Chapter 9, Part E
VII. Calculating the Probability of
Type II Errors
A common decision in business is whether to accept
a shipment or not, based upon a sample of the
entire shipment. This is called a “lot acceptance
decision” because you will either accept the entire
shipment or reject the lot.
Recall a type II error: We “fail to reject” Ho, when
the Ha was “true”.
A. An Example
The authors describe a scenario where a producer is
receiving a shipment of batteries that claim a life
of “at least 120 hours”.
Ho:   120 hours
Ha:  > 120 hours
If  = .05, then Z.05 = -1.645. If Ho is not rejected,
we would accept the shipment.
The Test
• Suppose we sample n=36 and =12 from past
tests.
• What is the sample mean that would cause us to
reject Ho?
• Set up the test statistic and solve for x-bar.
x   x  120
• So reject Ho if:
Z

 1.645
 


 n
 12 


 36 
Moving toward a type II error
• Solving the previous equation for x-bar = 116.71
hours.
• This means that you will reject a shipment if you
test 36 batteries and their sample mean lifespan is
less than 116.71 hours.
• Any sample mean above 116.71 hours and you
will “fail to reject” Ho, risking a type II error.
B. The Power of the Test
• A type II error in this scenario is if the true
shipment mean is less than 120 hours, but we
make the decision to accept Ho that it is at least
120 hours.
• This would only happen if <120 hours.
• What if the true =113 hours?
• What is the probability of accepting the shipment
is >=120 and thus making a type II error?
If =113 hours, and we fail to reject any shipment
with a mean over 116.71 hours, what is the
probability that we would “accept” Ho?
.0314
µ=113 116.71
Z
x   116.71  113

 1.86
  
 12 




 n
 36 
The Prob (x-bar>=116.71) =
.5-.4686 = .0314
So the probability of making a type II error, when
=113 hours, is the area in the tail which equals
=.0314.
See table 9.3 in the text for a variety of possible
population means less than 120 hours and the
corresponding probability of type II errors.
If <120 and we correctly reject Ho, there is a (1-)
probability of doing so.
(1-) is the “power” of the test, the probability of
correctly rejecting H0 when it is false.
See figure 9.16 in the text for a power curve for this
lot acceptance test.
C. Summary
1. Set up Ho and Ha.
2. Use  to establish the rejection rule for Z.
3. Using the rejection rule, solve for the x that
makes the rejection region for the test.
4. Using this result, state value of x that leads to
acceptance of Ho; this is the acceptance region for
the test.
5. Using the sampling distribution of x for any
value of µ from Ha and the acceptance region
from step 4. Compute the probability that x is in
the acceptance region. This is the  at the chosen
µ.
Do problem #61 for more practice and come to class
with plenty of questions.