chap 3 prob 4 estimate of tau(i) is y-bar(i,.)-y(.,.) in
 one case and y-bar(i,.)-ybar(v,.) in another. unless
 Y(v,.) is zero thee are not the same and thus not estimable

 tau(1)-tau(2) is estimable

chat 3 prob 14 
 a) plot time*pushes
 b) p-value is .88 accept means the same
 c) mean stdev
   38.2   .068
   38.17  .116
   38.19   .099
   38.17   .109

chap 3 prob 14d
  contrast ss is .004
chap 3 prob 17
 phi is sqrt (r/2v)*(delta/sigma) =
   sqrt(4/2*3)(.25/.2828)
 alpha is .05 with phi=.72 v(1)=2
 and v(2)=12-3=9 see power in table on p. 713 
  as lower than .24 (lenth calculator gives .093)

chap. 4 prob 3 b
  bonferroni
  use pt est=diff of above means
  w= 2.55
  sterr= sqrt(.01* (1/sample size  + 1/samples size)
 tukey
  as above with bonferroni but studentrange=3.402
 dunnett
 as above with bonferroni but w=2.154
 dunnett (onetailed)
 as above with bonferroni but w = 1.789
 scheffe
  as above with vonferroni but f is 2.298

chap 4 prob 11
 try 0,4,8,12 pushes
b) might use bonferroni
  y-bar(o,.)-y-bar(s,.) +-
  t(df=n-4,alpha/12)*sqrt(mse*2/r)
   where mse= sum of (y(it) - y-bar(i,.))squared / (n-4)
c) look to above bonferroni and the .01 from before
  2.58*sqrt(.01*2/r) = .1 gives r=13
 v plays a small role

chap 5 prob 1
  no pattern...all ok
cahp 6 prob 7
 a) .0001 for p-value
 b) estimate 'g3_v_1&2' cell -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
    2 2 2 2 2 /divisor=2;
  estimate=-32.25  sterr=6.4
c) estimate '13v15' cell 0 0 1 0 -1 0 0 0 0 0 0 0 0 0 0;
   estimate= 6.5  sterr=3.3
 d) mse=10.91 29*10.91/19.77=15.99
 e)t(df=n-v, alpha/2m) * sqrt (2/r*mse) < 8
   gives r=9, so 9*15                                 

chap 6 prob 8
 b) estimate=5.0 sterr=7.01