Tools › Statistics › Surveys and Sampling › Survey Analysis › Example
Cochran's sample size and the Krejcie–Morgan table at a 5 % margin
| Tool | Survey Analysis |
| Source | Cochran, W. G. (1977). Sampling Techniques, 3rd ed. Wiley, New York. Section 4.4; Krejcie, R. V., and Morgan, D. W. (1970). "Determining Sample Size for Research Activities." Educational and Psychological Measurement, Vol. 30, pp. 607–610. Table 1. |
| Level | Undergraduate |
Problem
How many respondents does a survey need for a 5 % margin of error at 95 % confidence when the proportion may be near 0.5, and how does the finite-population correction reduce that number for populations of 1,000, 500 and 100?
Inputs to Enter
- Target margin 5 %, planning p 50 %, confidence 95 %
- Population size blank, then 1000, 500, 100 (Krejcie–Morgan rows)
Answers
- n0 = 1.96² × 0.25/0.05² = 384.15, rounded up to 385
- With N: 1,000,000 → 384, 100,000 → 383, 1,000 → 278, 500 → 217, 100 → 80, 10 → 10 (Krejcie–Morgan Table 1, nearest integer)
- E = 3 % gives n0 = 1067.1; p = 0.2 at 5 % gives 245.9
Notes
The sample-size section runs with the default data; set the population size to read the finite-population rows. Krejcie and Morgan use χ²(1) = 3.841 = 1.96², so the two formulas coincide. Not a default-case entry.