Solution to the Take-Home Make-up, 11/20/07
===========================================

(1) Here we have 
> Nh = c(5500,3500, 1000)
  ph = c(.03, .06, .10)
  pqh = ph*(1-ph)*Nh/(Nh-1)
> ybar = sum(Nh*ph)/sum(Nh)
[1] .0475

(a) 2*1.96*sqrt((1-300/10000)*ybar*(1-ybar)/300)
[1] 0.04741225  ##SRS random sample CI width

## in (b), allocation is: 300*Nh/1e4 =  165 105  30
(b) 2*1.96*sqrt(sum((1-300/10000)*(Nh/1e4)^2*pqh/(300*Nh/1e4)))
[1] 0.04715945

(c) > nopt = 300*Nh*sqrt(pqh)/sum(Nh*sqrt(pqh))
## in (c), allocation is: nopt =  136.00 120.49  43.50
 > 2*1.96*sqrt(sum((1-nopt/Nh)*(Nh/1e4)^2*pqh/nopt))
[1] 0.04610368 ### Neyman alloc stratified sample CI width


(2) S_t^2 = sum_i (M_i*ybar_{Ui} - (K/N)*ybar_U})^2/(N-1)
          = sum_i M_i^2 ybar_{Ui}/(N-1) - Mbar*K*ybarU^2/(N-1)
and S_{yU}^2 = SST/(K-1). 

   Now we are given  Mbar = K/N=3.5,  and 
SSW/((K-N)*S_{yU}^2) = (SSW/SST)*(K-1)/(K-N) = 0.3 and 
sum_i M_i*(M_i-Mbar)* ybar_{Ui}^2 = 1.44*SST*(N-1)/(K-1).

Therefore
SSB = SST*(1-0.3*(K-N)/(K-1) 
and by definition
SSB =  sum_i M_i (ybar_{Ui}-ybar_U)^2 
    = sum_i M_i ybar_{Ui}^2 - K*ybarU^2

To solve the problem, note that
(N-1) * S_t^2 = sum_i M_i*(M_i-Mbar)*ybar_{Ui}^2
              + Mbar*sum_i M_i*(ybar_{Ui}-ybarU)^2
and we can now fill in the information:

(N-1) * S_t^2 =  1.44*SST*(N-1)/(K-1) + Mbar*SST*(1-0.3*(K-N)/(K-1) 
        =  SST*(1.44*(N-1)/(K-1) + (K/N)*(1-0.3*(K-N)/(K-1)))

or (a) S_t^2/S_{yU}^2 = (K-1)*S_t^2/SST
         = 1.44 + (K/N)*(K-1-.3*(K-N))/(N-1)
         = 1.44 + (K/N)*(.7*K+.3*N-1)/(N-1)
## After removing -1 terms due to large N, we obtain the
##   approximate ratio
 S_t^2/S_{yU}^2 = 1.44 + 3.5*(.7*3.5+.3) = 11.065

For (b): For single stage cluster sample, with N1 clusters,
variance in estimating population mean  
    = N^2*(1-N1/N)*S_t^2/(N1*K^2) 
 approximately = S_t^2/(N1*Mbar^2) = S_t^2/(N1*3.5^2).

For SRSWOR sample of N2 units, approx variance = S_{yU}^2/N2.

To equalize the two variances: we set 
S_t^2/(N1*3.5^2) = S_{yU}^2/N2

and because of the calculation done in part (a), we conclude 
that

11.065 = S_t^2/S_{yU}^2  = 3.5^2 * N1/N2

Since each of the N1 clusters sampled costs 25, and each 
of the N2 persons sampled costs 10, we find that the 
ratio of costs needed for the two plans is 

25*N1/(10*N2) = 2.5*11.065/3.5^2
 =  2.258163

### This means that the plan to sample clusters is more 
costly by a factor of 2.26.


(3)  Formula is :

 [ (300/20)* sum over i in S1 of 3*that_i plus
  1200/30 * sum over i in S2 of sampled incomes y_i]
  divided by the total number 3000 of units.

The variance of this estimator is obtained using formula (5.22) [using
S_t^2 = MSB*M, or (5.34)] for the first stratum 
and our standard SRS formula for the second,  as

[300^2*(1-20/300)*MSB_{y,1}*6/20 
   + (300/20)*(1-2/6)*(6^2/2)*(1/5)*SSW_{y,1}
   + 1200^2*(1-30/1200)*S_{y,2}^2/30]/3000^2

We note that   SST_{y,1} = 1799*S_{y,1}^2 = 16191,  and 

SSW_{y,1}/SST_{y,1} = (1800-300)*MSW_{y,1}/(1799*S_{y,1}^2) 
    = (1500/1799)*0.2 = 0.16676, so that
SSB_{y,1}/SST_{y,1} = 1 - 0.16676 = 0.83324  and 
MSB_{y,1} = 0.83324* 16191 /299 = 45.120
SSW_{y,1} = 0.16676* 16191 = 2700.011

Plugging everything in, we get 

SE  = (1/3000)*sqrt(300^2*(1-20/300)*45.120*6/20 + 
    (300/20)*(1-2/6)*(6^2/2)*.2*2700.011 + 1200^2*(1-30/1200)*16/30)
[1] 0.4693997
### Recalling units are $10000 this might be reasonable.


(4) NOTE: in this propblem the sample proportion of counties
with >500 farms is 160/300 = .53333, while we are told that the
national proportion of such counties = 1484/3078 = .48213.

As in all small-domain problems, we re-define the attribute
to be multiplied by the indicator of domain membership:

W_i = Y_i*I(i in D),   and we use ratio estimator over denom
which is total of V_i = I(i in D). Target is ratio of frame 
totals  sum W_i/ sum V_i.

Estimator = 50468635/160 = 315429
Standard Error = sqrt((1/n - 1/N)*(1/Vbar)^2*SampVar(W_i-Bhat*V_i))
       = sqrt((1/300-1/3078)*(3078/1484)^2*(1/299)*
               (2.733721e+13-160*315429^2))
       22231.26

So CI = 315429 + 1.96*c(-1,1)*22231.26 =  (271856, 359002)