7 The parametric function for comparing the first two (store) brands of oil filter with

Example 10.7 The parametric function for comparing the first two (store) brands of oil filter with

  the last three (national) brands is u5 1 2 (m 1m )2 1 1 2 3 (m 3 1m 4 1m 5 ) , from

  which

  2 1 2 1 2 1 2 1 2 1 g 2 c

  i 5 a b 1 a 1 1 1 2 5 2 b a2 3 b a2 3 b a2 3 b 6

  10.2 Multiple Comparisons in ANOVA

  With u5 ˆ 1 2 (x 1 .1x .) 2 1 2 3 (x 3 .1x 4 .1x 5 .) 5 .583 and MSE 5 .088 , a 95

  interval is

  ■ Sometimes an experiment is carried out to compare each of several “new”

  treatments to a control treatment. In such situations, a multiple comparisons tech- nique called Dunnett’s method is appropriate.

  EXERCISES Section 10.2 (11–21)

  11. An experiment to compare the spreading rates of five dif-

  Tukey’s pairwise comparisons

  ferent brands of yellow interior latex paint available in a

  Family error rate 5 0.0500

  particular area used 4 gallons (J 5 4) of each paint. The

  Individual error rate 5 0.00693

  sample average spreading rates (ft 2 gal) for the five brands

  Critical value 5 4.10

  were , x 1 5 462.0, x 2 5 512.8, x 3 5 437.5, x 4 5 469.3

  and x 5 532.1

  5 . The computed value of F was found to be

  Intervals for (column level mean) – (row level

  significant at level mean) a 5 .05 . With MSE 5 272.8 , use

  Tukey’s procedure to investigate significant differences in

  the true average spreading rates between brands.

  12. In Exercise 11, suppose x 3 . 5 427.5 . Now which true aver-

  age spreading rates differ significantly from one another?

  Be sure to use the method of underscoring to illustrate your

  conclusions, and write a paragraph summarizing your

  2 53.8 2 18.9 2 11.8 13. Repeat Exercise 12 supposing that 20.4 x

  2 . 5 502.8 in addition

  to . x 3 . 5 427.5

  a. Is it plausible that the variances of the five axial stiffness

  14. Use Tukey’s procedure on the data in Example 10.3 to iden-

  index distributions are identical? Explain.

  tify differences in true average bond strengths among the

  b. Use the output (without reference to our F table) to test

  five protocols.

  the relevant hypotheses.

  c. Use the Tukey intervals given in the output to determine

  15. Exercise 10.7 described an experiment in which 26 resistiv-

  which means differ, and construct the corresponding

  ity observations were made on each of six different concrete

  underscoring pattern.

  mixtures. The article cited there gave the following sample means: 14.18, 17.94, 18.00, 18.00, 25.74, 27.67. Apply

  17. Refer to Exercise 5. Compute a 95 t CI for u5

  Tukey’s method with a simultaneous confidence level of

  2 (m 1 1m 2 )2m 3 .

  95 to identify significant differences, and describe your

  18. Consider the accompanying data on plant growth after the

  findings (use MSE 5 13.929 ).

  application of five different types of growth hormone.

  16. Reconsider the axial stiffness data given in Exercise 8. ANOVA output from Minitab follows:

  Analysis of Variance for Stiffness

  a. Perform an F test at level

  b. What happens when Tukey’s procedure is applied?

  19. Consider a single-factor ANOVA experiment in which

  28.57 I53 , J 5 5, x 1 5 10, x 2 5 12 , and x 3 5 20 . Find a value

  20.83 of SSE for which f.F .05,2,12 , so that H 0 :m 1 5m 2 5m 3 is

  44.51 rejected, yet when Tukey’s procedure is applied none

  26.00 of the m i ’s can be said to differ significantly from one Pooled StDev 5 32.39 another.

  CHAPTER 10 The Analysis of Variance

  20. Refer to Exercise 19 and suppose x 1 5 10, x 2 5 15 , and

  a. Test the null hypothesis that true average survival time

  x 3 . 5 20 . Can you now find a value of SSE that produces such

  does not depend on an injection regimen against the

  a contradiction between the F test and Tukey’s procedure?

  alternative that there is some dependence on an injection

  21. The article “The Effect of Enzyme Inducing Agents on the

  regimen using a 5 .01 .

  Survival Times of Rats Exposed to Lethal Levels of Nitrogen

  b. Suppose that 100(1 2 a) CIs for k different paramet-

  Dioxide” (Toxicology and Applied Pharmacology, 1978:

  ric functions are computed from the same ANOVA data

  169–174) reports the following data on survival times for rats

  set. Then it is easily verified that the simultaneous confi-

  exposed to nitrogen dioxide (70 ppm) via different injection

  dence level is at least 100(1 2 ka) . Compute CIs with a

  regimens. There were J 5 14 rats in each group.

  simultaneous confidence level of at least 98 for

  x (min)

  6. p-Aminobenzoic Acid