Modeling Hydrologic Change: Statistical Methods - Chapter 8 pps

cumulative distribution FT is:8.3 When the alternative hypothesis suggests a one-tailed lower test, the critical value is the largest value of T for which α ≥ FT.. For a 1% level of sign

is described by the underlying probability distribution, which includes both theparameters and the function Change in either the meteorological or the hydrologicprocesses can induce change in the underlying population through either a change

in the probability function or the parameters of the function

A sequential series of hydrologic data that has been affected by watershed change

is considered nonstationary Some statistical methods are most sensitive to changes

in the moments, most noticeably the mean and variance; other statistical methodsare more sensitive to change in the distribution of the hydrologic variable Selecting

a statistical test that is most sensitive to detecting a difference in means when thechange in the hydrologic process primarily caused a change in distribution may lead

to the conclusion that hydrologic change did not occur It is important to hypothesizethe most likely effect of the hydrologic change on the measured hydrologic data sothat the most appropriate statistical test can be selected Where the nature of thechange to the hydrologic data is uncertain, it may be prudent to subject the data totests that are sensitive to different types of change to decide whether the change inthe hydrologic processes caused a noticeable change in the measured data

In this chapter, tests that are sensitive to changes in moments are introduced Inthe next chapter, tests that are appropriate for detecting change in the underlyingprobability function are introduced All of the tests follow the same six steps ofhypothesis testing (see Chapter 3), but because of differences in their sensitivities,the user should be discriminating in the selection of a test

8

A change in the variance of a hydrologic variable will be characterized by achange in spread of the data For example, an increase in variance due to hydrologicchange would appear in the elongation of the tails of a histogram A change in slope

of a frequency curve would also suggest a change in variance An increase or decrease

in the amplitude of a periodic function would be a graphical indication of a change

in variance of the random variable

Once a change has been detected graphically, statistical tests can be used toconfirm the change While the test does not provide a model of the effect of change,

it does provide a theoretical justification for modeling the change Detection is thefirst step, justification or confirmation the second step, and modeling the change isthe final step This chapter and Chapter 9 concentrate on tests that can be used tostatistically confirm the existence of change

8.3 THE SIGN TEST

The sign test can be applied in cases that involve two related samples The name ofthe test implies that the random variable is quantified by signs rather than a numericalvalue It is a useful test when the criterion cannot be accurately evaluated but can

be accurately ranked as being above or below some standard, often implied to meanthe central tendency Measurement is on the ordinal scale, within only three possibleoutcomes: above the standard, below the standard, or equal to the standard Symbolssuch as +, −, and 0 are often used to reflect these three possibilities

The data consist of two random variables or a single criterion in which thecriterion is evaluated for two conditions, such as before treatment versus aftertreatment A classical example of the use of the sign test would be a poll in whichvoters would have two options: voting for or against passage The first part of theexperiment would be to record each voter’s opinion on the proposed legislation,that is, for or against, + or − The treatment would be to have those in the randomsample of voters read a piece of literature that discusses the issue Then they would

be asked their opinion a second time, that is, for or against Of interest to thepollster is whether the literature, which may possibly be biased toward one decision,can influence the opinion of voters This is a before–after comparison Since thebefore-and-after responses of the individuals in the sample are associated with theindividual, they are paired The sign test would be the appropriate test for analyzingthe data

The hypotheses can be expressed in several ways, with the appropriate alternativepair of hypotheses selected depending on the specific situation One expression ofthe null hypothesis is

H0: P (X > Y) =P (X < Y) = 0.5 (8.1a)

in which X and Y are the criterion values for the two conditions In the polling case,

X would be the pre-treatment response (i.e., for or against) and Y would be the treatment response (i.e., for or against) If the treatment does not have a significanteffect, then the changes in one direction should be balanced by changes in the

post-other direction If the criterion did not change, then the values of X and Y are equal,

which is denoted as a tie Other ways of expressing the null hypothesis are

H0: P (+) =P (−) (8.1b)or

H0: E (X) =E (Y) (8.1c)

where + indicates a change in one direction while − indicates a change in the other

direction, and the hypothesis expressed in terms of expectation suggests that the two

conditions have the same central tendency

Both one- and two-tailed alternative hypotheses can be formulated:

H A: P (X > Y) > P (X < Y) or P (+) > P (−) (8.2a)

H A: P (X > Y) < P (X < Y) or P (+) < P (−) (8.2b)

H A: P (X > Y) ≠P (X < Y) or P (+) ≠P (–) (8.2c)

