use of more of the information in the data by the WSR test compared to the sign examination. The WSR examination incorporates the fact that the average rank for the four changes away from the mean is v.75 (= [1 + 5 + 7 + x]/4), less than the average rank of seven.fifty. This lower average rank of these four changes, along with the fact that there were only four changes abroad from the hateful, acquired the WSR examination to be pregnant. The sign test used only the number of changes toward the hateful, not the ranks of these changes, and was not pregnant. Although the sign test failed to reject the null hypothesis, its p-value of 0.0898 was non that dissimilar from 0.05.

In applying the WSR examination, ii types of ties can occur in the data. One type is that some observed values are the aforementioned as the hypothesized value or some paired observations are the aforementioned — that is, the differences are naught. If this type of tie occurs in an observational unit or pair, that unit of measurement or pair is deleted from the data set, and the sample size is reduced by i for every unit of measurement or pair deleted. Once more, this procedure is appropriate when there are just a few ties in the data. If there are many ties of this blazon, there is trivial reason to perform the test.

The other blazon of necktie occurs when two or more than differences have exactly the same nonzero value. This has an impact on the ranking of the differences. In this case, convention is that the differences are assigned the same rank. For instance, if two differences were tied every bit the smallest value, each would receive the rank of ane.v, the boilerplate of ranks 1 and 2. If three differences were tied as the smallest value, each would receive the rank of 2, the average of ranks 1, 2, and three. If in that location are few ties in the differences, the rank sum can still exist used equally the test statistic; nonetheless, the results of the test are now approximate. If there are many ties, an adjustment for the ties must exist fabricated (Hollander and Wolfe 1973), or one of the methods in the side by side chapter should exist used.

Instance 9.4

Let usa apply the WSR exam to the data in Example ix.2. The xiii measurements, the deviations from the truthful value of 41, and ranks of absolute differences are every bit follows:

Measures:	45	43	40	44	49	36	51	46	35	fifty	41	38	47
Differences:	+ 4	+2	−one	+three	+8	−v	+ten	+5	−6	+9	0	−three	+six
Ranks:	5	2	i	3.v	10	6.5	12	6.five	8.5	xi	—	three.five	8.5

Note that the average ranking procedure is used for the same values of absolute differences and the rank is not assigned to tenth observation.

Once again the investigator wishes to test whether the repeated measurements are significantly different from the value of 41 at the 0.05 significance level. We delete one ascertainment that has no rank. The test statistic (the sum of ranks for positive differences) is 58.5. Table B9 provides boundaries of the critical region. For n = 12 and α ≤ 0.05 under the two-sided comparison the boundaries are (13, 65). Since the test statistic is less than 65, nosotros fail to reject the null hypothesis in favor of the alternative hypothesis at the 0.05 significance level. This conclusion is consistent with the outcome of the sign exam in Case 9.2.

For a large sample, the normal approximation is used. If at that place are at to the lowest degree 16 pairs of observations used in the calculations, R _WSR will approximately follow a normal distribution. As we simply saw, the expected value of R_WSR, under the supposition that the null hypothesis is truthful, is n(due north + 1)/4, and its variance can exist shown to be n(n + ane)(twonorthward + i)/24. Therefore, the statistic

$\frac{\left|R_{w\mathrm{due\; south}r}-\left[\mathrm{north}\left(\mathrm{due\; north}+1\right)/4\right]\right|-0.5}{\sqrt{\mathrm{northward}\left(\mathrm{due\; north}+1\right)\left(2n+\mathrm{i}\right)/24}}$

approximately follows the standard normal distribution. The two vertical lines in the numerator indicate the absolute value of the difference — that is, regardless of the sign of the difference, it is now a positive value. The 0.five term is the continuity correction term, required because the signed rank sum statistic is not a continuous variable.

