Chapter 7
Hypothesis Testing with One
Sample
Larson/Farber 4th ed.
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Chapter Outline
•
•
•
•
•
7.1 Introduction to Hypothesis Testing
7.2 Hypothesis Testing for the Mean (Large Samples)
7.3 Hypothesis Testing for the Mean (Small Samples)
7.4 Hypothesis Testing for Proportions
7.5 Hypothesis Testing for Variance and Standard
Deviation
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Section 7.1
Introduction to Hypothesis Testing
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Section 7.1 Objectives
• State a null hypothesis and an alternative hypothesis
• Identify type I and type I errors and interpret the level
of significance
• Determine whether to use a one-tailed or two-tailed
statistical test and find a p-value
• Make and interpret a decision based on the results of
a statistical test
• Write a claim for a hypothesis test
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Hypothesis Tests
Hypothesis test
• A process that uses sample statistics to test a claim
about the value of a population parameter.
• For example: An automobile manufacturer
advertises that its new hybrid car has a mean mileage
of 50 miles per gallon. To test this claim, a sample
would be taken. If the sample mean differs enough
from the advertised mean, you can decide the
advertisement is wrong.
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Hypothesis Tests
Statistical hypothesis
• A statement, or claim, about a population parameter.
• Need a pair of hypotheses
• one that represents the claim
• the other, its complement
• When one of these hypotheses is false, the other must
be true.
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Stating a Hypothesis
Null hypothesis
• A statistical hypothesis
that contains a statement
of equality such as , =,
or .
• Denoted H0 read “H
subzero” or “H naught.”
Alternative hypothesis
• A statement of
inequality such as >, ,
or <.
• Must be true if H0 is
false.
• Denoted Ha read “H
sub-a.”
complementary
statements
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Stating a Hypothesis
• To write the null and alternative hypotheses, translate
the claim made about the population parameter from
a verbal statement to a mathematical statement.
• Then write its complement.
H0: μ ≤ k
Ha: μ > k
H0: μ ≥ k
Ha: μ < k
H0: μ = k
Ha: μ ≠ k
• Regardless of which pair of hypotheses you use, you
always assume μ = k and examine the sampling
distribution on the basis of this assumption.
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Example: Stating the Null and Alternative
Hypotheses
Write the claim as a mathematical sentence. State the null
and alternative hypotheses and identify which represents
the claim.
1. A university publicizes that the proportion of its
students who graduate in 4 years is 82%.
Solution:
H0: p = 0.82
Equality condition (Claim)
Ha: p ≠ 0.82
Complement of H0
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Example: Stating the Null and Alternative
Hypotheses
Write the claim as a mathematical sentence. State the null
and alternative hypotheses and identify which represents
the claim.
2. A water faucet manufacturer announces that the mean
flow rate of a certain type of faucet is less than 2.5
gallons per minute.
Solution:
H0: μ ≥ 2.5 gallons per minute
Ha: μ < 2.5 gallons per minute
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Complement of Ha
Inequality
(Claim)
condition
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Example: Stating the Null and Alternative
Hypotheses
Write the claim as a mathematical sentence. State the null
and alternative hypotheses and identify which represents
the claim.
3. A cereal company advertises that the mean weight of
the contents of its 20-ounce size cereal boxes is more
than 20 ounces.
Solution:
H0: μ ≤ 20 ounces
Ha: μ > 20 ounces
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Complement of Ha
Inequality
(Claim)
condition
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Types of Errors
• No matter which hypothesis represents the claim,
always begin the hypothesis test assuming that the
equality condition in the null hypothesis is true.
• At the end of the test, one of two decisions will be
made:
 reject the null hypothesis
 fail to reject the null hypothesis
• Because your decision is based on a sample, there is
the possibility of making the wrong decision.
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Types of Errors
