Analysis of Hydrologic DataEstimation of Design Discharge and Water Level Estimation of both flood discharges and high water levels are necessary for bank protection design. Careful estimation of discharge and water level is important for all sites with erodible banks. This section describes the methods of assessing flood discharge and water level at the site under consideration. The design discharge and water level are determined for selected probability of exceedance or return period. The design discharge and water level arising from floods should be selected after due consideration of the following: The maximum historical discharge as recorded at the site, or as calculated on the basis of recorded water level at the site, or as calculated on the basis of measured discharge at other points on the river from which corresponding site discharge can reasonably be inferred; the discharge derived from a frequency analysis using a probability of exceedance or return period which is appropriate to the importance and value of the protection work. The maximum historical water level as recorded at the site, or as inferred from observed or recorded water level at other points on the river from which level can reasonably be transferred to the site in question; the water level derived from a frequency analysis using a probability of exceedance or return period which is appropriate to the importance and value of the protection work. In estimating high flows, primary reliance should be placed on careful field investigations, local enquiries and searches of historical records. Data so obtained should be compared with recorded data for hydrometric stations, and supplemented by analytical procedure using stagedischarge curves: At most hydrometric gauging stations reasonably stable relationship exists between water level and discharge. At some sites, however, the stage discharge curve may be quite unstable because of aggradation or degradation at channel bed or backwater effect from downstream, and may change drastically during major floods. A persistent trend of rising or lowering of curve indicates progressive channel aggradation or degradation. The stage corresponding to design flood which exceeds any recorded flow obtained by extrapolating the stage-discharge relationships. The most commonly used method for estimating design discharge and water level examines the observed discharge and water level to arrive at suitable estimates. The method, known as frequency analysis, is founded on statistical analyses of discharge and water level records. For locations where records of stream flows are available, or where flows from another basin can be transported to the design location, design flood magnitude and water level can be estimated directly from those records by means of frequency analysis. Frequency Analysis Frequency of a hydrologic event, such as the annual peak flow is the probability that a value will be equaled or exceeded in any year. This is more appropriately called the exceedance probability, P(F). The reciprocal of the exceedance probability is the return period T in years, T 1 P ( F ) . The length of record should be sufficient to justify extrapolating the i.e., frequency relationship. For example, it might be reasonable to estimate a 50-year flood on the 1 basis of a 30-year record, but to estimate a 100-year flood on the basis of a 10-year record would normally be absurd (Neill 1973). Viessman and Lewis (1996) noted that as