gions are labeled as shown. The values of X and z at the boundary of the acceptance and
rejection regions are called the critical values.
At B, A(z) = 0.45. From the table of areas under the normal curve, z — -1.645. Thus, at
B, X = 3.50 + (-1.645)(0.101) = 3.334 h.
- Make a recommendation
Since the true sample mean of 3.37 h falls to the right of B, the null hypothesis stands.
Therefore, we must recommend that the modified method of production be disapproved
and the present method retained.
This recommendation does not necessarily imply that the industrial engineer's claim is
invalid. The decision must be based on probability rather than certainty, and the test re-
sults have failed to demonstrate a 95 percent probability that the modified method is supe-
rior to the present method. The difference between the assumed population mean of 3.50
h and the sample mean of 3.37 h can be ascribed to chance.
Related Calculations'. If the null hypothesis is rejected when, in fact, it is true, a
Type I error has been committed. The probability of committing this error is denoted by
a, and the acceptable value of a is termed the level of significance. In this case, if the null
hypothesis is correct, the sampling distribution of X is as shown in Fig. 24a. The hypoth-
esis will be rejected if X assumes a value to the left of B, and the probability of this event
is 1 - 0.95 = 0.05. Thus, the level of significance is 0.05.
Whether a null hypothesis is accepted or rejected depends largely on the level of sig-
nificance imposed. Therefore, selecting an appropriate level of significance is one of the
crucial problems that arise in statistical decision making. The selection must be based on
the amount of the loss that would result from a false decision.
PROBABILITY OFACCEPTING A FALSE
NULL HYPOTHESIS
With reference to the preceding calculation procedure, the time required to produce 1 unit
under the modified method has these characteristics: The standard deviation remains 0.64
h, but the arithmetic mean is (a) 3.40 h; (b) 3.20 h. Determine the probability that the in-
dustrial engineer's proposal will be vetoed despite its merit.
Calculation Procedure:- Compute the critical value of z in each case
Since cr remains 0.64 h and the sample size is still 40, CT^ remains 0.101 h. Refer to Fig.
24b and c, which gives the sampling distributions of X when LL = 3.40 h and LL = 3.20 h,
respectively. The null hypothesis will be accepted if X > 3.334, and it is necessary to cal-
culate the probability of this event.
In Fig. 246, at B', z = (3.334 - 3.40)70.101 = -0.653. In Fig. 24c, at B", z = (3.334 -
3.20)70101-1.327. - Compute the required probabilities
Refer to the table of areas under the normal curve. When z = -0.653, A(z) = 0.243. In Fig.
246, area to right of B' = 0.243 + 0.5 = 0.743. Thus, when LL = 3.40 h, there is a probabil-
ity of 74.3 percent that the proposal will be vetoed. Similarly, when z = 1.327, A(z) =
0.408. In Fig. 24c, area to right of B" = 0.5 - 0.408 = 0.092. Thus, when LL = 3.20 h, there
is a probability of 9.2 percent that the proposal will be vetoed.