Homework #3 -- Solutions

  *** see hwk053s.sas, .log, .lst ***

  a) plot the data -- There are some obvious troughs (low points) for 
   t under 10, and around t = 80, 150, 220 -- could be annual cycle.

  b) Periodogram shows some power at low frequencies, but strong peaks
   at f=0.014388 (period 69.5), and 2f=0.028777, which appear to be
   leaking from the fundamental (annual cycle) at 1/72 = 0.01389, and
   second harmonic 2/72 = 1/36 = .027778.

  c) First fit regression model with annual cycle and several harmonics;
   I chose 5.  (Five is rather small, ten would hopefully be more than
   enough, 36 a little excessive, but the absolute most since it's the
   Nyquist frequency.)  
   
   Testing H: only need annual fundamental and second harmonic vs.
   A: need more harmonics, by adding the Type I SS from the bottom 
   (or averaging the F-statistics), here the numerator SS is
    403 + 464 + 746 + 2 + 846 + 4 = 2465.  The F-statistic is
   F = (2465/6) / (142713/267) = 411/534 = 0.77.  Reject H if F is
   larger than F(6,267,0.05) = 2.80; do NOT reject.

  d) Periodogram white noise tests: Fisher's Kappa (SAS) or Gn (my
   notation) is 10.85.  Here n = 138 = (278-2)/2 and we would reject
   H: residuals are iid vs. A: not iid if Gn > g(.05) at n=138.
   Since 7.378 < g(.05) < 7.832, we would reject H at alpha = 0.05.

   Bartlett, Durbin, K-S test D = .254, so we would reject if
   sqrt(137)*D > 1.36 for a test a level alpha = 0.05, here we have
   sqrt(137)*D = 2.97 which screams rejection -- the residuals are
   not iid.

  e) The result of Fisher's test suggests that the peak at f=0.05036
   is real (notice that the next lower qerf is sizeable); there is 
   another at 0.10791 which could be its second harmonic.
   This could be interesting....