a general rule, frequency analysis is cautioned when working with shorter records and estimating frequencies of hydrologic events greater than twice the record length. Frequency analysis can be conducted in two ways: one is the analytical approach and the other is the graphical technique in which flood magnitudes are usually plotted against probability of exceedance. Here in the following sections, procedures are given mostly for discharge frequency analysis; the similar procedures can also be followed for water level frequency analysis. Analytical Frequency Analysis Analytical frequency analysis is based on fitting theoretical probability distributions to given data. Numerous distributions have been suggested on the basis of their ability to “fit” the plotted data from streams (Linsley et al. 1988). The Log-Pearson Type III (LP3) has been adopted for use in the United States federal agencies for flood analysis. The first asymptotic distribution of extreme values (EV1), commonly called Gumbel Distribution has been widely used and is recommended in the United Kingdom. EV1 Distribution was found to fit peak flow data for several rivers in Bangladesh (Bari and Saleque 1995). Extreme Value Distributions: Distributions of the extreme values selected from sets of samples of any probability distribution converge to any one of three forms of Extreme Value Distributions, called Type I, II, and III, respectively, when the number of selected extreme values is large. The three limiting forms are special cases of a single distribution called Generalized Extreme Value (GEV) Distribution. (Chow et al. 1988). The cumulative distribution function for the GEV is 1 xu F ( x) exp 1 (1) where , u, and are parameters to be determined. For EVI Distribution x is unbounded, while for EVII, x is bounded from below, and for EVIII, x is bounded from above. The EVI and EVII Distributions are also known as the Gumbel and Frechet Distributions, respectively. The Extreme Value Type I (EVI) cumulative distribution function is xu F ( x ) exp exp -x (2) The parameters are estimated by 6 s and u x 0.5772 (3) Eq (2) can be expressed as F ( x) e e y (4) where y is the reduced variate defined as y xu (5) Solving Eq (4) for y: 2 Annual peak discharges (m3/s) of the Old Brahmaputra River at Mymensingh for the period from 1964-98 Year Peak flow 1964 2830 1965 3230 1966 3490 1967 3000 1968 2810 1969 2770 1970 3250 1974 3820 1975 3060 1976 3210 1977 3550 Sample Size Max = 4910 Min = Skew. s = Peak flow 2770 2630 3340 2690 2470 2370 4780 3070 1930 3230 4910 Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 Peak flow 2180 2060 2900 1490 2060 1065 3187 2369 1973 3267 2867. 1 y ln ln F x (6) Noting that the probability of occurrence of an event x xT is the inverse of its return period T. a model is developed for frequency analysis of the annual peak flow data of Old Brahmaputra River at Mymensingh for 5. 10.53 and s = 804. When a fairly long record has a short gap. it may be justifiable to estimate the missing data by correlation with a nearby station.54. or xT u y T (8) Example 1 Using the EVI Distribution. Substituting in Eq (3) yields 3 . Cs = 1065 0. we can write 1 P ( x xT ) 1 P ( x xT ) 1 F ( xT ) T so F ( xT ) 1 1 T and substituting for F( xT ) into Eq (6) T yT ln ln T 1 (7) For a given return period xT is related to yT by Eq (5).54 Note that data for 1971. The latter approach is used in this example.372 Year 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 n = 32 Ave.53 804. 