Actual Truth of H0
Decision
Do not reject H0
Reject H0
H0 is true
Correct Decision
Type I Error
H0 is false
Type II Error
Correct Decision
• A type I error occurs if the null hypothesis is rejected
when it is true.
• A type II error occurs if the null hypothesis is not
rejected when it is false.
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Example: Identifying Type I and Type II
Errors
The USDA limit for salmonella contamination for
chicken is 20%. A meat inspector reports that the
chicken produced by a company exceeds the USDA
limit. You perform a hypothesis test to determine
whether the meat inspector’s claim is true. When will a
type I or type II error occur? Which is more serious?
(Source: United States Department of Agriculture)
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Solution: Identifying Type I and Type II
Errors
Let p represent the proportion of chicken that is
contaminated.
Hypotheses: H0: p ≤ 0.2
Ha: p > 0.2 (Claim)
Chicken meets
USDA limits.
H0: p ≤ 0.20
Chicken exceeds
USDA limits.
H0: p > 0.20
p
0.16
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0.18
0.20
0.22
0.24
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Solution: Identifying Type I and Type II
Errors
Hypotheses: H0: p ≤ 0.2
Ha: p > 0.2 (Claim)
A type I error is rejecting H0 when it is true.
The actual proportion of contaminated chicken is less
than or equal to 0.2, but you decide to reject H0.
A type II error is failing to reject H0 when it is false.
The actual proportion of contaminated chicken is
greater than 0.2, but you do not reject H0.
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Solution: Identifying Type I and Type II
Errors
Hypotheses: H0: p ≤ 0.2
Ha: p > 0.2 (Claim)
• With a type I error, you might create a health scare
and hurt the sales of chicken producers who were
actually meeting the USDA limits.
• With a type II error, you could be allowing chicken
that exceeded the USDA contamination limit to be
sold to consumers.
• A type II error could result in sickness or even death.
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Level of Significance
Level of significance
• Your maximum allowable probability of making a
type I error.
 Denoted by , the lowercase Greek letter alpha.
• By setting the level of significance at a small value,
you are saying that you want the probability of
rejecting a true null hypothesis to be small.
• Commonly used levels of significance:
  = 0.10
 = 0.05
 = 0.01
• P(type II error) = β (beta)
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Statistical Tests
• After stating the null and alternative hypotheses and
specifying the level of significance, a random sample
is taken from the population and sample statistics are
calculated.
• The statistic that is compared with the parameter in
the null hypothesis is called the test statistic.
Population
parameter
Test statistic
μ
x
p
σ2
p̂
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s2
Standardized test
statistic
z (Section 7.2 n  30)
t (Section 7.3 n < 30)
z (Section 7.4)
χ2 (Section 7.5)
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P-values
P-value (or probability value)
• The probability, if the null hypothesis is true, of
obtaining a sample statistic with a value as extreme or
more extreme than the one determined from the
sample data.
• Depends on the nature of the test.
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Nature of the Test
• Three types of hypothesis tests
 left-tailed test
 right-tailed test
 two-tailed test
• The type of test depends on the region of the
sampling distribution that favors a rejection of H0.
• This region is indicated by the alternative hypothesis.
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Left-tailed Test
• The alternative hypothesis Ha contains the less-than
inequality symbol (<).
H0: μ  k
Ha: μ < k
P is the area to
the left of the
test statistic.
z
-3
-2
-1
0
1
2
3
Test
statistic
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Right-tailed Test
• The alternative hypothesis Ha contains the greaterthan inequality symbol (>).
H0: μ ≤ k
Ha: μ > k
P is the area
to the right
of the test
statistic.
z
-3
-2
-1
0
1
2
3
Test
statistic
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Two-tailed Test
• The alternative hypothesis Ha contains the not equal
inequality symbol (≠). Each tail has an area of ½P.
H0: μ = k
Ha: μ  k
P is twice the
area to the right
of the positive
test statistic.
P is twice the