Let united states of america calculate the normal approximation to the pairs in Case 9.3. The expected value of R_WSR is 52.v (= [fourteen][xv]/4), and the standard fault is $\mathrm{fifteen.93}\left(=\sqrt{\left(14\right)\left(15\right)\left(29\right)/24}\right)$ . Therefore, the statistic's value is

What is the probability that Z is greater than i.82? This probability is plant from Table B4 to be 0.0344. This agrees very closely with the exact p-value of 0.0338. The verbal p-value is based on 554 of the xvi,384 possible signed rank sums having a value of 82 or greater, applying the same logic illustrated in Tables two and 3 to the case n = fourteen. Thus, even though n is less than 16, the normal approximation worked quite well in this case. The WSR test can be performed by the computer (see Programme Annotation 9.1 on the website).

The sign and Wilcoxon Signed Rank tests are both used most oft in the comparing of paired data, although they can be used with a single population to test that the median has a specified value. In the use of these tests with pre- and postintervention measurement designs, care must be taken to ensure that at that place are no extraneous factors that could take an touch during the study. Otherwise, the possibility of the confounding of the extraneous factor with the intervention variable is raised. In add-on, the research designer must consider whether or not reversion to the mean is a possibility. If extraneous factors or reversion to the hateful cannot exist ruled out, the research design should be augmented to include a control group to help business relationship for the consequence of these possibilities.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780123694928500145

Information Science: Theory and Applications

Sunil Mathur , in Handbook of Statistics, 2021

three One-sample methods

In large information, many times a single stream of data is nerveless on a unit of interest. The data might exist collected in batches, periodic, near-real-fourth dimension, or real-fourth dimension. As the analysis of big data evolves, it is necessary to build models that integrate different models for developing applications of certain needs. In the case of streaming information in real-time, the processing of real-fourth dimension information may be followed by batch processing. That gives rise to categories of data due to batches of data. Some of the tests bachelor in the literature are based on the empirical distribution function. The empirical distribution role is an guess of the population distribution function, which works for big data. It is defined as the proportion of sample observations that are less than or equal to x for all real numbers x.

Nosotros consider the classical one-sample location problem with univariate data. Let x ₁, x _two, …, x _n exist an independent random sample of size north from a continuous distribution with distribution role F. Let the hypothesized cumulative distribution role be denoted by F ₀(x) and the empirical distribution office be denoted by Due south _n (x) for all ten. The hypothesis to be tested is H ₀  : F  = F _o vs H _a   : F  ≠ F _o . If the cipher hypothesis is true then the divergence betwixt South _n (x) and F ₀(x) must be close to zero.

Thus, for large northward, the exam statistic

(1) $D_{\mathrm{north}}=\underset{x}{sup}\left|S_n\left(x\right)-F_X\left(x\right)\right|,$

will have a value shut to zippo under the nix hypothesis.

The test statistic, D _n , called the Kolmogorov-Smirnov ane-sample statistic (Gibbons and Chakraborti, 2014), does not depend on the population distribution function if the distribution part is continuous and hence D _n is a distribution-free test statistic. The goodness-of-fit exam for a sample was proposed by Kolmogorov (1933). The Kolmogorov-Smirnov test for two samples was-proposed by Smirnov (1939).

Here we define order statistic Ten ₍₀₎  =   −   ∞ and X _{(n  +   1)}  =   ∞, and

(2) ${\mathrm{South}}_{\mathrm{northward}}\left(x\right)=\frac{i}{n},\mathit{for}{X}_{\left(i\right)}\le \mathrm{ten}\le {\mathrm{10}}_{\left(i+\mathrm{i}\right)},i=0,\mathrm{i},.\dots \dots ,n.$

The probability distribution of the test statistic does not depend on the distribution function F _X (X) for a continuous distribution function F _Ten (10). The asymptotic distribution of the exam statistic D _northward is Chi-square.

The exact sampling distribution of the Kolmogorov-Smirnov examination statistic is known while the distribution of the Chi-square goodness-of-fit examination statistic is approximately Chi-foursquare for finite northward. Moreover, the Chi-square goodness-of-fit test requires that the expected number of observations in a cell must be greater than five while the Kolmogorov exam statistic does not require this status. On the other paw, the asymptotic distribution of the Chi-foursquare goodness-of-fit exam statistic does not require that the distribution of the population must be continuous merely the exact distribution of the Kolmogorov-Smirnov test statistic does require that F _X (X) must exist a continuous distribution. The ability of the Chi-square distribution depends on the number of classes or groups fabricated.

