How to Know if You Should Reject the Null Hypothesis in a Wilcoxon Rank Sum Test

Inferential Statistics III: Nonparametric Hypothesis Testing

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

6.3 Wilcoxon Signed-Rank Test

Although the sign test can be used to test both one-sample and two-sample paired data, the Wilcoxon signed-rank test is more than powerful than the sign test for these tasks because it makes utilise of the magnitudes of the differences rather than merely their signs.

The Wilcoxon signed rank test was developed past Frank Wilcoxon ⁱ in 1945. We will illustrate its use using two-sample paired data. Following our checklist from Section 5.2, the bones idea behind the Wilcoxon signed-rank test is:

■

Course null and alternative hypotheses and choose a degree of confidence. The nothing hypothesis is that the median of the population of differences betwixt the paired data is nothing. The alternative hypothesis is that it is not.

■

Compute the test statistic. Nosotros do this by first computing the differences between the paired data samples; then we rank the differences according to magnitude only, that is, without regard to their sign; next, we sum the ranks of the positive and negative differences; finally, we pick the minimum of the sums as our examination statistic.

■

Compare the test statistic to a critical value. If the test statistic is less than the critical value, so we reject the zilch hypothesis.

Again, nosotros will illustrate this procedure with an example. We volition revisit the instance we introduced in Section 5.7.ane for testing blood pressure information from patients suffering from hypertension. 2 sets of data take been gathered: before handling with a new drug and later on treatment. The researchers now doubt that their data are normally distributed and and then wish to endeavor a nonparametric hypothesis examination.

Our naught hypothesis is that the median of the population of differences between the two sets of blood pressure data is zero. The alternative hypothesis is that information technology is not zero. Denoting the before handling data every bit the control information and the after treatment data every bit the test information, our two samples are:

Control:	175.4	188.3	147.four	178.half dozen	173.2	156.9	165.7	173.4
Examination:	152.3	159.7	155.7	166.2	149.i	162.three	163.5	146.0

To compute our test statistic, we starting time compute the differences between the paired data, that is, exam minus control. These are (−23.1 −28.half dozen 8.3 −12.4 −24.1 v.4 −2.two −27.4).

Next, nosotros rank these differences by magnitude and assign ranks to each difference value. This procedure is illustrated in Tabular array vi.2. Note that the signs of the divergence values are ignored when ranking them, that is, they are ranked past magnitude but, but we remember the signs.

Tabular array 6.2. Wilcoxon signed-rank test: differences betwixt two paired samples ranked by magnitude, with corresponding rankings.

Differences	−23.1	−28.6	8.3	−12.4	−24.1	five.4	−ii.2	−27.iv
Ranked differences	−2.2	5.4	8.3	−12.4	−23.1	−24.1	−27.4	−28.vi
Ranks	1	2	iii	4	5	6	7	8

Next, nosotros sum the rankings for the positive and negative differences (i.east. using the remembered signs). Referring to the ranks in Table 6.2, the sum of ranks for the positive differences is

$T_{+}=\mathrm{ii}+3=5$ ,

and the sum of ranks for the negative differences is

$T_{-}=1+\mathrm{four}+5+\mathrm{vi}+\mathrm{vii}+8=31$ .

The minimum of these rank sums is $T_{+}$ , which is five, then this is our exam statistic.

Now we look upwards a critical value in Table A.iii (meet Appendix): for $\mathrm{due\; north}=\mathrm{eight}$ and a significance level of 0.05 in a 2-tailed test, we have a critical value of iii. We compare our calculated and critical values: if the calculated value is less than the critical value, then we reject the null hypothesis. Since five is non less than iii, nosotros cannot refuse the null hypothesis in this case, so nosotros cannot conclude that in that location is a significant divergence between the two samples. Therefore, based on this test, nosotros have not demonstrated that the new drug reduces blood force per unit area. Note the difference here with the parametric test on the same information (see Section v.7.one): in that case, we were able to evidence statistical significance (on the assumption that the distributions were normal). By and large, significance is more likely to be shown with parametric tests than with nonparametric ones.

Equally we pointed out earlier, it is possible to use both the sign test or the Wilcoxon signed-rank test in either the ane-sample case or the two-sample paired data case. To use the Wilcoxon signed-rank test in the i-sample instance, we but compute the differences by subtracting the expected median value beingness tested against from the sample. However, although both tests are applicable in both situations, the Wilcoxon signed-rank test is the preferred method as it makes use of the magnitudes of the differences rather than just the signs. It is important to call back that the Wilcoxon signed-rank examination does make a stronger assumption almost the information sample(s) existence tested: it requires that the distribution of the differences is symmetric. If this can be shown to be not the case or if we have good reason to doubt that it is the case, and then the sign examination should exist used.

The Intuition. Wilcoxon Signed-Rank Test

We accept seen how the Wilcoxon signed-rank test can be used to exam hypotheses nearly data that are not commonly distributed, but why does it piece of work similar this?

The best way to understand the process underlying this nonparametric exam is to think of it as converting the problem into one that tin exist addressed using a parametric approach. As noted before, the Wilcoxon signed-rank exam makes the assumption that the distribution of differences between the two samples is symmetric. This assumption is of import equally it allows us to convert the original nonparametric problem into i that tin can be addressed past a parametric test. Note that the assumption is not saying that the distributions of the two samples are symmetric – we are only talking near the differences computed past subtracting the paired values (or the values from the expected median in the one-sample instance).

Why exercise we demand to brand this assumption? An example of a distribution of differences is illustrated in Fig. vi.3A. If this is symmetric and the null hypothesis is truthful (i.east. the median of this distribution is nothing), and then there will be the aforementioned number of departure values in a higher place and below zero (this is the definition of a median). Therefore, the sums of the positive and negative ranks of these differences will be the aforementioned, that is, $T_{+}=T_{-}$ . What would be the expected value of these rank sums?

