Chapter 9: Introduction to Hypothesis Testing

390 CHAPTER 9 • INTRODUCTION TO HYPOTHESIS TESTING
FIGURE 9.6
Two-Tailed Test—
Golden Peanut
Example
H 0 : μ = 16
H 0 : μ = 16
α = 0.10
Rejection region
α/2 = 0.05
Decision Rule:
0.45
0.45
μ = 16
z = 0
p-value = 2(0.0005) = 0.0010
Rejection region
α/2 = 0.05
z = 3.32
P(z > 3.32)
= 0.0005
x
If p-value < α = 0.10, reject H 0.
Otherwise, do not reject H 0.
Because p-value = 0.0010 < α = 0.10, reject H 0.
However, because this is a two-tailed hypothesis test, the p-value is found by multiplying
the 0.0005 value by 2 (to account for the chance that our sample result could have
been on either side of the distribution). Thus
p-value 2(0.0005) 0.0010
Assuming an alpha 0.10 level, then because the
p-value 0.0010 α 0.10, we reject H 0
.
Figure 9.6 illustrates the two-tailed test for the Golden Peanut Company example.
SUMMARY Two-Tailed Test for a Hypothesis About a Population Mean, σ Known
To conduct a two-tailed hypothesis test when the population
standard deviation is known, you can perform the following
steps.
1. Specify the population parameter of interest.
2. Formulate the null and alternative hypotheses in terms
of the population mean, μ.
3. Specify the desired significance level, α.
4. Construct the rejection region.
Determine the critical values for each tail, z α/2
and z α/2
from the standard normal table. If needed, calculate x(/2)L
and x(/2)U
Define the two-tailed decision rule using one of the following:
If z z /2
, or if z z /2
reject H 0
; otherwise, do
not reject H 0
.
If x x (/2)
or x x (/2)U reject H 0
; otherwise, do
not reject H 0
.
If p-value α, reject H 0
; otherwise, do not reject H 0
.
5. Compute the test statistic, z( x ) ( / n)
, or
find the p-value.
6. Reach a decision.
7. Draw a conclusion.
TRY PROBLEM 9.2
EXAMPLE 9-6 Two-Tailed Hypothesis Test for µ, Known
The Wilson Glass Company The Wilson Glass Company has a contract to supply
plate glass for home and commercial windows. The contract specifies that the mean thickness
of the glass must be 0.375 inches. The standard deviation, σ, is known to be 0.05 inch.
Before sending the first shipment, Wilson managers wish to test whether they are meeting
the requirements by selecting a random sample of n 100 thickness measurements.