The Wilcoxon signed-rank test ( Wilcoxon, 1945) requires that the parent population should be symmetric. When data is nerveless in batches, the data might be symmetric at some point, particularly in the instance of seasonal and periodic information. Let united states of america consider a random sample 10 ₁, X ₂, …. . , Ten _n from a continuous cdf F which is symmetric virtually its median M. The null hypothesis can be stated as

(3) $H_0:\mathrm{Yard}=M_0$

The alternative hypotheses can be postulated accordingly. We notice that the differences D _i   = X _i   − M ₀ are symmetrically distributed near zero, and hence the number of positive differences will be equal to the number of negative differences. The ranks of the differences |  D ₁  |, |  D ₂  |, . ……………, |  D _{Due north}   | are denoted past Rank(.). Then, the exam statistic tin can be defined as

(4) $T^{+}=\sum _{i=\mathrm{one}}^n{a}_i\mathit{Rank}\left(\left|D_i\right|\right)$

(5) $T^{-}=\sum _{i=1}^n\left(1-a_i\right)\mathit{Rank}\left(\left|D_i\right|\right)$

where

(6) $a_i=\left\{\begin{array}{c}1,\mathit{if}{D}_i>0,\\ 0,\mathit{if}{D}_i<0.\end{array}\right.$

Since the indicator variables a _i are contained and identically distributed Bernoulli variates with P(a _i   =   i)   = P(a _i   =   0)   =   ½, therefore, under the null hypothesis

(seven) $E\left(T^{+}|H_0\right)=\sum _{i=1}^{n}E\left(a_i\right)\mathit{Rank}\left|D_i\right|=\frac{n\left(n+\mathrm{one}\right)}{4}.$

and

(8) $\mathit{Var}\left(T^{+}|H_0\right)=\sum _{i=\mathrm{i}}^n\mathit{var}\left(a_i\right){\left[\mathit{Rank}\left|D_i\right|\right]}^{\mathrm{two}}=\frac{n\left(\mathrm{due\; north}+1\right)\left(2n+\mathrm{one}\right)}{24}$

Some other common representation for the exam statistic T ⁺ is given as follows.

(9) $T^{+}=\sum _{1\le i\le j\le n}\sum T_{\mathit{ij}}$

where

(ten) $T_{\mathit{ij}}=\left\{\begin{array}{c}1,\mathit{if}{D}_i+D_j>0,\\ 0,\mathit{otherwise}.\end{array}\right.$

Similar expressions can be derived for T ⁻. The paired-samples can be divers based on the differences X ₁  − Y _one, X ₂  − Y _two,..…, X _{due north}   − Y _n of a random sample of n pairs (X ₁, Y _ane), (Ten _ii, Y ₂)..…, (X _n , Y _northward ). Now, these differences are treated as a single sample and the i-sample test procedure is practical. The null hypothesis to be tested will be

$H_0:\mathrm{Yard}={\mathrm{Yard}}_0$

where Chiliad ₀ is the median of the differences X ₁  − Y ₁, X _two  − Y ₂,..…, 10 _north   − Y _northward . These differences can exist treated as a unmarried sample with the hypothetical median M ₀. And so, the Wilcoxon signed-rank method described higher up for a unmarried sample tin can be applied to test the null hypothesis that the median of the differences is Grand ₀.

Since a good test must be non only fast in computing the test value but likewise should have the ability in finding out data hidden in big data. Wilcoxon signed-rank test fulfills that requirement, still, several other tests are bachelor in the literature that are competitors of the Wilcoxon signed-rank test.

Chattopadhyay and Mukhopadhyay (2019) used the kernel of degree yard (>   ane) to develop a one-sample nonparametric exam. Define a kernel (k   =   2):

(xi) ${\psi }^{\left(\mathrm{ii}\right)}\left(X_i,X_j\right)=\left\{\begin{array}{c}1\mathit{if}\frac{X_i+X_j}{\mathrm{ii}}\ge 0\\ 0\mathit{if}\frac{X_i+{\mathrm{Ten}}_j}{\mathrm{two}}<0\end{array}\right.$

This kernel is equivalent to U-Statistic of degree 2

(12) ${\mathrm{Southward}}_n^{\left(2\right)}={\left(\begin{array}{c}\mathrm{northward}\\ \mathrm{ii}\end{array}\right)}^{-1}\sum _{1\le i_{\mathrm{ane}}<i_2\le n}{\psi }^{\left(2\right)}\left({\mathrm{Ten}}_{i_{\mathrm{one}}},{\mathrm{10}}_{i_2}\right)$

Both the sign test and the Mann-Whitney exam involve U-statistics with a symmetric kernel of degree one, one, and one respectively.

