Modeling Hydrologic Change: Statistical Methods - Chapter 8 pps

Detection of Change in Moments 8.1 INTRODUCTION Hydrologic variables depend on many factors associated with the meteorological and hydrologic processes that govern their behavior. The measured values are often treated as random variables, and the probability of occurrence of all random variables is described by the underlying probability distribution, which includes both the parameters and the function. Change in either the meteorological or the hydrologic processes 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 methods are 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 the change 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 hypothesize the most likely effect of the hydrologic change on the measured hydrologic data so that the most appropriate statistical test can be selected. Where the nature of the change to the hydrologic data is uncertain, it may be prudent to subject the data to tests that are sensitive to different types of change to decide whether the change in the hydrologic processes caused a noticeable change in the measured data. In this chapter, tests that are sensitive to changes in moments are introduced. In the next chapter, tests that are appropriate for detecting change in the underlying probability function are introduced. All of the tests follow the same six steps of hypothesis testing (see Chapter 3), but because of differences in their sensitivities, the user should be discriminating in the selection of a test. 8.2 GRAPHICAL ANALYSIS Graphical analyses are quite useful as the starting point for detecting change and the nature of any change uncovered. Changes in central tendency can be seen from a shift of a univariate histogram or the movement up or down of a frequency curve. For a periodic or cyclical hydrologic processes, a graphical analysis might reveal a shift in the mean value, without any change in the amplitude or phase angle of the periodic function. 8 L1600_Frame_C08.fm Page 173 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC A change in the variance of a hydrologic variable will be characterized by a change in spread of the data. For example, an increase in variance due to hydrologic change 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 to confirm 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 the first step, justification or confirmation the second step, and modeling the change is the final step. This chapter and Chapter 9 concentrate on tests that can be used to statistically 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 of the test implies that the random variable is quantified by signs rather than a numerical value. 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 mean the central tendency. Measurement is on the ordinal scale, within only three possible outcomes: above the standard, below the standard, or equal to the standard. Symbols such as + , − , and 0 are often used to reflect these three possibilities. The data consist of two random variables or a single criterion in which the criterion is evaluated for two conditions, such as before treatment versus after treatment. A classical example of the use of the sign test would be a poll in which voters would have two options: voting for or against passage. The first part of the experiment 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 random sample 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 the pollster 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 the before-and-after responses of the individuals in the sample are associated with the individual, they are paired. The sign test would be the appropriate test for analyzing the data. The hypotheses can be expressed in several ways, with the appropriate alternative pair of hypotheses selected depending on the specific situation. One expression of the null hypothesis is H 0 : 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 post- treatment response (i.e., for or against). If the treatment does not have a significant effect, then the changes in one direction should be balanced by changes in the L1600_Frame_C08.fm Page 174 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC 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 H 0 : P ( + ) = P ( − ) (8.1b) or H 0 : 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 N 12 and N 21 are pertinent while the number of responses N 11 and N 22 are considered “ties” and are not pertinent to the calculation of the sample test statistic. If N = N 11 + N 22 + N 12 + N 21 but only the latter two are of interest, then the sample size of those affected is n = N 12 + N 21 , 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 Pre-test XY XN 11 N 12 YN 21 N 22 L1600_Frame_C08.fm Page 175 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC 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 one- tailed upper test, the critical values for 1% and 5% are 12 and 10. For a two-tailed test, the critical values for a 1% level of significance are 1 and 12, which means the null hypothesis is rejected if the computed value is equal to 0, 1, 12, or 13. For a 5% level of significance the critical values would be 2 and 11, which means that the null hypothesis is rejected if the computed value is equal to 0, 1, 2, 11, 12, or 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 using the 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 Tf(T) F(T) 0 0.00012 0.00012 1 0.00159 0.00171 2 0.00952 0.01123 3 0.03491 0.04614 4 0.08728 0.13342 5 0.15711 0.29053 6 0.20947 0.50000 7 0.20947 0.70947 8 0.15711 0.86658 9 0.08728 0.95386 10 0.03491 0.98877 11 0.00952 0.99829 12 0.00159 0.99988 13 0.00012 1.00000 FT n i pp n i nnT i T i T n () ( ) (.)=       −=       − == ∑∑ 105 00 L1600_Frame_C08.fm Page 176 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC For large sample sizes, the binomial mass function can be approximated with a normal distribution. The mean and standard deviation of the binomial distribution are µ = np (8.4) σ = [np (1 − p)] 0.5 (8.5) For a probability of 0.5, these reduce to µ = 0.5n and σ = 0.5n 0.5 . Using these with the standard normal deviate gives: (8.6) 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: (8.8) TABLE 8.2 Critical Values for the Sign Tests a nX .01 X .05 nX .01 X .05 nX .01 X .05 nX .01 X .05 10 1 b 2 b 20 4 6 b 30 8 10401214 11 1 2 21 4 6 31 8 10 41 12 15 b 12 1 2 22 5 6 32 9 11 b 42 13 15 13 2 b 3 23 5 7 33 9 11 43 13 15 14 2 3 24 6 b 73410 b 11 44 14 b 16 15 3 b 4 b 25 6 8 b 35 10 12 45 14 16 16 3 b 4266 8361012461417 b 17 3 4 27 7 8 37 11 13 47 15 17 18 3 5 28 7 9 38 11 13 48 15 18 b 19 4 5 29 8 b 93912 b 14 b 49 16 b 18 50 16 18 a For the upper tail critical values for probabilities of 5% and 1%, use X u.05 = n − X .05 and X u.01 = n – X .01 . b To obtain a conservative estimate, reduce by 1. This entry has a rejection probability slightly larger than that indicated, but the probability of the lower value is much less than the indicated rejection probability. For example, for n = 13 and a 1% rejection probability, X = 1 has a probability of 0.0017 and X = 2 has a probability of 0.0112. z xxn n = − = − µ σ 05 05 05 . . . z xn n = +−(.). . . 05 05 05 05 x n zn n zn=+ −= +−05 05 05 05 1 05 05 ( ) L1600_Frame_C08.fm Page 177 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC 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. For the 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 obtained with Equation 8.8. The rounded value can then be used with Equation 8.7 to estimate the actual rejection probability. Example 8.1 Mulch is applied to exposed soil surfaces at 63 sites, with each slope divided into two 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 storm events with the extent rated as high or low. The sign test is used to test the following hypotheses: H 0 : 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 mulch density showed the lower amount of eroded material on 41 sites. On 5 sites, the section 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 null hypothesis 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 and 16.9, respectively. These normal approximations are close to the actual binomial values of 14 and 17. 8.4 TWO-SAMPLE t-TEST In some cases, samples are obtained from two different populations, and it is of interest to determine if the population means are equal. For example, two laboratories may advertise that they evaluate water quality samples of some pollutant with an accuracy of ±0.1 mg/L; samples may be used to test whether the means of the two populations are equal. Similarly, tests could be conducted on engineering products to determine whether the means are equal. The fraction of downtime for two com- puter types could be tested to decide whether the mean times differ. A number of tests can be used to test a pair of means. The method presented here should be used to test the means of two independent samples. This test is frequently of interest in engineering research when the investigator is interested in comparing an experimental group to a control group. For example, an environmental engineer 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. L1600_Frame_C08.fm Page 178 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC Step 1: Formulate hypotheses. The means of two populations are denoted as µ 1 and µ 2 , the null hypothesis for a test on two independent means would be: H 0 : The means of two populations are equal. (8.9a) Mathematically, this is H 0 : µ 1 = µ 2 (8.9b) Both one-sided and two-sided alternatives can be used: H A1 : µ 1 < µ 2 (8.10a) H A2 : µ 1 > µ 2 (8.10b) H A3 : µ 1 ≠ µ 2 (8.10c) The selection of the alternative hypotheses should depend on the statement of the problem. 