Statistics for Environmental Engineers

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The test will be made at the 5% level with degrees of freedom v1 = к — 1 = 3 — 1 = 2 and v2 = N — к = 12 — 3 = 9. The relevant value is F2,9,0.05 = 4.26. The ratio computed for our experiment, F = 52/6 = 8.67 is greater than F2 9 a=0.05 = 4.26, so we conclude that or > ow. This provides sufficient evidence to conclude at the 95% confidence level that the means of the three treatments are not equal. We are entitled only to conclude that ца ^ Пв ^ Пс- This analysis does not tell us whether one treatment is different from the other two (i.e., nA ^ Пв but nB = Пс), or whether all three are different. To determine which are different requires the kind of analysis described in Chapter 20.


When ANOVA is done by a commercial computer program, the results are presented in a special ANOVA table that needs some explanation. For the example problem just presented, this table would be as given in Table 24.2. The “sum of squares” in Table 24.2 is the sum of the squared deviations in the numerator of each variance estimate. The “mean square” in Table 24.2 is the sum of squares divided by the degrees of freedom of that sum of squares. The mean square values estimate the within-treatment variance (sW) and the between-treatment variance (sb). Note that the mean square for variation between treatments is 52, which is the between-treatment variance computed above. Also, note that the within treatment mean square of 6 is the within-treatment variance computed above. The F ratio is the ratio of these two mean squares and is the same as the F ratio of the two estimated variances computed above.

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