72 and 73 are missing. 25 and 50 years return period peak flows are calculated. x = Std. For the given data x 2867. otherwise it is preferable to consolidate the various recorded sequences as if they formed a continuous record (Neill 1973). 62 and u x 0. 6 804 . yT and xT values are found as follows: T=10 years.27 + 627. 5 y5 ln ln 1. and it is applicable to many probability distributions used in hydrologic frequency analysis.25. Chow (1951) has shown that most frequency functions can be generalized to xT x K T s (9) where xT is a flood of specified probability or return period T.54 = 627. The K-T relationship can be expressed in mathematical terms or by a table. that is.62×1. such as LP3.27 x 2505. yT = 3. x is the mean of the flood series. given a value of T or 1 F ( xT ) 1 T . like the Normal and Pearson Type III Distributions. Similarly for other values of T. s is the standard deviation of the series. T=50 years.5 = 3446. it is convenient to use the reduced variate yT . Normal Distribution: From Eq (9) the frequency factor can be expressed as 4 . T=25 years. Chow (1951) proposed the frequency factor as in Eq (9).90.5 5 1 For T = 5 years. as well as coefficient of skewness for skewed distributions.27 F ( x) exp exp 626.20. the corresponding value of xT can be determined.62 The probability model is To determine the values of xT for various values of T. and K T is the frequency factor and is a function of return period and type of probability distribution. for example as in Lognormal and LP3 Distributions. In the event that the variable analyzed is y log x . Eq (7) gives and Eq (8) yields x5 = 2505. yT = 2.7 m3/s. yT = 3. and the required value of xT is found taking antilog of yT .5772 2505 . the same method is applied to the statistics for the logarithms of data using yT y K T s y . Thus an alternative method based on frequency factor is used for calculating the magnitudes of extreme events. xT = 3918 m3/s xT = 4513 m3/s xT = 4954 m3/s Frequency Analysis using Frequency Factors Calculating the magnitudes of extreme events by the method outlined in the above example requires that the probability distribution function be invertible. Some probability distribution functions are not readily invertible. for Normal Distribution K T is the same as the standard normal variable z.KT xT x z s (10) Thus. the frequency factor is equal to the standard normal variable z (Table 2). K T can be taken from Table 2. y and standard deviation y 3) Compute KT yT y K T s y yT y z sy So. The value of z and hence K T can be obtained from Table 2. is taken from Haan (1977). Lognormal Distribution: The recommended procedure for use of the Lognormal Distribution is to convert the data series to logarithms and compute: 1) y i log xi 2) s Compute the mean. When Cs =0. 5) Finally compute xT anti log yT Table 3 gives values of the frequency factors for the LP3 Distribution for various values of return period and coefficient of skewness.33 years.5772 ln ln T 1 (12) When xT . Cs. Eq (9) (in population term) gives K T 0 and Eq (12) gives T=2. This is the return period of the mean of the EVI Distribution. 5 . The values computed from the above equation are equivalent to an infinite sample size in Table 4. given in Table 4. y and standard deviation Compute coefficient of skewness Cs 4) n ( y i y ) 3 (n 1)( n 2) s y Compute sy 3 yT y K T s y (11) where K T is taken from Table 3. Extreme Value I (EVI) Distribution: Chow (1951) derived the following expression for frequency factor for the EVI Distribution T 6 KT 0. Table of frequency factors for the EVI Distribution. 