Page 210 - Statistics for Environmental Engineers

l1592_frame_Ch23 Page 210 Tuesday, December 18, 2001 2:44 PM

The c i are the costs to collect and analyze each specimen. The optimal sample size per stratum is:

n i = n-----------------------------------------------------------------
w i s i / c i
w i s i / c i + … + w n s n s
s / c n s
This says that the sample size in stratum i will be large if the stratum is large, the variance is large, or
the cost is low. If sampling costs are equal in all strata, then:

n i = n----------------------------------------
w i s i
w i s i + … + w n s n s
s
Using these equations requires knowing the total sample size, n. This might be constrained by budget,
or it might be determined to meet an estimation error criterion for the population mean, or to have a
specified variance (Gilbert, 1987).
The sample size needed to estimate the overall mean with a specified margin of error (E) and an appr-
oximate probability (1 − α)100% = 95% of exceeding that error is:

4
n = ------ 2∑ w i s i 2
E

Example 23.10

Using the data from Example 23.9 (Table 23.4), design a stratified sampling plan to estimate the
mean with a margin of error of 1.0 unit with 95% confidence. There are three strata, with variances
2 2 2
s 1 = 35.4, s 2 = 180, and s 3 = 12, and having weights w 1 = 0.5, w 2 = 0.25, and w 3 = 0.25. Assume
equal sampling costs in the three strata. The total sample size required is:

4
n = ---- 2 0.5 35.4( ) + 0.25 180) + 0.25 12) = 263
(
(
1
The allocation among strata is:

n 1 = 263 ---------------------------------------------------------------------------------- =
w i s i
(
0.5 35.4) + 0.25 180) + 0.25 12) 4w i s i
(
(
giving
(
(
n 1 = 4 0.5) 35.4) = 71
n 2 = 4 0.25) 180) = 180
(
(
(
(
n 3 = 4 0.25) 12) = 12
This large sample size results from the small margin of error (1 unit) and the large variance in
stratum 2.
Example 23.11

The allocation of the n = 40 samples in Example 23.9 gave a 95% confidence interval of ±2.7.
2 2 , which gives n 1 =
The allocation according to Example 23.10 is n i = 40(w i s i /65.7) = 0.61w i s i
11, n 2 = 28, and n 3 = 2. Because of rounding, this adds to n = 41 instead of 40.
© 2002 By CRC Press LLC