chapter 1

Chapter 10: The Analysis of Varianceˆ 2.63 + 2.13 + 2.41 + 2.49θ = , t. 025,25= 2. 06042 22222c = 1 + −.25+ −.25+ −.25+ −.25= 1.39. = 2.58 −. 165Σ i, MSE = .108, and( ) ( ) ( ) ( ) ( ) 25(.108)( 1.25). 165 2.060= .165 ± .309 = ( −.144,.474)interval for θ is6, so a 95% confidence± . Thisinterval does include zero, so 0 is a plausible value for θ .40. µ µ = µ µ = µ = µ −σ1=2 3, 4 5 1, so µ µ − 21σ5232 J αα4α5= − σ Φ =i52== . ThenI∑σ226 ⎡ 233( 5σ ) 2( −5σ ) ⎤⎢ += 1 .63222 ⎥5 ⎢⎣σ σ ⎥⎦≈ . , so ≈ . 52inspection of figure (10.6), power 48= , α α 2= σβ .and =1. 28α 1 2 3=5= ,Φ , ν 4 , ν 251=2=. By241. This is a random effects situation. : 0H σ states that variation in laboratories doesn’t0 A=contribute to variation in percentage. H o will be rejected in favor of H a iff ≥ F .= 4.0705,3,8= .7114. Thus = = 3. 96. SST = 86,078.9897 – 86,077.2224 = 1.7673, SSTr = 1.0559, and SSE1.05593f , which is not ≥ 4. 07.71148.05. Variation in laboratories does not appear to be present., so H o cannot be rejected at level42.a. µi= true average CFF for the three iris colors. Then the hypotheses areH : µ = µ = µ0 1 2 3vs. H : aat least two ' is= 61.31,22( 204.7) ( 134.6) ( 169.0)µ differ. SST = 13,659.67 – 13,598.362⎛⎞SSTR = ⎜⎟+ + −13,598.36= 23.00 The8 5 6⎝⎠ANOVA table follows:Source Df SS MS FTreatments 2 23.00 11.50 4.803Error 16 38.31 2.39Total 18 61.31Because F = .63 < 4.803 < F 6. 23 , .01 < p-value < .05, so we reject. 05,2,163.01,2, 16=H o . There are differences in CFF based on iris color.309