Question

The following data are from a completely randomized design.

Answer 1

Answer:

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.