area to the left of
the negative test
statistic.
z
-3
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-2
-1
Test
statistic
0
1
2
Test
statistic
3
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Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine
whether the hypothesis test is a left-tailed, right-tailed,
or two-tailed test. Sketch a normal sampling distribution
and shade the area for the P-value.
1. A university publicizes that the proportion of its
students who graduate in 4 years is 82%.
Solution:
H0: p = 0.82
½ P-value
½ P-value
Ha: p ≠ 0.82
area
area
Two-tailed test
Larson/Farber 4th ed.
-z
0
z
z
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Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine
whether the hypothesis test is a left-tailed, right-tailed,
or two-tailed test. Sketch a normal sampling distribution
and shade the area for the P-value.
2. A water faucet manufacturer announces that the
mean flow rate of a certain type of faucet is less than
2.5 gallons per minute.
Solution:
P-value
H0: μ ≥ 2.5 gpm
area
μ
<
2.5
gpm
Ha:
Left-tailed test
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-z
0
z
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Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine
whether the hypothesis test is a left-tailed, right-tailed,
or two-tailed test. Sketch a normal sampling distribution
and shade the area for the P-value.
3. A cereal company advertises that the mean weight of
the contents of its 20-ounce size cereal boxes is
more than 20 ounces.
Solution:
P-value
H0: μ ≤ 20 oz
area
Ha: μ > 20 oz
Right-tailed test
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0
z
z
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Making a Decision
Decision Rule Based on P-value
• Compare the P-value with .
 If P  , then reject H0.
 If P > , then fail to reject H0.
Claim
Decision
Claim is H0
There is enough evidence to
There is enough evidence to
support the claim
There is not enough evidence
to reject the claim
There is not enough evidence
to support the claim
Fail to reject H0 reject the claim
Reject H0
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Claim is Ha
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Example: Interpreting a Decision
You perform a hypothesis test for the following claim.
How should you interpret your decision if you reject
H0? If you fail to reject H0?
1. H0 (Claim): A university publicizes that the
proportion of its students who graduate in 4 years is
82%.
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Solution: Interpreting a Decision
• The claim is represented by H0.
• If you reject H0 you should conclude “there is
sufficient evidence to indicate that the university’s
claim is false.”
• If you fail to reject H0, you should conclude “there is
insufficient evidence to indicate that the university’s
claim (of a four-year graduation rate of 82%) is
false.”
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Example: Interpreting a Decision
You perform a hypothesis test for the following claim.
How should you interpret your decision if you reject
H0? If you fail to reject H0?
2. Ha (Claim): Consumer Reports states that the mean
stopping distance (on a dry surface) for a Honda
Civic is less than 136 feet.
Solution:
• The claim is represented by Ha.
• H0 is “the mean stopping distance…is greater than or
equal to 136 feet.”
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Solution: Interpreting a Decision
• If you reject H0 you should conclude “there is enough
evidence to support Consumer Reports’ claim that the
stopping distance for a Honda Civic is less than 136
feet.”
• If you fail to reject H0, you should conclude “there is
not enough evidence to support Consumer Reports’
claim that the stopping distance for a Honda Civic is
less than 136 feet.”
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Steps for Hypothesis Testing
1. State the claim mathematically and verbally. Identify
the null and alternative hypotheses.
H0: ? Ha: ?
2. Specify the level of significance. This sampling distribution
is based on the assumption
α= ?
that H0 is true.
3. Determine the standardized
sampling distribution and
draw its graph.
z
0
4. Calculate the test statistic
and its standardized value.
Add it to your sketch.
z
0
Test statistic
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Steps for Hypothesis Testing
5. Find the P-value.
6. Use the following decision rule.
Is the P-value less
