### Chapter09

```Chapter 9
Hypothesis
Testing with
Single
Samples
Slide 9-1
Learning Objectives
• Understand the logic of hypothesis testing,
and know how to establish null and alternate
hypotheses.
• Understand Type I and Type II errors.
• Use large samples to test hypotheses about
a single population mean and about a single
population proportion.
• Test hypotheses about a single population
mean using small samples when  is
unknown and the population is normally
distributed.
Slide 9-2
Method of Indirect Proof
X
Either X or Y is true but not both
Y
X
X is demonstrated not to be true
Y
Y is true by default
Y
Slide 9-3
Hypothesis Testing
A process of testing hypotheses about
parameters by setting up null and
alternative hypotheses, gathering
sample data, computing statistics from
the samples, and using statistical
the hypotheses.
Slide 9-4
Steps in Testing Hypotheses
1. Establish hypotheses: state the null and
2.
3.
4.
5.
6.
7.
8.
alternative hypotheses.
Determine the appropriate statistical test
and sampling distribution.
Specify the Type I error rate (
State the decision rule.
Gather sample data.
Calculate the value of the test statistic.
State the statistical conclusion.
Make a managerial decision.
Slide 9-5
Null and Alternative Hypotheses
• The Null and Alternative Hypotheses are
mutually exclusive. Only one of them can
be true.
• The Null and Alternative Hypotheses are
collectively exhaustive. They are stated to
include all possibilities. (An abbreviated
form of the null hypothesis is often used.)
• The Null Hypothesis is assumed to be true.
• The burden of proof falls on the Alternative
Hypothesis.
Slide 9-6
Null and Alternative
Hypotheses: Example
• A soft drink company is filling 12 oz.
cans with cola.
• The company hopes that the cans are
averaging 12 ounces.
Ho:   12 oz
Ha:   12 oz
Slide 9-7
Rejection and Nonrejection Regions
Rejection Region
Rejection Region
Nonrejection Region
=12 oz
Critical Value
Critical Value
Slide 9-8
Type I and Type II Errors
• Type I Error
– Rejecting a true null hypothesis
– The probability of committing a Type I error
is called , the level of significance.
• Type II Error
– Failing to reject a false null hypothesis
– The probability of committing a Type II
error is called .
– Power is the probability of rejecting a false
null hypothesis, and equal to 1- 
Slide 9-9
Decision Table
for Hypothesis Testing
Null True
Null False
Fail to
reject null
Correct
Decision
Type II error
( )
Reject null
Type I error
()
Correct Decision
(Power)
Slide 9-10
One-tailed and Two-tailed Tests
• One-tailed Tests
Ho:   12
Ho:   12
Ha:   12
Ha:   12
• Two-tailed Test
Ho:   12
Ha:   12
Slide 9-11
One-tailed Tests
Ho:   12
Ho:   12
Ha:   12
Ha:   12
Rejection Region
Nonrejection Region
=12 oz
Critical Value
Rejection Region
Nonrejection Region
=12 oz
Critical Value
Slide 9-12
Two-tailed Tests
Ho:   12
Ha:   12
Rejection
Region
Rejection
Region
Nonrejection Region
=12 oz
Critical Values
Slide 9-13
CPA Net Income Example:
Two-tailed Test
Ho :   \$74,914
If Z  Zc  196
. , reject Ho.
If Z  Zc  196
. , do not reject Ho.
Ha :   \$74,914

.025
2

.025
2
Rejection
Region
Nonrejection Region
Rejection
Region
X 

n

78, 646  74, 914
 2.75
14,530
112
Z = 2.75  Zc = 1.96, reject Ho
=0
Zc  196
.
Z
Zc 196
.
Slide 9-14
CPA Net Income Example:
Critical Value Method (Part 1)
Lower
Ho:   \$74,914

X  Z n
Ha:   \$74,914
14,530
c
c

.025
2

.025
2
Rejection
Region
Nonrejection Region
72,223
Zc  196
.
 74,914  196
.
=0
 72,223
Rejection
Region
Upper
X
77,605
Zc 196
.
112
c
   Zc

n
14,530
 74,914  196
.
112
 77,605
Slide 9-15
CPA Net Income Example:
Critical Value Method (Part 2)

2

.025
2
Rejection
Region
Nonrejection Region
72,223
Zc  196
.
.025
Rejection
Region
77,605
Zc 196
.
=0
If X  77,223 or X  77,605, reject Ho.
If 77,223  X  77,605, do not reject Ho.
Since X  78,646 
X
c
 77,605, reject Ho.
Slide 9-16
Demonstration Problem 9.1 (Part 1)
Ho :   4.30
Ha :   4.30
Rejection
Region
=.05
Nonrejection Region
Zc  1645
.
