85085b15cB5d85eμ = 0.678M₂ = 0.414Me = 1.316Me = 1.382SourceBetween groupResidualTotalSS3.4092307122701214SSE =MSR = 3.4082/3 - 101364MSE3-7/22/16= 0.2320Y = 0.947522 Y₁² = 0.07² +124² ++ 1.11² + 1.79² = 25.0765SST EE(Y₁-F²EE Y² -n Y² - 25.0765 - 20 (0.9475)² = 7.1214SSR = En (√₁₂-F)² = 5 (0.678-0-7474)² + 5 (0.414-0.9476)² +5 (10316-0.9474) + 5 (1-382-0-94+1)= 3.4092=SST - SSR = 7.1214 - 3.4092 = 3.7122Ho. The meanH₁ He is not true.·r-1.3.・ny. -r. = 16M²₁ -1=1996% confidence internal of unMA ± to.rs, N-r [. 5²/MA.•√35200.678 ± 2.120- (0.22131-1347).ms.1.13640.2320F = 4.8983 > Fo. 05, 3,16 = 3.24.It means that at least two of the meanseach group is not significant different MAM = μc = μ0.678 0.414 I 0.6458.= (-0.3818.7.0.9098).Therefore, we areF-M³R = 1.1364MSE= 4.8983Increase T-0.3818 and 0.3088.95% confidence interval forMA-MB.-0.22/3+1=1347 -0.414 ± 2.120 (0.2320 ( 33 + 5)0.2320reject He at 5% level of significant.sigaficant different.are95% confident that the true difference between the meanhamstring extensibility for treatment A and B. is between A study was conducted to examine the effect of heat and duration of stretching on theextensibility of hamstring muscles and their electromyographic responses to passive stretch inchildren with hypertonia and severe mental retardation. Twenty subjects were randomlyassigned to the following four treatments with five subjects per treatment: (A) 10-secondstretching, (B) 30-second stretching, (C) hot pack followed by 10-second stretching and (D) hotpack followed by 30-second stretching. The outcome measure is the increase in hamstringextensibility (Y) and the data are given in Table 4.Table 4TreatmentABCD0.07-0.191.110.571.240.51.282.19Y0.620.641.081.250.340.482.031.11Consider a one-way ANOVA to study the effects of the four treatments. Denote μA to μD as themeans of increase in hamstring extensibility under treatments A to D respectively.the 95% confidence interval for µ^ — µB.1.120.641.081.79(f) Estimate and comment on the contrast of the treatment means : L =With a 90% confidence interval.stensibiliMA + MB2eqμc + UD2

μA=0.678μB=0.414μC=1.316μD=1.382 L=μA+μB2-μC-μD2

Answered: A study was conducted to examine the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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