The alternative hypothesis selected would depend on the intent of the analysis

To conduct the test, a sample of N is collected and measurements on both X and

Y made, that is, pre-test and post-test Given that two conditions are possible, X and

Y, for each test, the analysis can be viewed as the following matrix:

Only the responses where a change was made are of interest The treatment would

have an effect for cells (X, Y) and (Y, X) but not for cells (X, X) and (Y, Y) Thus, the

number of responses N12 and N21 are pertinent while the number of responses N11

and N 22 are considered “ties” and are not pertinent to the calculation of the sample

test statistic If N=N11+N22+N12+N21 but only the latter two are of interest, then

the sample size of those affected is n=N12+N21, where n≤N The value of n, not

N, is used to obtain the critical value

The critical test statistic depends on the sample size For small samples of about

20 to 25 or less, the critical value can be obtained directly from the binomial

distribution For large samples (i.e., 20 to 25 or more), a normal approximation is

applied The critical value also depends on the alternative hypothesis: one-tailed

lower, one-tailed upper, and two-tailed

For small samples, the critical value is obtained from the cumulative binomial

distribution for a probability of p = 0.5 For any value T and sample size n, the

Post-test

cumulative distribution F(T) is:

(8.3)

When the alternative hypothesis suggests a one-tailed lower test, the critical value

is the largest value of T for which α ≥ F(T) For a one-tailed upper test, the critical value is the value of T for which α ≥ F(n − T) For a two-tailed test, the lower and upper critical values are the values of T for which α/2 ≥ F(T) and α/2 ≥ F(n − T) Consider the case where n = 13 Table 8.1 includes the mass and cumulative

mass functions as a function of T For a 1% level of significance, the critical value for a one-tailed lower test would be 1 since F(2) is greater than α For a 5% level

of significance, the critical value would be 3 since F(4) is greater than α For a tailed upper test, the critical values for 1% and 5% are 12 and 10 For a two-tailedtest, the critical values for a 1% level of significance are 1 and 12, which means thenull hypothesis is rejected if the computed value is equal to 0, 1, 12, or 13 For a5% level of significance the critical values would be 2 and 11, which means thatthe null hypothesis is rejected if the computed value is equal to 0, 1, 2, 11, 12, or

one-13 Even though α is set at 5%, the actual rejection probability is 2(0.01123) =0.02246, that is, 2.2%, rather than 5% Table 8.2 gives critical values computed usingthe binomial distribution for sample sizes of 10 to 50

TABLE 8.1 Binomial Probabilities for Sign Test

with n ==== 13; f(T) ==== Mass Function and

F(T ) ==== Cumulative Mass Function

i T

For large sample sizes, the binomial mass function can be approximated with anormal distribution The mean and standard deviation of the binomial distribution are

Equation 8.6 gives a slightly biased estimate of the true probability A better estimate

can be obtained by applying a continuity correction to x, such that z is

(8.7)

Equation 8.7 can be rearranged to solve for the random variable x, which is the

upper critical value for any level of significance:

0 5

0 5 0 5

.

n

=( + )− . .

0 5 0 5

0 5 0 5

x=0 5 n+0 5 zn0 5. −0 5 =0 5 (n+zn0 5. −1)

Enter Equation 8.8 with the standard normal deviate z corresponding to the rejection

probability to compute the critical value Equation 8.8 yields the upper bound Forthe one-sided lower test, use −z Values obtained with this normal transformation

are approximate It would be necessary to round up or round down the value obtainedwith Equation 8.8 The rounded value can then be used with Equation 8.7 to estimatethe actual rejection probability

Example 8.1

Mulch is applied to exposed soil surfaces at 63 sites, with each slope divided intotwo equal sections The density of mulch applied to the two sections is different(i.e., high and low) The extent of the erosion was qualitatively assessed during stormevents with the extent rated as high or low The sign test is used to test the followinghypotheses:

H0: The density of mulch does not affect the amount of eroded soil

H A: Sites with the higher density of mulch experienced less erosion

The statement of the alternative hypothesis dictates the use of a one-sided test

Of the 63 sites, 46 showed a difference in eroded soil between the two sections,with 17 sites not showing a difference Of the 46 sites, the section with higher mulchdensity showed the lower amount of eroded material on 41 sites On 5 sites, thesection with higher mulch density showed the higher amount of erosion Therefore,the test statistic is 5 For a sample size of 46, the critical values for 1% and 5%levels of significance are 14 and 17, respectively (see Table 8.2) Therefore, the nullhypothesis should be rejected To use the normal approximation, the one-sided 1%

and 5% values of z are −2.327 and −1.645, which yield critical values of 14.6 and16.9, respectively These normal approximations are close to the actual binomialvalues of 14 and 17

In some cases, samples are obtained from two different populations, and it is ofinterest to determine if the population means are equal For example, two laboratoriesmay advertise that they evaluate water quality samples of some pollutant with anaccuracy of ±0.1 mg/L; samples may be used to test whether the means of the twopopulations are equal Similarly, tests could be conducted on engineering products

to determine whether the means are equal The fraction of downtime for two puter types could be tested to decide whether the mean times differ

com-A number of tests can be used to test a pair of means The method presentedhere should be used to test the means of two independent samples This test isfrequently of interest in engineering research when the investigator is interested incomparing an experimental group to a control group For example, an environmentalengineer might be interested in comparing the mean growth rates of microorganisms

in a polluted and natural environment The procedure presented in this section can

be used to make the test

Step 1: Formulate hypotheses The means of two populations are denoted as

µ1 and µ2, the null hypothesis for a test on two independent means would be:

H0: The means of two populations are equal (8.9a)Mathematically, this is

Step 2: Select the appropriate model For the case of two independent samples,

the hypotheses of step 1 can be tested using the following test statistic:

(8.11)

in which and are the means of the samples drawn from populations

1 and 2, respectively; n1 and n2 are the sample sizes used to compute and

, respectively; t is the value of a random variable that has a t distribution

with degrees of freedom (υ) of ν = n1+ n2− 2; and S p is the square root ofthe pooled variance that is given by

(8.12)

in which and are the variances of the samples from population 1 and 2,respectively This test statistic assumes that the variances of the two popu-lations are equal, but unknown

Step 3: Select the level of significance As usual, the level of significance

should be selected on the basis of the problem However, values of either5% or 1% are used most frequently

Step 4: Compute an estimate of test statistic Samples are drawn from the two

populations, and the sample means and variances computed Equation 8.11can be computed to test the null hypothesis of Equation 8.9

Step 5: Define the region of rejection The region of rejection is a function of

the degrees of freedom (ν = n1 + n2 − 2), the level of significance (α), andthe statement of the alternative hypothesis The regions of rejection for thealternative hypotheses are as follows:

Step 6: Select the appropriate hypothesis The sample estimate of the t statistic

from step 4 can be compared with the critical value (see AppendixTable A.2), which is based on either tα or tα /2 obtained from step 5 If thesample value lies in the region of rejection, then the null hypothesis should

be rejected

Example 8.2

A study was made to measure the effect of suburban development on total nitrogenlevels in small streams A decision was made to use the mean concentrations beforeand after the development as the criterion Eleven measurements of the total nitrogen(mg/L) were taken prior to the development, with a mean of 0.78 mg/L and a standarddeviation of 0.36 mg/L Fourteen measurements were taken after the development,with a mean of 1.37 mg/L and a standard deviation of 0.87 mg/L The data are used

to test the null hypothesis that the population means are equal against the alternativehypothesis that the urban development increased total nitrogen levels, which requiresthe following one-tailed test:

where µb and µa are the pre- and post-development means, respectively Rejection

of H0 would suggest that the nitrogen levels after development significantly exceedthe nitrogen levels before development, with the implication that the developmentmight have caused the increase

Based on the sample data, the pooled variance of Equation 8.12 is

The computed value of the test statistic is

(8.16)

which has ν = 11 + 14 − 2 = 23 degrees of freedom From Table A.2, with a 5%

level of significance and 23 degrees of freedom, the critical value of t is −1.714.Thus, the null hypothesis is rejected, with the implication that the developmentcaused a significant change in nitrogen level However, for a 1% significance level,the critical value is −2.500, which leads to the decision that the increase is notsignificant This shows the importance of selecting the level of significance beforeanalyzing the data

8.5 MANN–WHITNEY TEST

For cases in which watershed change occurs as an episodic event within the duration

of a flood series, the series can be separated into two subseries The Mann–Whitney

U-test (Mann and Whitney, 1947) is a nonparametric alternative to the t-test for two

independent samples and can be used to test whether two independent samples havebeen taken from the same population Therefore, when the assumptions of the

parametric t-test are violated or are difficult to evaluate, such as with small samples,

the Mann–Whitney U-test should be applied This test is equivalent to the Wilcoxon–

Mann–Whitney sum test described in many textbooks as a t-test on the

rank-transformed data (Inman and Conover, 1983)

The procedure for applying the Mann–Whitney test follows

1 Specify the hypotheses:

H0: The two independent samples are drawn from the same population

H A: The two independent samples are not drawn from the same lation

popu-The alternative hypothesis shown is presented as a two-sided hypothesis;one-sided alternative hypotheses can also be used:

H A: Higher (or lower) values are associated with one part of the series.For the one-sided alternative, it is necessary to specify either higher orlower values prior to analyzing the data

2 The computed value (U) of the Mann–Whitney U-test is equal to the lesser

of U a and U b where

U a = n a n b + 0.5 n b (n b + 1) − S b (8.17a)

U b = n a n b + 0.5 n a (n a + 1) − S a (8.17b)

in which n a and n b are the sample sizes of subseries A and B, respectively

The values of S a and S b are computed as follows: the two groups are

114

2 104

combined, with the items in the combined group ranked in order fromsmallest (rank = 1) to the largest (rank = n = n a + n b) On each rank, include

a subscript a or b depending on whether the value is from subseries A or

B S a and S b are the sums of the ranks with subscript a and b, respectively. Since S a and S b are related by the following, only one value needs to becomputed by actual summation of the ranks:

5 Obtain the critical value (Zα /2 for a two-sided test or Zα for a one-sided

test) from a standard-normal table (Table A.1)

6 Reject the null hypothesis if the computed value of Z (step 4) is greater than Zα /2 or less than −Zα /2 (Zα for an upper one-sided test or −Zα for a

lower, one-sided test)

In some cases, the hydrologic change may dictate the direction of change in theannual peak series; in such cases, it is appropriate to use the Mann–Whitney test as

a one-tailed test For example, if channelization takes place within a watershed, thepeaks for the channelized condition should be greater than for the natural watershed

To apply the U-test as a one-tailed test, specify subseries A as the series with the

smaller expected central tendency and then use U a as the computed value of the test

statistic, U = U a (rather than the lesser of U a and U b ); the critical value of Z is −Zα

(rather than −Zα /2) and the null hypothesis is accepted when Z > − Zα

Example 8.3

The 65-year annual maximum series for the Elizabeth River appears (Figure 7.2) tochange in magnitude about 1951 when the river was channelized To test if thechannelization was accompanied by an increase in flood peaks, the flood series wasdivided into two series, 1924 to 1951 (28 years) and 1952 to 1988 (37 years) Statisticalcharacteristics for the periods are given in Table 8.3

Since the direction of change was dictated by the problem (i.e., the peak charges increased after channelization), the alternative hypothesis is

dis-H A: The two independent samples are not from the same population andthe logarithms of the peaks for 1924–1951 are expected to be less thanthose for 1952–1988

Analysis of the annual peak discharges yields U a = 200 and U b= 836 Since a

one-tailed test is made, U = U a Using Equation 8.18, the computed value of Z is −4.213.For a one-tailed test and a level of significance of 0.001, the critical value is −3.090

Since Z < Zα, the null hypothesis is rejected The results of the Mann–Whitney test suggest that channelization significantly increased the peak discharges

U-8.5.1 R ATIONAL A NALYSIS OF THE M ANN –W HITNEY T EST

When applying the Mann–Whitney test, the flows of the annual maximum series aretransformed to ranks Thus, the variation within the floods of the annual series isreduced to variation of ranks The transformation of measurements on a continuousvariable (e.g., flow) to an ordinal scale (i.e., the rank of the flow) reduces the importance

of the between-flow variation However, in contrast to the runs test, which reducesthe flows to a nominal variable, the importance of variation is greater for the Mann–Whitney test than for the runs test Random variation that is large relative to thevariation introduced by the watershed change into the annual flood series will morelikely mask the effect of the change when using the Mann–Whitney test compared

to the runs test

This can be more clearly illustrated by examining the statistics U a and U b ofEquations 8.17 The first two terms of the equation represent the maximum possible

total of the ranks that could occur for a series with n a and n b elements The third

terms, S a and S b, represent the summations of the actual ranks in sections A and B

of the series Thus, when S a and S b are subtracted, the differences represent

devia-tions If a trend is present, then one would expect either S a or S b to be small and the

other large, which would produce a small value of the test statistic U If the flows

exhibit random variation that is large relative to the variation due to watershed

change, then it can introduce significant variation into the values of S a and S b, therebymaking it more difficult to detect a trend resulting from watershed change In asense, the test is examining the significance of variation introduced by watershedchange relative to the inherent natural variation of the flows

