Suppose an ANOVA has been performed on a completely randomized design containing

Question

Suppose an ANOVA has been performed on a completely randomized design containing six treatment levels. The mean for group 3 is 15.85 and the sample size for group 3 is eight. the mean for group 6 is 17.21 and the sample size for group 6 is seven. MSE is .3352. The total number of observations is 46. compute the signifcant difference for the means of these two group by using the Tukey-Kramer produre. let a=.05

Answer: HSD=0.896, group 3&6 signifantly different    solve it step by step with details

Explanation / Answer

here for a=6 treament and (46-6=40) degree of freedom for error critical value for q at 0.05 level =4.2320

therfore HSD =(q/(2)1/2)*(MSE*(1/n1+1/n2))1/2 =(4.2320/(2)1/2)*(0.3352*(1/8+1/7))1/2 =0.896

as absolute mean diffierence between group 3 and 6 =|x3-x6| =|15.85-17.21| =1.36

as  absolute mean diffierence is greater then HSD ; therefore group 3&6 are signifantly different   

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