A full general examination statistic (Chattopadhyay and Mukhopadhyay, 2019) based on the kernel of one thousand (<   n) tin be defined as:

(13) $S_n^{\left(k\right)}={\left(\begin{array}{c}n\\ k\end{array}\right)}^{-\mathrm{ane}}\sum _{1\le i_{\mathrm{one}}<\dots <i_{\mathrm{grand}}\le \mathrm{north}}{\psi }^{\left(k\right)}\left(X_{i_1},\dots ,X_{i_{\mathrm{thousand}}}\right)$

where

(14) ${\psi }^{\left(m\right)}\left({\mathrm{10}}_{i_{\mathrm{ane}}},\dots ,{\mathrm{Ten}}_{i_k}\right)=\left\{\begin{array}{c}\mathrm{ane}\mathit{if}{\overline{\mathrm{Ten}}}_{i_{\mathrm{thou}}}\ge 0\\ 0\mathit{if}{\overline{\mathrm{Ten}}}_{i_k}<0\end{array}\right.$

and ${\overline{X}}_{i_k}=\frac{1}{k}{\sum }_{j=\mathrm{i}}^g{X}_{i_j,}$ 1   ≤ i ₁  <   …   < i _{one thousand}   ≤ n for n   >   1000.

The cipher hypothesis, every bit given by argument (iii), tin be tested at a level α using the following criterion:

Reject the null hypothesis if S _n ^(chiliad)  > c _α , where P _{H o} (S _n ^(k)  > c _α )   ≤ α, and c _α is the critical region.

In big data scenarios, confront recognition has received significant attending due to increasing attention to security at public places such as airports, rail stations, and similar places. A single sample is generally received from an ID card or e-passport, captured in a very stable environment while probe images are captured in a highly unstable surround ordinarily from surveillance cameras. The probe images may include noise, blur, arbitrary pose, and illumination, which makes the comparison with the standard database image difficult and hence makes the recognition difficult. The performance of available computational methods based on main component analysis, linear discriminant analysis, sparse representation, kernel-based and similar methods in face recognition, yet, is heavily influenced past the number of training samples per person.

Since at that place is only one sample available for such problems, we attempt to increase the number of samples artificially using synthetic sample generation from a 3D model of the bachelor epitome. The new dataset with multiple artificially generated samples can be used every bit a gallery set and the probe prepare contains images from surveillance cameras in an unconstrained environment. Now one can select 2D facial points and the landmark points in the 3D model in the gallery set and discover median points at each landmark signal. Similarly, select those points in the probe set up and run the i-sample test, such every bit Eq. (9) at each landmark betoken. The larger similarities at landmark points volition betoken toward similarities of probe and gallery and probe images.

The problem is generally faced when the interest is in identifying an individual at busy common places such every bit airports, train stations, and public coming together places. That volition involve matching the gallery set with a probe fix containing millions for images. In order to achieve that task speedily and efficiently, one tin set up three layers of batch processing. The beginning layer of processing involves eliminating the information which has more than two standard deviations of variations in major landmarks. Thus, the probe data containing a wider face up or too small face volition get eliminated. In the 2nd layer, matching of finer landmarks such as ear length and width, olfactory organ length is done. The data having two standard deviations of variations in those landmarks is eliminated. Thus, the remaining data will be a lot easier to handle with the batch processing method (Fig. 2).

Fig. ii. Converting 2D facial model to 3D model and increasing virtual sample size for gallery information.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/S0169716120300481

kennadaunded.blogspot.com

Source: https://www.sciencedirect.com/topics/mathematics/wilcoxon-signed-rank-test

Share this post