2.3 Selecting Simple Random Samples 21
- Then referring to the systematic list, we see that the index 7 corre-
sponds to the sample {A, C, D, E}.
Now this method is feasible when N and n are small like 6 and 4
above, since the number of combinations is only 15. But as N and n
get larger, the number of combinations gets out of hand very quickly.
So a simpler alternative is to consider the sampling without replace-
ment approach. In this approach, the individual patients get ordered.
One ordering that we could have is as follows:
1 is A
2 is B
3 is C
4 is D
5 is E
6 is F
Now we divide [0, 1) into six equal intervals and assign the uniform
random number as follows:
If 0.0000 ≤ U < 0.1667, then the index is 1.
If 0.1667 ≤ U < 0.3333, then the index is 2.
If 0.3333 ≤ U < 0.5000, then the index is 3.
If 0.5000 ≤ U < 0.6667, then the index is 4.
If 0.6667 ≤ U < 0.8333, then the index is 5.
If 0.8333 ≤ U < 1.0000, then the index is 6.
Example: From a table of uniform random numbers, suppose the fi rst
number to be 00439 for 0.00439, since 0.00439 is in the interval [0,
0.1667], we choose index 1 corresponding to patient A. Now A is taken
out so we rearrange the indexing.
1 is B
2 is C
3 is D