variable; then P and p are also continuous. Since the sample is very large, the sampling
distribution ofp is approximately normal, and it is shown in Fig. 25a. The null hypothesis
is to be rejected if.p assumes a value greater than that corresponding to 90 percent of all
possible samples. In Fig. 25<z, locate B such that area to the left of B = 0.90; then area
from A to B = 0.90 - 0.50 = 0.40. At B 9 A(z) = 0.40. From the table of areas under the nor-
mal curve, z = 1,282. Thus, at B 9 p = 0.03 + (1.282)(0.0108) = 0.044.
- State the decision rule
If the proportion of defectives in the sample is 4.4 percent or less, accept the shipment; if
the proportion is greater, reject the shipment.
By setting the limiting proportion of defectives in the sample at 4.4 percent as com-
pared with the limiting proportion of 3 percent in the population, allowance is being made
for the random variability of sample results.
PROBABILITY OF ACCEPTING AN
UNSATISFACTORY SHIPMENT
With reference to the preceding calculation procedure, what is the probability that a ship-
ment in which the incidence of defectives is 5 percent will nevertheless be accepted?
Acceptance region Rejection regionFIGURE 25. Sampling distribution of proportion of defective units.