In addition to the variation from the watershed change, the importance of thetemporal location of the watershed change is important In contrast to the runs test,the Mann–Whitney test is less sensitive to the location of the trend It is more likely

to detect a change in flows that results from watershed change near one of the ends

TABLE 8.3

Annual Peak Discharge Characteristics for Elizabeth River,

New Jersey, Before and After Channelization

Characteristics of Series Logarithms Period

Mean (ft 3 /s)

Standard Deviation (ft 3 /s) Mean

Standard Deviation Skew

of the series than the runs test If watershed change occurs near an end of the series,

then either N a or N b of Equations 8.17 would be small, but the sum of the ranks

would also be small The summation terms S a and S b decrease in relative proportion

to the magnitudes of the first two terms Thus, a change near either end of a series

is just as likely to be detected as a change near the middle of the series

The Mann–Whitney test is intended for the analysis of two independent samples.When applying the test to hydrologic data, a single series is separated into two partsbased on a watershed-change criterion Thus, the test would not strictly apply if thewatershed change occurred over an extended period of time Consequently, it may

be inappropriate to apply the test to data for watershed change that occurred graduallywith time Relatively abrupt changes are more appropriate

8.6 THE t-TEST FOR TWO RELATED SAMPLES

The t-test for independent samples was introduced in Section 8.4 The t-test described

here assumes that the pairs being compared are related The relationship can arise

in two situations First, the random sample is subjected to two different treatments

at different times and administered the same test following each treatment The twovalues of the paired test criterion are then compared and evaluated for a significant

difference As an example, n watersheds could be selected and the slope of the main

channel computed for maps of different scales The same computational method isused, but different map scales may lead to different estimates of the slope Thelengths may differ because the map with the coarser scale fails to show the samedegree of meandering The statistical comparison of the slopes would seek to deter-mine if the map scale makes a significant difference in estimates of channel slope.The second case where pairing arises is when objects that are alike (e.g., identicaltwins) or are similar (e.g., adjacent watersheds) need to be compared In this case,the pairs are given different treatments and the values of the test criterion compared.For example, pairs of adjacent watersheds that are similar in slope, soils, climate, andsize, but differ in land cover (forested vs deforested) are compared on the basis of acriterion such as base flow rate, erosion rate, or evapotranspiration (ET) A significantdifference in the criterion would indicate the effect of the difference in the land cover

The hypotheses for this test are the same as those for the t-test for two

in which n is the number of pairs and

(8.20)

where D i is the difference between the ith pair of scores and is the average

difference between scores The test statistic is:

with (8.21)

This test assumes that the difference scores are normally distributed A log formation of the data may be used for highly skewed data If the difference scoresdeviate significantly from normality, especially if they are highly asymmetric, theactual level of significance will differ considerably from the value used to find thecritical test statistic

trans-The computed test statistic is compared with the tabled t value for level of

significance α and degrees of freedom υ The region of rejection depends on thestatement of the alternative hypothesis, as follows:

Example 8.4

Two different methods of assigning values of Manning’s roughness coefficient (n)

are used on eight small urban streams The resulting values are given in columns 2and 3 of Table 8.4 The two-tailed alternative hypothesis is used since the objective

is only to decide if the methods give, on the average, similar values The value of

is equal to −0.027/8 = −0.003375 Results of Equation 8.20 follow:

i

i i

i i

i i

The test statistic is:

(8.24)

with 7 degrees of freedom For a two-tailed test and a level of significance of 20%,the critical value is ±1.415 (see Table A.2) Therefore, little chance exists that thenull hypothesis should be rejected In spite of differences in the methods of estimat-ing the roughness coefficients, the methods do not give significantly different values

on the average

Example 8.5

The paired t-test can be applied to baseflow discharge data for two watersheds of

similar size (see Table 8.5) It is generally believed that increases in the percentage

of urban/suburban development cause reductions in baseflow discharges becauseless water infiltrates during storm events It is not clear exactly how much landdevelopment is necessary to show a significant decrease in baseflow The baseflows

of two pairs of similarly sized watersheds will be compared The baseflow (ft3/sec)based on the average daily flows between 1990 and 1998 were computed for eachmonth Since baseflow rates depend on watershed size, both pairs have similar areas,with percentage differences in the area of 7.50% and 3.16% Such area differencesare not expected to have significant effects on the differences in baseflow