4) Finally compute xT anti log yT Log-Pearson Type III (LP3) Distribution: The recommended procedure for use of the LP3 Distribution is to convert the data series to logarithms and compute: 1) y i log xi 2) 3) Compute the mean. 4394 + 2.592×804.542825 3.5 + 2. Determining the probability to assign a data point is commonly referred to as determining probability position. Year 1964 1965 1966 1967 1968 1969 1970 1974 1975 1976 1977 Peak flow 2830 3230 3490 3000 2810 2770 3250 3820 3060 3210 3550 Ave.98 Table 2 is entered and z = 2.509202 3.02 and F x P Z z 0.511883 3. which is a plot of event magnitude versus probability.506505 3.532×0.313867 3.6425 y T y K T s y y 50 x 10 4390 m3/s . the data are plotted on specially designed probability paper that linearizes the distribution function.4394 +1.679427 3.007 for n=32 years).98.514149 Skew. the value of 1.9303 Lognormal Distribution: For T = 50 year.532. Graphical Frequency Analysis The frequency of an event can be obtained by use of probability plot. LogPearson Type III and EVI Distributions.313867 3. xT x K T s . so.442479 3. 3. The plotted data are then fitted with a straight line for interpolation and extrapolation purposes.392696 3.477121 3.550228 3.54 = 4953 m3/s.487138 3.451786 3.503382 3.054 is obtained by interpolation corresponding to the tabular value of P Z z 0. so.691081 Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 0.6425. 3.462397 3. Table 4 gives K 50 x 3. y= y = log Q 3.338456 3.Example 2 For illustration the 5 and 50 years return period annual maximum discharges (m 3/s) for the Old Brahmaputra River near Mymensingh is calculated using the Lognormal.509202 3.374565 3.419955 3.429752 3.054×0.1326 = 3.7118 5150 m3/s K 50 Log-Pearson Type III (LP3) Distribution: For Cs = -0.485721 3.4394 Year 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 Peak flow 2770 2630 3340 2690 2470 2370 4780 3070 1930 3230 4910 Std. 50 Extreme Value I (EVI) Distribution: Eq (12) gives K 50 2.582063 3.7118.9303. 6 .448706 3. yT y K T s y y 50 .027349 3. 50 2867.295127 3.523746 3. Note that the value of frequency factor can be obtained from Table 3 with Cs = 0.592 (however. sy = Log Q 3.285557 3.173186 3.374748 3.1326 = 3. 3.442479 3. 1/T = 0.1326 Peak flow 2180 2060 2900 1490 2060 1065 3187 2369 1973 3267 Log Q 3. As a check that a probability distribution fits a set of hydrologic data. Cs = -0. x50 10 3. 578 0. Rank Peak Plotting flow position* m Q P 1 4910 0.453 0.111 5 3490 0.920 0.142 6 3340 0.5156 0. P( X x m ) xm m n However this simple formula (known as California’s formula) produces a probability of 100%.049 3 3820 0.640 0. Several plotting position formulas are given below.983 . Also plotted data are compared with best-fit EVI Distribution.0797 4 3550 0.422 0.7958 0. that is.017 2 4780 0. Example 3 As an example.827 0. which implies that the largest sample value is the largest possible value.671 Rank m 23 24 25 26 27 28 29 30 31 32 Peak flow Q 2470 2370 2369 2180 2060 2060 1973 1930 1490 1065 Plotting position P 0. is P m n 1 When m is ranked from lowest to highest.4) (m-3/8)/(n+1/4) (3m-1/3n+1) The technique in all cases is to arrange the data in increasing or decreasing order of magnitude and to assign order number m to the ranked values.733 0.Plotting Positions: Plotting position refers to the probability value assigned to each piece of data to be plotted. To overcome this limitation other formulas have been proposed.235 9 3230 0.5)1/n m/(n+1) (m-0. If n is the total number of values to be plotted and m is the rank of a value in a list ordered by descending magnitude.173 7 3267 0. P is P(Xx).267 10 3230 0.