Stratified Random Sampling, Optimal Allocation Example
An advertising firm, interested in determining how much to emphasize television advertising in a
certain county, decides to conduct a sample survey to estimate the average number of hours per week
that households within the county watch television. The county contains two towns, town A and town
B, and a rural area. Town A is built around factories, and most households contain factory workers
with school-aged children. Town B is an upscale suburb of a city in a neighboring county and contains
older residents with few children at home. There are 1550 households in Town A, 620 in town B, and
930 in the rural area.
Since there are differences between the two towns in the prevalence of school-aged children, we
believe that the towns represent natural strata differing on the variable of interest, number of hours per
week that a household watches television. It is possible that the rural area will also differ from each
town on this characteristic. Hence we have a population of size N = 3100, with three strata of sizes N1
= 1550, N2 = 620, N3 = 930, respectively. A prior survey suggests that the stratum variances are
approximately S12  25 , S 22  225 , and S 32  100 . The advertising firm finds that obtaining an
observation from a rural household costs more than obtaining an observation from either Town A or
Town B. This is due to costs of traveling from one rural household to another. The cost per
observation in each town is estimated to be c1  c2  $9.00 , and the cost per observation in the rural
area is c3  $16.00 . We want our sample size to be n = 58. How should the sample be allocated
among the three strata?
Since the cost of sampling differs among the strata, and the variances differ among the strata, we have
N 1 S1 1550 5
N S
62015  3100.00 , N 3 S 3  93010  2325.00 .

 2583.33 , 2 2 
3
3
4
c1
c2
c3
Hence, the sample sizes should be
 N 1 S1 


 c 
1 
 2583.33 
n1  
n
58  18.7097  19 ,
 3 Ni Si 
 8008.33 


 i 1 c 
i 

 N2S2 


 c 
2 
 3100 
n2  
n
58  22.4516  22 , and
 3 N i Si 
 8008.33 


 i 1 c 
i 

 N 3 S3 


 c 
3 
 2325.00 
n3  
n
58  16.8387  17 .
 3 N i Si 
 8008.33 


 i 1 c 
i 

Now assume that the overall sample size has not yet been determined; only the relative sample sizes
for the strata have been determined using the optimal allocation method. We want to determine the
overall sample size by specifying that we want a 95% confidence interval estimate of the population
mean with a margin of error of e = 2 hours.
2
1 3 n  Ni  2 v
   Si  n .
n i 1 ni  N 
If we assume that the stratum sample sizes will be large enough so that the normal approximation
z 2 v
applies, then we may find the overall sample size from n  22 . For this example we have
e
2
2
2
 58  1550 
 58  620 
 58  930 
v   
 25   
 225   
 100  73.5121 , and
 19  3100 
 22  3100 
 17  3100 
z 2 v
1.962 73.5121  70.6010 . We round up to n = 71. For the individual strata, we have
n  22 
4
e
n1  23 , n2  27 , n3  21 .
If the finite population corrections may be ignored, then we have V  y str  