than or equal to the
level of significance?
No
Fail to reject H0.
Yes
Reject H0.
7. Write a statement to interpret the decision in the
context of the original claim.
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Section 7.1 Summary
• Stated a null hypothesis and an alternative hypothesis
• Identified type I and type I errors and interpreted the
level of significance
• Determined whether to use a one-tailed or two-tailed
statistical test and found a p-value
• Made and interpreted a decision based on the results
of a statistical test
• Wrote a claim for a hypothesis test
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Section 7.2
Hypothesis Testing for the Mean
(Large Samples)
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Section 7.2 Objectives
• Find P-values and use them to test a mean μ
• Use P-values for a z-test
• Find critical values and rejection regions in a normal
distribution
• Use rejection regions for a z-test
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Using P-values to Make a Decision
Decision Rule Based on P-value
• To use a P-value to make a conclusion in a hypothesis
test, compare the P-value with .
1. If P  , then reject H0.
2. If P > , then fail to reject H0.
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Example: Interpreting a P-value
The P-value for a hypothesis test is P = 0.0237. What is
your decision if the level of significance is
1. 0.05?
Solution:
Because 0.0237 < 0.05, you should reject the null
hypothesis.
2. 0.01?
Solution:
Because 0.0237 > 0.01, you should fail to reject the
null hypothesis.
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Finding the P-value
After determining the hypothesis test’s standardized test
statistic and the test statistic’s corresponding area, do one
of the following to find the P-value.
a. For a left-tailed test, P = (Area in left tail).
b. For a right-tailed test, P = (Area in right tail).
c. For a two-tailed test, P = 2(Area in tail of test statistic).
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Example: Finding the P-value
Find the P-value for a left-tailed hypothesis test with a
test statistic of z = -2.23. Decide whether to reject H0 if
the level of significance is α = 0.01.
Solution:
For a left-tailed test, P = (Area in left tail)
P = 0.0129
-2.23
0
z
Because 0.0129 > 0.01, you should fail to reject H0
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Example: Finding the P-value
Find the P-value for a two-tailed hypothesis test with a
test statistic of z = 2.14. Decide whether to reject H0 if
the level of significance is α = 0.05.
Solution:
For a two-tailed test, P = 2(Area in tail of test statistic)
1 – 0.9838
= 0.0162
0.9838
0
2.14
P = 2(0.0162)
= 0.0324
z
Because 0.0324 < 0.05, you should reject H0
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Z-Test for a Mean μ
• Can be used when the population is normal and  is
known, or for any population when the sample size n
is at least 30.
• The test statistic is the sample mean x
• The standardized test statistic is z
x 
  standard error  
z
x
 n
n
• When n  30, the sample standard deviation s can be
substituted for .
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Using P-values for a z-Test for Mean μ
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Determine the standardized test
statistic.
z
4. Find the area that corresponds
to z.
Larson/Farber 4th ed.
x 
 n
Use Table 4 in
Appendix B.
44
Using P-values for a z-Test for Mean μ
In Words
In Symbols
5. Find the P-value.
a. For a left-tailed test, P = (Area in left tail).
b. For a right-tailed test, P = (Area in right tail).
c. For a two-tailed test, P = 2(Area in tail of test
statistic).
Reject H0 if P-value
6. Make a decision to reject or
is less than or equal
fail to reject the null hypothesis.
to . Otherwise,
fail to reject H0.
7. Interpret the decision in the
context of the original claim.
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Example: Hypothesis Testing Using Pvalues
In an advertisement, a pizza shop claims that its mean
delivery time is less than 30 minutes. A random
selection of 36 delivery times has a sample mean of
28.5 minutes and a standard deviation of 3.5 minutes. Is
there enough evidence to support the claim at  = 0.01?
Use a P-value.
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Solution: Hypothesis Testing Using Pvalues
•
•
•
•
H0: μ ≥ 30 min
Ha: μ < 30 min
 = 0.01
Test Statistic:
z
x 
 n
28.5  30

