Chapter 8 Handout
Introduction to Hypothesis Testing:
One Population Value
Chapter 8 Summary
Hypothesis Testing for One Population Value:
1. Population Mean (m)
a.
b.
 (population standard deviation) is given (known):
 Use z/standard normal/bell shaped distribution
 (pop std dev) is not given but s (sample std dev) is
given
 Use student’s t distribution
2. Population proportion ()
 Use z/standard normal/bell shaped distribution
3. Population variance (2)
 Use 2 (Chi-Square) distribution
PS: population standard deviation = 
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.
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: m  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.
The Alternative Hypothesis, H1
or HA
•
Is the opposite of the null hypothesis
e.g. The average # TV sets in US homes is
less than 3 (H1: m < 3)
•
•
Challenges the Status Quo
Never contains the ‘=‘ sign
•
The Alternative Hypothesis may or may
not be accepted
Identify the Problem
Steps:
 State the Null Hypothesis (H0: m  3)
 State its opposite, the Alternative
Hypothesis (H1: m < 3)
 Hypotheses
are mutually exclusive &
exhaustive
 Sometimes it is easier to form the
alternative hypothesis first.
Hypothesis Testing Process
Assume the
population
mean age is 50.
(Null Hypothesis)
Is X  20  m  50?
Population
The Sample
Mean Is 20
No, not likely!
REJECT
Null Hypothesis
Sample
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
m = 50.
... if in fact this were
the population mean.
20
m = 50
H0
Sample Mean
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
Level of Significance, a and
the Rejection Region
a
H0: m  3
H1: m < 3
H0: m  3
H1: m > 3
Rejection
Regions
0
0
H0: m  3
H1: m  3
0
Critical
Value(s)
a
a/2
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)

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
Correct
Error
Error
Correct Reject
H0
1-a
Type II
Error (b )
Type I
Error
(a )
Power
(1 - b)
a & b Have an
Inverse Relationship
Reduce probability of one error
and the other one goes up.
b
a
Factors Affecting
Type II Error, b
•
True Value of Population Parameter

•
Significance Level a

•
b
Increases When a Decreases
Population Standard Deviation 

•
Increases When Difference Between Hypothesized
Parameter & True Value Decreases

b 
Increases When  Increases
Sample Size n
Increases When n Decreases
a
b
n
3 Methods for Hypotheses Tests
Refer to Figure 8-6 (page 299) for a hypothesis test for
means (m) with  (pop. std. dev.) is given:
Method 1: Comparing Xa (X critical) with X
Method 2: Z test, i.e., comparing Za (Z critical) with Z (or
Z statistics or Z calculated)
Method 3: Comparing a (significance level) with p-value
You can modify those three methods for other cases.
For example, if  is unknown, you must use
student’s t distribution. If you would like to use
Method 2, please compare t a (t critical) with t (or t
statistics or t calculated). Refer to Figure 8-8 (page
303).
You always get:
• Za (Z critical) from Z distribution
• ta (t critical) from student’s t distribution
• .2a (2critical) from 2distribution
You always get:
• Z or Z calculated or Z statistics from sample
(page 299 and Figure 8-6)
• t or t calculated or t statistics from sample
(Figure 8-8, page 299)
• .2 or 2 calculated or 2 statistics from
sample (Figure 8-19, page 322)
Z-Test Statistics ( Known)
• Convert Sample Statistic (e.g., X ) to
Standardized Z Variable
Z
X  mX
X

X m

Test Statistic
n
• Compare to Critical Z Value(s)

If Z test Statistic falls in Critical Region,
Reject H0; Otherwise Do Not Reject H0
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

Hypothesis Testing: Steps
Test the Assumption that the true mean #
of TV sets in US homes is at least 3.
1.
State H0
H0 : m  3
2.
State H1
H1 : m < 3
3.
Choose a
a = .05
4.
Choose n
n = 100
5.
Choose Method: Z Test (Method 2)
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
The true mean # of TV set
is less than 3 in the US
households.
One-Tail Z Test for Mean
( Known)
•
Assumptions
Population Is Normally Distributed
 If Not Normal, use large samples
 Null Hypothesis Has =, , or  Sign Only

•
Z Test Statistic:
z
x  mx
x

xm

n
Rejection Region
H0: m  
H1: m < 0
H0: m  0
H1: m > 0
Reject H0
Reject H 0
a
a
0
Must Be Significantly
Below m = 0
Z
0
Z
Small values don’t contradict H0
Don’t Reject H0!
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  to
be 15 grams. Test at the
a0.05 level.
368 gm.
H0: m  368
H1: m > 368
Finding Critical Values:
One Tail
What Is Z Given a = 0.05?
.50
-.05
.45
Z = 1
a = .05
0 1.645 Z
Critical Value
= 1.645
Standardized Normal
Probability Table (Portion)
Z
.04
.05
.06
1.6 .4495 .4505 .4515
1.7 .4591 .4599 .4608
1.8 .4671 .4678 .4686
1.9 .4738 .4744 .4750
Example Solution: One Tail
H0: m  368
H1: m > 368
Test Statistic:
a = 0.025
n = 25
Critical Value: 1.645
Reject
.05
0 1.645 Z
Z
X m

 1.50
n
Decision:
Do Not Reject Ho at a = .05
Conclusion:
No Evidence True Mean
Is More than 368
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
1.0000
- .9332
.0668
.9332
0 1.50
From Z Table:
Lookup 1.50
Z
Z Value of Sample
Statistic
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
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
 to be 15 grams. Test at the
a0.05 level.
368 gm.
H0: m  368
H1: m  368
Example Solution: Two Tail
H0: m  386
H1: m  386
Test Statistic:
a = 0.05
n = 25
Critical Value: ±1.96
Reject
.025
.025
-1.96
0 1.96
Z
X m
372.5  368
Z

 1.50

15
n
25
Decision:
Do Not Reject Ho at a = .05
Conclusion:
No Evidence that True
Mean Is Not 368
Two tail hypotheses tests =
Confidence Intervals
_
For X = 372.5oz,  = 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  m  378.38
If this interval contains the Hypothesized mean (368), we
do not reject the null hypothesis. It does. Do not reject Ho.
t-Test:  Unknown
Assumptions
Population is normally distributed
 If not normal, only slightly skewed & a large
sample taken

Parametric test procedure
t test statistic
X m
t
S
n
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.
 is not given,
368 gm.
H0: m  368
H1: m > 368
Example Solution: One Tail
H0: m  368
H1: m > 368
Test Statistic:
X  m
372 . 5  368
t 

 1 . 80
S
15
n
36
a = 0.01
n = 36, df = 35
Critical Value: 2.4377
Reject
.01
0 2.4377 Z
Decision:
Do Not Reject Ho at a = .01
Conclusion:
No Evidence that True
Mean Is More than 368
Proportions
• Involves categorical variables
• Fraction or % of population in a category
• If two categorical outcomes, binomial
distribution

Either possesses or doesn’t possess the characteristic
• Sample proportion (p)
number of successes
p X

n
samplesize
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 n = 500 were surveyed with x =
25 responses.
•Solution: Test at the a = .05 significance
level.
Z Test for Proportion:
Solution
H0:   .04
H1:   .04
Test Statistic:
p-
a = .05
Z 
n = 500, x = 25
p = x/n = 25/500 = 0.05
Critical Values:  1.96
Reject
(1 -  )
n
Decision:
Reject
.025
.025
0
.05-.04
=
= 1.14
.04 (1 - .04)
500
Z
Do not reject Ho at a = .05
Conclusion:
We do not have sufficient
evidence to reject the company’s
claim of 4% response rate.