If Z  1645
. , reject H0.
If Z  1645
. , do not reject H0.
0
X   4.156  4.30
Z

 142
.
s
0.574
n
32
Z  142
.  1645
.
,
do not reject H0.
Slide 9-17
Demonstration Problem 9.1 (Part 2)
Ho:   4.30
Ha:   4.30
=.05
Nonrejection Region
Zc  1645
.
s
XcZ n
 4.30  ( 1.645)
Rejection
Region
0.574
32
xc  4.133
0
4.30
 4.133
If X  4.133, reject H0.
X  4156
.
 4133
. , do not reject H0.
If X  4.133, do not reject H0.
Slide 9-18
Demonstration Problem 9.1 (Part 3)
Ho:   4.30
Ha:   4.30
Rejection
Region
=.05
Nonrejection Region
0
If p - value <  , reject Ho.
If p - value   , do not reject Ho.
X   4.156  4.30

 1.42
s
0.574
n
32
P ( Z  1.42) .0778
Z
Since p - value = .0778 >  = .05,
do not reject Ho.
Slide 9-19
Two-tailed Test: Small Sample,
 Unknown,  = .05 (Part 1)
Weights in Pounds of a Sample of 20 Plates
22.6
27.0
26.2
25.8
22.2
26.6
25.3
30.4
23.2
28.1
23.1
28.6
27.4
26.9
24.2
23.5
24.5
24.9
26.1
23.6
X  2551
. , S = 2.1933, and n = 20
Slide 9-20
Two-tailed Test: Small Sample,
 Unknown,  = .05 (Part 2)
Ho:   25
Ha:   25
Rejection Regions

2
df  n  1  19

.025
2
.025
Nonrejection Region
t
c
t
 2.093
c
 2.093
Critical Values
Slide 9-21
Two-tailed Test: Small Sample,
 Unknown,  = .05 (Part 3)
If t  2.093, reject Ho.
Rejection Regions
If t  2.093, do not reject Ho.

.025
2

.025
2
Non Rejection Region
t
c
t
 2.093
Critical Values
c
 2.093
X   2551
.  250
.
t

 104
.
S
21933
.
n
20
Since t  104
.  2.093, do not reject Ho.
Slide 9-22
Demonstration Problem 9.2 (Part 1)
Size in Acres of 23 Farms
445
463
466
561
489
466
477
560
474
557
557
505
502
433
553
449
545
477
438
511
545
500
590
X  498.78, S = 46.94,and n = 23
Slide 9-23
Demonstration Problem 9.2 (Part 2)
Ho :   471
Rejection Region
Ha :   471
 .05
df  n  1  22
Nonrejection Region
.
t 1717
c
Critical Value
Slide 9-24
Demonstration Problem 9.2 (Part 3)
If t  1717
. , reject Ho.
If t  1717
. , do not reject Ho.
Rejection Region
 .05
Nonrejection Region
.
t 1717
c
Critical Value
X   498.78  471
t

 2.84
S
46.94
n
23
Since t  2.84  1.717, reject Ho.
Slide 9-25
Z Test of Population Proportion
p  P
Z
PQ
n
where: p = sample proportion
P = population proportion
n  P  5, and
nQ  5
Q = 1-P
Slide 9-26
Proportion: Manufacturer Example
(Part 1)
Rejection Regions
Ho : P  .08
Ha : P  .08

.05
2
Nonrejection Region
Z
c
 1645
.
Z
c

.05
2
 1645
.
Critical Values
Slide 9-27
Proportion: Manufacturer Example
(Part 2)
If Z  1645
. , reject Ho.
Rejection Regions
If Z  1645
. , do not reject Ho.

.05
2
Nonrejection Region
Z
c
 1645
.
Z
Critical Values
c

.05
2
 1645
.
33
p 
.165
200
p  P
.165.08
Z

 4.43
P Q
(.08)(.92)
n
200
Since Z  4.43  1645
. , reject Ho.
Slide 9-28
Demonstration Problem 9.3 (Part 1)
Ho: P .17
Ha: P .17
Rejection Region
 .05
Nonrejection Region
Z
c
 1645
.
Critical Value
Slide 9-29
Demonstration Problem 9.3 (Part 2)
Rejection Region
 .05
Nonrejection Region
Z
Critical Value
c
 1645
.
If Z  1645
. , reject Ho.
If Z  1645
. , do not reject Ho.
115
p 
.209
550
p  P
.209 .17
Z

 2.44
P Q
(.17)(.83)
n
550
Since Z = 2.44  1645
. , reject Ho.
Slide 9-30
```