3.5 36
 2.57
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• P-value
0.0051
-2.57
0
z
• Decision: 0.0051 < 0.01
Reject H0
At the 1% level of significance,
you have sufficient evidence to
conclude the mean delivery time
is less than 30 minutes.
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Example: Hypothesis Testing Using Pvalues
You think that the average franchise investment
information shown in the graph is incorrect, so you
randomly select 30 franchises and determine the
necessary investment for each. The sample mean
investment is $135,000 with a
standard deviation of $30,000. Is
there enough evidence to support
your claim at  = 0.05? Use a
P-value.
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Solution: Hypothesis Testing Using Pvalues
•
•
•
•
H0: μ = $143,260
Ha: μ ≠ $143,260
 = 0.05
Test Statistic:
z

x 

n
135, 000  143, 260
30, 000
 1.51
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• P-value
P = 2(0.0655)
= 0.1310
0.0655
-1.51
0
z
• Decision: 0.1310 > 0.05
Fail to reject H0
At the 5% level of significance,
there is not sufficient evidence to
conclude the mean franchise
investment is different from
$143,260.
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Rejection Regions and Critical Values
Rejection region (or critical region)
• The range of values for which the null hypothesis is
not probable.
• If a test statistic falls in this region, the null
hypothesis is rejected.
• A critical value z0 separates the rejection region from
the nonrejection region.
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Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution
1. Specify the level of significance .
2. Decide whether the test is left-, right-, or two-tailed.
3. Find the critical value(s) z0. If the hypothesis test is
a. left-tailed, find the z-score that corresponds to an area
of ,
b. right-tailed, find the z-score that corresponds to an area
of 1 – ,
c. two-tailed, find the z-score that corresponds to ½ and
1 – ½.
4. Sketch the standard normal distribution. Draw a vertical
line at each critical value and shade the rejection region(s).
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Example: Finding Critical Values
Find the critical value and rejection region for a twotailed test with  = 0.05.
1 – α = 0.95
Solution:
½α = 0.025
z0
-z0 = -1.96
½α = 0.025
0 z0 =z01.96
z
The rejection regions are to the left of -z0 = -1.96 and
to the right of z0 = 1.96.
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Decision Rule Based on Rejection
Region
To use a rejection region to conduct a hypothesis test,
calculate the standardized test statistic, z. If the
standardized test statistic
1. is in the rejection region, then reject H0.
2. is not in the rejection region, then fail to reject H0.
Fail to reject Ho.
Fail to reject H0.
Reject H0.
z < z0
Reject Ho.
z0
z
0
Fail to reject H0
Left-Tailed Test
Reject H0
z < -z0 z0
0
0
z0
z > z0
z
Right-Tailed Test
Reject H0
z
z0 z > z0
Two-Tailed Test
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Using Rejection Regions for a z-Test for a
Mean μ
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
In Symbols
State H0 and Ha.
Identify .
3. Sketch the sampling distribution.
4. Determine the critical value(s).
Use Table 4 in
Appendix B.
5. Determine the rejection region(s).
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Using Rejection Regions for a z-Test for a
Mean μ
In Words
6. Find the standardized test
statistic.
7. Make a decision to reject or fail
to reject the null hypothesis.
8. Interpret the decision in the
context of the original claim.
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In Symbols
z
x 
or if n  30
 n
use   s.
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
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Example: Testing with Rejection Regions
Employees in a large accounting firm claim that the
mean salary of the firm’s accountants is less than that of
its competitor’s, which is $45,000. A random sample of
30 of the firm’s accountants has a mean salary of
$43,500 with a standard deviation of $5200. At
α = 0.05, test the employees’ claim.
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Solution: Testing with Rejection Regions
•
•
•
•
H0: μ ≥ $45,000
Ha: μ < $45,000
 = 0.05
Rejection Region:
• Test Statistic
x   43,500  45, 000
z

 n
5200 30
0.05
-1.645 0
-1.58
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z
 1.58
• Decision: Fail to reject H0
At the 5% level of significance,
there is not sufficient evidence
to support the employees’ claim
that the mean salary is less than
$45,000.
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Example: Testing with Rejection Regions
The U.S. Department of Agriculture reports that the
mean cost of raising a child from birth to age 2 in a rural
area is $10,460. You believe this value is incorrect, so
you select a random sample of 900 children (age 2) and
find that the mean cost is $10,345 with a standard
deviation of $1540. At α = 0.05, is there enough
evidence to conclude that the mean cost is different
from $10,460? (Adapted from U.S. Department of Agriculture
Center for Nutrition Policy and Promotion)
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Solution: Testing with Rejection Regions
•
•
•
•
H0: μ = $10,460
Ha: μ ≠ $10,460
 = 0.05
Rejection Region:
0.025
-1.96
-2.24
Larson/Farber 4th ed.
• Test Statistic
x   10,345  10, 460
z

