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<br />. <br /> <br />SOME STATISTICAL TOOLS IN HYDROLOGY <br /> <br />27 <br /> <br />. <br /> <br />of the equation constants, and of the <br />standard error. <br />2. It allows testing of the coefficients for sig- <br />nificant difference from zero. <br />3. Results can be presented in a clear, concise <br />manner which most hydrologists can under <br />stand. <br />4. Results are unique for the model and sample <br />used; different investigators would get the <br />same results. <br />Disadvantages of an analytical method are: <br /> <br />1. Computation is time consuming, especially <br />for several variables and complicated mod- <br />els, use of computers reduces the actual <br />computation time but requires consider- <br />able time to prepare the data. <br />2. The existence of wild points is masked. as <br />would be the existence of a group of points <br />much different from the majority (unless <br />departures of all points from the estimates <br />are computed). <br />3. The model selected may not be the appro- <br />priate one. <br />In general a graphical method should be used <br />for exploratory work and the final conclusions <br />should be based on a computed relation. <br /> <br />Determining Equations of Graphical <br />Relations <br /> <br />Graphical analyses are often adequate for <br />certain problems. The results may be reported <br />by furnishing copies of the graphs, but interpre- <br />tation of graphs in more than two variables is <br />difficult for someone not familiar with the pro- <br />cedure. For instance, consider the three-variable <br />relation of figure 18. What is the expected runoff <br />corresponding to a water content of snow of 20 <br />inches and a precipitation at Three Creek of 10 <br />inches? It is 40,000 acre-feet from the left curve <br />plus 14,000 from the right curve, a total of <br />54,000 acre-feet. A better method of presenta- <br />tion would be as a family of curves. Another <br />way would be to write the equation of the graph- <br />ical relation. The equation for the relation of <br />figure 18 is <br /> <br />. <br /> <br />R= -22 + 1.68+4,.4P <br /> <br />12<8<28 <br />4<P<11. <br /> <br />The limits of definition to the right of the equa- <br />tion tell the reader that he applies the equation <br />to values of 8 and P outside those limits at his <br />own risk. <br />Another advantage of defining the equations <br />of graphical relations appears when it is desired <br />to compare relations of the same type but <br />developed from different data. For example, <br />Riggs (1965) related base flow discharges of <br />nine small streams to drainage area and per- <br />centage of basin cleared. He made eight different <br />relations, each based on measurements of the <br />same streams but at different times. Interest was <br />in the variability of the effect of the percentage <br />of basin cleared; this variability was apparent <br />when the equations of the relations were defined <br />and the regression coefficients of the percentage <br />of basin cleared were compared. <br />Still another use for the equation of a graphi- <br />cal relation is to reduce a relation to its simplest <br />terms. This reduction is a desirable procedure <br />if tbe graphical analysis uses compound vari- <br />ables. The graphical regression of figure 20 is <br />the result of an exploratory study of data for a <br />basin in Western United States and indicates <br />that l11AF may be estimated quite reliably from <br />drainage area and mean flow in cubic feet per <br />second per square mile. Because drainage area <br />is used twice, the actual effect of drainage area <br />should be assessed. We begin by writing the <br />equation of the relation (by a method described <br />subsequently), which is <br /> <br />log MAF=1.00+log A+1.0210g Q, <br /> <br />where A is drainage area in square miles and <br />Q is mean flow in cubic feet per second per <br />square m:ile. Let q be mean flow in cubic feet <br />per second. Then <br />Ii qlA. <br /> <br />Substituting this for Ii in the first equation gives <br /> <br />log MAF= 1.00 + log A + 1.02 log (iliA) <br /> <br />= 1.00+log A+1.02Iog q-1.02 log A. <br /> <br />Thus the net regression coefficient of log A is <br />-0.02 which is negligible and, if eliminated <br />from the relation, leaves <br /> <br />log MAF=1.00+1.0210g q, <br /> <br />or <br /> <br />MAF=10q1.02. <br />