10- 1
Chapter
Ten
McGraw-Hill/Irwin
© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.
10- 2
Chapter Ten
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
ONE
Define a hypothesis and hypothesis testing.
TWO
Describe the five step hypothesis testing procedure.
THREE
Distinguish between a one-tailed and a two-tailed test of
hypothesis.
FOUR
Conduct a test of hypothesis about a population mean.
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Chapter Ten continued
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
FIVE
Conduct a test of hypothesis about a population proportion.
SIX
Define Type I and Type II errors.
SEVEN
Compute the probability of a Type II error.
10- 4
A statement
about the
value of a
population
parameter
developed for
the purpose of
testing.
The mean monthly
income for systems
analysts is $6,325.
What is a
Hypothesis?
Twenty percent of all
customers at Bovine’s
Chop House return for
another meal within a
month. What is a Hypothesis?
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Hypothesis testing
Based on
sample
evidence and
probability
theory
Used to determine
whether the hypothesis is
a reasonable statement
and should not be
rejected, or is
unreasonable and should
be rejected
What is Hypothesis Testing?
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Step 1: State null and alternate hypotheses
Step 2: Select a level of significance
Step 3: Identify the test statistic
Step 4: Formulate a decision rule
Step 5: Take a sample, arrive at a decision
Do not reject null
Reject null and accept alternate
Hypothesis Testing
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Step One: State the null and alternate hypotheses
Null Hypothesis H0
A statement about the
value of a population
parameter
Alternative Hypothesis H1:
A statement that is
accepted if the sample data
provide evidence that the
null hypothesis is false
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Step One: State the null and alternate
hypotheses
Three
possibilities
regarding
means
H 0: m = 0
H1: m = 0
H 0: m < 0
H 1: m > 0
H0 : m > 0
H 1: m < 0
The null
hypothesis
always contains
equality.
3 hypotheses about means
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Type I Error
Level of Significance
Rejecting the null
hypothesis when it
is actually true (a).
The probability of
rejecting the null
hypothesis when it is
actually true; the level of
risk in so doing.
Type II Error
Accepting the null
hypothesis when it
is actually false (b).
Step Two: Select a Level of
Significance.
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Step Two: Select a Level of Significance.
Null
Hypothesis
Ho is true
Ho is false
Researcher
Accepts
Rejects
Ho
Ho
Correct
Type I
error
decision
(a)
Type II
Correct
Error
decision
(b)
Risk
table
10- 11
Test statistic
A value, determined
from sample
information, used to
determine whether
or not to reject the
null hypothesis.
Examples: z, t, F, c2
z Distribution as a
test statistic
X −m
z=
/ n
The z value is based on the
sampling distribution of X,
which is normally
distributed when the sample
is reasonably large (recall
Central Limit Theorem).
Step Three: Select the test statistic.
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Step Four: Formulate the decision rule.
Critical value: The dividing point between the region
where the null hypothesis is rejected and the region
where it is not rejected.
Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
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Decision Rule
Reject the null
hypothesis and
accept the
alternate
hypothesis if
Computed -z < Critical -z
or
Computed z > Critical z
Decision Rule
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Step Five: Make a decision.
Movie
10- 15
One-Tailed Tests of Significance
The
alternate
hypothesis,
H1, states a
direction
H1: The mean
speed of trucks
traveling on I95 in Georgia
is less than 60
miles per hour.
(µ<60)
H1: The mean yearly
commissions earned by
full-time realtors is more
than $35,000. (µ>$35,000)
H1: Less than 20
percent of the
customers pay
cash for their
gasoline purchase.
(p<.20)
One-Tailed Tests of Significance
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Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
One-Tailed Test of Significance
.
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Two-Tailed Tests of Significance
No direction is specified in the alternate hypothesis H1.
H1: The mean
amount spent by
customers at the
Wal-mart in
Georgetown is
not equal to $25.
(µ ne $25).
H1: The mean
price for a
gallon of
gasoline is not
equal to $1.54.
(µ ne $1.54).
Two-Tailed Tests of Significance
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Two-Tailed Tests of Significance
Regions of
Nonrejection
and Rejection
for a TwoTailed Test,
.05 Level of
Significance
Region of
Region of
Do not
rejection
rejection
reject
[Probability=.025]
[Probability =.95]
-1.96
Critical value
0
[Probability=.025]
1.96
Critical value
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Test for the population
mean from a large sample
with population standard
deviation known
X −m
z=
/ n
Testing for the Population Mean: Large
Sample, Population Standard Deviation
Known
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The processors of Fries’ Catsup
indicate on the label that the
bottle contains 16 ounces of
catsup. The standard deviation
of the process is 0.5 ounces. A
sample of 36 bottles from last
hour’s production revealed a
mean weight of 16.12 ounces per
bottle. At the .05 significance
level is the process out of
control? That is, can we
conclude that the mean amount
per bottle is different from 16
ounces?
Example 1
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Step 5
Make a decision and
interpret the results.
Step 4
State the decision rule.
Reject H0 if z > 1.96
or z < -1.96
Step 3
Identify the test statistic. Because
we know the population standard
deviation, the test statistic is z.
Step 1
State the null and the
alternative hypotheses
H0: m = 16
H1: m = 16
Step 2
Select the significance level.
The significance level is .05.
EXAMPLE 1
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Step 5: Make a
decision and
interpret the results.
z=
X −m

=
n
16 .12 − 16 .00
0.5
oComputed
z of 1.44
< Critical z of 1.96,
Do not reject the null
hypothesis.
= 1.44
36
We cannot
conclude the
mean is different
from 16 ounces.
Example 1
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Testing for the
Population Mean:
Large Sample,
Population Standard
Deviation Unknown
Here  is unknown,
so we estimate it
with the sample
standard deviation s.
As long as the
sample size n > 30, z
can be approximated
using
X −m
z=
s/ n
Testing for the Population Mean: Large Sample,
Population Standard Deviation Unknown
10- 24
Roder’s Discount Store
chain issues its own
credit card. Lisa, the
credit manager, wants to
find out if the mean
monthly unpaid balance
is more than $400. The
level of significance is set
at .05. A random check
of 172 unpaid balances
revealed the sample
mean to be $407 and the
sample standard
deviation to be $38.
Should Lisa conclude
that the population
mean is greater than
$400, or is it reasonable
to assume that the
difference of $7 ($407$400) is due to chance?
Example 2
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Step 5
Make a decision
and interpret the
results.
Step 4
H0 is rejected if
z > 1.65
Step 3
Because the sample is large
we can use the z
distribution as the test
statistic.
Step 1
H0: µ < $400
H1: µ > $400
Step 2
The significance
level is .05.
Example 2
Step 5
Make a decision
and interpret the
results.
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z=
oComputed
z of 2.42
> Critical z of 1.65,
Reject H0.
Lisa can conclude that
the mean unpaid balance
is greater than $400.
X −m
s
n
=
$407 − $400
$38
172
= 2.42