 n
1540 900
0.025
0
1.96
z
 2.24
• Decision: Reject H0
At the 5% level of significance,
you have enough evidence to
conclude the mean cost of
raising a child from birth to age
2 in a rural area is significantly
different from $10,460.
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Section 7.2 Summary
• Found P-values and used them to test a mean μ
• Used P-values for a z-test
• Found critical values and rejection regions in a
normal distribution
• Used rejection regions for a z-test
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Section 7.3
Hypothesis Testing for the Mean
(Small Samples)
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Section 7.3 Objectives
• Find critical values in a t-distribution
• Use the t-test to test a mean μ
• Use technology to find P-values and use them with a
t-test to test a mean μ
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Finding Critical Values in a t-Distribution
1. Identify the level of significance .
2. Identify the degrees of freedom d.f. = n – 1.
3. Find the critical value(s) using Table 5 in Appendix B in
the row with n – 1 degrees of freedom. If the hypothesis
test is
a. left-tailed, use “One Tail,  ” column with a negative
sign,
b. right-tailed, use “One Tail,  ” column with a
positive sign,
c. two-tailed, use “Two Tails,  ” column with a
negative and a positive sign.
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Example: Finding Critical Values for t
Find the critical value t0 for a left-tailed test given
 = 0.05 and n = 21.
Solution:
• The degrees of freedom are
d.f. = n – 1 = 21 – 1 = 20.
• Look at α = 0.05 in the
“One Tail, ” column.
• Because the test is lefttailed, the critical value is
negative.
Larson/Farber 4th ed.
0.05
-1.725 0
t
64
Example: Finding Critical Values for t
Find the critical values t0 and -t0 for a two-tailed test
given  = 0.05 and n = 26.
Solution:
• The degrees of freedom are
d.f. = n – 1 = 26 – 1 = 25.
• Look at α = 0.05 in the
“Two Tail, ” column.
• Because the test is twotailed, one critical value is
negative and one is positive.
Larson/Farber 4th ed.
0.025
-2.060 0
0.025
2.060
t
65
t-Test for a Mean μ (n < 30,  Unknown)
t-Test for a Mean
• A statistical test for a population mean.
• The t-test can be used when the population is normal
or nearly normal,  is unknown, and n < 30.
• The test statistic is the sample mean x
• The standardized test statistic is t.
x 
t
s n
• The degrees of freedom are d.f. = n – 1.
Larson/Farber 4th ed.
66
Using the t-Test for a Mean μ
(Small Sample)
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Identify the degrees of freedom
and sketch the sampling
distribution.
d.f. = n – 1.
4. Determine any critical value(s).
Use Table 5 in
Appendix B.
Larson/Farber 4th ed.
67
Using the t-Test for a Mean μ
(Small Sample)
In Words
5. Determine any rejection
region(s).
6. Find the standardized test
statistic.
7. Make a decision to reject or
fail to reject the null
hypothesis.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed.
In Symbols
x 
t
s n
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
68
Example: Testing μ with a Small Sample
A used car dealer says that the mean price of a 2005
Honda Pilot LX is at least $23,900. You suspect this
claim is incorrect and find that a random sample of 14
similar vehicles has a mean price of $23,000 and a
standard deviation of $1113. Is there enough evidence to
reject the dealer’s claim at α = 0.05? Assume the
population is normally distributed. (Adapted from Kelley
Blue Book)
Larson/Farber 4th ed.
69
Solution: Testing μ with a Small Sample
•
•
•
•
•
H0: μ ≥ $23,900
Ha: μ < $23,900
α = 0.05
df = 14 – 1 = 13
Rejection Region:
• Test Statistic:
t
-3.026
Larson/Farber 4th ed.
s
n

