chapter 1

Chapter 10: The Analysis of VarianceΣcx± tc. The 99% t confidence interval isi i .005, I ( J −1)MSE2( Σc)i•.( 21 1 11314412− Σc) iΣ c ix i •= x1+2+ + −5−6= 4.164 •x4 •x4 •x•x•x2 •, = . 1719 ,JMSE = .273, t 2. 845 . The resulting interval is. 005,20=( 2.845) (.273)(.1719) = −4.16± .62 = ( − 4.78, 3.54)− 4.16±−Jii. The intervalin the answer section is a Scheffe’ interval, and is substantially wider than the t interval.27.a. Let µi= true average folacin content for specimens of brand I. The hypotheses to betested are H : µ = µ = µ = µ0 1 2 3 4vs. H : aat least two µ ' is differ .ΣΣx 2 ij=1246.88 and2x2••n=( 168.4)242= 1181.61222( 57.9) ( 37.5) ( 38.1) ( 34.9)Σ xi•= + + +Ji7 5 6SSTr = 1205 .10 −1181.61= 23.49 .62, so SST = 65.27.= 1205.10, soSource Df SS MS FTreatments 3 23.49 7.83 3.75Error 20 41.78 2.09Total 23 65.27With numerator df = 3 and denominator = 20,F = .10 < 3.75 < F 4.94 , so . 01 < p − value < . 05. 05,3,203.01,3, 20=, and since thep-value < .05, we reject H o . At least one of the pairs of brands of green tea has differentaverage folacin content.304