Next Page Previous Page Home Tools & Aids Search Handbook
7. Product and Process Comparisons
7.2. Comparisons based on data from one process
7.2.4. Does the proportion of defectives meet requirements?

7.2.4.2.

Sample sizes required

Derivation of formula for required sample size when testing proportions The method of determining sample sizes for testing proportions is similar to the method for determining sample sizes for testing the mean. Although the sampling distribution for proportions actually follows a binomial distribution, the normal approximation is used for this derivation.
Problem formulation We want to test the hypothesis
    H0:p=p0
    Ha:pp0
with p denoting the proportion of defectives.

Define δ as the change in the proportion defective that we are interested in detecting

    δ=|p1p0|.
Specify the level of statisitical significance and statistical power, respectively, by

    P(reject H0|H0 is true with any pp0)α

    P(reject H0|H0 is false with any pδ)1β.

Definition of allowable deviation If we are interested in detecting a change in the proportion defective of size δ in either direction, the corresponding confidence interval for p can be written as p^δpp^+δ.
Relationship to confidence interval For a 100(1α) % confidence interval based on the normal distribution, where z1α/2 is the critical value of the normal distribution which is exceeded with probability α/2, is δ=z1α/2p0(1p0)N+z1βp1(1p1)N.
Minimum sample size Thus, the minimum sample size is
  1. For a two-sided interval

    N(z1α/2p0(1p0)+z1βp1(1p1)δ)2.

  2. For a one-sided interval

    N(z1αp0(1p0)+z1βp1(1p1)δ)2.

    The mathematical details of this derivation are given on pages 30-34 of Fleiss, Levin, and Paik.

Continuity correction Fleiss, Levin, and Paik also recommend the following continuity correction,
    N=N+1/δ,
with N denoting the sample size computed using the above formula.
Example of calculating sample size for testing proportion defective Suppose that a department manager needs to be able to detect any change above 0.10 in the current proportion defective of his product line, which is running at approximately 10 % defective. He is interested in a one-sided test and does not want to stop the line except when the process has clearly degraded and, therefore, he chooses a significance level for the test of 5 %. Suppose, also, that he is willing to take a risk of 10 % of failing to detect a change of this magnitude. With these criteria:
  1. z0.95=1.645,z0.90=1.282
  2. δ=0.10(p1=0.20)
  3. p0=0.10
and the minimum sample size for a one-sided test procedure is N(z1αp0(1p0)+z1βp1(1p1)δ)2=(1.6450.1×0.9+1.2820.2×0.80.1)2102. With the continuity correction, the minimum sample size becomes 112.
Home Tools & Aids Search Handbook Previous Page Next Page