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<br />BIOLOGICAL METHODS <br /> <br />slides suspended in streams varies from stream to <br />stream. A simple analysis such as this could <br />precede a more in-depth biological study of the <br />comparative productivity of the streams. Data <br />from such a study are presented in Table 7. <br /> <br />TABLE 7. PERIPHYTON <br />PRODUCTIVITY DATA <br /> <br />Stream <br /> <br />Slide <br /> <br />Biomass <br />(mg dry wt.) <br />26 <br />20 <br />14 <br />25 <br />34 <br />28 <br />Lost <br />23 <br />31 <br />35 <br />40 <br />28 <br /> <br />2 <br /> <br />1 <br />2 <br />3 <br />4 <br />1 <br />2 <br />3 <br />4 <br />1 <br />2 <br />3 <br />4 <br /> <br />3 <br /> <br />In testing with the analysis of variance, as <br />with other methods, a null hypothesis should be <br />formulated. In this case the null hypothesis <br />could be: <br />Ho: There are no differences in the <br />biomass of organisms attached to the <br />slides that may be attributed to differ- <br />ences among streams. <br />In utilizing the analysis of variance, the test <br />for whether there are differences among streams <br />is made by comparing two types of variances, <br />most often called "mean squares" in this con- <br />text. Two mean squares are computed: one <br />based upon the means for streams; and one that <br />is free of the effect of the means. In our <br />example, a mean square for streams is computed <br />with the use of the averages (or totals) from the <br />streams. The magnitude of this mean square is <br />affected both by differences among the means <br />and by differences among slides of the same <br />stream. The mean square for slides is computed <br />that has no contribution due to stream differ- <br />ences. If the null hypothesis is true, then differ- <br />ences among streams do not exist and, therefore, <br />they make no contribution to the mean square <br />for streams. Thus, both mean squares (for <br />streams and for slides) are estimates of the same <br />variance, and with repeated sampling, they <br />would be expected to average to the same value. <br /> <br />If the null hypothesis (lio) is true, the ratio of <br />these values is expected to equal one. If Ho is <br />not true, i.e., if there are real differences due to <br />the effect of streams, then the mean square for <br />streams is affected by these differences and is <br />expected to be the larger. The ratio in the <br />second case is expected to be greater than one. <br />The ratio of these two variances forms an F-test. <br />The analysis of variance is presented in Table <br />8. <br />The computations are: <br />(85 + 85 + 134)2 <br />C = 11 8401.45 <br /> <br />. <br /> <br />~ Xi j2 = 262 + 202 + . . . + 402 + 282 = 8936 <br />i j <br /> <br />Total SS = 8936 - 8401.45 = 534.55 <br />Xi.2 852 852 1342 <br />~ (-)=-+-+- = 8703 58 <br />i ri 4 3 4 . <br /> <br />Streams SS = 8703.58 - 8401.45 = 302.13 <br />Slides w/i streams SS = Total SS - Streams SS <br />= 534.55 - 302.13 <br />= 232.42 <br /> <br />The mean squares (MS column) are computed <br />by dividing the sums of squares (SS column) by <br />its corresponding degrees of freedom (df <br />column). (Nothing is usually learned in this <br />context by computing a total MS.) The F-test is <br /> <br />TABLE 8. F-TEST USING PERIPHTON DATA <br /> <br />Source df SS <br />Total N-l* ~ . Xi j2 - C <br /> 1 J <br /> 2 <br />Streams t-l ~ ~-C <br /> ri <br />Slides w/i streams ~ (fj-l) Total SS - Stream SS <br /> i <br /> <br />*The symbols are defined as: N = total number of observations <br />(slides); t = nwnber of streams; ri = number of slides in stream i; <br />Xij = an observation (biomass of a slide); Xi. = sum of the <br />observations for stream i;and C = correction for mean = <br />(~ Xi .)2 <br />.. J <br />1J <br />N <br /> <br />Source df SS <br />Total 10 534.55 <br />Streams 2 302.13 <br />Slides w/i <br />streams 8 232.42 <br /> <br />MS F <br /> <br />151.065 5.20* <br />29.055 <br /> <br />*Significant at the 0.05 probability level. <br /> <br />. <br /> <br />16 <br />