(0.547 0.298 11 3210 0. P(Xx).12) (m-0.889 0. for large n.3604 0. probability plotting analysis of the annual maximum discharges (m 3/s) of the Old Brahmaputra near Mymensingh is performed. the exceedance probability of the mth largest value is.5)/n 1 . Plotting position formulas California Hazen Beard Weibull Gringorten Chegodayev Blom Tukey (m/n) (m-0.951 0.765 0.329 Sample size n = 32 Rank m 12 13 14 15 16 17 18 19 20 21 22 Peak flow Q 3187 3070 3060 3000 2900 2830 2810 2770 2770 2690 2630 7 Plotting position P 0.484 0. when the rank is from highest to lowest.858 0.3)/(n+0. The most efficient formula for computing plotting positions for unspecified distribution and the one now commonly used for most sample data.44)/(n+0.391 0.702 0. A value of 100% cannot be plotted on many probability paper (Haan 1977).610 0. P is an estimate of the probability of values being equal to or less than the ranked value.204 8 3250 0. P(Qq) = 0. for m=1. P(Qq) = 0.44)/(32+0. P(Qq) = 0. Goodness-of-fit Tests The goodness of fit of a probability distribution can be tested by comparing the theoretical and sample values of the relative frequency or the cumulative frequency function. the probability distribution must be defined.Ave = 2867.02 K50 = 2.37198 * Gringorten. as shown below to smallest (m=n=32). the χ 2 – test is used and with cumulative frequency function the Kolmogorov-Smirnov test is used. To describe the test. A confidence level is chosen for the test. P = (m−0. Several points on the best-fit EVI line are calculated using Eq (9) as follows: T=5 years. The critical distribution function is tabulated (in Table 5) from Haan (1977).017. the theoretical value of the relative frequency function (also called incremental probability function) is p(xi) = F(xi) . it is often expressed as 1-α.44)/(n+1−0. Gringorten’s plotting formula (b=0. fs(xi) = ni/n.88) = (1−0. 2 8 . the observed number of occurrences in interval i. Q100 = 5390 m3/s A straight line is drawn through the calculated points to obtain the best-fit EVI Distribution line. Q25 = 4511 m3/s T =50 years. In the case of the relative frequency function.01 K100 = 3.592.44) was used since data are being fitted to EVI Distribution.5443.04 K25 = 2.88) First the data are ranked from largest (m=1). where α is termed the significance level. Q5 = 3446 m3/s T =25 years. the sample value of the relative frequency of interval i is.044. Chi-Square Test: The test statistic is given by k n f x p x 2 2 s i i i 1 p xi (13) where k is the number of intervals.12 = 0. Q50 = 4952 m3/s T =100 years.12) = 0.137. The plotted points represent the empirical distribution obtained using 32 observed peak flows.20 K5 = 0. For example.531 Std dev = 804. A distribution with υ = k-l-1 degrees of freedom (l is the number of parameters used in fitting the proposed distribution) is the distribution for the sum of squares of υ 2 2 2 independent standard normal random variables zi. In this example the plotted points show good-fit with EVI Distribution. P(Qq) = 0.719. and np(xi) is the corresponding expected number of occurrences in interval i. It may be noted that nfs(xi) = ni. Skew = 0.56/32.44)/(n+1−0. the exceedance probability (Q 4910 m3/s) = (m−0. Similarly all the plotting positions are calculated and plotted on EVI paper (Fig 6).F(xi-1). Mymensingh Example 4 Chi square test is used to determine whether EVI Distribution adequately fits the Old Brahmaputra river annual peak flow data.25. The observed cumulative frequency found by summing up the relative frequencies. ni in each class is counted. 