Question

Consider the following computer output for an experiment. Source DF SS MS F P-value ? ? ? Factor 5 ? Error 27.48 ? ? Total 29 66.24 (a) How many replicates did the experimenter use? (b) Fill in the missing information in the ANOVA table. Round your answers to three decimal places (e.g. 98.765), except P-value. DF(Error) = SS(Factor) MS(Error) MS(Factor) F = Round your answer to five decimal places (e.g. 98.76543). P-value (c) Compute an estimate for o. Round your answer to three decimal places (e.g. 98.765) =

Answer 1

sal Source DF SS MS P-value 7.752 6.77030 0.00045 Factar 38.76 Errar 24 27.48 1.145 Total 66.24 Here anona b) abone variation between the block (whose degree of freedom = 4-1) is missing r-1 it means that is uro so 9-1=0 replication DF (Total) =n-1=29 is one {IF(TO 2n=30 DF (Total) DF (Fachar) + DF (Errar) 29 = 5 + DE (Ervas) DF (Errar ) = 24 Esschun) +48] → Sstoral ) = SS (Factor) + SS (Errcar) 66.24 ss (factar) + SSCErvar) SS (factar) = 66.24 - 27.48 = 38-76 Mserrar) ss cerras) 27.48 DF (crocar 24 → MS (facteur) = SS (Factor) 7.752 38.76 DF (factar) F = MS (factas) 7.752 6.77030 Ms (errar) 1.us P-value for this scalu leaving degree of freedom (524) can be computed by using t-table D-value = 0.00045 that (ss ( ory] - In-y) where 0-1 = degree of freedom (factar) 10-15 v-1=DE (factan) - number = 6 v=6 Ssceran) 27.48 = 1.145 30-6 C know of factas A ny