Effigy half dozen.iii. The intuition behind the Wilcoxon signed-rank test: (A) shows the (symmetric) distribution of differences betwixt paired values from the two samples – the examination statistics T ₊ or T ₋ will be equal if the nil hypothesis is true; (B) shows the distribution of rank sums, which is normal according to the central limit theorem – the computed examination statistic is not beyond the disquisitional value for the 0.05 significance level, so we cannot reject the zip hypothesis.

Nosotros can compute this using knowledge of the properties of an arithmetic series. The ranks for a sample of northward numbers are simply the arithmetic series $1+\mathrm{two}+\dots +n$ . The sum of this serial is $\frac{\mathrm{northward}(n+1)}{\mathrm{two}}$ , then we would look the sums of either the positive or negative ranks to be $\frac{n(n+\mathrm{one})}{\mathrm{four}}$ (i.east. half of the overall sum) if the nada hypothesis were true. This value $\frac{n(n+1)}{\mathrm{iv}}$ is the hateful of the distribution of rank sums. Importantly, if the sample size n is large enough, and so this distribution volition exist normal according to the central limit theorem. The central limit theorem equally we described it in Section 4.5 referred to the sample mean, but the same principle applies to the sum (which is just the mean multiplied by the sample size). Over again, using knowledge of the backdrop of arithmetic serial and some basic algebra, the standard mistake of the (normal) distribution of rank sums can be calculated as $\sqrt{\frac{\mathrm{due\; north}(n+1)(2\mathrm{north}+\mathrm{one})}{24}}$ . This is illustrated in Fig. six.3B.

Now we have converted our nonparametric trouble into a parametric one, and we tin proceed in a like way to our procedure for the Student's t-test (encounter Section 5.5). Nosotros compute our actual sum of ranks (i.e. from our data – for our blood pressure example, $T_{+}$ was five) and compare this to the (now known) population distribution of rank sums. For our example, nosotros accept a population distribution of rank sums that has the following mean and standard error:

$\begin{array}{ccc}\hfill \mu & \hfill =\hfill & \frac{n(\mathrm{due\; north}+\mathrm{one})}{\mathrm{iv}}=\frac{\mathrm{eight}(\mathrm{viii}+\mathrm{ane})}{4}=18,\hfill \\ \hfill \text{southward.e.}& \hfill =\hfill & \sqrt{\frac{n(n+1)(2n+1)}{24}}=\sqrt{\frac{\mathrm{eight}(8+\mathrm{ane})(2\times \mathrm{eight}+1)}{24}}=\mathrm{vii.14}.\hfill \end{array}$

Following the procedure outlined in Section v.5, we compute our examination statistic using Eq. (v.1):

$t=\frac{\overline{x}-\mu }{s/\sqrt{n}}=\frac{5-\mathrm{xviii}}{\mathrm{7.xiv}}=-1.82.$

This tells united states that the rank sum we got from our data is i.82 standard errors away from the expected rank sum. How likely is information technology that we would get this figure (or a more extreme one) if the nada hypothesis were true?

For a one-sample Pupil's t-exam with a sample size of 8 (i.e. seven degrees of freedom), the critical value from Table A.one for 0.05 significance is ii.365 (this is too shown in Fig. 6.3B). The magnitude of our test statistic (1.82) is not greater than the disquisitional value (2.365), so we cannot reject the nix hypothesis. This is the same result as we got using the simplified procedure outlined in a higher place. Note that the critical values in Table A.iii 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 accept high cholesterol. Data accept been gathered from 9 patients who had loftier cholesterol. The data consist of their cholesterol levels (in mmol/Fifty) before and after taking the drug. The data are shown in the table.

O6.B

Cholesterol levels (mmol/L)
Before	6.4	4.2	3.eight	three.six	four.ane	3.7	5.0	four.4	iv.7
After	iii.4	ii.6	3.1	3.three	3.3	iii.3	2.6	4.3	iv.9

There is good reason to doubtable that the data are not unremarkably distributed. Perform a Wilcoxon signed-rank test to determine if there was a difference in cholesterol level as a effect of taking the drug. Conspicuously state what your hypotheses are, evidence all working, and utilize a 95% caste of confidence in your test.

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Nonparametric Methods

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

14.2 I Sample

In Section 4.5 we considered an culling approach to analyzing some income information that had a single extreme observation. This approach was based on the fact that the median is non 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 so-called sign test was based on the proportion of successes. In other words, the exam was performed on a set of data that were converted from the ratio to the nominal calibration.

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

The Wilcoxon signed rank test is used to examination that a distribution is symmetric well-nigh some hypothesized value, which is equivalent to the test for location. We illustrate with a test of a hypothesized median, which is performed as follows:

1.

Rank the magnitudes (accented 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").

three.

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 $n\left(\mathrm{northward}+\mathrm{i}\right)∕2$ , so 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 test statistic is the minimum of $T\left(+\right)$ and $T\left(-\right)$ , the critical region consists of the left tail of the distribution, containing a probability of at well-nigh $\alpha ∕2$ . For small samples $\left(n\le \mathrm{v}0\right)$ , critical values are plant in Appendix Table A.ix. If $\mathrm{due\; north}$ is large, the sampling distribution of $T$ is approximately normal with

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

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

Example xiv.2

Example 4.vii, particularly the information in Table four.5, concerned a test for the mean family income of a neighborhood whose results were unduly influenced by an farthermost outlier. A test for the median was used to overcome the influence of that observation. We now use that example to illustrate the Wilcoxon signed rank test. The hypothesis of interest is

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

with a ii-tailed alternative,

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

Solution

The deviations of the observed values from thirteen.0 (the specified $H_0$ value) are given in Tabular array fourteen.two in the cavalcade labeled "Diff," followed by the signed ranks corresponding to the differences. Note that several ties are given boilerplate ranks, and that zero is arbitrarily given a positive sign. A quick inspection shows that in that location are fewer negative signed ranks so we outset compute $T\left(-\right)$ :

