I (J 2 1) MSE

I2 1 I (J 2 1) MSE

  The computations are often summarized in a tabular format, called an ANOVA

  table , as displayed in Table 10.2. Tables produced by statistical software customarily include a P-value column to the right of f.

  Table 10.2 An ANOVA Table

  Source of

  Sum of

  Variation

  df Squares

  Mean Square f

  Treatments

  I2 1 SSTr MSTr 5 SSTr y(I 2 1) MSTrMSE

  Error

  I (J 2 1) SSE MSE 5 SSE y[I(J 2 1)]

  Total

  IJ 2 1 SST

  418 Chapter 10 the analysis of Variance

  ExamplE 10.4

  According to the article “Evaluating Fracture Behavior of Brittle Polymeric

  Materials Using an IASCB Specimen” (J. of Engr. Manuf., 2013: 133–140) , researchers have recently proposed an improved test for the investigation of frac- ture toughness of brittle polymeric materials. This new fracture test was applied to the brittle polymer polymethylmethacrylate (PMMA), more popularly known as Plexiglas, which is widely used in commercial products. The test was performed by applying asymmetric three-point bending loads on PMMA specimens. The location of one of the three loading points was then varied to determine its effect on fracture load. In one experiment, three loading point locations based on different distances from the center of the specimen’s base were selected, resulting in the following fracture load data (kN):

  Let m i denote true average fracture load when distance i is used (i 5 1, 2, 3). The null hypothesis asserts that these three m i ’s are identical, whereas the alternative hypothesis says that not all the m i ’s are the same. Before using the F test at signifi- cance level .01, we should check the plausibility of underlying assumptions. The three sample standard deviations are .322, .167, and .278, respectively. Sure enough, the largest of these three is no more than twice the smallest. So the assumption of equal variances is plausible. Figure 10.4 shows a normal probability plot of the 12 residuals obtained by subtracting the mean of each sample from the four sample observations. They don’t come much straighter than this! It is reasonable to assume that the three fracture load distributions are normal.

  Figure 10.4 Normal probability plot of the residuals from Example 10.4

  Squaring each of the 12 observations and adding gives oox 2 5 (2.62) ij 2 1…1

  (5.06) 2 5 181.7376. The values of the three sums of squares are

  SST 5 181.7376 2 (45.74) 2 y12 5 181.7376 2 174.3456 5 7.3920 SSTr 5 1 [(11.86) 2 1 (14.72) 2 1 (19.16) 4 2 ] 2 174.3456 5 6.7653

  SSE 5 7.3920 2 6.7653 5 .6267

  10.1 Single-Factor aNOVa 419

  The accompanying ANOVA table from Minitab summarizes the computations. With a P-value of .000, the null hypothesis can be rejected at any sensible signifi- cance level, and in particular at the chosen level .01. There is compelling evidence for concluding that true average fracture load is not the same for all three distances.

  ExErciSES section 10.1 (1–10)

  1. In an experiment to compare the tensile strengths of

  Strength and Stiffness of Second-Growth Douglas-Fir

  I5 5 different types of copper wire, J 5 4 samples of

  Dimension Lumber” (Forest Products J., 1991: 35–43),

  each type were used. The between-samples and within-

  except that the sample sizes there were larger]:

  samples estimates of s 2 were computed as MSTr 5

  2673.3 and MSE 5 1094.2, respectively. Use the F test

  at level .05 to test H 0 :m 1 5m 2 5m 3 5 m 4 5m 5 versus

  H a : at least two m i ’s are unequal.

  2. Suppose that the compression-strength observations on

  the fourth type of box in Example 10.1 had been 655.1, 748.7, 662.4, 679.0, 706.9, and 640.0 (obtained by add-

  Use this data and a significance level of .01 to test the

  ing 120 to each previous x 4j ). Assuming no change in the

  null hypothesis of no difference in mean modulus of

  remaining observations, carry out an F test with a 5 .05.

  elasticity for the three grades.

