Chapter Topics
•Hypothesis Testing Methodology
•Z Test for the Mean (s Known)
• p-Value Approach to Hypothesis Testing
•Connection to Confidence Interval Estimation
•One Tail Test
• t Test of Hypothesis for the Mean
•Z Test of Hypothesis for the Proportion
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Chap. 8 - 1
What is a Hypothesis?
A hypothesis is an
assumption about the
population parameter.

A parameter is a
Population mean or
proportion

The parameter must be
identified before
analysis.
I assume the mean GPA
of this class is 3.5!
© 1984-1994 T/Maker Co.
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Chap. 8 - 2
The Null Hypothesis, H0
•
States the Assumption (numerical) to be tested
e.g. The average # TV sets in US homes is at
least 3 (H0:   3)
•
Begin with the assumption that the null
hypothesis is TRUE.
(Similar to the notion of innocent until proven guilty)
•Refers to the Status Quo
•Always contains the ‘ = ‘ sign
•The Null Hypothesis may or may not be rejected.
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Chap. 8 - 3
The Alternative Hypothesis, H1
•
Is the opposite of the null hypothesis
e.g. The average # TV sets in US homes is
less than 3 (H1:  < 3)
•
•
Challenges the Status Quo
Never contains the ‘=‘ sign
•
The Alternative Hypothesis may or may
not be accepted
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Chap. 8 - 4
Identify the Problem
Steps:
 State the Null Hypothesis (H0:   3)
 State its opposite, the Alternative
Hypothesis (H1:  < 3)
 Hypotheses
are mutually exclusive &
exhaustive
 Sometimes it is easier to form the
alternative hypothesis first.
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Chap. 8 - 5
Hypothesis Testing Process
Assume the
population
mean age is 50.
(Null Hypothesis)
Is X  20    50?
Population
The Sample
Mean Is 20
No, not likely!
REJECT
Null Hypothesis
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Sample
Chap. 8 - 6
Reason for Rejecting H0
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... Therefore, we
reject the null
hypothesis that
 = 50.
... if in fact this were
the population mean.
20
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 = 50
Sample Mean
H0
Chap. 8 - 7
Level of Significance, a
•
Defines Unlikely Values of Sample
Statistic if Null Hypothesis Is True

•
Called Rejection Region of Sampling
Distribution
Designated a (alpha)

Typical values are 0.01, 0.05, 0.10
•
Selected by the Researcher at the Start
•
Provides the Critical Value(s) of the Test
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Chap. 8 - 8
Level of Significance, a and
the Rejection Region
a
H0:   3
H1:  < 3
H0:   3
H1:  > 3
Rejection
Regions
0
0
H0:   3
H1:   3
Critical
Value(s)
a
a/2
0
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Chap. 8 - 9
Errors in Making Decisions
•
Type I Error
Reject True Null Hypothesis
 Has Serious Consequences
 Probability of Type I Error Is a


•
Called Level of Significance
Type II Error
Do Not Reject False Null Hypothesis
 Probability of Type II Error Is b (Beta)

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Chap. 8 - 10
Result Possibilities
H0: Innocent
Jury Trial
Actual Situation
Hypothesis Test
Actual Situation
Verdict
Innocent Guilty Decision H0 True H0 False
Innocent
Do Not
Reject
H0
Guilty
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Correct
Error
Error
Correct Reject
H0
1-a
Type II
Error (b )
Type I
Error
(a )
Power
(1 - b)
Chap. 8 - 11
a & b Have an
Inverse Relationship
Reduce probability of one error
and the other one goes up.
b
a
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Chap. 8 - 12
Z-Test Statistics (s Known)
• Convert Sample Statistic (e.g., X ) to
Standardized Z Variable
Z
X  X
sX

X 
s
Test Statistic
n
• Compare to Critical Z Value(s)

If Z test Statistic falls in Critical Region,
Reject H0; Otherwise Do Not Reject H0
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Chap. 8 - 13
p Value Test
•
Probability of Obtaining a Test Statistic
More Extreme  or ) than Actual
Sample Value Given H0 Is True
•
Called Observed Level of Significance

•
Smallest Value of a H0 Can Be Rejected
Used to Make Rejection Decision
If p value  a Do Not Reject H0
 If p value < a, Reject H0

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Chap. 8 - 14
Hypothesis Testing: Steps
Test the Assumption that the true mean #
of TV sets in US homes is at least 3.
1.
State H0
H0 :   3
2.
State H1
H1 :  < 3
3.
Choose a
a = .05
4.
Choose n
n = 100
5.
Choose Test:
Z Test (or p Value)
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Chap. 8 - 15
Hypothesis Testing: Steps
(continued)
Test the Assumption that the average # of
TV sets in US homes is at least 3.
6. Set Up Critical Value(s)
Z = -1.645
7. Collect Data
100 households surveyed
8. Compute Test Statistic
Computed Test Stat.= -2
9. Make Statistical Decision
Reject Null Hypothesis
10. Express Decision
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The true mean # of TV set
is less than 3 in the US
households.
Chap. 8 - 16
One-Tail Z Test for Mean
(s Known)
•
Assumptions
Population Is Normally Distributed
 If Not Normal, use large samples
 Null Hypothesis Has  or  Sign Only