$\begin{array}{lll}\hfill T\left(-\right)=\mathrm{four}.5+\mathrm{ix}+\mathrm{seven}+2.\mathrm{v}=23.& & \hfill \\ \hfill \end{array}$

The total sum of $n$ ranks is $\left(n\right)\left(n+\mathrm{i}\right)∕2$ ; hence, it follows that $T\left(+\right)+T\left(-\right)=\left(20\right)\left(21\right)∕\mathrm{ii}=210$ . Thus $T\left(+\right)=\mathrm{ii}10-23=18\mathrm{vii}$ . The test statistic is the smaller of the ii, $T=T\left(-\right)=23$ . From Appendix Table A.9, using $\mathrm{due\; north}=20$ and $\alpha =0.01$ , nosotros see that the critical value is 37. We reject $H_0$ 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 14.2. Deviations from the Median and Signed Ranks

Obs	Diff	Rank Signed	Obs	Diff	Rank Signed
1	iv.1	18	11	two.7	xiv
2	$-\text{ 0.3}$	$-\text{ 4.5}$	12	eighty.iv	20
iii	3.five	16	13	ane.9	13
four	i.0	11	xiv	0.0	i
5	1.two	12	fifteen	0.8	10
6	$-\text{ 0.7}$	$-\text{ 9}$	xvi	3.ii	xv
7	0.2	ii.five	17	0.6	8
8	0.3	4.5	xviii	$-\text{ 0.ii}$	$-\text{ two.five}$
9	4.9	19	19	0.iv	6
10	$-\text{ 0.five}$	$-\text{ 7}$	20	iii.6	17

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

$\begin{array}{ll}\hfill \mu & =\left(\mathrm{two}0\right)\left(2\mathrm{i}\right)∕\mathrm{four}=10\mathrm{five},\text{and}\\ \hfill {\sigma }^2& =\left(20\right)\left(\mathrm{two}\mathrm{ane}\right)\left(\mathrm{iv}1\right)∕\mathrm{two}4=7\mathrm{i}7.5;\end{array}$

hence $\sigma =\mathrm{two}\mathrm{half-dozen}.79$ . These values are used to compute the exam statistic

$\begin{array}{l}\hfill z=\frac{2\mathrm{three}-105}{26.79}=-3.0\mathrm{vi}.\end{array}$

Using Appendix Tabular array A.one, we find a (two-tailed) $p$ value of approximately 0.002; hence, the null hypothesis is readily rejected. However, the sample is rather small; hence, the $p$ value calculated from the large sample approximation should not be taken besides literally.

The $p$ value obtained for the sign test in Section 4.5 was 0.012. Thus, for $\alpha =0.0\mathrm{one}$ the Wilcoxon signed rank test rejected the null hypothesis while the sign test did non. ³

Some texts recommend discarding zero differences, such as the one arbitrarily assigned a positive value in Table fourteen.2. This discard is done before the ranking, and the test statistic computed using the remaining observations with the correspondingly smaller sample size. See Higgins (2004) for a discussion.

A pop awarding of the signed rank test is for comparison means from paired samples. In this application the differences between the pairs are computed equally is done for the paired $t$ test (Section 5.iv). The hypothesis to be tested is that the distribution of differences is symmetric about 0.

Example xiv.three

To decide the consequence of a special nutrition on activity in small children, 10 children were rated on a scale of 1 to xx for degree of activity during lunch hour past a school psychologist. Later vi weeks on the special nutrition, the children were rated over again. The results are give in Table fourteen.3. Nosotros test the hypothesis that the distribution of differences is symmetric about 0 against the alternative that it is non.

Tabular array xiv.3. Result of Diet on Activity

Child	Before Rating	Afterward Rating	$\mathfrak{|}\text{ Difference}\mathfrak{|}$	Signed Rank
1	19	11	8	$-\text{ 10}$
2	fourteen	15	one	$+\text{ one}$
3	20	17	three	$-\text{ 3.5}$
4	half dozen	12	half dozen	$+\text{ eight}$
5	12	8	4	$-\text{ 5}$
6	iv	nine	5	$+\text{ vi.5}$
7	10	7	iii	$-\text{ 3.five}$
8	13	6	seven	$-\text{ 9}$
9	15	x	5	$-\text{ 6.5}$
ten	9	xi	2	$+\text{ 2}$

Solution

The sum of the positive ranks is $T\left(+\right)=\mathrm{ane}\mathrm{seven}.5$ ; hence $T\left(-\right)=\mathrm{v}\mathrm{five}-17.5=\mathrm{three}7.\mathrm{v}$ . Using $\alpha =0.05$ , the rejection region is for the smaller of $T\left(+\right)$ and $T\left(-\right)$ to be less than 8 (from Appendix Table A.ix). Using $T\left(+\right)$ equally our test statistic, we cannot refuse the nix hypothesis, so nosotros conclude that there is insufficient evidence to conclude that the nutrition affected the level of activity.

The Randomization Approach for Example fourteen.iii

Because this data contains ties and is a small sample, nosotros might request an verbal $p$ value computed using a randomization test. How should the randomization be done? That the values are paired by child is an inherent characteristic of this information, and we must maintain it. When nosotros randomize, the just possibility is that the earlier-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, we would need to list all the $2^{10}=\mathrm{one}02\mathrm{four}$ possible sets where the signed ranks in Table 14.3 are free to contrary their signs. For each of these hypothetical (or pseudo) data sets, we compute the pseudo-value of T. In 33.ii% of the sets, the pseudo-T is at or below our observed value of 17.5. Hence, our $p$ value is 0.3320, which agrees with the value from the SAS Arrangement'south Proc UNIVARIATE.

Example Study 14.ane

Gumm et al. (2009) studied the preferences of female zebrafish for males with several possible fin characteristics. Each zebrafish can be long fin, brusque fin, or wildtype. Do females have a preference for a particular fin type? In each trial, a female zebrafish (the focal individual) 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 end, behind a divider, was a male of a contrasting fin type. The males are referred to as the stimulus fish. The researchers recorded the amount of fourth dimension each female spent in the vicinity of each stimulus fish, yielding ii measurements for each trial.

