5 An experiment was carried out to compare five different brands of automobile oil fil-

Example 10.5 An experiment was carried out to compare five different brands of automobile oil fil-

  ters with respect to their ability to capture foreign material. Let m i denote the true average amount of material captured by brand i filters under (i 5 1, c, 5) con- trolled conditions. A sample of nine filters of each brand was used, resulting in the

  following sample mean amounts: x 1 5 14.5, x 2 5 13.8, x 3 5 13.3, x 4 5 14.3 , and

  x 5 . 5 13.1 . Table 10.3 is the ANOVA table summarizing the first part of the analysis.

  Table 10.3 ANOVA Table for Example 10.5 Source of Variation

  df Sum of Squares

  Mean Square f

  Treatments (brands)

  Since , F .05,4,40 5 2.61 H 0 is rejected (decisively) at level .05. We now use Tukey’s pro-

  cedure to look for significant differences among the m i ’s. From Appendix Table A.10, Q .05,5,40 5 4.04 (the second subscript on Q is I and not I21 as in F), so w5

  4.04 1.0889 5 .4 . After arranging the five sample means in increasing order, the two smallest can be connected by a line segment because they differ by less than .4.

  CHAPTER 10 The Analysis of Variance

  However, this segment cannot be extended further to the right since

  13.8 2 13.1 5 .7 .4 . Moving one mean to the right, the pair x 3 and x 2 cannot be

  underscored because these means differ by more than .4. Again moving to the right, the next mean, 13.8, cannot be connected to any further to the right. The last two means can be underscored with the same line segment.

  x 5 x 3 x 2 x 4 x 1

  13.1 13.3 13.8 14.3 14.5 Thus brands 1 and 4 are not significantly different from one another, but are signif-

  icantly higher than the other three brands in their true average contents. Brand 2 is significantly better than 3 and 5 but worse than 1 and 4, and brands 3 and 5 do not differ significantly.

  If x 2 5 14.15 rather than 13.8 with the same computed w, then the configura-

  tion of underscored means would be

  x 5 x 3 x 2 x 4 x 1 ■

  Example 10.6

  A biologist wished to study the effects of ethanol on sleep time. A sample of 20 rats, matched for age and other characteristics, was selected, and each rat was given an oral injection having a particular concentration of ethanol per body weight. The rapid eye movement (REM) sleep time for each rat was then recorded for a 24-hour period, with the following results:

  Treatment (concentration of ethanol)

  32.76 x 5 1107.5 x 5 55.375

  Does the data indicate that the true average REM sleep time depends on the con- centration of ethanol? (This example is based on an experiment reported in “Relationship of Ethanol Blood Level to REM and Non-REM Sleep Time and Distribution in the Rat,” Life Sciences, 1978: 839–846.)

  The x i .s differ rather substantially from one another, but there is also a great deal of variability within each sample, so to answer the question precisely we must carry out the

  ANOVA. With g gx 2 ij 5 68,697.6 and correction factor x 2 .. (IJ) 5 (1107.5) 2 20 5

  61,327.8 , the computing formulas yield SST 5 68,697.6 2 61,327.8 5 7369.8

  Table 10.4 is a SAS ANOVA table. The last column gives the P-value as .0001. Using a significance level of .05, we reject the null hypothesis

  10.2 Multiple Comparisons in ANOVA

  H 0 :m 1 5m 2 5m 3 5m 4 , since P-value .0001 < .05 a. True average REM

  sleep time does appear to depend on concentration level.

  Table 10.4 SAS ANOVA Table

  Analysis of Variance Procedure Dependent Variable: TIME

  DF Squares

  Square

  F Value Pr F

  Corrected Total

  There are I54 treatments and 16 df for error, from which Q .05,4,16 5 4.05 and w 5 4.05 193.05 5 17.47 . Ordering the means and underscoring yields

  x 4 x 3 x 2 x 1

  The interpretation of this underscoring must be done with care, since we seem to have concluded that treatments 2 and 3 do not differ, 3 and 4 do not differ, yet 2 and

  4 do differ. The suggested way of expressing this is to say that although evidence allows us to conclude that treatments 2 and 4 differ from one another, neither has been shown to be significantly different from 3. Treatment 1 has a significantly higher true average REM sleep time than any of the other treatments.

  Figure 10.4 shows SAS output from the application of Tukey’s procedure.

  Alpha df 5 0.05 5 16 MSE 5 92.9625

  Critical Value of Studentized Range 5 4.046 Minimum Significant Difference 5 17.446

  Means with the same letter are not significantly different.

  Tukey Grouping

  C B 47.920

  Figure 10.4 Tukey’s method using SAS

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