For the first pair of watersheds (USGS gage station numbers 1593500 and1591000) the monthly baseflows are given in columns 2 and 3 of Table 8.5 The twowatersheds have drainage areas of 37.78 mi2 and 35.05 mi2 For the larger watershed,the percentages of high density and residential area are 1.25% and 29.27%, respec-tively For the smaller watershed, the corresponding percentages are 0% and 1.08%

A similar analysis was made for a second pair of watersheds (USGS gage station

TABLE 8.4

Comparison of Roughness Coefficients Using the Two-Sample

t-Test for Related Samples: Example 8.5

numbers 1583600 and 1493000) that have areas of 21.09 mi2 and 20.12 mi2, tively The larger watershed has a percentage of high-density development of 1.04%and of residential development of 20.37%, while the smaller watershed has corre-sponding percentages of 0.04% and 1.77%, respectively The first pair of watersheds

respec-is compared first The mean and standard deviation of the difference scores in column

4 of Table 8.5 are −2.733 and 3.608, respectively Using Equation 8.21 the computedvalue of the statistic is:

with (8.25)

The critical t values for levels of significance of 5%, 2.5%, and 1% are 1.796, 2.201,

and 2.718, respectively Thus, the rejection probability is approximately 1.5% Thenull hypothesis of equal means would be rejected at 5% but accepted at 1%.For the second pair of paired watersheds, the difference scores (column 7 ofTable 8.5) have a mean and standard deviation of 2.708 and 3.768, respectively The

computed value of the t statistic is:

Since the critical t values are negative because this is a one-tailed lower test, the

null hypothesis cannot be rejected The mean baseflow discharges are not cantly different in spite of the differences in watershed development

signifi-TABLE 8.5

Two-Sample t-Test of Baseflow Discharge Rates

Watershed Pair 1 Watershed Pair 2

Yιιιι

Gage No.

1591000 (3)

Difference Scores

Yιιιι

Gage No.

1493000 (6)

Difference Scores

8.7 THE WALSH TEST

The Walsh test is a nonparametric alternative to the two-sample t-test for related

samples It is applicable to data measured on an interval scale The test assumes thatthe two samples are drawn from symmetrical populations; however, the two popu-lations do not have to be the same population, and they do not have to be normaldistributions If the populations are symmetrical, then the mean of each populationwill equal the median

The null hypothesis for the Walsh test is:

H0: The average of the difference scores (µd) is equal to zero (8.27a)Both one-tailed and two-tailed alternative hypotheses can be tested:

The test statistic for the Walsh test is computed based on the differences betweenthe paired observations Table 8.6 gives the test statistics, which differ with thesample size, level of significance, and tail of interest The critical value is alwayszero, so for a one-tailed lower test (Equation 8.27a), one of the test statistics incolumn 4 of Table 8.6 should be used For a one-tailed upper test (Equation 8.27b),one of the test statistics of column 5 should be used For a two-tailed test (Equation8.27c), the test statistics of both columns 4 and 5 are used

The procedure for conducting Walsh’s test is as follows:

1 Arrange the data as a set of related pairs: (x i , y i ), i = 1, 2, …, n.

2 For each pair, compute the difference in the scores, d i : d i = x i − y i (note

that the values of d i can be positive, negative, or zero)

3 Arrange the difference scores in order of size, with the most negativevalue given a rank of 1 to the algebraically largest value with a rank of

n Tied values of d i are given sequential ranks, not average ranks

There-fore, the d values are in order such that d1≤ d2≤ d3≤ … ≤ d n (Note that

the subscript on d now indicates the order according to size, not the

number of the pair of step 2.)

4 Compute the sample value of the test statistic indicated in Table 8.6

5 The critical test statistic is always zero

One-Tailed

Tailed One-Tailed: Accept µµµµd < 0 If One-Tailed: Accept µµµµd>>>> 0 If

.031

.125 062

6 Compare the computed value (W L and/or W U) to the critical value of zero,

as indicated in the table The decision to accept or reject is indicated by:

be rejected for a level of significance of 4.7% but would be accepted for the smallerlevels of significance It may be of interest to note that the computed values for thethree smaller levels of significance are equal to the critical value of zero While this

is not sufficient for rejection, it may suggest that the baseflows for the more highlydeveloped watershed tend to be lower In only 1 month (October) was the baseflowfrom the more developed watershed higher than that for the less developed water-shed This shows the importance of the selection of the level of significance, as well

as the limitations of statistical analysis

Application of Walsh Test to Baseflow Discharge Data

Difference Scores

Y i

Gage No.

1493000 (7)