12.1

1. ANOVA tests the null hypothesis that the population means are all equal.

2. Experiments are best for establishing causation.

3. ANOVA is used to compare means (and assumes that the variances are equal).

4. A two-way ANOVA is used to compare the means based on two factors.

12.3 Answers will vary.

12.5 xij=μi+∈ij, i=1, 2,3, j=1,2,...,20.∈ij~N(0,σ); I=3,ni=20. Parameters: μ1, μ2, μ3, and σ.

12.7

1. Yes, 80<2(68).

2. The estimates for μ1, μ2, and μ3 are 279, 245, and 258. The estimate for σ is 75.516.

12.9 The Normal quantile plot shows that the residuals are Normally distributed.

12.11

1. This sentence describes between-group variation.

2. The sums of squares in an ANOVA table will add; that is, SST=SSG+SSE.

3. σ, not sp, is a parameter.

4. A small P means the means are not all the same, but the distributions may still overlap quite a bit.

12.13 a1=0.5, a2=0.5, a3=−0.5, a4=−0.5.

12.15 Because there are only two groups, if the ANOVA test establishes that there are differences in means, then we already know that the two means we have must be different. In this case, contrasts and multiple comparison provide no further useful information.

12.17 The power would be larger. For larger differences between alternative means, λ gets bigger, increasing our power to see these differences.

12.1

1. 3 and 16. In Table E, F>3.24.

3. 0.025<P-value<0.05.

4. Because the P-value is small, we reject H0; however, this does not say that all pairs of group means are different, only that at least one mean is different.

12.3 Generally, the one with the largest difference between means and the smallest standard deviation will be the most significant. If this is not clear, a ratio between the two can be used. So part (b) can be ruled out because it has a larger sigma than part (a), with the same means. Of the remaining two, we can see that part (c) has a bigger max difference relative to its sigma (10 to 3) than part (a) does (5 to 2), so part (c) will be the most significant.

12.5

1. F=340/50=6.80 with 2 and 60 df. P=0.0022.

2. F=11/4.75=2.32 with 7 and 40 df. P=0.0442.

12.7

1. Response: egg cholesterol level. Populations: chickens with different diets or drugs. I=3, n1=n2=n3=25, N=75.

2. Response: rating on seven-point scale. Populations: the three groups of students. I=3, n1=31, n2=18, n3=45, N=94.

3. Response: quiz score. Populations: students in each TA group. I=3, n1=n2=n3=14, N=42.

12.9 For all three situations, we have H0: μ1=μ2=μ3. Ha: not all of the μi are equal. DFG=I−1=DFE=N−I, and DFT=N−1. The degrees of freedom for the F test are DFG and DFE. (a) DFG 2, DFE 72, DFT 74; F(2, 72). (b) DFG 2, DFE 91, DFT 93; F(2, 91). (c) DFG 2, DFE 39, DFT 41; F(2, 39).

12.11

1. This sounds like a fairly well-designed experiment, so the results should at least apply to this farmer’s breed of chicken.

2. It would be good to know what proportion of the total student body falls in each of these groups—that is, is anyone overrepresented in this sample?

3. How well a TA teaches one topic (power calculations) might not reflect that TA’s overall effectiveness.

12.13

1. Yes.

2. sp2=32.6084; sp=5.71.

3. Yes, there appears to be a difference in average gains in performance; the estimated common standard deviation is 5.71, which is a value similar to the individually reported standard deviations; this suggests that the variability in the means is not driven by group deviations.

12.15

1. Both drugs cause an increase in activity level; Drug B appears to have a greater effect.

2. Yes; 17.2<27.75, sp=3.487.

3. DFG=4, DFE=20.

4. 2.25<F<2.87, 0.05<P-value<0.10.

12.17

1. 4 and 178.

2. 5+146=151 athletes actually participated.

3. For example, the individuals could have been outliers in terms of their ability to withstand the water-bath pain. In either case of low or high outliers, their removal would lessen the standard deviation for their sport and move that sports mean.

12.19

1. Source	df	SS	MS	F
   ELF-EMF	4	662.86	165.715	5.70
   Error	55	1599.004	29.0728
   Total	59

2. P-value=0.0007. There is evidence that at least one population mean is different.

3. 5.392.

12.21 1.37.

12.23 (−0.1453,5.3453).

12.25

1. Yes, 0.824<2(0.657); sp=0.7683.

2. df=2, 767, P-value<0.001.

