Question

please show answers clearly

You may need to use the appropriate technology to answer this question To test whether the mean time needed to mix a batch of material is the same for machines produced by the manufacturers, a chemical company obtained the following data on the time in minutes) needed to mix the material Manufacturer (a) Use these data to test whether the population mean times forming a batch of materialer for the the manufacturer -0.05. State the null and alternative hypotheses O Hol Not all the population means are equal OH " NA O NA least two of the potations are At least two of the potion means are different ON H OH- - 2 Now the population means are equal

Answer 1

treatment 1 treatment 2 treatment 3 Total 276 23.0000 20.0000 grand mean Xgrand- Total=EX = Average=X=EX-/n= Total=ex} = s={x?-(Ex)/n:)/(n-1) = 22.0000 1966 27.0000 2930 4.6667 1618 10.0000 6.0000 Total observation number of groups between groups degree of freedom dfe =k-1 = within groups degree of freedom dfwg=N-K = Total degree of freedom dfTotal = treatment 1 treatment 2 treatment 3 total 22.0000 27.0000 20.0000 10.000 4.667 6.000 n*(X--Xgrand) 4.00 64.00 36.00 104.0000 (n-1)*s? 30.00 14.00 18.00 62.0000 104.0000 sum of square between groups -SSG = sum of square within groups ESSE = total sum of square =SST = En*(X--Xgrand) = :(n-1)*2= SSG+SSE = 62.0000 166.0000 Summary table Source of variation treatments error total MS 52.0000 6.8889 Fp vlaue 7.548 0.0119 104.000 62.000 166.000

Ho: μ1=μ2=μ3

Ha: not all the population means are equal

test statistic =

7.55

p value =0.012

reject HO , there is sufficient evidence,,,,,,,,,,,,,,

b)

Fisher's (LSD) for group i and j =                  (tN-k)*(sp*√(1/ni+1/nj)           =

4.20

x1-x3

2.00

there is not a significant difference between the means for treatment 1 and 3