o occupants ol e. where the treatments are three levels of longitudinal spacing

Question

o occupants ol e. where the treatments are three levels of longitudinal spacing cussed two insignificant factors, which are omitted here) There were ninc replicates energy absorption of the pillar ve, The response is the head injury criterion (HIC), which is a unitless quandity that measures the impac One way ANOVA: Spacing Source Spacing Error Total DF MS 50946.6 25473.3 5.071 0.015 24 120550.9 26 171497.4 5023.0 The treatment means were Treatment 930.87 873.14 979.41 an Can you conclude that the longitudinal spacing affects the absorption of impoct energy a. se the Tukey-Kramer method to determine which pairs of treatment means, if any, are 5% level c. Use the Bonferroni d. Which method is more powerful in this case, the Tukey-Kramer method or the method? method to determine which pairs of treatment means, if any, are different at the 5% level Bonferroni a. Use the Bonferroni method to determine which means, if any, differ from the mean of the control grou b. Use the Tukey-Kramer method to determine which means, if any. differ from the mean of the control c. Which is the more powerful method to find the treatments whose mean differs from that of the con 3B) 5% level. the 5% level. the Bonferroni method or the Tukey-Kramer method?

Explanation / Answer

3a. Assume alpha=0.05. Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.05, here, p value is less than 0.05, therefore, reject null hypothesis [H0: there is no difference in absorption of energy for different longitudinal spacing] and conclude that spacing affect the absorption of impact of energy.

3b. For the analysis, select experimentwise error rate, alpha=0.05. There are p=3 treatments [df(spacing)=#of treatments-1], v=24 [number of degrees of freedom associated with a MSE], nt=9 observations per treatment, s=sqrt MSE=sqrt 5023=70.8731. The critical value of the Studentized range q0.05(3,24)=3.532. Substitute the values in the following formula.

w=q0.05(3,24)(s/sqrt nt)=3.532(70.8731/sqrt 9)=83.44

The 3 sample means are ranked as follows:

873.14 930.87 979.41

B A C

Place a bar over those treatment means that differ by less than w. Any pair of treatment means not connected by bar implies a difference in corresponding population means.

Place a bar over BA and AC. Thus, BC differ significantly.

3c. Error rat eis alpha=0.05. For p=3 treatment means, the number of pairwise comparison is g=p(p-1)/2=3

Here, the adjusted alpha level is alpha*=alpha/g=0.05/3=0.017. The critical value for Student's t statistic with v=24 [df associated with MSE] is t0.017/2, df=24=2.56., s=70.8731 [s=sqrt MSE]. Substitute the values in following formula.

B=t0.0085(s)sqrt[1/ni+1/nj] [where, ni and nj are number of observations in sample for treatment i and j respectively]

=2.56*70.8731*sqrt[1/9+1/9]

=85.53

The 3 sample means are ranked as follows:

873.14 930.87 979.41

B A C

Place a bar over those treatment means that differ by less than B. Any pair of treatment means not connected by bar implies a difference in corresponding population means.

Place a bar over BA and AC. Thus, BC differ significantly.

3.d Since, the ANOVA is balanced (each treatment has equal number of observations-9] therefore, Tukey Crammer method is better. Bonferroni method would have been superior if ANOVA were unbalanced, that is sample sizes for treatmnets are unequal.

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