3. We want to do this because it is a linear combination of other means.

4. ψ=μ2−μ1; we test H0: ψ=0; Ha: ψ>0. We find c=0.6; t=6.07, P-value<0.0001.

5. ψ=μ2−0.5(μ1+μ3); we test H0: ψ=0; Ha: ψ>0. We find c=0.585, t=5.99, P-value<0.0001.

12.27

1. 1μ2−0.5μ1−0.5μ4.

2. 1/3μ1+1/3μ2+1/3μ4−1μ3.

12.29

1. ψ1=μ1−1/2(μ2+μ4) and ψ2=μ3−μ2−(μ5−μ4).

2. c1=−3.9 and c2=2.35; SEc1=2.1353 and SEc2=3.487.

3. The first contrast is significant (t1=−1.826,P-value=0.0414), but the second is not (t2=−0.674,P-value=0.2540).

12.31 Answers will vary.

12.33

1. ψ=μ7−0.25(μ1+μ2+μ3+μ4).

2. H0: ψ=0; Ha: ψ>0.

3. t=1.894, P-value=0.0302.

12.35 The power would be larger. For larger differences between alternative means, λ gets bigger, increasing our power to see these differences.

12.37

1. See accompanying table. Pooling is appropriate, 0.621669<2(0.572914).

   Level of food	N	Score Mean	Std dev
   Comfort	22	4.8873	0.5729
   Control	20	5.0825	0.6217
   Organic	20	5.5835	0.5936

2. While the distributions are not Normal, there are no outliers or extreme departures from Normality that would invalidate the results. We can likely proceed with the ANOVA.

12.39

1. H0: μ1=μ2=μ3, Ha: not all of the μi are equal, F=8.89, P-value=0.000. There are significant differences in the number of minutes that the three groups are willing to volunteer. The Comfort group is willing to donate significantly more minutes than the Organic group. The Control group is in the middle, not significantly different from either the Comfort or Organic group.

2. The residual plot shows a possible violation of constant variance. The Normal quantile plot looks fine and shows a roughly Normal distribution.

12.41

1. F can be made very small, and the P-value can be close to 1.

2. F increases, and the P-value decreases.

12.43

1. Observational.

2. Yes, the mean for SC appears higher than the others.

3. Yes, the largest s is less than twice the smallest.

4. Yes, it is reasonable to use the one-way ANOVA procedure, as long as the data are quantitative.

5. MSG=271.57 MSE=3.136;F=86.587;P-value=0.

12.45 Using the Bonferroni method, coffeehouse 1 is different from coffeehouses 2 and 4; coffeehouse 2 is different from coffeehouses 4 and 5; coffeehouse 3 is different from coffeehouses 4 and 5; coffeehouse 4 is different from coffeehouse 5.

12.47

1. H0: μ1=μ2=μ3, Ha: not all of the μi are equal, F=5.31,P-value=0.0067.

2. The Bonferroni shows that group 2 is not significantly different from either group 1 or group 3, but group 3 is significantly different from (larger than) group 1.

3. This is not appropriate. The regression assumes that group 2 (coded as 2) would have twice the effect of group 1 (coded as 1), and group 3 (coded as 3) would have three times the effect of group 1, etc. This is likely not true.

12.49

1. Portals and Transactions have higher integration features than Presence.

2. An observational study. They are not imposing a treatment on the winery.

3. Yes; 2.346<2(2.097).

4. Our inference is based on sample means, which will be approximately Normal, given the sample sizes.

5. F(2, 190); P-value<0.001.

12.51

1. See accompanying table.

   	n	x¯	s
   Ctrl	35	−1.009	11.501
   Grp	34	−10.785	11.139
   Indiv	35	−3.709	9.078

2. Yes, the largest s is less than twice the smallest s; 11.501<2(9.078)=18.156.

3. All three distributions are roughly Normal.

12.53

1. All weight loss values are divided by 2.2.

2. F=7.77, df=2, 101, P-value=0.0007. The results are identical with the transformed data.

12.55

1. The variation in sample size is some cause for concern, but there can be no extreme outliers in a 1-to-7 scale.

2. Yes; 1.26<2(1.03).

3. F(4, 405), P-value<0.0002.

4. Hispanic Americans are highest, Japanese are in the middle, and the other three are the lowest.

12.57

1. df=2, 117.

2. P-value<0.001.

3. Because the bargainer was the same person each time, the results would certainly not be generalizable.

12.59

1. See accompanying table.

   	n	x¯	s	SE
   Control	15	0.2189	0.01159	0.002992
   Low dose	15	0.2159	0.01151	0.002972
   High dose	15	0.2351	0.01877	0.004847

2. All three distributions appear to be reasonably close to Normal, and the standard deviations are suitable for pooling.

3. F=7.72(df=2, 42), P-value=0.001.

4. For Bonferroni, t**=2.4937 and MSD=0.01308. The high-dose mean is significantly different from the other two.

5. High doses of kudzu isoflavones increase BMD.

12.61

1. See accompanying table. Yes; 27.364<2(16.594).

2. F=7.98(df=2,27), P-value=0.002.

   	n	x¯	s
   Control	10	601.1	27.364
   Low jump	10	612.5	19.329
   High jump	10	638.7	16.594

12.63

1. ψsex=μ1−μ3; csex=−1.255, t=−3.6; P-value<0.001.

2. ψnat=μ1−μ2; cnat=0.005, t=0.014; P-value>0.25.

3. The conclusions from the contrasts were limited to only comparing sex among Hispanics and nationalities among males. Including Anglo females would alleviate this limited inference.

12.65

1. See accompanying table. Pooling is not appropriate, the largest s is more than twice the smallest s; 8.6603>2(2.8868)=5.7736.

   	n	Mean	s
   ECM1	3	65.00%	8.66%
   ECM2	3	63.33%	2.89%
   ECM3	3	73.33%	2.89%
   MAT1	3	23.33%	2.89%
   MAT2	3	6.67%	2.89%
   MAT3	3	11.67%	2.89%

2. F=137.94 (df=5, 12), P-value<0.0005.

12.67 H0: μLC=μLF, Ha: μLC≠ μLF. sp2=732.192, sp=27.059. t=4.171, df=63, P-value<0.0005 (0.0001 from software) (0.0001 from software). F=17.396. Software gives the P-value=0.0001. We note that 4.1712=17.397; t2 is within rounding error of F.

12.69

1. The plot shows granularity, but otherwise, independence is not violated.

2. Yes; 1.6<2(0.93)=1.86.

3. The individual Normal quantile plots for each shampoo show a lot of granularity due to using integer values, but most look at least roughly Normal.

4. The Normal quantile plot for the residuals shows that the residuals are Normally distributed.

12.71

1. ψ1=1/3(μPyr1+μPyr2+μKeto)−μPlacebo, ψ2=0.5(μPyr1+μPyr2)−μKeto, ψ3=μPyr1−μPyr2.

2. sp=1.1958. c1=−12.51, SEC1=0.2355; c2=1.269, SEC2=0.1413; c3=0.191, SEC3=0.1609.

3. H0: ψi=0, Ha: ψ1≠0; t1=−53.17, P-value1 <0.0005; t2=8.98, P-value2<0.0005; t3=1.19, P-value3=0.2359. The Placebo mean is significantly higher than the average of the other three, while the Keto mean is significantly lower than the average of the two Pyr means. The difference between the Pyr means is not significant (meaning the second application of the shampoo is of little benefit).

12.73 The means increase by 5%, but everything else remains the same.

12.75

1. Answers will vary with choice of Ha and desired power. For example, with the alternative μ1=μ2=4.4, μ3=5, and σ=1.2, three samples of size 75 will produce power 0.78.

12.77 The design can be similar, although the types of music might be different. Bear in mind that spending at a casual restaurant will likely be less than at the restaurants examined in Exercise 12.58; this might also mean that the standard deviations could be smaller. A pilot study might be necessary to get an idea of the size of the standard deviations. Decide how big a difference in mean spending you would want to detect; then do some power computations.

12.79 Answers will vary. Students should note that this is a challenging data set because coins can only range between 0 and 100, but most chose values on 5-coin increments, and many invested all the coins. Depending on answers, if assumptions are not satisfied for CLT, then a nonparametric rank test (Chapter 15) or a bootstrap analysis (Chapter 16) can be used.

12.81

1. ψ1=μC−0.25(μ30×1+μ30×2+μ60×1+μ60×2), ψ2=0.5(μ30×1+μ30×2)−0.5(μ60×1+μ60×2), ψ3=0.5(μ60×1−μ60×2)−0.5(μ30×1−μ30×2),

2. c1=14.65, c2=6.1, c3=0.5. SEC1=4.209, SEC2=SEC3=3.784.

3. t1=3.481, P-value=0.0007;t2=1.612, P-value=0.1097;t3=0.132, P-value=0.8952. The first two are significant; the third is not.