23, 000  23,900
1113 14
 3.026
• Decision: Reject H0
0.05
-1.771 0
x 
t
At the 0.05 level of
significance, there is enough
evidence to reject the claim
that the mean price of a 2005
Honda Pilot LX is at least
$23,900
70
Example: Testing μ with a Small Sample
An industrial company claims that the mean pH level of
the water in a nearby river is 6.8. You randomly select
19 water samples and measure the pH of each. The
sample mean and standard deviation are 6.7 and 0.24,
respectively. Is there enough evidence to reject the
company’s claim at α = 0.05? Assume the population is
normally distributed.
Larson/Farber 4th ed.
71
Solution: Testing μ with a Small Sample
•
•
•
•
•
H0: μ = 6.8
Ha: μ ≠ 6.8
α = 0.05
df = 19 – 1 = 18
Rejection Region:
0.025
-2.101 0
0.025
2.101
t
• Test Statistic:
t
x 
s
n

6.7  6.8
0.24
19
 1.816
• Decision: Fail to reject H0
At the 0.05 level of
significance, there is not
enough evidence to reject
the claim that the mean pH
is 6.8.
-1.816
Larson/Farber 4th ed.
72
Example: Using P-values with t-Tests
The American Automobile Association claims that the
mean daily meal cost for a family of four traveling on
vacation in Florida is $118. A random sample of 11 such
families has a mean daily meal cost of $128 with a
standard deviation of $20. Is there enough evidence to
reject the claim at α = 0.10? Assume the population is
normally distributed. (Adapted from American Automobile
Association)
Larson/Farber 4th ed.
73
Solution: Using P-values with t-Tests
• H0: μ = $118
• Ha: μ ≠ $118
TI-83/84set up:
Calculate:
Draw:
• Decision: 0.1664 > 0.10
Fail to reject H0. At the 0.10 level of significance, there
is not enough evidence to reject the claim that the
mean daily meal cost for a family of four traveling on
vacation in Florida is $118.
Larson/Farber 4th ed.
74
Section 7.3 Summary
• Found critical values in a t-distribution
• Used the t-test to test a mean μ
• Used technology to find P-values and used them with
a t-test to test a mean μ
Larson/Farber 4th ed.
75
Section 7.4
Hypothesis Testing for Proportions
Larson/Farber 4th ed.
76
Section 7.4 Objectives
• Use the z-test to test a population proportion p
Larson/Farber 4th ed.
77
z-Test for a Population Proportion
z-Test for a Population Proportion
• A statistical test for a population proportion.
• Can be used when a binomial distribution is given
such that np  5 and nq  5.
• The test statistic is the sample proportion p̂ .
• The standardized test statistic is z.
z
Larson/Farber 4th ed.
pˆ   pˆ
 pˆ
pˆ  p

pq n
78
Using a z-Test for a Proportion p
Verify that np ≥ 5 and nq ≥ 5
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
In Symbols
State H0 and Ha.
Identify .
3. Sketch the sampling distribution.
4. Determine any critical value(s).
Larson/Farber 4th ed.
Use Table 5 in
Appendix B.
79
Using a z-Test for a Proportion p
In Words
In Symbols
5. Determine any rejection
region(s).
6. Find the standardized test
statistic.
7. Make a decision to reject or
fail to reject the null
hypothesis.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed.
p̂  p
z
pq n
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
80
Example: Hypothesis Test for
Proportions
Zogby International claims that 45% of people in the
United States support making cigarettes illegal within
the next 5 to 10 years. You decide to test this claim and
ask a random sample of 200 people in the United States
whether they support making cigarettes illegal within the
next 5 to 10 years. Of the 200 people, 49% support this
law. At α = 0.05 is there enough evidence to reject the
claim?
Solution:
• Verify that np ≥ 5 and nq ≥ 5.
np = 200(0.45) = 90 and nq = 200(0.55) = 110
Larson/Farber 4th ed.
81
Solution: Hypothesis Test for Proportions
•
•
•
•
• Test Statistic
pˆ  p
0.49  0.45
z

