z
Decision rule using the p-value:
Reject
H
0
if
p
-value < significance level
339

Calculating the p-value
1.
Calculate the z-score
2.
Using the Normal chart, find the probability of
getting (
more than
/
less than
) this score,
based
on the direction of the equality expression in
the ALTERNATE hypothesis
3.
If it is a two-tailed test, multiple the result by 2.
z
NOTE:
The final calculated p-value can be
between 0.00 and 1 (0% and 100%)
Calculating the p-value - Details
z
If
H
A
:
ȝ
>
#
–
Find probability
Z
>
Z-value
z
If
H
A
:
ȝ
<
#
–
Find probability
Z
<
Z-value
z
If
H
A
:
ȝ
¡
#
a.
If Z-value is
positive
, Find probability
Z
>
Z-value
b.
If Z-value is
negative
, Find probability
Z
<
Z-value
–
Multiply that answer by 2

p
-Value in Hypothesis Testing - Example
Recall the last problem where the
hypothesis and decision rules were set
up as:
H
0
:
P
¢
200
H
1
:
P
> 200
Reject H
0
if Z > Z
D
where Z = 1.55 and Z
D
=2.33
Reject H
0
if p-value <
D
0.0606 is not < 0.01
Conclude: Fail to reject H
0
339
What does it mean when p-value <
D
?
(a)
.10
, we have
some evidence
that
H
0
is not true.
(b)
.05
, we have
strong evidence
that
H
0
is not true.
(c)
.01
, we have
very strong evidence
that
H
0
is not true.
(d)
.001
, we have
extremely strong evidence
that
H
0
is not
true.
339

LYING
with
Statistics
z
Create Hypothesis at 0.05
z
Run the test and find INSIGNIFICANT
z
Find p-value = 0.07
z
Rerun at 0.10 to state SIGNIFICANT
THIS is
BAD
statistics
z
NOTE:
Not bad to run a test and state
p-value (even if you do not reject Null)
?
Testing for the Population Mean:
Population Standard Deviation Unknown
z
When the population standard deviation (
ı
)
is
unknown, the sample standard deviation (
s
) is used in
its place
z
The
t
-distribution is used as test statistic, which is
computed using the formula:
341

Testing for the Population Mean: Population
Standard Deviation Unknown - Example
The McFarland Insurance Company Claims Department reports the
mean
cost to
process a claim is
$60
. An industry comparison showed this amount to be larger
than most other insurance companies, so the company instituted cost-cutting
measures. To evaluate the effect of the cost-cutting measures, the Supervisor
of the Claims Department selected a random sample of
26
claims processed
last month. The sample information is reported below.
At the
0.01
significance level is it reasonable a claim is
now less than $60
?
342
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 1: State the null hypothesis and the alternate hypothesis.
H
0
:
P
¡
$60
H
1
:
P
< $60
(note: keyword in the problem “now
less
than”)
Step 2: Select the level of significance.
Į
= 0.01 as stated in the problem
Step 3: Select the test statistic.
Use
t
-distribution since
ı
is unknown
342

t
-Distribution Table (portion)
343
Testing for a Population Mean with a
Known Population Standard Deviation- Example
Step 5: Make a decision and interpret the result.

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- Spring '11
- Leany
- Statistics, Normal Distribution, Null hypothesis, Statistical hypothesis testing, Hypothesis and Testing