An experiment was carried out to investigate the effect of species (factor a, with i = 4) and grade (factor b, with j = 3) on breaking strength of wood specimens. one observation was made for each species—grade combination—resulting in ssa = 449.0, ssb = 425.6, and sse = 127.4. assume that an additive model is appropriate. (a) test h0: α1 = α2 = α3 = α4 = 0 (no differences in true average strength due to species) versus ha: at least one αi ≠ 0 using a level 0.05 test. calculate the test statistic. (round your answer to two decimal places.) f =

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1  

step-by-step explanation:

what "following statement"?

step-by-step explanation:

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional