Statistics for Business and
Economics
Chapter 6
Inferences Based on a Single Sample:
Tests of Hypothesis
Learning Objectives
1. Distinguish Types of Hypotheses
2. Describe Hypothesis Testing Process
3. Explain p-Value Concept
4. Solve Hypothesis Testing Problems Based
on a Single Sample
5. Explain Power of a Test
Statistical Methods
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
Hypothesis
Testing
Hypothesis Testing
Concepts
Hypothesis Testing
Population


 


I believe the
population mean
age is 50
(hypothesis).

Random
sample
Mean 
X = 20
Reject
hypothesis!
Not close.
What’s a Hypothesis?
A belief about a population
parameter
I believe the mean GPA of
this class is 3.5!
• Parameter is
population mean,
proportion, variance
• Must be stated
before analysis
© 1984-1994 T/Maker Co.
Null Hypothesis
1.
2.
3.
4.
5.
What is tested
Has serious outcome if incorrect decision made
Always has equality sign: , , or 
Designated H0 (pronounced H-oh)
Specified as H0:   some numeric value
•
•
Specified with = sign even if  or 
Example, H0:   3
Alternative Hypothesis
1. Opposite of null hypothesis
2. Always has inequality sign: ,, or 
3. Designated Ha
4. Specified Ha:  ,, or  some value
•
Example, Ha:  < 3
Identifying Hypotheses
Steps
Example problem: Test that the population mean
is not 3
Steps:
• State the question statistically (  3)
• State the opposite statistically ( = 3)
—
Must be mutually exclusive & exhaustive
• Select the alternative hypothesis (  3)
—
Has the , <, or > sign
• State the null hypothesis ( = 3)
What Are the Hypotheses?
Is the population average amount of TV
viewing 12 hours?
• State the question statistically:  = 12
• State the opposite statistically:   12
• Select the alternative hypothesis: Ha:   12
• State the null hypothesis: H0:  = 12
What Are the Hypotheses?
Is the population average amount of TV
viewing different from 12 hours?
• State the question statistically:   12
• State the opposite statistically:  = 12
• Select the alternative hypothesis: Ha:   12
• State the null hypothesis: H0:  = 12
What Are the Hypotheses?
Is the average cost per hat less than or equal
to $20?
• State the question statistically:   20
• State the opposite statistically:   20
• Select the alternative hypothesis: Ha:   20
• State the null hypothesis: H0:   20
What Are the Hypotheses?
Is the average amount spent in the bookstore
greater than $25?
• State the question statistically:   25
• State the opposite statistically:   25
• Select the alternative hypothesis: Ha:   25
• State the null hypothesis: H0:   25
Basic Idea
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... therefore, we
reject the
hypothesis that
 = 50.
... if in fact this were
the population mean
20
 = 50
H0
Sample Means
Level of Significance
1. Probability
2. Defines unlikely values of sample statistic if
null hypothesis is true
•
Called rejection region of sampling
distribution
3. Designated (alpha)
•
Typical values are .01, .05, .10
4. Selected by researcher at start
Rejection Region
(One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1–

Nonrejection
Region
Critical
Value
Ho
Value
Sample Statistic
Observed sample statistic
Rejection Region
(One-Tail Test)
Sampling Distribution
Level of Confidence
Rejection
Region

1–
Nonrejection
Region
Ho
Value
Critical
Value
Observed sample statistic
Sample Statistic
Rejection Regions
(Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1–
1/2 
1/2 
Nonrejection
Region
Critical
Value
Ho
Value
Sample Statistic
Critical
Value
Observed sample statistic
Rejection Regions
(Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1–
1/2 
Nonrejection
Region
Ho
Value
Critical
Critical
Value
Value
Observed sample statistic
Sample Statistic
Rejection Regions
(Two-Tailed Test)
Sampling Distribution
Level of Confidence
Rejection
Region
1/2 
Rejection
Region
1–
1/2 
Nonrejection
Region
Ho
Value
Critical
Critical
Value
Value
Observed sample statistic
Sample Statistic
Decision Making Risks
Errors in
Making Decision
1. Type I Error
•
•
•
Reject true null hypothesis
Has serious consequences
Probability of Type I Error is (alpha)
— Called level of significance
2. Type II Error
•
•
Do not reject false null hypothesis
Probability of Type II Error is (beta)
Decision Results
H0: Innocent
Jury Trial
Actual Situation
Verdict
Innocent
Guilty
Innocent
Guilty
H0 Test
Actual Situation
Decision
Correct
Error
Accept
H0
Error
Correct
Reject
H0
H0 True
H0
False
1–
Type II
Error
()
Type I Power
Error () (1 – )
 &  Have an
Inverse Relationship
You can’t reduce both
errors simultaneously!


Factors Affecting 
1. True value of population parameter
•
Increases when difference with hypothesized
parameter decreases
2. Significance level, 
•
Increases when decreases
3. Population standard deviation, 
•
Increases when  increases
4. Sample size, n
•
Increases when n decreases
Hypothesis Testing Steps
H0 Testing Steps
•
State H0
•
Set up critical values
•
State Ha
•
Collect data
•
Choose 
•
Compute test statistic
•
Choose n
•
Make statistical decision
•
Choose test
•
Express decision
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
Two-Tailed Z Test
of Mean ( Known)
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
Two-Tailed Z Test
for Mean ( Known)
1. Assumptions
•
•
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
2. Alternative hypothesis has  sign
3. Z-Test Statistic
Z
X  x
x

X 

n
Two-Tailed Z Test
for Mean Hypotheses
H0:= 0 Ha: ≠ 0
Reject H 0
Reject H
/2
/2
0
Z
Two-Tailed Z Test
Finding Critical Z
What is Z given  = .05?
.500
- .025
.475

=1

 /2 = .025

Standardized Normal
Probability Table (Portion)
Z
.05
.06
.07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
-1.96 0 1.96 Z

1.9 .4744 .4750 .4756
Two-Tailed Z Test Example
Does an average box of cereal
contain 368 grams of cereal?
A random sample of 25 boxes
showed x = 372.5. The
company has specified  to
be 25 grams. Test at the .05
level of significance.
368 gm.
Two-Tailed Z Test Solution
•
•
•
•
•
H0:  = 368
Ha:   368
  .05
n  25
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Test Statistic:
X   372.5  368
Z

 1.50

15
n
25
Decision:
Do not reject at  = .05
Conclusion:
No evidence average
is not 368
Two-Tailed Z Test Thinking
Challenge
You’re a Q/C inspector. You want to find out if
a new machine is making electrical cords to
customer specification: average breaking
strength of 70 lb. with  = 3.5 lb. You take a
sample of 36 cords & compute a sample mean
of 69.7 lb. At the .05 level of significance, is
there evidence that the machine is not meeting
the average breaking strength?
Two-Tailed Z Test Solution*
•
•
•
•
•
H0:  = 70
Ha:   70
 =.05
n = 36
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Test Statistic:
X   69.7  70
Z

 .51

3.5
n
36
Decision:
Do not reject at  = .05
Conclusion:
No evidence average
is not 70
One-Tailed Z Test
of Mean ( Known)
One-Tailed Z Test
for Mean ( Known)
1. Assumptions
•
•
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
2. Alternative hypothesis has < or > sign
3. Z-test Statistic
Z
X  x
x

X 

n
One-Tailed Z Test
for Mean Hypotheses
H0:= 0 Ha: < 0
H0:= 0 Ha: > 0
Reject H 0
Reject H0


0
Must be significantly
below 
Z
0
Z
Small values satisfy H0 .
Don’t reject!
One-Tailed Z Test
Finding Critical Z
What Is Z given  = .025?
.500
- .025
.475

=1

 = .025

Standardized Normal
Probability Table (Portion)
Z
.05
.06
.07
1.6 .4505 .4515 .4525
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
0 1.96 Z

1.9 .4744 .4750 .4756
One-Tailed Z Test
Example
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  to
be 25 grams. Test at the .05
level of significance.
368 gm.
One-Tailed Z Test Solution
•
•
•
•
•
H0:  = 368
Test Statistic:
Ha:  > 368
X   372.5  368
Z

 1.50

15
 = .05
n = 25
n
25
Critical Value(s):
Decision:
Reject
Do not reject at  = .05
.05
0 1.645 Z
Conclusion:
No evidence average is
more than 368
One-Tailed Z Test Thinking
Challenge
You’re an analyst for Ford. You
want to find out if the average miles
per gallon of Escorts is at least 32
mpg. Similar models have a
standard deviation of 3.8 mpg. You
take a sample of 60 Escorts &
compute a sample mean of 30.7
mpg. At the .01 level of
significance, is there evidence that
the miles per gallon is at least 32?
One-Tailed Z Test Solution*
•
•
•
•
•
H0:  = 32
Ha:  < 32
 = .01
n = 60
Critical Value(s):
Reject
.01
-2.33 0
Z
Test Statistic:
X   30.7  32
Z

 2.65

3.8
n
60
Decision:
Reject at  = .01
Conclusion:
There is evidence average
is less than 32
Observed Significance
Levels: p-Values
p-Value
1. Probability of obtaining a test statistic more
extreme (or than actual sample value,
given H0 is true
2. Called observed level of significance
•
Smallest value of  for which H0 can be
rejected
3. Used to make rejection decision
•
•
If p-value  , do not reject H0
If p-value < , reject H0
Two-Tailed Z Test
p-Value Example
Does an average box of cereal
contain 368 grams of cereal? A
random sample of 25 boxes
showed x = 372.5. The
company has specified  to be
25 grams. Find the p-Value.
368 gm.
Two-Tailed Z Test
p-Value Solution
X   372.5  368
Z

 1.50

15
n
25
0
1.50

Z
Z value of sample
statistic (observed)
Two-Tailed Z Test
p-Value Solution
p-value is P(Z  -1.50 or Z  1.50)
1/2 p-Value
1/2 p-Value
.4332
-1.50

0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic (observed)
Two-Tailed Z Test
p-Value Solution
p-value is P(Z  -1.50 or Z  1.50) = .1336
1/2 p-Value
.0668
-1.50

1/2 p-Value
.0668
0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
Two-Tailed Z Test
p-Value Solution
(p-Value = .1336)  ( = .05).
Do not reject H0.
1/2 p-Value = .0668
1/2 p-Value = .0668
Reject H0
Reject H0
1/2  = .025
1/2  = .025
-1.50
0
1.50
Test statistic is in ‘Do not reject’ region
Z
One-Tailed Z Test
p-Value Example
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 
to be 25 grams. Find the pValue.
368 gm.
One-Tailed Z Test
p-Value Solution
X   372.5  368
Z

 1.50

15
n
25
0
1.50

Z
Z value of sample
statistic
One-Tailed Z Test
p-Value Solution
p-Value is P(Z  1.50)

Use
alternative
hypothesis
to find
direction
p-Value
.4332

0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
One-Tailed Z Test
p-Value Solution
p-Value is P(Z  1.50) = .0668

p-Value
.0668
Use
alternative
hypothesis
to find
direction
.4332

0
From Z table:
lookup 1.50
1.50


.5000
- .4332
.0668
Z
Z value of sample
statistic
One-Tailed Z Test
p-Value Solution
(p-Value = .0668)  ( = .05).
Do not reject H0.
p-Value = .0668
Reject H0
 = .05
0
1.50
Test statistic is in ‘Do not reject’ region
Z
p-Value
Thinking Challenge
You’re an analyst for Ford. You
want to find out if the average
miles per gallon of Escorts is at
least 32 mpg. Similar models
have a standard deviation of 3.8
mpg. You take a sample of 60
Escorts & compute a sample mean
of 30.7 mpg. What is the value of
the observed level of significance
(p-Value)?
p-Value
Solution*
p-Value is P(Z  -2.65) = .004.
p-Value < ( = .01). Reject H0.

Use
alternative
hypothesis
to find
direction
p-Value
.004
.5000
- .4960
.0040
.4960
-2.65


Z value of sample
statistic
0

Z
From Z table:
lookup 2.65
Two-Tailed t Test
of Mean ( Unknown)
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
t Test for Mean
( Unknown)
1. Assumptions
• Population is normally distributed
• If not normal, only slightly skewed & large
sample (n  30) taken
2. Parametric test procedure
3. t test statistic
X 
t
S
n
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2

 /2 = .05


 /2 = .05
Critical Values of t Table
(Portion)
v
t .10
t .05
t .025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
-2.920 0 2.920 t
3 1.638 2.353 3.182

Two-Tailed t Test
Example
Does an average box of
cereal contain 368 grams of
cereal? A random sample
of 36 boxes had a mean of
372.5 and a standard
deviation of 12 grams. Test
at the .05 level of
significance.
368 gm.
Two-Tailed t Test
Solution
•
•
•
•
•
H0:  = 368
Ha:   368
 = .05
df = 36 - 1 = 35
Critical Value(s):
Reject H0
Reject H0
.025
.025
-2.030
0 2.030
t
Test Statistic:
X   372.5  368
t

 2.25
S
12
n
36
Decision:
Reject at  = .05
Conclusion:
There is evidence population
average is not 368
Two-Tailed t Test
Thinking Challenge
You work for the FTC. A
manufacturer of detergent claims that
the mean weight of detergent is 3.25
lb. You take a random sample of 64
containers. You calculate the sample
average to be 3.238 lb. with a standard
deviation of .117 lb. At the .01 level
of significance, is the manufacturer
correct?
3.25 lb.
Two-Tailed t Test
Solution*
•
•
•
•
•
H0:  = 3.25
Ha:   3.25
  .01
df  64 - 1 = 63
Critical Value(s):
Reject H 0
Reject H0
.005
.005
-2.656
0 2.656
t
Test Statistic:
X   3.238  3.25
t

 .82
S
.117
n
64
Decision:
Do not reject at  = .01
Conclusion:
There is no evidence
average is not 3.25
One-Tailed t Test
of Mean ( Unknown)
One-Tailed t Test
Example
Is the average capacity of
batteries at least 140 amperehours? A random sample of 20
batteries had a mean of 138.47
and a standard deviation of
2.66. Assume a normal
distribution. Test at the .05 level
of significance.
One-Tailed t Test
Solution
•
•
•
•
•
H0:  = 140
Ha:  < 140
 = .05
df = 20 - 1 = 19
Critical Value(s):
Reject H0
.05
-1.729
0
t
Test Statistic:
X   138.47  140
t

 2.57
S
2.66
n
20
Decision:
Reject at  = .05
Conclusion:
There is evidence population
average is less than 140
One-Tailed t Test
Thinking Challenge
You’re a marketing analyst for WalMart. Wal-Mart had teddy bears on sale
last week. The weekly sales ($ 00) of
bears sold in 10 stores was:
8 11 0 4 7 8 10 5 8 3
At the .05 level of significance, is there
evidence that the average bear sales per
store is more than 5 ($ 00)?
One-Tailed t Test
Solution*
•
•
•
•
•
H0:  = 5
Ha:  > 5
 = .05
df = 10 - 1 = 9
Critical Value(s):
Reject H0
.05
0 1.833
t
Test Statistic:
X   6.4  5
t

 1.31
S
3.373
n
10
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
average is more than 5
Z Test of Proportion
Data Types
Data
Quantitative
Discrete
Continuous
Qualitative
Qualitative Data
1. Qualitative random variables yield responses
that classify
•
e.g., Gender (male, female)
2. Measurement reflects number in category
3. Nominal or ordinal scale
4. Examples
•
•
Do you own savings bonds?
Do you live on-campus or off-campus?
Proportions
1. Involve qualitative variables
2. Fraction or percentage of population in a
category
3. If two qualitative outcomes, binomial
distribution
• Possess or don’t possess characteristic
^
4. Sample Proportion (p)
x number of successes
pˆ  
n
sample size
Sampling Distribution
of Proportion
1. Approximated by
Normal Distribution
npˆ  3 npˆ 1  pˆ 
–
Excludes 0 or n
2. Mean
 P̂  p
Sampling Distribution
^
P(P )
.3
.2
.1
.0
^
P
.0
.2
.4
.6
.8
1.0
3. Standard Error
 pˆ 
p0  (1  p0 )
where p0 = Population Proportion
n
Standardizing Sampling
Distribution of Proportion
Z 
p^   p^
 p^

p^  p0
p0 (1  p0)
n
Sampling
Distribution
 P^
Standardized Normal
Distribution
z = 1
 P^
^
P
 Z= 0
Z
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
One-Sample Z Test
for Proportion
1. Assumptions
• Random sample selected from a binomial
population
• Normal approximation can be used if
npˆ 0  15 and nqˆ0  15
2.
Z-test statistic for proportion
p̂  p0
Z
p0 q0
n
Hypothesized population
proportion
One-Proportion Z Test
Example
The present packaging system
produces 10% defective
cereal boxes. Using a new
system, a random sample of
200 boxes had 11 defects.
Does the new system produce
fewer defects? Test at the .05
level of significance.
One-Proportion Z Test
Solution
•
•
•
•
•
H0: p = .10
Ha: p < .10
 = .05
n = 200
Critical Value(s):
Reject H0
.05
-1.645 0
Z
Test Statistic:
11
 .10
pˆ  p0 200
Z