Nosotros would prefer to utilize a paired t examination to compare the preference of females for 1 blazon of fin versus the other. However, the authors state:

The data were not usually distributed subsequently all attempts at transformations and thus nonparametric statistics were used for within treatment assay. Total time spent with each stimulus was compared within treatments with a Wilcoxon-Signed Rank test.

The results of their analysis are summarized as follows:

Treatment	Wilcoxon Signed Rank Test
wildtype female person: wildtype vs. long fin male	$\mathrm{north}=19,z=-\mathrm{one}.\mathrm{viii}\mathrm{ane},p=0.86$
wildtype female: short fin vs. wildtype male	$n=20,z=-0.\mathrm{i}31,p=0.90$
long fin female person: wildtype vs. long fin male	$n=20,z=-2.427,p=0.0\mathrm{two}$
short fin female: short fin vs. wildtype male	$n=\mathrm{ii}0,z=-0.08,p=0.45$

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

The preference for males with longer fins was observed but in females that as well 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 decision-making female person association preferences.

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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 ane-sided hypothesis (supervene upon α by α/2 for a two-sided alternative hypothesis). Suppose the level, α, and the power, γ, for a particular alternative Δ _A are specified. Allow n denote the number of pairs or blocks to be selected. Under these conditions, the recommended number of pairs is given by

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

Annotation that it does depend on τ which, in applications, would have to be guessed or estimated in a pilot study. As in the 2-sample location problem, if the underlying density of the errors is causeless to exist normal with standard difference σ then $\tau =\sqrt{\pi /3}\sigma$ . For LS, the formula for north would be the same, except τ would exist replaced by σ.

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Nonparametric Statistics

Kandethody M. Ramachandran , Chris P. Tsokos , in Mathematical Statistics with Applications in R (3rd 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 tabular array with the normal approximation using several sets of data of various sample sizes. Likewise, if the sample size is very small, compare the results from the Wilcoxon signed rank examination with a small sample t-test.

(ii)

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

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RANK-BASED AND NONPARAMETRIC METHODS

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

fifteen.six.3 Wilcoxon Signed Rank Examination

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

$H_0:F_1\left(x\right)=F_2\left(x\right),$

the hypothesis that two dependent groups accept identical distributions. To apply it, first form difference scores every bit was washed in conjunction with the paired T-examination in Chapter 11 and discard any difference scores that are equal to aught. It is assumed that there are n divergence scores non equal to zero. That is, for the ith pair of observations, compute

$D_i=X_{i1}-X_{i2},$

i = ane, …, n and each D_i value is either less than or greater than zero. Side by side, rank the |D_i| values and permit U_i denote the result for |D_i|. So, for instance, if the Di values are 6, − two, 12, 23, −eight, then U1 = 2, because afterward taking absolute values, six has a rank of 2. Similarly, U₂ = 1, because later taking absolute values, the second value, −2, has a rank of one. Side by side set

$R_i=U_i,$

if Di > 0; otherwise

$R_i=-U_i,$

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

If the sample size (n) is less than or equal to forty and there are no ties amongst the |Di| values, the test statistic is West, the sum of the positive Ri values. For example, if R_one =4,R₂ = −3, R_iii = 5,R₄ = ii, and R_five = −one, then

$\mathrm{West}=4+5+2-11.$

A lower critical value, cL, is read from Table 12 in Appendix B. So for α = .05 and north = five, the critical value corresponds to α/two = .025 and is 0, so reject if West ≤ 0. The upper critical value is

${}^{c}U=\frac{\mathrm{north}(n+1)}{2}-{}^c\mathrm{50}.$

In the illustration, because cL, = 0,

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

meaning that yous reject if W ≥ 15. Considering W = xi is between ane and 15, neglect to reject.

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

$W=\frac{\sum R_i}{\sqrt{\sum R_i^2}}.$

If at that place are no ties, this last equation simplifies to

$W=\frac{\sqrt{6}\sum R_i}{\sqrt{\mathrm{northward}(n+1)(2n+1)}}.$

For a two-sided test, reject 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 ii dependent groups have different distributions. Although the signed rank examination can accept more power than the sign test, a criticism is that it does not provide certain details about how the groups differ. For instance, in the cork ho-hum case, rejecting indicates that the distribution of weights differs for the north versus east side of a tree, but how might we elaborate on what this difference is? One possibility is to approximate p, the probability that the weight from the north side is less than the weight from the east side. So despite lower power, i might fence that the sign test provides a useful perspective on how groups compare.

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Nonparametric Methods

Donna 50. Mohr , ... Rudolf J. Freund , in Statistical Methods (Quaternary Edition), 2022

Exercises

ane.

In 11 test runs a brand of harvesting machine operated for 10.1, 12.two, 12.4, 12.4, nine.iv, eleven.2, 14.viii, 12.6, 10.i, 9.two, and xi.0   h on a tank of gasoline.

(a)

Utilise the Wilcoxon signed rank test to determine whether the car lives up to the manufacturer'due south claim of an average of 12.five   h on a tank of gasoline. (Use $\alpha =0.05$ .)

(b)

For the sake of comparison, use the one-sample $t$ test and compare results. Comment on which method is more advisable.

2.

Twelve adult males were put on a liquid diet in a weight-reducing plan. Weights were recorded before and afterward the diet. The data are shown in Table 14.ten. Apply the Wilcoxon signed rank exam to define whether the plan was successful. Do you think the use of this test is advisable for this set of data? Comment.

Table 14.x. Information for Practise two.

	SUBJECT
	1	ii	3	4	5	6	7	8	9	10	11	12
Before	186	171	177	168	191	172	177	191	170	171	188	187
Afterwards	188	177	176	169	196	172	165	190	165	180	181	172

3.

The examination scores shown in Tabular array 14.11 were recorded by two unlike professors for two sections of the same class. Using the Mann–Whitney exam and $\alpha =0.05$ , decide whether the locations of the two distributions are equal. Why might the median be a better measure of location than the mean for these information?