6 EVI probability plot for annual peak flows of Old Brahmaputra. Thirty two peak flow observation are divided into six class intervals.Return Period (Years) Exceedence Probability Discharge (m3/s) Fig. 9 . The observed or sample values of relative frequency fs(xi) is calculated with n = 32. for the second class interval fs(x2) = 8/32 = 0. The number or frequency of observations. For example. 5 and s = 804.35 is less than the critical value of 6.3815 0. x = 2867.25.37214 0.03575 0.32904 The value of 0. The theoretical cumulative frequencies corresponding to the upper limit of each of class interval is calculated by finding reduced variate y from Eq (5) and the F(x) by Eq (4). The test statistic D.3463 For a confidence level of 90%.06068 0.1866 0.25 0.36479 0. Kolmogorov-Smirnov Test: The theoretical and sample values of the cumulative frequency are compared with the Kolmogorov-Smirnov (S-K) test. Since the computed Chi square value of 2. this is compared with the critical χ 2 value to be obtained from tabular values as shown below.2033 0.62.03575 0. u = 2505.54 m3/s). The calculation is repeated for other class intervals and summed to obtain χ 2 = 2.32904 0.91174 0.1875 0.96875 3.75 1. Obs frequency Obs cum frequency Reduced variate Expected cum frequency Expected relative frequency Chi square Lower limit 1000 Upper limit 1750 ni fs(xi) Fs(xi) yi F(xi) p(xi) 2 2 0.00 Total Computed Chi square χ2 2. the data fits EVI Distribution adequately. the test requires that the value of D computed from the sample distribution be less than the tabulated value of D (Table 6) at 10 .0625 0.02948 4000 4750 1 0. the parameters α and u are calculated as before (α = 627.25.32904 is entered under the expected relative frequency corresponding to the class interval 1750-2500 in the table below.45676 4750 5500 1 0.97242 0. χ 2 = 6.4375 0.0084 0.3125 -0.03125 0.To fit EVI Distribution.7716 0.35.27. the critical Chi square for υ = k-l-1 = 6-21 = 3 degree of freedom.5765 0.64050 1750 2500 8 0. This is the computed χ 2 value.73693 0.99157 0.0625 -1.9375 2.60757 2500 3250 14 0.24465 32 1. from Table 5. To test the goodness of fit. Class limit Num of obs.0 4. for the second class interval y2 x2 u e and F x 2 e y2 0.17481 0.36734 3250 4000 6 0. For example. which is based on deviations of the sample distribution function P(x) from the completely specified continuous hypothetical distribution function Po(x).36479 p(x2) = P(1750 ≤ X ≤ 2500) = F(2500) – F(1750) = 0. such that: D max P( x) Po ( x) Developed by Kolmogorov (Kite 1988) in 1933.03125 1.01915 0. For Brahmaputra. This is the state of maximum velocity in the channel. Split this into discrete of equal class interval. To estimate the dominant discharge the following steps are followed: 1. therefore. Dominant Discharge Analysis The dominant discharge is the flow doing most geomorphic work and it is. F Q 11 . Frequency. but it is difficult to measure in the field. Quoting return periods for bankful discharge is a tricky business because over a dozen methods are available. and a wide variety of field procedures exist for this measurement. Obtain long-term (30 year plus) distribution of flows for gauging station. Bankful discharge is assumed to be a major determinant of the size and shape of a river channel. the channel forming discharge. It probably does not correspond to bankful flow on any river. let us try initially 5.the required confidence level. The dominant or channel forming flow represents an alternative benchmark criterion to bankful flow when analyzing channel form and process. but the frequency of its occurrence seems to vary with climatic regimes. Kolmogorov-Smirnov test for Gumbel’s Extremal Distribution gives better result in Bangladesh Bankful Discharge The bankful discharge of a river may be defined as the discharge which is contained within the banks of the river. F 2. and therefore of maximum competence for the transport of sediment load. Find mid point of each class. and check sensitivity of results to this choice of interval.000 m3/s class interval. Multiply the sediment transport rate for each discharge class with the frequency of that class to obtain the total sediment load transported by that flow during the period. Total Qs Q Mode = Qd 6. Obtain the most reliable sediment rating curve for the gauging station. plot this as a histogram. From the histogram. Qs Qsi Q Qi 5.3. Qs Q 4. Ideally this should be for total load. identify the mode. Determine the magnitude of Qd = dominant discharge and use the flow duration curve to establish its return period. This corresponds to the dominant discharge. Use the sediment rating curve to find the sediment transport rate (tons/sec) for the mid-point discharge of each flow class. but a suspended load curve may be used provided that suspended load makes up most of the total load. as is usually the case. 12 Exceedance Probability . Table 2 Cumulative probability of the Standard Normal Distribution 13 . Table 3 Frequency factors for Pearson Type III Distribution 14 . 15 . Table 3 Continued 16 . 91 3 2.91 1.14 5 3.32 1 3.32 6 3.117 2.02 9 2.86 6 0.78 2 0.78 0 0.02 3 1.01 3 2.85 1 0.26 7 2.75 7 1.52 0 3.78 5 2.75 2 2.80 1 0.41 3 0.97 2 0.02 0 2.116 3.96 7 0.23 5 7 2.18 3 3.18 7 2.82 4 3.42 9 3.10 9 3.55 4 0.59 2 2.22 0 2.359 5.41 7 1.35 7 3.47 8 1.46 7 0.04 4 2.80 2 2.54 1 1.51 6 1.51 3.30 2 2.83 8 1.02 6 2.21 9 3.78 8 2.65 3 3.44 4 9 3.36 7 3.76 2 1.Table 4 Frequency factors for Extreme Value I Distribution Return Period Sampl e size (n) 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 α 5 10 15 20 25 50 75 100 1000 0.39 3 2.41 0 2.88 9 2.478 5.63 5 1.13 7 6.16 9 3.46 6 1.80 6 1.07 1 2.44 6 1.34 1 3.006 5.76 7 2.77 3 1.86 6 2.936 .46 3 6 3.18 8 2.89 1 1.40 9 1.83 7 2.78 0 1.77 0 3.97 9 2.96 3 2.43 0 1.78 8 0.261 4.911 17 3.88 8 5 1.10 4 2.72 1 3.56 4.80 7 0.20 5 2.79 6 1.727 5.40 5 1.34 9 0.19 8 2.49 1 3.265 9 0.30 1 3.43 7 1.17 3.40 1 1.03 8 2.81 8 2.05 9 2.30 3 2.79 3 3.77 9 1.30 5 1.49 5 1.39 3 3.81 2 2.21 2 2.71 9 1.41 3 1.12 6 2.37 6 0.00 7 2.24 1 2.62 2.84 7 1.79 7 1.04 8 2.12 5 3.92 2 1.00 2 1.83 1 1.15 5 3.82 0 0.23 0 2.78 5 1.28 3 2.842 5.44 6 3.20 0 3.24 1 3.86 9 3.13 5 3.00 5 3.72 9 0.08 8 3 3.19 3 2.86 6 2.83 6.08 6 2.99 8 2.45 5 1.40 0 3.94 3 3.59 8 3.42 3 1.77 7 2.63 2 2.38 7 3.82 9 0.57 5 2.26 8 3.35 4 2.576 5.25 3 2.85 2 2.15 2 2.81 3 1.70 3 1. 16 2.64 9.0 34.8 24.67 9.3 23.7 60.84 9.06 0.831 1.5 32.5 20.5 33.35 5.4 32. 