The following data are from a completely randomized design.

Treatment

A B C

162 142 126
142 156 122
165 124 138
145 142 140
148 136 150
174 152 128

Sample mean 156 142 134
Sample variance 164.4 131.2 110.

a. Compute the sum of squares between treatments.
b. Compute the mean square between treatments.
c. Compute the sum of squares due to error.
d. Compute the mean square due to error.
e. Set up the ANOVA table for this problem.
f. At the α = . 05 level of significance, test whether the means for the three treatments are equal.

a)1488

b) 744

c) 2030

d) 135.333

e) see attached table

f) There is sufficient evidence to reject the claim of equal means.  

Step-by-step explanation:

a) Determine the value of total-group variability SS(tot):

SS(tot)=∑x^2(tot)-(x(tot))^2/N=376766-2592^2/18 ≅ 3518

Determine the value of the sum of squares between groups:  

SS(bet) = ∑_(all groups) (∑x_i)^2/n_i -(∑x(tot))^2/N

            =936^2/6+852^2/6+804^2/6-2592^2/6≅ 1488

b) d.f(BET) is the number of groups k decreased by 1.  

d.f(BET) = k - 1= 3-1 = 2

MS(BET) is SS(BET) divided by d.f(BET):

MS(BET)=SS(BET)/ d.f(BET)

               =1488/2

               =744

c) The value of the sum of squares within groups (due to error) is then the value of the total-group variability decreased by the value of the sum of squares between groups (Note: using this calculation you then immediately obtain that the sum SS(tot) =SS(bet)+ SS(W) holds):  

SS(W) = SS(tot) - SS(bet) = 3518 - 1488 = 2030

d)  d.f.w is the total sample size decreased by the number of groups k.  

  d.f.w = N — k = 18 — 3 = 15

MSw is SSw divided by d.f.w:  

MSw =SSw/d.f.w=2030/15=135.333

e) see attached table

f) The value of the test statistic F is then MS(BET) divided by MSw:

F=MS(BET)/MSw=744/135.3333≅5.5

The degrees of freedom are the same as those for between groups and within groups:  

d.f.N=d.f(BET)=2

d.f.D=d.f.w=15

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table 4 containing the F-value in the row dfn = 2 and dfd = 15:  

0.01<P<0.025

If the P-value is less than the significance level, reject the null hypothesis.  

P<0.05==> Reject H_o

There is sufficient evidence to reject the claim of equal means.  

2x^2+5x+17

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