A randomized block design has three levels of factor A and five levels of factor

Question

A randomized block design has three levels of factor A and five levels of factor B where six replicates for each combination are examined. The results include the following sum of square terms. Use Table 4 SST = 1558 SSA = 1008 SSB = 400 SSAB = 30 a. Construct an ANOVA table. (Leave no cells blank be certain to enter "0" wherever required. Round "p-value" to 3 decimal places and all other answers except "df" and "SS" to 2 decimal places.) b. At the 1% significance level, can you conclude that there is interaction between factor A and factor B? Yes No c. At the 1% significance level, can you conclude that the factor A means differ? Yes No d. At the 1% significance level, can you conclude that the factor B means differ? Yes No

Explanation / Answer

df total = 5 *3 * 6 -1 = 89

df total = df row + df column + df interaction + df error

MS = SS/df

F = MS /MS error

p-value = 1 -F.dist(F, df ,df error)

if p-value is less then alpha , we reject the null

a) p-value for interaction = 0.0263 > 0.01

we fail to reject the null

hence No

b)

p-value = 0 < 0.01

hence Yes

c)

p-value = 0 < 0.01

hence Yes

Please rate my solution

SS df MS F p-value rows 400 4 100 62.5 0 columns 1008 2 504 315 0 interaction 30 8 3.75 2.34375 0.026335 error 120 75 1.6 total 1558 89

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