Solutions to Sample Final Problems 1--9. 
========================================
(See below also for solutions to Fall 2007 
Supplement of 4 sample problems.)
-----------------------------------------

(1) Ignoring fpc, and using M=4, K=4N, 
design effect = (N^2/n) S_t^2/((4N)^2 S_y^2/(4n))
which after equating fpc's and terms like (N/(N-1)) 
to 1,  is equal to 4 * SSB /SST  = 2.8

(2)  We find the average of the 10 groups' total 
monthly rentals = $96000, and the standard deviation =
sqrt((9.4464e10 - 10(9.6e4)^2)/9) = 1.6e4.
Then the overall average-estimator  \bar{\tau}/80 has
observed value 1200, mean \bar{Y} and standard error 
estimated by sqrt(1/10)*(1/80)*1.6e4 = 63.25.
So the CI for \bar{Y} is 1200 + 1.96*c(-1,1)*63.25 =
(1076.03, 1323.97).


(3) In this problem, do not ignore fpc's but ignore 
factor M/(M-1) = 1000/999.

In both parts: \hat{t}_i = 1000* block mean given in Table,
and the sample variance estimator 
s_t^2 = 1000^2 * var(c(15.2, 13.5, 11.7, 14.4)) = 2.26e6

Part (a): estimator \bar{y}_{unb} 
                  = (15.2+13.5+11.7+14.4)/4 = 13.7
while  SE = 1e-4*sqrt((10^2/4)*(1-4/10)*2.26e6 + 
            (10/4)*(1000^2/20)*(1-20/1000)*(10.2+8.6+13.3+11.1))
 = 0.626  ## which is a reasonable value

Part (b): The unbiased estimator for SSB in this problem is:
(9/1000)*(s_t^2 - (1/4)*(1000^2/20)*(1-20/1000)*(10.2+8.6+13.3+11.1))
 = 15577.2
The unbiased estimator for SSW is (999/4)*10*(10.2+8.6+13.3+11.1) = 107982
So the solution is to estimate 
   SST/(1e4-1) = (15577+107982)/1e4 =  12.356
showing that only a little more than 1/8 of the overall variance in 
Y is due to between stratum variance.


(4) Here all units within each stratum have the same 
inclusion probability, resp. 
c((35/2500)*2/4, (50/4500)*1/4, 20/3000) = 
c(0.007, 0.00278, 0.00667). 

So the HT estimate is  101500/(.007) + 118750/(.00278) + 
248000/(.00667) =  94397237  .

[Compare (40000/200)*(101500+118750+248000) = 93650000 which 
is an SRS-looking estimator, just to check the order of 
magnitude is right.]


(5) > nh <- c(.23, .24, .25, .28)*c(4200, 3000, 1900, 2400)
> round(nh*1000/sum(nh))
[1] 341 254 168 237
> sum(c(.23, .24, .25, .28)^2*c(4200, 3000, 1900, 2400)^2 *
     (1-nh/(80000*c(.23,.24,.25,.28)))/nh)
[1] 2723.844   ### sqrt of this is the SE


(6) Ignoring fpc, want 2.576*sqrt(.55*.45/n) <= .05, 
or n=657.

(7) > sum(c(5,4,2,3,2,1)*(0:5)) ## 30
> (sum(c(5,4,2,3,2,1)*(0:5)^2) - 17*(30/17)^2)/16 ## 2.691
(a) Estimator is (175/17)*30 = 308.82, with 
    estimated SE = sqrt((175^2/17)*(1-17/175)*2.691) = 66.16
(b) Estimator is regression estimator based on 
x_i= I[at least 1 cav], y_i=#cav. Then the regression 
estimators are 
\hat{B}_1 = (30 - 17*(30/17)*(12/17))/(12 - 17*(12/17)^2) 
     = 2.5
\hat{B}_0 = 30/17 - 2.5*(12/17) = 0
and \hat{t}_{y,reg} = \hat{B}_1 *120 = 2.5*120 = 300
with estimated SE = sqrt((175^2/17)*(1-17/175)*
    (sum((0:5 - c(0, rep(1,5))*2.5)^2*c(5,4,2,3,2,1))/16))
  =  46.203 .