  3. The lumen output was determined for each of I 5 3 dif-

  6. The article “Origin of Precambrian Iron Formations”

  ferent brands of lightbulbs having the same wattage, with

  (Econ. Geology, 1964: 1025–1057) reports the follow-

  J5 8 bulbs of each brand tested. The sums of squares

  ing data on total Fe for four types of iron formation

  were computed as SSE 5 4773.3 and SSTr 5 591.2.

  (15carbonate, 2 5silicate, 35magnetite, 4 5 hematite).

  State the hypotheses of interest (including word defini-

  tions of parameters), and use the F test of ANOVA

  (a 5 .05) to decide whether there are any differences in

  true average lumen outputs among the three brands for

  this type of bulb by obtaining as much information as

  possible about the P-value.

  4. It is common practice in many countries to destroy

  (shred) refrigerators at the end of their useful lives. In

  this process material from insulating foam may be

  Carry out an analysis of variance F test at significance

  released into the atmosphere. The article “Release of

  level .01, and summarize the results in an ANOVA table.

  Fluorocarbons from Insulation Foam in Home

  7. An experiment was carried out to compare electrical

  Appliances During Shredding” (J. of the Air and

  Waste Mgmt. Assoc., 2007: 1452–1460) gave the follow-

  resistivity for six different low-permeability concrete bridge deck mixtures. There were 26 measurements on

  ing data on foam density (gL) for each of two refrigera-

  concrete cylinders for each mixture; these were obtained

  tors produced by four different manufacturers:

  28 days after casting. The entries in the accompanying

  ANOVA table are based on information in the article

  “In-Place Resistivity of Bridge Deck Concrete

  Does it appear that true average foam density is not the

  Mixtures” (ACI Materials J., 2009: 114–122) . Fill in

  same for all these manufacturers? Carry out an appropriate

  the remaining entries and test appropriate hypotheses.

  test of hypotheses by obtaining as much P-value infor-

  Sum of

  mation as possible, and summarize your analysis in an

  Source

  df Squares

  Mean Square f

  ANOVA table.

  5. Consider the following summary data on the modulus of

  Mixture

  elasticity (3 10 6 psi) for lumber of three different grades

  Error

  [in close agreement with values in the article “Bending

  Total 5664.415

  420 Chapter 10 the analysis of Variance

  8. A study of the properties of metal plate-connected

  Wheat

  trusses used for roof support (“Modeling Joints Made

  Barley

  with Light-Gauge Metal Connector Plates,” Forest

  Maize

  Products J., 1979: 39–44) yielded the following obser-

  Oats

  vations on axial-stiffness index (kipsin.) for plate lengths

  4, 6, 8, 10, and 12 in:

  Does this data suggest that at least two of the grains dif- fer with respect to true average thiamin content? Use a

  level a 5 .05 test.

  10. In single-factor ANOVA with I treatments and J observa-

  tions per treatment, let m 5 (1 yI)om i .

  a. Express E (X..) in terms of m. [Hint: X.. 5 (1 yI)oX

  i? ] b. Determine E(X i? 2 ). [Hint: For any rv Y, E(Y 2 )5

  V Does variation in plate length have any effect on 2 (Y) 1 [E(Y)] .] true average axial stiffness? State and test the rel-

  c. Determine E (X.. 2 ).

  evant hypotheses using analysis of variance with

  d. Determine E(SSTr) and then show that

  a5 .01. Display your results in an ANOVA table.

  J

  I2 1 o i

  [Hint:

  ) oox 2 5,241,420.79.] 2m

  ij 5

  E (MSTr) 5 s 2

  1 (m

  9. Six samples of each of four types of cereal grain grown

  e. Using the result of part (d), what is E(MSTr) when

  in a certain region were analyzed to determine thiamin

  H is true? When H is false, how does E(MSTr)

  content, resulting in the following data (mgg):

  compare to s 2 ?