Stat 217 Final Exam Name: May 1, 2002 Problem 1. Three brands of batteries are under study. It is suspected that the lives (in weeks) of the three brands are different. Five batteries of each brand are tested with the following results: Table 1: Weeks of Life Brand 1 Brand 2 Brand 3 100 76 108 96 80 100 92 75 96 96 84 98 92 82 100 a. (5 pts) Are the lives of these brands of batteries different? Table 2: ANOVA Table for Problem 1 Source df SS MS F p-value brand 2 1196.13 598.06 38.34.0001 error 12 187.2 15.6 total 14 1383.3 There is a treatment effect since the p-value is very small. b. (5 pts) Estimate the components of the appropriate statistical model; that is, mean responses and error variance. The model is y ij = µ + τ j + ɛ ij where i = 1, 2, 3, 4, 5 and j = 1, 2, 3 (treatment or brand index). The assumption is that E(y ij ) = µ + τ j and Var(y ij ) = σ 2. ˆµ 1 = ȳ 1 = 95.2, ˆµ 2 = ȳ 2 = 79.4, ˆµ 3 = ȳ 3 = 100.4 The estimate of the error variance is ˆσ 2 = MSE = 15.6. 1
c. (5 pts) Construct a 95 percent interval estimate on the mean life of battery brand 2. Construct a 95 percent interval estimate on the mean difference between the lives of battery brands 2 and 3. A 95% CI for µ 2 is given by ȳ 2 ± t 0.025 (error df) MSE 79.4 ± 3.85 = (75.55, 83.25) n = 79.4 ± t 0.025 (12) 15.6 5 A 95% CI for the difference µ 3 µ 2, which is a treatment contrast (C = 0µ 1 µ 2 + µ 3 ), is given by ȳ 3 ȳ 2 ± t 0.025 (15 3) s 2 2 15.6 2 C n = 100.4 79.4 ± t 0.025(12) 5 21 ± 2.18 6.24 = (15.55, 26.46) d. (5 pts) Which brand would you select for use? If the manufacturer will replace without change any battery that fails in less than 85 weeks, what percentage would he expect to replace? Since brand 2 differs from both brands 1 and 3, and brand 3 batteries appear to have the longest life, we would select brand 3 batteries. Given this choice, as the estimate of mean life for brand 3 batteries is 100.4 weeks, the probability of getting less than 85 week life is the area under the student s t(12) curve to the left of 85 100.4 = 8.718 which is negligible. 15.6/5 2
Problem 2: In an attempt to study the effectiveness of two corn varieties (V 1 and V 2 ), two corn-producing counties (C) in Iowa were chosen at random. Four farms (F ) were then randomly selected within each county. Seeds from both varieties were sent to the four farms and planted in random plots. At harvest, the number of bushels of corn per acre was recorded for each variety on each farm. a. (5 pts) Write a mathematical model for this design. This is a nested factorial design that reads: y ijkl = µ + v i + C j + V C ij + F k(j) + V F ik(j) + ɛ l(ijk) with i = 1, 2, j = 1, 2, k = 1, 2, 3, 4 and l = 1 from what we can infer from the problem statement. Hence, the error is non-estimable. b. (10 pts) Determine the EMS column for this experiment. F R R R 2 2 4 1 Source i j k l EMS v i 0 2 4 1 8φ v + 4σV 2 C + σ2 V F + σ2 C j 2 1 4 1 8σC 2 + 2σ2 F + σ2 V C ij 0 1 4 1 4σV 2 C + σ2 V F + σ2 F k(j) 2 1 1 1 2σF 2 + σ2 V F ik(j) 0 1 1 1 σv 2 F + σ2 ɛ l(ijk) 1 1 1 1 σ 2 c. (5 pts) Assuming that only counties (C) showed significantly different average yields, and MS C = 130 and MS F = 10 (farms or error), find the percentage of the variance in this experiment that can be attributed to county differences. We do not have sufficient information to compute the percentage of total variance attributed to county differences. The total variance in this model is equal to σc 2 + σ2 CV + σ2 F + σ2 V F + σ2. There was no penalty for missing this question. 3
