Chapter 6 Statistical Inference 235
Let’s apply this formula to our example: m 0550 , s515, n 52 5, and we
set a5 0. 0 5 so that the probability of Type I error is 5%. The acceptance
region is therefore
5506 1.96 315 /" 25
5506 5. 88
51 44.12,55.88^2
Any value that is less than 44.12 or greater than 55.88 will cause us to reject
the null hypothesis. Because 45 falls in the acceptance region, we accept
the null hypothesis and do not conclude that the new process decreases the
number of defective resistors in a batch.
p Values
The p value is the probability of a value as extreme as the observed value.
We can calculate that by examining the z test statistic. For the manufactur-
ing example, the z test statistic is
z 5
x2m 0
s/!n
5
45250
15 /" 25
52 1. 67
The probability of a standard normal value of less than 2 1. 6 7 is 0.0478.
To calculate the p value, we need to take into account the terms of the al-
ternate hypothesis. In this case, the alternative hypothesis was that the new
process made no difference (either positive or negative) in the number of
defects. Thus, we need to calculate the probability of an extreme value 1.67
units from 0 in either direction. Because the standard normal distribution
is symmetric, the probability of a value being ,21. 6 7 is equal to the prob-
ability of a value being .1. 6 7, so we can simply double the probability,
resulting in a p value of 230. 047850. 0956.
This was an example of a two-tailed test, in which we assume that extreme
values can occur in either direction. We can also construct a one-tailed test,
in which we consider differences in only one direction. A one-tailed test
could have these hypotheses.
H 0 : The mean number of defective resistors in the new process is 50.
Ha: The mean number of defective resistors is , 50.
We use a one-tailed test if something in the new process would absolutely
rule out the possibility of an increase in the number of defective resistors.
