Homework 1 St 711           Due date listed on web.
(just place it on the front desk when you come to class)

A proper randomization scheme would make the probability of the first 
two units getting the same treatment the same as the probability that 
the last two get the same. In assigning 10 cows to 2 treatments, five 
per treatment, the probability that the second cow gets the same 
treatment as the first is 4/9 in a proper randomization scheme, since 
there are 9 equally likely possibilities remaining and 4 of them match 
the first choice.  

Suppose my treatment units are 10 cows on which I will run my 
experiment, getting one observation per cow. I have 2 treatments, 
perhaps 2 ways of doing the morning milking of a cow, and I will 
assign them as follows.  For cow 1, I toss a coin.  If it comes
up heads, I use treatment 1 and if tails, treatment 2.  I do the 
same on all subsequent cows until I have used one of the 
treatments 5 times. Then I will do the rest of the trials 
with the other treatment.  

1. What is the probability of tossing exactly 4 heads (and 4 tails) in 
8 tosses of a fair coin? (you will see later why I asked this) 

2. Run this program to see how the treatments are assigned in 100 repeats 
of this "randomization" scheme.  Note that the numeric value for cowj 
will be the treatment assigned to cow j. 

Data random; array cow(10); 
keep cow1-cow10 match910; 

do trial = 1 to 100; 
sum1=0; sum2=0; 
do unit = 1 to 10; 
  if ranuni(1635409) > 0.5 then do; 
  cow(unit)=1; sum1=sum1+1; end; 
  else do; cow(unit)=2; sum2=sum2+1; end; 
  if sum1 = 5 then do i=unit to 9; cow(i+1)=2; unit=10; end; 
  if sum2 = 5 then do i=unit to 9; cow(i+1)=1; unit=10; end;  
end; match910 = (cow9=cow10); output; 
end; 

proc print data=random(obs=20); run; 
proc means sum data=random; var match910; run; 

(A) How many times (out of 100) did cows 9 and 10 get the same treatment?

Note: You can print everything out and count by hand but since 
X=(A=B); is a logical function returning X=1 when A and 
B are equal and X=0 otherwise, the program uses PROC MEANS 
with this idea to count up how many times the condition holds. 

(B) How many times (out of 100) did cows 1 and 2 get the same treatment?  

Do the answers to (A) and (B) seem close to each other?  Why is that of
interest? 

3. What is the probability that I will have to toss the coin when I 
get to cow 9? Use the fact that the only way this could happen is 
if I had tossed exactly 4 heads and 4 tails up to that point. 
If I have stopped tossing when I get to cow 9, what is the implication
for the cow 9 and 10 treatments?

4. Using the above, compute the probability that cows 9 and 10 will get 
get the same treatment.  How many, then, out of 100 do you expect to 
get the same treatment?  Comment on how that compares to what you 
observed. How does it compare to 100(4/9)=44.4 that we would expect 
under a proper randomization scheme (see the introduction).

5. In a binomial situation with 100 trials where p is the
probability of some event (like cows 9 and 10 getting the same 
treatment) and f is the observed proportion of times that event 
occurred, p(1-p)/100 is the variance of f and Z=(f-p)/s 
is approximately N(0,1) where s is the square root of p(1-p)/100. 
Using the program results, find the Z that goes with f=proportion 
of 100 trials in which cows 9 and 10 got the same treatment. For p 
use the probability that cows 9 and 10 get the same treatment when 
coin tossing is used.  Is |Z|>1.96?

6. In the program, 1635409 is referred to as a "seed" for the random 
number generator.  Maybe our results just occured by accident.  Change
the seed to 49153 and comment on how the 2 counts (first 2 same, last 2 
same) change.  You might also run a few more seeds just to see how 
volatile these numbers are (just write up the 49153 result).

7. Assigning treatments as described here is not uncommon.  Briefly 
summarize in a few sentences what you have learned about the validity of 
this randomization scheme. 

Note: there may be nothing particularly important about the order
in which the cows enter the milking barn, but then again there may.  The
last cows may be in some ways weaker or more lethargic than those that
enter first (which may or may not affect the milking).  The point of 
randomization is that it guards against any such gradient whereas if 
we do not randomize, we must guarantee that we've thought of all 
possible such effects and none of them matters. 

Optional (not collected)  You might redo question 5 letting p be the expected 
probability under a proper randomization.  In that way you are doing a 
computer experiment to test the hypothesis that the scheme is a valid 
randomization scheme.  You might also try, analytically or using the 
computer, to figure out how many cows it would take to make the probability 
of the last two getting the same treatment some prechosen value like 0.99.