Chapter 9: More Complicated Experimental Designs9.1 a. The F-test from the ANOVA table tests 2-sided alternatives:Test Ho: µAttend= µDidN otvs Ho: µAttend6= µDidN otThe ANOVA table is given here:Source DF SS MS F p-valuePair 5 1319.42 263.88Treatment 1 420.08 420.08 71.40 0.0001Error 5 29.42 5.88Total 11 1768.92Reject Hoand conclude there is significant evidence that the mean scores of studentsattending Head Start are significantly different from the mean scores of students whodo not attend Head Start.b. RE(RCB,CR)=(b−1)MSB+b(t−1)M SE(bt−1)MSE=(6−1)(263.88)+(6)(2−1)(5.88)((6)(2)−1)(5.88)= 20.94 ⇒It would take approximately 21 times as many observations (126) per treatment in acompletely randomized design to achieve the same level of precision in estimating thetreatment means as was accomplished in the randomized complete block design.9.2 The F-test from the ANOVA table tests 2-sided alternativesHo: µAttend= µDidN otvs Ho: µAttend6= µDidN ot.The paired t-test can test one-sided alternatives:Ho: µAttend≤ µDidN otvs Ho: µAttend≥ µDidN ot¯yAttend= 78.33 ¯yDidN ot= 66.50t =¯DSD/√n=11.833.43/√6= 8.45p-value = Pr(t5≥ 8.45) = 0.0002 ⇒Yes, there is significant evidence that the mean scores attending Head Start are greaterthan the mean scores of students who do not attend.Note (8 .45)2= 71.409.3 a. Blocks are Investigators and Treatments are Mixturesb. Randomly assign the four Mixtures to the each of the Investigatorsc. The randomized complete block design guarantees that each investigator measureseach of the four mixtures, whereas in a completely randomized design, it is possiblethat some of the investigators may not measure some of the mixtures. This may causea bias towards some of the mixtures if a particular investigator tends to always givehigh readings no matter which mixture is measured.709.4 a. yij= µ + βj+ αi+ ²ij; i = 1, 2, 3, 4 j = 1, 2, 3, 4, 5yijis measurement of jth investigator on ith mixtureαiis the ith mixture effectβjis the jth investigator effectb. ˆµ = ¯y..= 2463.75ˆα1= ¯y1.−¯y..= 2351.0−2463.75 = −112.75, ˆα2= ¯y2.−¯y..= 2653.2−2463.75 = 189.45ˆα3= ¯y3.−¯y..= 2444.2−2463.75 = −19.55, ˆα4= ¯y4.−¯y..= 2406.6−2463.75 = −57.15ˆβ1= ¯y.1− ¯y..= 2462.5 − 2463.75 = −1.25,ˆβ2= ¯y.2− ¯y..= 2468.25 − 2463.75 = 4.5ˆβ3= ¯y.3− ¯y..= 2469.25 − 2463.75 = 5.5,ˆβ4= ¯y.4− ¯y..= 2456.0 − 2463.75 = −7.75ˆβ5= ¯y.5− ¯y..= 2462.75 − 2463.75 = −1c. F = 1264.73 with p-value < 0.0001 ⇒Reject Ho: µ1= µ2= µ3= µ4and conclude there is significant evidence of adifference in the means for the four mixtures.d. Mixture 2 with the highest mean response would appear to be possibly the best mix-ture.A multiple comparison procedure could be used to confirm that the other three mix-tures have significantly lower means.e. RE(RCB,CR)=(b−1)MSB+b(t−1)M SE(bt−1)MSE=(5−1)(113.12)+(5)(4−1)(68.86)((5)(4)−1)(68.86)= 1.14 ⇒It would take 1.14 times as many observations (approximately 6) per treatment in acompletely randomized design to achieve the same level of precision in estimating thetreatment means as was accomplished in the randomized complete block design.9.5 a. yij= µ + αi+ βj+ ²ij; i = 1, 2, 3, j = 1, 2, 3, 4, 5, 6, 7yijis score on test of jth subject hearing the ith music typeαiis the ith music type effectβjis the jth subject effectˆµ = 21.33, ˆα1= −0.47, ˆα2= −1.19, ˆα3= 1.67ˆβ1= 0,ˆβ2= −3,ˆβ3= 3.33,ˆβ4= −1.33,ˆβ5= 1,ˆβ6= 3.67,ˆβ7= −3.67b. F =SST RT/dfT RTSSError/dfError=30.952/228.38/12= 6.54 with df=2,12.Therefore, p-value = Pr(F2,12≥ 6.54) = 0.0120 ⇒ Reject Ho: µ1= µ2= µ3.We thus conclude that there is significant evidence of a difference in mean typingscores for the three types of music.c. An interaction plot of the data is given here:71NNNNNNNHHHHHHHCCCCCCCInteraction PlotsSubjectTest Scores1 2 3 4 5 6 716 18 20 22 24 26 28 30 32NNo MusicHHard RockCClassicalBased on the interaction plot, the additive model may be inappropriate because thereis some crossing of the three lines. However, the plotted points are means of a singleobservation and hence may be quite variable in their estimation of the populationmeans µij. Thus, exact parallelism is not required in the data plots to ensure thevalidity of the additive model.d. t = 3, b = 7 ⇒ RE(RCB, CR) =(7−1)(24.889)+(7)(3−1)(2.365)((7)(3)−1)(2.365)= 3.86 ⇒It would take 3.86 times as many observations (approximately 27) per treatment ina completely randomized design to achieve the same level of precision in estimatingthe treatment means as was accomplished in the randomized complete block design.Since RE was much larger than 1, we would conclude that the blocking was effective.9.6 The model conditions appear to be satisfied:The normal probability plots and box plots of the residuals do not indicate nonnormality.Plot of residuals versus estimated mean does not indicate nonconstant varianceInteraction plot indicates a potential interaction between subjects and type of music, butthe indications are fairly weak.9.7 a. yij= µ + αi+ βj+ ²ij; i = 1, 2, 3, 4 j = 1, 2, 3, 4, 5yijis the increase in productivity of worker having jth level of attitude and attendingworkshop type ithαiis the ith workshop type effectβjis the jth attitude effectˆµ = 50.25, ˆα1= −7.45, ˆα2= −3.65, ˆα3= 0.35, α4= 10.75ˆβ1= −9.75,ˆβ2= −8.5,ˆβ3= −4.75,ˆβ4= 0.5,ˆβ5= 22.5b. F =SST RT/dfT RTSSError/dfError=922.55/355.7/12= 114.12 with df=3,12.Therefore, p-value = P r(F3,12≥ 114.12) < 0.0001 ⇒ Reject Ho: µ1= µ2= µ3= µ4.We thus conclude that there is significant evidence of a difference in the mean increasein productivity for the four types of workshops.72c. A profile plot of the data is given here:AAAAABBBBBCCCCCDDDDDProfile PlotAttitude LevelIncrease in Productivity1 2 3 4 530 40 50 60 70 80 90Workshop AWorkshop BWorkshop CWorkshop DBased on the profile plot, the additive model appears to be appropriate because thefour lines are relatively parallel. Note further that the plotted points are means of asingle observation and hence may be quite variable in their estimation of the populationmeans µij. Thus, exact parallelism is not required in the data plots to ensure thevalidity of the additive model.d. t = 4, b = 5 ⇒ RE(RCB, CR) =(5−1)(696.375)+(5)(4−1)(4.6417)((5)(4)−1)(4.6417)= 32.37 ⇒It would take 32.37 times as many observations
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