Chapter 9
Part C
III. One-Tailed Tests
B. P-values
• Using p-values is another approach to conducting
a hypothesis test, yielding the same result.
• In general: a p-value is the probability of
obtaining a sample result that is at least as unlikely
as what is observed.
• For example, suppose you calculate a z-score of
negative 2.80.
A diagram
The standard normal probability table tells us that
there is a .0026 probability of observing a sample
mean less than or equal to what was observed.
P-value is
P=.0026
.4974
Z=-2.80
0
Z
How do you use it?
•If you’re testing the hypothesis at the 99%
level, then =.01.
•If p-value< , reject Ho.
=.01
P-value is
P=.0026
0
Z=-2.80 Z=-2.33
Z
Result of the test.
• So in this example, .0026<.01, so we reject Ho.
• Had we been using our previous methodology of
comparing the test statistic to the critical value, we
would get the same decision.
• Since Z=-2.80 is greater (absolute value) than
Z=.01=-2.33, we would reject Ho.
An Example
Ho:   10
Ha:  < 10
n=36, x =8, s=5, =.01, Z=.01=-2.33
Test statistic, Z=(-2)/(5/6)=-2.4
p-value for a Z=-2.4 is .0082 and since this is less
than =.01, reject the null Ho.
MAKE SURE YOU CAN DO THIS!
IV. Two-Tailed Tests about :
(large sample)
• The big important gold star difference when you
do a two-tailed test is that the rejection range is
split equally in each tail of the sampling
distribution.
• If you get a test statistic that is either too low, or
too high, you will reject Ho.
A. An Example
• A lathe is set to cut bars of steel into perfect
lengths of 6 centimeters. If the bars are anything
but 6 cm, our customers will incur substantial
costs in resizing the bars and will find other
suppliers.
Ho:  = 6 cm
Ha:   6 cm
The rejection range
If you are testing this hypothesis at the =.05 level
of significance, then .025 (/2) is in each tail.
.025
.025
.4750
.4750
=6
x
Critical Z and Test Statistic
• Find the z-score that is associated with .4750, and
you’ll find Z=.025=±1.96.
• In a sample of 121 bars, you find a sample mean
of 6.08 cm with a standard deviation of .44 cm.
Z
6.08  6
.44 /

121
 2.0
• With a test statistic of 2.0, you would reject Ho
and conclude, with 95% confidence, that the bars
are not exactly 6 cm.
B. P-values for 2-tailed tests
• Because the significance level () is split between
each tail of the sampling distribution, so is the pvalue.
• The p-value is reported as 2 times the area in one
tail.
• If 2*(area in tail) < , reject Ho.
Example
• In the previous example, Z=2.0. The area in the
tail, beyond Z=2.0 is (.5-4772)=.0228
• The p-value = 2(.0228) = .0456, and since this is
less than =.05, reject Ho.
/2=.025 p=.0228
/2=.025
0
Z
C. Confidence Intervals
Recall, that with a (1- ) confidence coefficient, a
confidence interval is constructed as: x  z 
 /2
If you have a 2-tailed hypothesis test like:
Ho:  = 0
Ha:    0
n
If you take a sample, and your confidence interval
includes the hypothesized value of 0, then you
cannot reject the null.
If your 0 is outside of the confidence interval, reject
the null.
Back to our example
Our sample mean was 6.08 cm, with a standard error
of (.44/11)=.04.
So our 95% confidence interval would look like:
6.08  1.96(.04)
or a range between 6.0016 and 6.1584. Since 6 cm
is NOT in the interval, we reject Ho.