Table fourteen.11. Data for Exercise three.

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 Practice xi in Chapter v suggests that the information may not exist normally distributed. Redo the problem using the Mann–Whitney test. Compare the results with those obtained by the pooled $t$ exam.

5.

Eight human molar teeth were sliced in half. For each tooth, one randomly chosen half was treated with a compound designed to ho-hum loss of minerals; the other half served as a control. All tooth halves were then exposed to a demineralizing solution. The response is percent of mineral content remaining in the tooth enamel. The information are given in Tabular array 14.12.

(a)

Perform the Wilcoxon signed rank test to make up one's mind 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. Information for Exercise 5.

	Mineral Content
Control	66.ane	79.3	55.3	68.8	57.eight	71.8	81.3	54.0
Treated	59.1	58.9	55.0	65.nine	54.1	69.0	60.two	55.5

6.

Iii teaching methods were tested on a grouping 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 six-calendar 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 divergence in distributions of test scores for the different teaching methods using the Kruskal–Wallis exam.

(b)

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

Table fourteen.xiii. Data for Practice 6.

METHOD
ane	two	3
94	82	89
87	85	68
ninety	79	72
74	84	76
86	61	69
97	72
	lxxx

seven.

Hail damage to cotton wool, in pounds per planted acre, was recorded for 4 counties for three years. The data are shown in Tabular array 14.14. Using years as blocks utilize the Friedman test to determine whether at that place was a difference in hail damage among the iv counties. If a deviation exists, determine the nature of this departure with a multiple comparison test. Besides discuss why this test was recommended.

Table 14.xiv. Data for Practise vii.

Canton	Year
	ane	2	3
P	49	141	82
B	13	64	8
C	175	30	7
R	179	9	seven

8.

To be as fair as possible, most county fairs utilize more one gauge for each blazon of event. For example, a pie-tasting contest may take 2 judges testing each entered pie and ranking information technology according to preference. The Spearman rank correlation coefficient may be used to determine the consistency betwixt the judges (the interjudge reliability). In ane such competition there were 10 pies to be judged. The results are given in Table xiv.15.

(a)

Calculate the Spearman correlation coefficient betwixt the two judges' rankings.

(b)

Test the correlation for significance at the 0.05 level.

Table fourteen.15. Ranking of pies by judges.

Pie	Judge A	Guess B
i	4	5
2	7	vi
iii	5	4
4	8	9
5	10	8
6	i	one
7	2	3
8	9	10
9	3	2
10	6	7

nine.

An agronomics experiment was conducted to compare four varieties of sugariness potatoes. The experiment was conducted in a completely randomized design with varieties every bit the treatment. The response variable was yield in tons per acre. The information are given in Tabular array 14.16. Examination for a difference in distributions of yields using the Kruskal–Wallis test.

Table fourteen.16. Yield of sweet potatoes.

Diverseness A	Variety B	Variety C	Variety D
8.3	nine.1	10.i	vii.viii
9.four	ix.0	10.0	8.2
9.1	8.i	nine.half dozen	viii.1
nine.1	viii.2	9.3	vii.9
9.0	viii.8	ix.eight	vii.vii
8.9	8.iv	9.5	8.0
eight.nine	8.3	9.4	8.1

10.

In a study of educatee behavior, a school psychologist randomly sampled four students from each of five classes. He then gave each student one of iv unlike tasks to perform and recorded the time, in seconds, necessary to complete the assigned chore. The data from the study are listed in Tabular array 14.17. Using classes as blocks apply the Friedman examination to determine whether in that location is a divergence in tasks. Employ a level of significance of 0.ten. Explicate your results.

Table fourteen.17. Time to perform assigned job.

Class	Job
	i	2	three	4
ane	43.2	45.eight	45.4	44.7
2	48.3	48.seven	46.nine	48.8
three	56.6	56.i	55.3	54.vi
4	72.0	74.1	89.5	82.7
5	88.0	88.half dozen	91.v	88.ii

11.

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

Table 14.18. Bird counts for 20-Five years.

Year	Road			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	80	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 use the square root transformation, but another alternative is to use a nonparametric method. Perform the analysis using the Friedman test. Compare results with those obtained in Exercise 10.10. Which method appears to provide the nigh useful results?

12.

The ratings by respondents on the visual bear on of air current farms (Table 12.24 for Exercise 12.16) are on an ordinal scale that makes rankings possible. Use a nonparametric test to compare the ratings from residents of Gigha to those of Kintyre. How does the interpretation of these results compare to the estimation of the analysis in Do 12.xvi?

13.

The data in Tabular array v.ane give the price-to-earnings ratio (PE) for samples of stocks on the NYSE and NASDAQ exchanges.

Is there evidence that the typical PE values differ on the ii exchanges? (Rather than depending on a logarithmic transformation to reach normality, as in Example v.1, utilise a technique that does not require normality.)

fourteen.

Compare the variability in the test scores for the 3 pedagogy methods given in Table fourteen.v. To do this, implement a nonparametric version of Levene's test past first calculating the absolute differences of each value from its group median. Compare the typical magnitudes of the absolute differences using a nonparametric test from this affiliate. What do you conclude?

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Estimating Measures of Location and Scale

Rand Wilcox , in Introduction to Robust Estimation and Hypothesis Testing (Third Edition), 2012

3.9 The Hodges–Lehmann Estimator

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

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

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

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Estimating Measures of Location and Calibration

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

3.ix The Hodges–Lehmann Estimator

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

The Walsh averages of n observations refers to all pairwise averages, $({\mathrm{Ten}}_i+X_j)/2$ , for all $i\le j$ . The Hodges–Lehmann computer is the median of all Walsh averages, namely,

${\hat{\theta }}_{\mathrm{HL}}={\mathrm{med}}_{i\le j}\frac{X_i+X_j}{2}.$

As noted in Section two.2, there are conditions where this mensurate of location, as well as R-estimators in general, has good backdrop. Just at that place are general conditions under which they perform poorly (e.m., Bickel and Lehmann, 1975; Huber, 1981, p. 65; Morgenthaler and Tukey, 1991, p. 15).