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; n 1 and n 2 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 = n 1 + n 2 − 2; and S p is the square root of the 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 either 5% or 1% are used most frequently. t XX S nn p = − +       12 12 05 11 . X 1 X 2 X 1 X 2 S nSn S nn p 2 11 2 22 2 12 11 2 = −+− +− ()() S 1 2 S 2 2 L1600_Frame_C08.fm Page 179 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC Step 4: Compute an estimate of test statistic. Samples are drawn from the two populations, and the sample means and variances computed. Equation 8.11 can 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 = n 1 + n 2 − 2), the level of significance ( α ), and the statement of the alternative hypothesis. The regions of rejection for the alternative 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 Appendix Table A.2), which is based on either t α or t α /2 obtained from step 5. If the sample 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 nitrogen levels in small streams. A decision was made to use the mean concentrations before and 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 standard deviation 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 alternative hypothesis that the urban development increased total nitrogen levels, which requires the following one-tailed test: H 0 : µ b = µ a (8.13) H A : µ b < µ a (8.14) where µ b and µ a are the pre- and post-development means, respectively. Rejection of H 0 would suggest that the nitrogen levels after development significantly exceed the nitrogen levels before development, with the implication that the development might have caused the increase. Based on the sample data, the pooled variance of Equation 8.12 is (8.15) If H A is then reject H 0 if µ 1 < µ 2 t < −t α µ 1 > µ 2 t > t α µ 1 ≠ µ 2 t < −t α /2 or t > t α /2 S p 2 22 11 1 0 36 14 1 0 87 11 14 2 0 4842= −+− +− = ()(.)()(.) . L1600_Frame_C08.fm Page 180 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC 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 development caused 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 not significant. This shows the importance of selecting the level of significance before analyzing 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 have been 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 rank-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: H 0 : The two independent samples are drawn from the same population. H A : The two independent samples are not drawn from the same popu- lation. 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 or lower 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 t = − +     =− 078 137 0 4842 1 11 1 14 2 104 . . L1600_Frame_C08.fm Page 181 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC combined, with the items in the combined group ranked in order from smallest (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 be computed by actual summation of the ranks: (8.17c) 3. The level of significance ( α ) must be specified. As suggested earlier, α is usually 0.05. 4. Compute the test statistic value of U of step 2 and the value of Z: (8.18) in which Z is the value of a random variable that has a standard normal dis- tribution. 5. Obtain the critical value (Z α /2 for a two-sided test or Z α 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 Z α /2 or less than −Z α /2 (Z α for an upper one-sided test or −Z α for a lower, one-sided test). In some cases, the hydrologic change may dictate the direction of change in the annual 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, the peaks 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 −Z α (rather than −Z α /2 ) and the null hypothesis is accepted when Z > − Z α . Example 8.3 The 65-year annual maximum series for the Elizabeth River appears (Figure 7.2) to change in magnitude about 1951 when the river was channelized. To test if the channelization was accompanied by an increase in flood peaks, the flood series was divided into two series, 1924 to 1951 (28 years) and 1952 to 1988 (37 years). Statistical characteristics for the periods are given in Table 8.3. Since the direction of change was dictated by the problem (i.e., the peak dis- charges increased after channelization), the alternative hypothesis is H A : The two independent samples are not from the same population and the logarithms of the peaks for 1924–1951 are expected to be less than those for 1952–1988. SS nn a b += +05 1.( ) Z Unn nn n n a b a b a b = − ++ 05 112 05 . (( )/) . L1600_Frame_C08.fm Page 182 Tuesday, September 24, 2002 3:24 PM © 2003 by CRC Press LLC [...]