(8) NOTE that this is a sampling design with random 
sample size ! We know that the inclusion prob's pi_i are
resp. .05, .15, .10 for individuals in the respective 
precincts, so the estimator is
> (35/.05 + 60/.15 + 30/.10)/3000 ## =  1400/3000 = 46.7%
## So the CV is the rMSE/mean of the estimator, est'd by:
> sqrt(35*.95/.05^2+60*.85/.15^2+30*.9/.10^2)/(3000*1400/3000) 
= .0965

(9) (a) 
> (3/2.088e-4 + 12/4.175e-4 + 25/5.219e-4 + 77/6.263e-4)/60
[1] 3565.942

## var not available without data on sum 
##    of (HH #5yr+ persons)^2

Since we know that the sum of squares of numbers of 
persons in samples HH in town for 5+ yrs were respectively 
     3    ,    18  ,   45   ,   251

we can calculate that the with-replacement SE is 
estimated by :

sqrt((1/(60*59))*(3/(2.088e-4)^2+18/(4.175e-4)^2+45/(5.219e-4)^2+
             251/(6.263e-4)^2 - 60*3565.942^2))

= 246.0

=======================================================
Solutions to 4 Fall 2007 Supplementary Sample Problems
======================================================

#1. Estimator = (N_1*y1bar + N_2* y2bar)/(N_1*x1bar + N_2* x2bar)
       = (12+2*20)/(1+2*3) = 52/7 = Bhat

Estimated Variance = (1/xbar_U)^2*((N_1/N)^2*Shat^2_{y-Bx,1}*
       (1/n_1-1/N_1) + (N_2/N)^2*Shat^2_{y-Bx,2}*
       (1/n_2-1/N_2))  witb Bhat substituted for B

= (3/7)^2*((1/9)*(.98/200)*(2+(52/7)^2*.36-52/7)+(4/9)*(.985/300)*
       (4+(52/7)^2*.64-2*52/7))   =  .00800

#2. As indicated in class,  log(y1bar) is the estimator, with
   log(y1bar) - log(Ybar_{U1})  approx = (1/Ybar_{U1})*(y1bar-Ybar_{U1})
 
which means that the estimated variance is given by

 (1/y1bar)^2(1/n_1-1/N_1)*S^2_{y,1} = (1/12^2) *(.98/200)*2 = 6.8056e-5

#3. (a) Horvitz-Thompson estimator of total = sum_{i in s} y_i w_i = 
         1000*(14*1+6*4+13*8) = 142000

(b) Vhat = sum_{i,j in s} (pi_{ij}-pi_i *pi_j)*y_i*y_j/(pi_i*pi_j*pi_{ij})
        =  sum_{i in s} (1-pi_i)*y_i^2/pi_i^2 = 1000*(999*(25+4+9+16) + 
               3999*(4+9+1)+7999*(64+25)) =  82184300 =  28668^2

The Sen-Yates-Grundy formula with these pi_{ij} would give 0 !
But it does not apply, because with these pi_{ij} 
sum_{j: j ne i } pi_{ij} = pi_i*(sum_k pi_k - pi_i) ne pi_i*(n-1)

#4.  Estimated response rates for the 4 strata are respectively 
4/5, 5/6, 3/4, 1/2.  The adjusted weights are the original 
weights*r_i times the reciprocals of these class-estimated 
response rates, so the adjusted estimator in (a) becomes 
(5/4)*130 + (6/5)*170 + (4/3)*120 + 2*150 =  826.5. The same 
estimator could be used in (b), but under the adjustment class 
model with responders and nonresponders assumed to have the same
average y-attribute values, we could also estimate in (b) by
(130/1200)*1400 + (170/2500)*2400 + (120/1500)*1900 + (150/2000)*2500
  =  654.4 which we can guess might be a little more accurate.