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Nonparametric Tests

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

9.3 The Wilcoxon Signed Rank Test

Another much more recently developed test that can be used to examine whether or non there is reversion toward the mean in the data in Example ix.1 is the Wilcoxon Signed Rank (WSR) exam. An American statistician, Frank Wilcoxon, who worked in the chemical manufacture, developed this test in 1945. Unlike the sign test which tin can be used with nonnumeric data, the WSR examination requires that the differences in the paired data come from a continuous distribution.

To apply the WSR test to examine whether or not there is reversion toward the mean, nosotros set the data as follows: The data for the 14 boys with an extreme twenty-four hours 1 value are shown in Tabular array ix.2. In this tabular array, the differences between day 1 and twenty-four hours 2 values are shown as either a change in the direction of the mean (+) or away from the hateful (−). If the day i and day ii values for a boy are the same, then we cannot assign a sign, and such a pair would exist excluded from the analysis. The accented differences are ranked from smallest to largest, and the ranks are summed separately for those changes in the management of mean and for those changes away from the hateful. We use R _WSRto represent the signed rank sum statistic for the positive differences — in this case, those changes toward the mean.

Tabular array ix.two. Days i and ii caloric intakes for the 14 boys with the more than extreme caloric intakes on 24-hour interval 1.

					Rank
ID	Twenty-four hours one	Day 2	Change (+) Toward the Mean	Change (−) Away from the Hateful	+	−
13	1,053	two,484	1,431		13
14	4,322	ii,926	1,396		12
16	i,753	i,054		699		10
27	three,532	3,289	243		3
30	ii,842	two,849		vii		1
33	i,505	ane,925	420		four
41	iii,076	two,431	645		9
fifty	ane,292	810		482		7
51	3,049	two,573	476		6
101	i,277	2,185	ane,092		11
118	one,781	i,844	63		2
130	2,773	three,236		463		five
149	1,645	two,269	624		8
150	1,723	3,163	1,440		xiv
				Sum of Ranks	82	23

We now consider the logic behind the testing of R _WSR. When there are north observations or pairs of data, the sum of the ranks is the sum of the integers from 1 to n and that sum is due north(n + 1)/2. The average rank for an observation is therefore (north + 1)/two.

The null hypothesis is that the differences have a median of zero and the alternative hypothesis that the median is not equal to zero for a two-sided examination or greater (or smaller) than zero for a one-sided test. If the null hypothesis is true, the distribution of the differences will exist symmetric, and in that location should exist n/2 positive differences and northward/2 negative differences. Therefore, if the null hypothesis is true, the sum of the ranks for positive (or negative) differences, R_WSR, should be (north/2) times the boilerplate rank: (n/ii)(n + 1)/2 = n(n + 1)/4.

The test statistic is the sum of the ranks of positive (or negative) differences, R _WSR. For a modest sample, Table B9 (due north < 30) provides boundaries for the disquisitional 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 event from the enumeration of possible outcomes as shown in Tabular array 9.3.

Table nine.3. Positive ranks for a sample of size 5 for 0, 1, and 2 positive ranks.

Number of Positive Ranks	Possible Ranks	Sum of Positive Ranks	Sum of Negative Ranks
0		0	xv
1	i	1	14
	2	2	13
	iii	3	12
	iv	4	eleven
	5	v	10
ii	one, ii	iii	12
	1, 3	4	11
	i, 4	five	10
	1, v	half-dozen	9
	2, iii	5	10
	2, four	six	9
	2, 5	seven	8
	3, 4	7	8
	3, 5	8	7
	4, 5	9	6

In Tabular array nine.iii, there is no need to show the sum of ranks for 3, iv, 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, at that place are 5 negative ranks with a sum of 15. Only the sum of 5 positive ranks must also be 15. When there is 1 positive rank, there are 4 negative ranks with the indicated sums. Only these are also the sum for the possibilities with four positive ranks. The same reasoning applies for 2 and 3 positive ranks.

Based on Table nine.3, nosotros can course Tabular array 9.4, which shows all the possible values of the sum and their relative frequency of occurrence. Using Table 9.four, we encounter that the smallest rejection region for a 2-sided test is 0 or 15, and this gives the probability of a Type I fault of 0.062. Thus, in Tabular array B9, there is no rejection region shown for a sample size of five and a significance level of 0.05. If the test of interest were a one-sided examination, so it would be possible to have a Type I error probability less than 0.05.

Table 9.iv. All possible sums and their relative frequency.

Sum	Frequency	Relative Frequency
0 or 15	1	0.031
1 or 14	1	0.031
2 or 13	1	0.031
3 or 12	2	0.063
iv or 11	two	0.063
v or ten	three	0.094
6 or 9	3	0.094
vii or 8	three	0.094

Example nine.3

Permit us return to the information prepared for the 14 pairs in Table 9.2. We shall perform the test at the 0.05 significance level, the same level used in the sign test. Since this is a one-sided test, we read the boundary in a higher place α ≤ 0.05 under i-sided comparisons shown at the bottom of the table, which is equivalent to α ≤ 0.10 under two-sided comparisons. Using the row n = 14, the critical values are (25, 80). Since our test statistic is 82, greater than 80, we decline the zilch hypothesis of no regression toward the mean in favor of the alternative that there is regression toward the mean.

This consequence is inconsistent with the result of the sign test in Example 9.1 and reflects the greater ability of the WSR exam. This greater ability is due to the utilize of more than of the information in the information by the WSR test compared to the sign test. The WSR examination incorporates the fact that the average rank for the four changes away from the mean is 5.75 (= [1 + 5 + vii + ten]/4), less than the average rank of vii.50. This lower average rank of these four changes, forth with the fact that there were only four changes away from the mean, caused the WSR examination to be significant. The sign examination used merely the number of changes toward the hateful, not the ranks of these changes, and was non significant. Although the sign exam failed to reject the null hypothesis, its p-value of 0.0898 was not that dissimilar from 0.05.