32 (1941).1 12.3 71.60 3.554 .3 13.7 46.3 44.3 17.455 1.6 12.207 5 6 7 8 9 16.8 16.26 6.1 20.77 4.8 31.1 6.4 20.66 2.39 0.94 4.9 7.5 96.2995 x.9 9.7 71.2 26.0 33.3 25.43 8.84 5.290 x.1 10.38 9.81 9.21 1.26 8.61 2.1 12.4 31.3 42.4 23.1 30 40 50 60 53.8 15.8 14.103 .7 24.4 14.59 10.7 22.8 18.21 11.6 34.90 9.7 33.9 23.25 5.0 17.7 13.2 11.4 29.8 13.8 20.82 4.1 9.2 100.99 7.1 22.4 53.81 6.0 106.0002 .87 5.41 7.676 .90 1.0 13.297 0.8 36.4 34.2 4.9 40.54 10.0 17.3 18.5 79.8 44.8 14.17 2.3 10.0 29.8 21.3 15.7 14.7 7.9 39.210 x.6 56.57 4.9 12.6 15.26 6.34 10.83 3.8 95.2 27.26 9.18 2.56 3.0 33.67 3.34 2.8 43.5 101.0 22.03 8.2025 x.9 11.6 16.3 49.3 64.73 10 11 12 13 14 25.6 37.9 113.6 15.30 7.6 43.6 107.31 10.2 74.4 42.1 43.6 12.17 1.2 10.0100 .49 4.35 11.20 .61 6.70 6.0 41.7 66.3 89.57 4.3 18.25 7.58 8.63 9.3 39. Thompson.4 34.5 19.1 77.7 38.6 90.3 27.39 2.4 13.6 30.3 27.3 13.0201 .4 40.3 16.9 48.2 16.1 16.1 55.35 7.115 .299 x.7 29.23 8.3 59.575 1.45 4.2 11.14 5.9 20.6 14.24 1.6 32.9 63.7 6.211 .3 26.8 9.3 5.5 12. Table 6 Kolmogorov-Smirnov Distribution Sampl e size (n) 1 Significance Level .1 124.3 51.5 19.5 13.3 22.3 30.88 10.3 128.1 80. v x.8 34.6 36.0158 .6 38.6 28.3 67.3 35.0039 .7 37.0 50.8 45.5 13.1 129.3 11.2 67.2005 1 2 3 4 7.91 7.63 7.1 28.9 40.1 29.2 36.2 12.216 .5 13.5 20.8 12.23 5.7 37.5 77.2975 x.23 5.8 26.3 45.4 31.4 8.5 36.15 1.0 11.5 32.49 2.1 18.5 92.8 70.1 19.8 63.6 88.55 9.2 16.15 .1 98.1 40.7 26.9 11. by permission of the author and publisher.5 16.3 17.0 22.37 3.0 11.5 70 80 90 100 104.3 6.3 23.9 48.275 Χ2a.7 12.7 37.7 23.11 5.05 3.2 59.584 1.4 16.1 135.24 1.4 47.0 17.5 26.6 15.7 21.3 43.6 118.35 6.5 34.3 Source: Catherine M.7 18.5 24.9 35.07 3.3 12.2 18.0010 .6 40.5 21.6 22.2 23.3 16.3 124.9 6.3 18.0 48.3 11.3 20.872 1.78 1.7 10.5 18.0 59.3 41.0 23.3 51.60 5.4 83.3 24.7 28.072 .34 1.2 19.2 29.3 28.995 .92 0.3 20.57 3.6 29.412 .352 .1 18.8 55.8 67.925 0.0000 .6 51.900 0.3 23.8 21.5 15.0 11.71 4.4 112.7 37.950 18 0.2 18.4 28.32 2.2 27.1 15.3 82.4 35.05 .102 0.4 51.0 27. Table of percentage points of the 2 distribution.201 x.4 32.8 14.56 8.3 13.5 61.44 9.205 x.3 49.2 116.2 19.7 14.7 12.6 14.01 .26 7.484 0.6 109.0 14.1 69.79 3.5 21.2 38.20 2. Vol.3 99.5 11.8 57.3 24.975 0.70 .5 28.3 22.2 24.2 65.5 16.0 12.36 0.3 85.225 x.74 7.58 6.01 7.9 26.4 4.2 31.6 74.33 .3 29.6 8.5 45.07 5.2 45.6 90.89 6.4 16.2 35.6 47.25 3.69 2.8 14.6 12.84 20 21 22 23 24 40.11 4.8 79.5 17. Biometrika.02 7.04 10.91 5.2 88.9 43.Table 5 χ2 Distribution DOF x.10 .1 21.1 41.4 69.3 11.2 45.63 4.8 24.9 25.989 1.7 26.9 19.6 33.3 15.250 x.3 61.07 15 16 17 18 19 32.8 32.7 42.0 27.0 11.7 14.30 10.8 38.6 30.0506 .89 25 26 27 28 29 46.40 5.1 37.0 35.7 76.0 26.6 41.3 21.2 44.1 3.6 118.8 28.3 140.73 3.1 13.04 7.96 8.711 0.39 10.8 30.7 16.295 x.9 52.1 39.64 2.34 8.2 14.9 30.24 10.3 79.65 2.9 13.3 19.0 52.09 .9 17.57 5.1 13.3 49.3 73.63 2.3 19.01 5. 418 .17 .266 .436 .328 .264 .21 .565 .286 .231 .514 .22 .708 .642 .70 1.274 . 19 .14 .342 .521 .577 .725 .302 .29 .301 .564 .375 .669 6 7 8 9 10 .776 .244 .726 .22 1.304 .339 .252 .237 .370 .266 .349 .19 .358 . 1952.543 .14 1.21 .380 .360 .438 .246 .309 .326 .510 .258 .325 .381 .32 .14 .259 .450 .Z.264 .381 .307 .391 .829 .292 .295 .338 .457 .474 .23 .494 .295 .405 .322 .313 .318 .404 16 17 18 19 20 .684 .18 60 70 80 90 100 Asymptotic Formula 1.242 .284 .22 .470 .468 .929 . Birnbaum.21 .274 .272 .352 .624 .16 .486 .410 .842 .338 .25 .734 .23 .432 .597 .361 .2 3 4 5 .391 .388 .446 .368 .409 .250 .19 .294 .314 .63 n n n n n Source: Journal American Statistical Association 47:425-441.27 .24 .20 .618 .278 .563 .411 .19 .486 11 12 13 14 15 .352 25 30 35 40 50 .18 .433 .36 1.W.19 .15 .21 .361 .283 .