Question

Timed assignment. please hurry.

You may need to use the appropriate technology to answer this question. The following data are from a completely randomized design. In the following calculations, use a = 0.05. Treatment Treatment Treatment 69 47 51 77 58 130.00 84.67 91.33 (a) Use analysis of variance to test for a significant difference among the means of the three treatments. State the null and alternative hypotheses. Ở Họ: H = A2= 43 H: Not all the population means are equal. O HoiH, H₂ = 43 HM, & H₂ H3 He: At least two of the population means are equal. H: At least two of the population means are different. Ho: Not all the population means are equal. HiM = 42 = 13. Ho: My & H₂ H3 Hill, 4₂ 43 Find the value of the test statistic. (Round your answer to two decimal places.)

Answer 1

treatment 1 treatment 2 treatment 3 Total 214 173 205 152 12 744 62.0000 232 51.0000 58.0000 grand mean Xgrand Total=EX = Average=X=Ex/n= Total=ex? = s? =(x-(EX:)?/n:)/(n-1) = 10794 77.0000 23970 84.6667 13730 130.0000 91.3333 Ź Total observation number of groups between groups degree of freedom dfbg =k-1 = within groups degree of freedom dfwg=N-k = Total degree of freedom dfTotal = . 4 treatment 1 treatment 2 treatment 3 total 51.0000 77.0000 58.0000 130.000 84.667 91.333 n*(x-Xgrand) 484.00 900.00 64.00 1448.0000 (n-1)*s? 390.00 254.00 274.00 918.0000 1448.0000 sum of square between groups =SSG = sum of square within groups -SSE = total sum of square =SST = En *(X-Xgrand)? :(n-1)*2= SSG+SSE = 918.0000 2366.0000 Summary table Source of variation treatments error total MS 724.0 102.0 Fp vlaue 7.10 0.0141 1448.000 918.000 2366.000 11

Ho:μ1=μ2=μ3

Ha:Not all the means are equal.

value of the sample F statistic=

7.10

p value=

0.014

reject Ho , there is sufficient evidence,,,,,,,,,,,,

b)

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

16.15

	(xi-xj )
μ1-μ2	26.00
μ1-μ3	7.00
μ2-μ3	19.00

there is a signiifcant difference between the means of treatment 1 and 2

there is a signiifcant difference between the means of treatment 1 and 3