TEST SOLUTIONS, STAT 440                                   11/7/05
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#1. Ybar = .4*5+.4*9+.2*16 = 8.8,
(a) SSB/N = .4*(5-8.8)^2 + .4*(9-8.8)^2 + .3*(17-8.8)^2 = 16.16
    SSW/N = .4*3 + .4*6 + .2*12 = 6, so SST/N = Sy2 = 22.16
(b) MSE= (1/100)*22.16*(1-.5^2) ### = 0.1662  for regression estimator
Then
    CV = sqrt(MSE)/Ybar = sqrt(.1662)/8.8 = .0463    

#2. > lg2 <- (agsrs$FARMS87 >= 500)
> sum(agsrs$ACRES87[lg2])
[1] 52873455
> sum(agsrs$ACRES87[lg2]^2)
[1] 2.735729e+13

NB A brief argument (general for small-domain estimation of ratios) 
shows that the regression estimation is identical to the ratio
estimator. (This will be done in class.)  

(a)-(b) Ratio estimator = Regression estimator = 52873455/175 
           = 302134
SE = (3078/1800)*(sqrt((1-300/3078)*
           (2.735729e+13 - 52873455^2/175)/(299*300))
   =  18299.91
So CI = 302134 + c(-1,1)*1.96*18299.9 = (266266.2, 338001.8)

NB The corresponding estimator based on plain SRS estimation of 
the total is :  52873455*(3078/300)/1800 = 301378.7
and its (larger!) SE is:  
(3078/1800)*sqrt((1-300/3078)*
     (2.735729e+13 - 52873455^2/300)/(299*300)) = 23037.39

#3. We have s_i^2 = (1/2)*d_i^2, so that the 2-stage cluster sampling
MSE formula (for total, then divided by N^2) is:  
  (1/150)* (3.6e9 *(1-150/10000) + (1/10000)*(16/2)*(1-2/4)* (0.5)*5.4e11 )
> (1/150)* (3.6e9 *(1-150/10000))
[1] 23640000
> (1/150)*(1/10000)*(16/2)*(1-2/4)*0.5*5.4e11
[1] 720000
## Thus the two terms are respectively 2.364e7 + 0.72e6 = 2.436e7
So the CI = 
> 3.8e6*2/150 + 1.96*c(-1,1)*sqrt(2.436e7)
[1] 40992.9 60340.4

#4. (a) Formula for optimal nh's is:
> nh <- 2000*c(1000,3000,5000)*sqrt(c(.01,.05,.12)*(1-c(.01,.05,.12)))/
     (sqrt(c(40,20,10))*sum(c(1000,3000,5000)*sqrt(c(.01,.05,.12)*
     (1-c(.01,.05,.12))*c(40,20,10))))
> nh
[1]   3.620154  33.642827 118.233730
 ## So we sample respectively 4,34,118 in the three strata.
> nh <- round(nh)
## With this number sampled, the precision for ybar is:
> SEybar <- sqrt(sum(c(1/9,3/9,5/9)^2*(1-nh/c(1000,3000,5000))*
            c(.01,.05,.12)*(1-c(.01,.05,.12))/nh))
[1] 0.02139862
(b) The overall proportion is: p <-(.01*1+.05*3+.12*5)/9 = 0.08444.
Now we match sqrt((1-n/9000)*p*(1-p)/n) = 0.021594, or
> n <- 1/((0.02139862^2/(p*(1-p))+1/9000)) ### =  165.7, or 166.
## Roughly, sample proportions 1/9, 3/9, 5/9 in the three strata, giving
##   overall cost:  > sum(c(1,3,5)*(166/9)*c(40,20,10))
[1]  2766.667  ### Thus the cost becomes considerably larger with SRS!