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<br />32
<br />
<br />TECHNIQUES OF WATER-RESOURCES INVESTIGATIONS
<br />
<br />sample may be resolved into component varia-
<br />tions due to independent factors, It is closely
<br />related to correlation but is applicable to
<br />problems where some of the factors can be
<br />described only by classes, not as numerical
<br />variates.
<br />The analysis depends on the additive
<br />charaoteristic .of variances. Its purpose is to
<br />test whether several means are alike or not.
<br />The basic features of the process are (1) the
<br />measurement of variance among experimen tal
<br />data by the sum of the squared deviations of
<br />the observations from their mean, (2) the par-
<br />titioning of the total sum of squared deviations
<br />into independent parts, each part associated
<br />with some physical feature of the experiment,
<br />(3) the estimation of parameters in the distribu-,
<br />tion6 postulated to underlie the data, and (4)
<br />tests of significance regarding these parameters.
<br />Results of the test give the probability of there
<br />being a significant difference between the effoots
<br />of a factor or factors at different levels,
<br />A very simple example of an analysis of
<br />variance concerns whether the mean runoffs
<br />for two periods of record at a gaging station
<br />are estimates of the same population mean. The
<br />annual runoffs are given below:
<br />
<br />Period. 1 Pftiod I
<br />17.3..____________..__ 6. 4
<br />21.9_____________,____ 15.2
<br />13.6.-,.____,;.._____, 9.7
<br />10.8________._________ 4. 4
<br />19.7uu__________n__ 9.9
<br />20.7.-_..__,_,___,____ 11.9
<br />16.3______.___________ 11.9
<br />16.2______.___________ 15.4
<br />12.5______.u__u___u 9,4
<br />11,3_,u.____.________ 7.0
<br />14.0________________,_ 16.0
<br />16.5________________,. 17.0
<br />15.3________._________ 11.2
<br />19.2_________________, 13.2
<br />13.0____,_______.___,_ 11.5
<br />
<br />238.3
<br />
<br />170.1
<br />
<br />Computations are as follows:
<br />
<br />Grand total=T=238.3+170,1=408.4.
<br />Total number of items=N=30.
<br />Number of items in each period=n=15.
<br />T'IN= (408.4)'/30=5,559.7.
<br />
<br />Sum of squares of all individuals
<br />=L;Yi'.=6,067.3.
<br />(Sum of 6quares of sums)/n=L;1'1ln
<br />= [(238.3) ,+ (170,1)'1/15= 5,714.7.
<br />
<br />.
<br />
<br />Between-periods sum of squares= L; 1'1ln
<br />- T'IN=5,714.7 -5,559.7= 155.0,
<br />
<br />Within-periods sum of squares = L;Y;'
<br />- L;1'1ln=6,067,3-5,714.7=352.6.
<br />
<br />Total sum of squares=L;Yl,-T'IN=6,067.3
<br />-5,559.7=507.6.
<br />
<br />The analysis of variance table is
<br />
<br />Source
<br />
<br />Sum of Degrees of Mean Average
<br />squares freedom square mea.n
<br />square
<br />
<br />Between
<br />periods _ _ _ 155.0 1 **155 a'+nol
<br />Within
<br />periods _ _ _ 352, 6 28 12,6 a'
<br />TotaL__ 507.6 29 --------- .
<br />
<br />"Statistical slgn:1fiunce above the O.011evel.
<br />
<br />The degrees of freedom, D.F., are one less
<br />than the number of periods, p, for the between-
<br />periods sum of squares and N -1 for the total.
<br />Thus the degrees of freedom associated with the
<br />within-periods source is N - p. Mean square is
<br />obtained by dividing the sum of squares by
<br />D.F.
<br />The last column in the analysis of variance
<br />table shows expected values of the mean
<br />squares. If the means for the periods are alike,
<br />the term nul would be zero, Estimates of the
<br />ratio [u'+nulllu' may be greater than one,
<br />because of chance or because there is a real
<br />difference. This ratio has the F distribution and
<br />can be tested statistically. The ratio in the
<br />above table is 155/12.6= 12.3. The value of F
<br />for 1 and 28 degrees of freedom and a proba-
<br />bility of 0.01 is F',...O.0l=7.6 from a table of
<br />F distribution. Because the sample ratio
<br />exceeds the tabular ratio, we conclude that
<br />there is a real difference between periods; that
<br />is, the probability is less than 0.01 that such
<br />a difference in means would have occurred by
<br />
<br />.
<br />
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