Problem 3. A 3 3 factorial experiment is to be conducted with one observation per treatment combination. Past evidence suggests that the effect of the three-factor interaction is zero. a. (5 pts) Write down the mathematical model assumed here. y ijkl = µ + A i + B j + AB ij + C k + AC ik + BC jk + ABC ijk + ɛ l(ijk) with i = 1, 2, 3, j = 1, 2, 3, k = 1, 2, 3 and l = 1. In effect, as the three-way interaction is negligible, is written as y ijk = µ + A i + B j + AB ij + C k + AC ik + BC jk + ABC ijk where the three way-interaction term serves as error. b. (5 pts) How many degrees of freedom are associated with the error sum of squares? As the three way-interaction term serves as the error term, there are 2 2 2 = 8 degrees of freedom for error. c. (10 pts) If all factors are at fixed levels, what is an appropriate test statistic for a formal test on factor B? Carrying out the EMS calculation for all effects, we can see that the test statistic for B main effects is given by F B = MSB MSABC. 4
Problem 4: a. (10 pts) Work out a confounding scheme for a 3 3 factorial confounding in 3 blocks of nine each. Show the ANOVA table outline. There are four interactions of 2 degrees-of-freedom each. We ll select one to define the three blocks, say I = AB 2 C 2. This gives the defining contrast L = (x 1 + 2x 2 + 2x 3 ) mod 3. The principal block is defined by L = 0 mod 3. Obviously treatment combo 000 belongs to the principal block. Also, 021 and 220. To obtain the remaining six, recall that the elements of the principal block form a group with respect to addition mod 3. Thus, 021 + 220 = 211 mod 3 also belongs to the principal block, etc. The principal block consists of 000, 021, 220, 211, 012, 202, 110, 220, 101. You obtain the elements of blocks L = 1 mod 3 and L = 2 mod 3 in a similar manner as explained in class. The ANOVA outline reads: Source of Variation Degrees of freedom Blocks (AB 2 C 2 ) 2 A 2 B 2 C 2 AB 4 AC 4 BC 4 Error (ABC + AB 2 C + ABC 2 ) 6 Total 26 b. (10 pts) Repeat a. in four replications confounding different interactions in each replication. Show the ANOVA table outline. A different component of the three way interaction will be confounded with blocks in each replication of the design. As there are four reps, we will use all four two-degree-of-freedom interaction components. The defining contrasts are: L 1 = (x 1 + 2x 2 + 2x 3 ) mod 3, L 2 = (x 1 + x 2 + x 3 ) mod 3, L 3 = (x 1 + 2x 2 + x 3 ) mod 3, and L 4 = (x 1 + x 2 + 2x 3 ) mod 3. The ANOVA outline reads: 5
Source of Variation Degrees of freedom Replications 3 Blocks (AB 2 C 2 + ABC + AB 2 C + ABC 2 ) 8 Reps blocks 24 A 2 B 2 C 2 AB 4 AC 4 BC 4 ABC 8 Error 46 Total 107 Problem 5. Material is analyzed for weight in grams from three vendors (A, B, C), by three different inspectors (I, II, III) and using three different scales (1, 2, 3). The experiment yielded the following results: Scale Inspector 1 2 3 I A = 16 B = 10 C = 11 II B = 15 C = 9 A = 14 III C = 13 A = 11 B = 13 a. (5 pts) Identify the design used and write the mathematical model for it. This is a Latin square design with mathematical model: y ijk = µ + β i + τ j + γ k + ɛ ijk where β stands for the inspector effect or first restriction on randomization, τ is the treatment effect (vendor) and γ stands for the scale effect or second restriction on randomization. b. (15 pts) Analyze the data and draw the appropriate conclusions. There is significant vendor effect at level 0.05. It appears that introducing the inspector restriction on randomization does not have a significant effect. Scale, on the other hand, is highly significant. Carrying out multiple comparisons obtains more specific results identifying vendor C to be significantly different from both A and B. 6
Table 3: ANOVA Table for Problem 5 Source df SS MS F p-value inspector 2 0.22 0.11 1.5 vendor 2 10.88 5.44 49.02 scale 2 32.88 16.44 148.0067 error 2 0.22 0.11 total 8 44.22 7