In applying the WSR test, two types of ties can occur in the data. One blazon is that some observed values are the same as the hypothesized value or some paired observations are the same — that is, the differences are zero. 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 1 for every unit of measurement or pair deleted. Again, this process is appropriate when there are only a few ties in the data. If in that location are many ties of this type, in that location is little reason to perform the examination.

The other type of tie occurs when two or more 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 aforementioned rank. For example, if 2 differences were tied as the smallest value, each would receive the rank of 1.five, the average of ranks 1 and two. If 3 differences were tied as the smallest value, each would receive the rank of 2, the average of ranks 1, 2, and iii. If in that location are few ties in the differences, the rank sum can still be used as the test statistic; still, the results of the test are at present 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 adjacent affiliate should exist used.

Instance 9.4

Let us use the WSR test to the data in Example 9.2. The 13 measurements, the deviations from the truthful value of 41, and ranks of absolute differences are equally follows:

Measures:	45	43	xl	44	49	36	51	46	35	50	41	38	47
Differences:	+ 4	+2	−1	+3	+eight	−five	+ten	+five	−6	+9	0	−iii	+6
Ranks:	5	two	one	three.v	10	6.v	12	6.five	8.v	xi	—	3.5	8.five

Notation that the boilerplate 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 ane ascertainment that has no rank. The test statistic (the sum of ranks for positive differences) is 58.5. Table B9 provides boundaries of the disquisitional region. For n = 12 and α ≤ 0.05 under the ii-sided comparing the boundaries are (13, 65). Since the test statistic is less than 65, we neglect to refuse the null hypothesis in favor of the alternative hypothesis at the 0.05 significance level. This conclusion is consistent with the upshot of the sign test in Example 9.2.

For a large sample, the normal approximation is used. If there are at least 16 pairs of observations used in the calculations, R _WSR will approximately follow a normal distribution. Every bit we only saw, the expected value of R_WSR, under the assumption that the null hypothesis is truthful, is n(n + 1)/4, and its variance can be shown to exist due north(north + 1)(2n + 1)/24. Therefore, the statistic

$\frac{\left|R_{w\mathrm{south}r}-\left[n\left(\mathrm{north}+1\right)/\mathrm{four}\right]\right|-\mathrm{0.five}}{\sqrt{n\left(n+\mathrm{i}\right)\left(2n+\mathrm{i}\right)/24}}$

approximately follows the standard normal distribution. The 2 vertical lines in the numerator indicate the accented 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 us calculate the normal approximation to the pairs in Example 9.three. The expected value of R_WSR is 52.5 (= [fourteen][15]/iv), and the standard error is $\mathrm{xv.93}\left(=\sqrt{\left(\mathrm{xiv}\right)\left(15\right)\left(29\right)/24}\right)$ . Therefore, the statistic's value is

What is the probability that Z is greater than 1.82? This probability is constitute 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 16,384 possible signed rank sums having a value of 82 or greater, applying the same logic illustrated in Tables 2 and 3 to the case n = 14. Thus, even though due north is less than xvi, the normal approximation worked quite well in this instance. The WSR exam can be performed past the reckoner (see Program Note 9.i on the website).

The sign and Wilcoxon Signed Rank tests are both used most often in the comparison of paired information, although they can be used with a unmarried population to test that the median has a specified value. In the apply 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 have an impact during the study. Otherwise, the possibility of the confounding of the extraneous cistron with the intervention variable is raised. In addition, the research designer must consider whether or non reversion to the mean is a possibility. If inapplicable factors or reversion to the mean cannot be ruled out, the research design should be augmented to include a command grouping to assist account for the effect of these possibilities.

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Data Science: Theory and Applications

Sunil Mathur , in Handbook of Statistics, 2021

3 1-sample methods

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

We consider the classical i-sample location problem with univariate information. Let x _one, x _ii, …, 10 _north be an independent random sample of size north from a continuous distribution with distribution function F. Let the hypothesized cumulative distribution role be denoted by F ₀(x) and the empirical distribution function be denoted past Due south _n (10) for all x. The hypothesis to be tested is H ₀  : F  = F _o vs H _a   : F  ≠ F _o . If the nada hypothesis is true then the difference between S _n (x) and F ₀(ten) must be close to zero.

Thus, for large n, the test statistic

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

will have a value shut to zero under the null hypothesis.

The test statistic, D _northward , called the Kolmogorov-Smirnov one-sample statistic (Gibbons and Chakraborti, 2014), does non depend on the population distribution function if the distribution function is continuous and hence D _n is a distribution-costless exam statistic. The goodness-of-fit exam for a sample was proposed by Kolmogorov (1933). The Kolmogorov-Smirnov test for 2 samples was-proposed by Smirnov (1939).

Here we define order statistic 10 ₍₀₎  =   −   ∞ and Ten _{(n  +   ane)}  =   ∞, and

(two) $S_{\mathrm{northward}}\left(\mathrm{10}\right)=\frac{i}{n},\mathit{for}{X}_{\left(i\right)}\le x\le X_{\left(i+1\right)},i=0,\mathrm{one},.\dots \dots ,\mathrm{northward}.$

The probability distribution of the exam statistic does not depend on the distribution office F _X (X) for a continuous distribution function F _X (Ten). The asymptotic distribution of the test statistic D _n is Chi-foursquare.

The exact sampling distribution of the Kolmogorov-Smirnov test statistic is known while the distribution of the Chi-foursquare goodness-of-fit exam statistic is approximately Chi-square for finite northward. Moreover, the Chi-foursquare goodness-of-fit test requires that the expected number of observations in a cell must be greater than five while the Kolmogorov examination statistic does non require this status. On the other hand, the asymptotic distribution of the Chi-square goodness-of-fit test statistic does not crave that the distribution of the population must exist continuous but the exact distribution of the Kolmogorov-Smirnov test statistic does require that F ₁₀ (X) must exist a continuous distribution. The power of the Chi-square distribution depends on the number of classes or groups made.