•
Z Test Statistic:
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z
x  x
sx

x
s
n
Chap. 8 - 17
Rejection Region
H0:   
H1:  < 0
H0:   0
H1:  > 0
Reject H0
Reject H 0
a
a
0
Must Be Significantly
Below  = 0
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Z
Z
0
Small values don’t contradict H0
Don’t Reject H0!
Chap. 8 - 18
Example: One Tail Test
Does an average box of
cereal contain more than
368 grams of cereal? A
random sample
of 25 boxes
_
showed X = 372.5. The
company has specified s to
be 15 grams. Test at the
a0.05 level.
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368 gm.
H0:   368
H1:  > 368
Chap. 8 - 19
Finding Critical Values:
One Tail
What Is Z Given a = 0.05?
.50
-.05
.45
sZ = 1
a = .05
0 1.645 Z
Critical Value
= 1.645
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Standardized Normal
Probability Table (Portion)
Z
.04
.05
.06
1.6 .5495 .5505 .5515
1.7 .5591 .5599 .5608
1.8 .5671 .5678 .5686
1.9 .5738 .5744 .5750
Chap. 8 - 20
Example Solution: One Tail
H0:   368
H1:  > 368
Test Statistic:
a = 0.025
n = 25
Critical Value: 1.645
Reject
.05
0 1.645 Z
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Z
X 
s
 1.50
n
Decision:
Do Not Reject at a = .05
Conclusion:
No Evidence True Mean
Is More than 368
Chap. 8 - 21
p Value Solution
p Value is P(Z  1.50) = 0.0668
Use the
alternative
hypothesis
to find the
direction of
the test.
p Value
.0668
.9332
0 1.50
From Z Table:
Lookup 1.50
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1.0000
- .9332
.0668
Z
Z Value of Sample
Statistic
Chap. 8 - 22
p Value Solution
(p Value = 0.0668)  (a = 0.05).
Do Not Reject.
p Value = 0.0668
Reject
a = 0.05
0
1.50
Z
Test Statistic Is In the Do Not Reject Region
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Chap. 8 - 23
Example: Two Tail Test
Does an average box of
cereal contains 368 grams of
cereal? A random sample of
25 boxes showed X = 372.5.
The company has specified
s to be 15 grams. Test at the
a0.05 level.
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368 gm.
H0:   368
H1:   368
Chap. 8 - 24
Example Solution: Two Tail
H0:   386
H1:   386
Test Statistic:
a = 0.05
n = 25
Critical Value: ±1.96
Reject
.025
.025
-1.96
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0 1.96
Z
X 
372.5  368
Z

 1.50
s
15
n
25
Decision:
Do Not Reject at a = .05
Conclusion:
No Evidence that True
Mean Is Not 368
Chap. 8 - 25
Connection to
Confidence Intervals
_
For X = 372.5oz, s = 15 and n = 25,
The 95% Confidence Interval is:
372.5 - (1.96) 15/ 25 to 372.5 + (1.96) 15/ 25
or
366.62    378.38
If this interval contains the Hypothesized mean
(368), we do not reject the null hypothesis.
It does. Do not reject.
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Chap. 8 - 26
t-Test: s Unknown
Assumptions
Population is normally distributed
 If not normal, only slightly skewed & a large
sample taken

Parametric test procedure
t test statistic
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X 
t
S
n
Chap. 8 - 27
Example: One Tail t-Test
Does an average box of cereal
contain more than 368 grams
of cereal? A random sample of
36 boxes showed X = 372.5,
and s  15. Test at the a0.01
level.
s is not given,
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368 gm.
H0:   368
H1:  > 368
Chap. 8 - 28
Example:Z Test for Proportion
•Problem: A marketing company claims
that it receives 4% responses from its
Mailing.
•Approach: To test this claim, a random
sample of 500 were surveyed with 25
responses.
•Solution: Test at the a = .05 significance
level.
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Chap. 8 - 29
Z Test for Proportion:
Solution
H0: p  .04
H1: p  .04
Test Statistic:
p - ps
Z 
p (1 - p)
n
a = .05
n = 500
Critical Values:  1.96
Reject
.025
0
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Decision:
Reject
.025
.04 -.05
=
= 1.14
.04 (1 - .04)
500
Z
Do not reject at a = .05
Conclusion:
We do not have sufficient
evidence to reject the company’s
claim of 4% response rate.
Chap. 8 - 30