... Series 80 96 220 230 276 300 316 340 376 420 540 570 626 630 650 676 716 82 0 990 1040 1230 1310 1430 1450 187 0 2496 2600 X Score Y Score 13 14 17 18 21 22 25 26 29 30 33 34 37 38 41 42 45 46 49 50 53 54 57 58 61 62 623 Tx 65 1522 Ty L1600_Frame_C 08. fm Page 206 Tuesday, September 24, 2002 3:24 PM 8- 3 8- 4 8- 5 8- 6 8- 7 8- 8 8- 9 8- 1 0 8- 1 1 8- 1 2 b On a single plot of frequency versus flood magnitude, show the histograms... (X) and Post-Urbanization (Y) of Elizabeth River Watershed, Elizabeth, New Jersey Discharge (cfs) X 1924–1949 Y 1950–1 988 1290 980 741 1630 82 9 903 4 18 549 686 1320 85 0 614 1720 1060 1 680 760 1 380 1030 82 0 1020 9 98 3500 1100 1010 83 0 1030 452 2530 1740 186 0 1270 2200 1530 795 1760 80 6 1190 952 1670 82 4 702 1490 1600 80 0 3330 1540 2130 3770 2240 3210 2940 2720 2440 3130 4500 289 0 2470 1900 1 980 2550 3350... 0.360 0. 187 0.261 0.097 0.127 0. 183 0.1 58 0.207 0.2 68 8- 1 5 Use the Walsh test to decide whether the deforested watersheds of Problem 8- 1 3 tend to have higher discharges than the forested watershed 8- 1 6 Use the Walsh test to decide if the two methods of measuring infiltration in Problem 8- 1 4 give dissimilar estimates 8- 1 7 Use the Wilcoxon test to decide if the deforested watersheds of Problem 8- 1 3 tend... L1600_Frame_C 08. fm Page 194 Tuesday, September 24, 2002 3:24 PM TABLE 8. 9 Detection of Change in Cover Density Annual Maximum for Watershed Year 1 2 Difference 1 983 1 984 1 985 1 986 1 987 1 988 1 989 1990 1991 1992 1993 4.7 6.7 6.3 3.7 5.2 6.3 4.3 4.9 5.9 3.5 6.7 5.1 6 .8 5.6 3.2 4.6 5.2 2.4 2.7 3.4 1.2 4.0 −0.4 −0.1 0.7 0.5 0.6 1.1 1.9 2.2 2.5 2.3 2.7 rp rn 2 1 5 3 4 6 7 8 10 9 11 Sp = 63 Sn = 3 8. 8.1 TIES Tied... −0.027) 2 = 0.0009699 8 (8. 22) The sampling variation is computed using Equation 8. 19:  0.0009699  Sm =    8( 8 − 1)  © 2003 by CRC Press LLC 0.5 = 0.004162 (8. 23) L1600_Frame_C 08. fm Page 186 Tuesday, September 24, 2002 3:24 PM TABLE 8. 4 Comparison of Roughness Coefficients Using the Two-Sample t-Test for Related Samples: Example 8. 5 Roughness with Channel Reach 1 2 3 4 5 6 7 8 Method 1 Method 2... distribution? 8- 2 1 Using the rainfall phosphate concentration of Problem 8- 8 , test to determine whether the standard deviation is greater than 0.1 mg/L 8- 2 2 Using the BOD measurements for the business district (Problem 8- 1 1), test to determine whether the standard deviation is significantly less than 25 mg/L © 2003 by CRC Press LLC L1600_Frame_C 08. fm Page 2 08 Tuesday, September 24, 2002 3:24 PM 8- 2 3 Is it... scores per side Note: Table 8. 12 illustrates the assignment of scores for N from 4 to 11 TABLE 8. 12 Assignment of Scores for Siegel–Tukey Test N 4 5 6 7 8 9 10 11 4123 43125 541236 7431256 85 412367 87 4312569 9 8 5 4 1 2 3 6 7 10 11 8 7 4 3 1 2 5 6 9 10 © 2003 by CRC Press LLC 3214 52134 632145 6521347 763214 58 9652134 78 10 7 6 3 2 1 4 5 8 9 10 9 6 5 2 1 3 4 7 8 11 L1600_Frame_C 08. fm Page 202 Tuesday, September... discharges 8- 1 4 Two methods for measuring infiltration rates are used on each of 12 watersheds The issue is to determine whether the methods provide similar values a Use the paired t-test to decide b Apply the two-sample t-test (Section 8. 4) and compare the results with the paired t-test Discuss any differences detected in the results A B 0.221 0. 284 0.314 0.363 0.265 0.3 38 0.166 0.231 0.1 28 0.196 0.272... of the residential BOD measurements (Problem 8- 1 1) is significantly smaller than the variance of the values for the business district? Use the two-sample F-test 8- 2 4 Is it reasonable to conclude that the discharges for the forested watersheds (Problem 8- 1 3) have a smaller variance than the variance of the deforested watersheds? Use the two-sample F-test 8- 2 5 Is it reasonable to conclude that the discharges... tend to have higher discharges than the forested watershed 8- 1 8 Use the Wilcoxon test to decide if the two methods of measuring infiltration in Problem 8- 1 4 give dissimilar estimates 8- 1 9 The standard deviation of 58 annual maximum discharges for the Piscataquis River is 4095 ft3/sec Test whether this is significantly different from 3500 ft3/sec 8- 2 0 A sample of five measurements follows: x = {0.16, −0.74, . 0.03491 0.04614 4 0. 087 28 0.13342 5 0.15711 0.29053 6 0.20947 0.50000 7 0.20947 0.70947 8 0.15711 0 .86 6 58 9 0. 087 28 0.95 386 10 0.03491 0. 988 77 11 0.00952 0.9 982 9 12 0.00159 0.99 988 13 0.00012 1.00000 FT n i pp n i nnT i T i T n (). 8 b 35 10 12 45 14 16 16 3 b 4266 83 61012461417 b 17 3 4 27 7 8 37 11 13 47 15 17 18 3 5 28 7 9 38 11 13 48 15 18 b 19 4 5 29 8 b 93912 b 14 b 49 16 b 18 50 16 18 a For the upper tail critical. 0.049 −0.0 08 0.000064 3 0.0 68 0. 083 −0.015 0.000225 4 0.054 0.070 −0.016 0.000256 5 0.044 0.055 −0.011 0.000121 6 0.066 0.057 0.009 0.000 081 7 0. 080 0.071 0.009 0.000 081 8 0.037 0.045 −0. 088 0.000064 ∑