The Wilcoxon signed-rank test ( Wilcoxon, 1945) requires that the parent population should be symmetric. When data is collected in batches, the data might be symmetric at some betoken, peculiarly in the case of seasonal and periodic data. Allow the states consider a random sample X _ane, X _two, …. . , Ten _n from a continuous cdf F which is symmetric almost its median M. The naught hypothesis can be stated every bit

(3) $H_0:G={\mathrm{Thou}}_0$

The alternative hypotheses can be postulated appropriately. Nosotros detect that the differences D _i   = X _i   − K ₀ are symmetrically distributed about cypher, and hence the number of positive differences will exist equal to the number of negative differences. The ranks of the differences |  D ₁  |, |  D ₂  |, . ……………, |  D _Northward   | are denoted by Rank(.). Then, the test statistic tin can be defined as

(4) $T^{+}=\sum _{i=1}^n{a}_i\mathit{Rank}\left(\left|D_i\right|\right)$

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

where

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

Since the indicator variables a _i are independent and identically distributed Bernoulli variates with P(a _i   =   one)   = P(a _i   =   0)   =   ½, therefore, under the zero hypothesis

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

and

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

Another common representation for the test statistic T ⁺ is given as follows.

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

where

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

Similar expressions can be derived for T ⁻. The paired-samples can be defined based on the differences X ₁  − Y ₁, X ₂  − Y ₂,..…, X _n   − Y _north of a random sample of n pairs (Ten ₁, Y _one), (X ₂, Y ₂)..…, (Ten _n , Y _north ). Now, these differences are treated as a unmarried sample and the one-sample exam procedure is applied. The null hypothesis to be tested will be

$H_0:M=M_0$

where M ₀ is the median of the differences X ₁  − Y _one, X _two  − Y ₂,..…, X _{due north}   − Y _n . These differences can be treated equally a unmarried sample with the hypothetical median K ₀. Then, the Wilcoxon signed-rank method described in a higher place for a single sample can exist applied to test the nix hypothesis that the median of the differences is K ₀.

Since a expert examination must exist not only fast in calculating the test value just as well should accept the power in finding out information hidden in big data. Wilcoxon signed-rank examination 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 caste k (>   1) to develop a i-sample nonparametric test. Define a kernel (k   =   2):

(xi) ${\psi }^{\left(\mathrm{two}\right)}\left({\mathrm{10}}_i,X_j\right)=\left\{\begin{array}{c}\mathrm{i}\mathit{if}\frac{{\mathrm{10}}_i+X_j}{2}\ge 0\\ 0\mathit{if}\frac{{\mathrm{Ten}}_i+X_j}{2}<0\end{array}\right.$

This kernel is equivalent to U-Statistic of degree 2

(12) $S_n^{\left(2\right)}={\left(\begin{array}{c}n\\ 2\end{array}\right)}^{-\mathrm{i}}\sum _{\mathrm{ane}\le i_1<i_2\le \mathrm{north}}{\psi }^{\left(2\right)}\left({\mathrm{10}}_{i_1},X_{i_{\mathrm{ii}}}\right)$

Both the sign test and the Isle of mann-Whitney test involve U-statistics with a symmetric kernel of degree i, 1, and 1 respectively.

A general exam statistic (Chattopadhyay and Mukhopadhyay, 2019) based on the kernel of k (<   n) can be defined as:

(13) ${\mathrm{Due\; south}}_{\mathrm{northward}}^{\left(\mathrm{thousand}\right)}={\left(\begin{array}{c}n\\ k\end{array}\right)}^{-1}\sum _{1\le i_1<\dots <i_{\mathrm{one\; thousand}}\le \mathrm{due\; north}}{\psi }^{\left(k\right)}\left({\mathrm{10}}_{i_{\mathrm{ane}}},\dots ,{\mathrm{10}}_{i_k}\right)$

where

(14) ${\psi }^{\left(k\right)}\left({\mathrm{10}}_{i_1},\dots ,{\mathrm{10}}_{i_k}\right)=\left\{\begin{array}{c}1\mathit{if}{\overline{X}}_{i_k}\ge 0\\ 0\mathit{if}{\overline{X}}_{i_{\mathrm{1000}}}<0\end{array}\right.$

and ${\overline{X}}_{i_{\mathrm{1000}}}=\frac{\mathrm{ane}}{k}{\sum }_{j=1}^k{X}_{i_j,}$ one   ≤ i ₁  <   …   < i _k   ≤ northward for due north   >   k.

The zip hypothesis, every bit given past argument (3), can exist tested at a level α using the post-obit benchmark:

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

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

Since there is only one sample available for such problems, nosotros try to increase the number of samples artificially using constructed sample generation from a 3D model of the available epitome. The new dataset with multiple artificially generated samples tin exist used as a gallery ready and the probe set contains images from surveillance cameras in an unconstrained surround. Now one can select 2D facial points and the landmark points in the 3D model in the gallery set up and discover median points at each landmark signal. Similarly, select those points in the probe set and run the one-sample exam, such equally Eq. (9) at each landmark bespeak. The larger similarities at landmark points will point toward similarities of probe and gallery and probe images.

The problem is generally faced when the interest is in identifying an individual at decorated mutual places such equally airports, train stations, and public coming together places. That will involve matching the gallery set with a probe set containing millions for images. In order to accomplish that job quickly and efficiently, ane can set up three layers of batch processing. The first layer of processing involves eliminating the data which has more than two standard deviations of variations in major landmarks. Thus, the probe data containing a wider face or too pocket-size face will get eliminated. In the second layer, matching of finer landmarks such every bit ear length and width, olfactory organ length is washed. The data having 2 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. 2. Converting 2nd facial model to 3D model and increasing virtual sample size for gallery data.

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

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