 2.12
p0 q0
.10  .90
200
n
Decision:
Reject at  = .05
Conclusion:
There is evidence new
system < 10% defective
One-Proportion Z Test
Thinking Challenge
You’re an accounting manager. A
year-end audit showed 4% of
transactions had errors. You implement
new procedures. A random sample of
500 transactions had 25 errors. Has
the proportion of incorrect
transactions changed at the .05 level
of significance?
One-Proportion Z Test
Solution*
•
•
•
•
•
H0: p = .04
Ha: p  .04
 = .05
n = 500
Critical Value(s):
Reject H 0
Reject H 0
.025
.025
-1.96 0 1.96 Z
Test Statistic:
25
 .04
pˆ  p0 500
Z

 1.14
p0 q0
.04  .96
500
n
Decision:
Do not reject at  = .05
Conclusion:
There is evidence
proportion is not 4%
Calculating Type II Error
Probabilities
Power of Test
1. Probability of rejecting false H0
• Correct decision
2. Designated 1 - 
3. Used in determining test adequacy
4. Affected by
•
•
•
True value of population parameter
Significance level 
Standard deviation & sample size n
Finding Power
Step 1

Hypothesis:
H0: 0  368
Ha: 0 < 368
n

15
25

Reject H0
 = .05
Do Not
Draw
Reject H0
0 = 368
X
Finding Power
Steps 2 & 3

Hypothesis:
H0: 0  368
Ha: 0 < 368
n


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’ Situation:
a = 360 (Ha)

Specify

Draw
1-
a = 360

X
X
Finding Power
Step 4

Hypothesis:
H0: 0  368
Ha: 0 < 368
n


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’ Situation:
a = 360 (Ha)

Specify
X

15
 368  1.64
n
25
 363.065

X L  0  Z
Draw

1-
a = 360
363.065
X
Finding Power
Step 5

n
Hypothesis:
H0: 0  368
Ha: 0 < 368


Reject H0
Do Not
Draw
Reject H0
15
25
 = .05
0 = 368
‘True’ Situation:
a = 360 (Ha)


Specify
Z Table

Draw
X

15
 368  1.64
n
25
 363.065
X L  0  Z
 = .154

1- =.846
a = 360
363.065
X
Power Curves
Power
H0:  0
Power H0:  0
Possible True Values for a
Power
Possible True Values for a
H0:  =0
Possible True Values for a
 = 368 in
Example
Chi-Square (c2) Test
of Variance
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
2
(c )
Chi-Square
Test
for Variance
1. Tests one population variance or standard
deviation
2. Assumes population is approximately
normally distributed
3. Null hypothesis is H0: 2 = 02
4. Test statistic
c 
2
(n 1)  S 2
2
0
Sample variance
Hypothesized pop. variance
Chi-Square (c2) Distribution
Population
Select simple random
sample, size n.
Compute s 2


Sampling Distributions
for Different Sample
Sizes
Compute c2 = (n-1)s 2 /2
0
Astronomical number
of c2 values
1
2
3 c2
Finding Critical Value
Example
What is the critical c2 value given:
Ha: 2 > 0.7
Reject
n=3
 =.05?
 = .05
df = n - 1 = 2
c2 Table
(Portion)
0
DF .995
1
...
2 0.010
5.991
c2
Upper Tail Area
…
.95
…
… 0.004
…
… 0.103
…
.05
3.841
5.991
Finding Critical Value
Example
What is the critical c2 value given:
Ha: 2 < 0.7
n=3
What do you do
 =.05?
if the rejection
region is on the
left?
Finding Critical Value
Example
What is the critical c2 value given:
Ha: 2 < 0.7
Upper Tail Area
Reject H0
for Lower Critical
n=3
Value = 1-.05 = .95

=
.05
 =.05?
df = n - 1 = 2
c2 Table
(Portion)
0 .103
DF .995
1
...
2 0.010
c2
Upper Tail Area
…
.95
…
… 0.004
…
… 0.103
…
.05
3.841
5.991
Chi-Square (c2) Test
Example
Is the variation in boxes of
cereal, measured by the
variance, equal to 15
grams? A random sample
of 25 boxes had a standard
deviation of 17.7 grams.
Test at the .05 level of
significance.
2
(c )
Chi-Square
Test
Solution
•
•
•
•
•
H0: 2 = 15
Ha: 2  15
 = .05
df = 25 - 1 = 24
Critical Value(s):
 /2 = .025
0 12.401
39.364
c2
Test Statistic:
c2 
(n  1) S 2
 02
(25  1) 17.7 2

152
= 33.42
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
2 is not 15
Conclusion
1. Distinguished Types of Hypotheses
2. Described Hypothesis Testing Process
3. Explained p-Value Concept
4. Solved Hypothesis Testing Problems Based
on a Single Sample
5. Explained Power of a Test