pq n
(0.45)(0.55) 200
H0: p = 0.45
Ha: p ≠ 0.45
 = 0.05
Rejection Region:
0.025
-1.96
0.025
0
1.96
1.14
Larson/Farber 4th ed.
z
 1.14
• Decision: Fail to reject H0
At the 5% level of significance,
there is not enough evidence to
reject the claim that 45% of
people in the U.S. support
making cigarettes illegal within
the next 5 to 10 years.
82
Example: Hypothesis Test for
Proportions
The Pew Research Center claims that more than 55% of
U.S. adults regularly watch their local television news.
You decide to test this claim and ask a random sample of
425 adults in the United States whether they regularly
watch their local television news. Of the 425 adults, 255
respond yes. At α = 0.05 is there enough evidence to
support the claim?
Solution:
• Verify that np ≥ 5 and nq ≥ 5.
np = 425(0.55) ≈ 234 and nq = 425 (0.45) ≈ 191
Larson/Farber 4th ed.
83
Solution: Hypothesis Test for Proportions
•
•
•
•
H0: p ≤ 0.55
Ha: p > 0.55
 = 0.05
Rejection Region:
• Test Statistic
pˆ  p
255 425  0.55
z

pq n
(0.55)(0.45) 425
0.05
0 1.645
2.07
Larson/Farber 4th ed.
z
 2.07
• Decision: Reject H0
At the 5% level of significance,
there is enough evidence to
support the claim that more
than 55% of U.S. adults
regularly watch their local
television news.
84
Section 7.4 Summary
• Used the z-test to test a population proportion p
Larson/Farber 4th ed.
85
Section 7.5
Hypothesis Testing for Variance and
Standard Deviation
Larson/Farber 4th ed.
86
Section 7.5 Objectives
• Find critical values for a χ2-test
• Use the χ2-test to test a variance or a standard
deviation
Larson/Farber 4th ed.
87
Finding Critical Values for the χ2-Test
1. Specify the level of significance .
2. Determine the degrees of freedom d.f. = n – 1.
3. The critical values for the χ2-distribution are found in Table 6
of Appendix B. To find the critical value(s) for a
a. right-tailed test, use the value that corresponds to d.f. and
.
b. left-tailed test, use the value that corresponds to d.f. and
1 – .
c. two-tailed test, use the values that corresponds to d.f. and
½ and d.f. and 1 – ½.
Larson/Farber 4th ed.
88
Finding Critical Values for the χ2-Test
Right-tailed
Left-tailed


1–α
1–α
χ2
 02
χ2
 02
Two-tailed
1

2
 L2
Larson/Farber 4th ed.
1

2
1–α

2
R
χ2
89
Example: Finding Critical Values for χ2
Find the critical χ2-value for a left-tailed test when
n = 11 and  = 0.01.
Solution:
• Degrees of freedom: n – 1 = 11 – 1 = 10 d.f.
• The area to the right of the critical value is
1 –  = 1 – 0.01 = 0.99.
0.01
 02 022.558
χ2
From Table 6, the critical value is  02  2.558.
Larson/Farber 4th ed.
90
Example: Finding Critical Values for χ2
Find the critical χ2-value for a two-tailed test when
n = 13 and  = 0.01.
Solution:
• Degrees of freedom: n – 1 = 13 – 1 = 12 d.f.
• The areas to the right of the critical values are
1
  0.005
2
1
1    0.995
2
1
  0.005
2
1
  0.005
2
2
2 2

 L3.074  R R 28.299
2
L
χ2
From Table 6, the critical values are  L2  3.074 and
 R2  28.299
Larson/Farber 4th ed.
91
The Chi-Square Test
χ2-Test for a Variance or Standard Deviation
• A statistical test for a population variance or standard
deviation.
• Can be used when the population is normal.
• The test statistic is s2.
• The standardized test2 statistic
s
 2  (n  1)
2
follows a chi-square distribution with degrees of
freedom d.f. = n – 1.
Larson/Farber 4th ed.
92
Using the χ2-Test for a Variance or
Standard Deviation
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Determine the degrees of
freedom and sketch the sampling
distribution.
d.f. = n – 1
4. Determine any critical value(s).
Use Table 6 in
Appendix B.
Larson/Farber 4th ed.
93
Using the χ2-Test for a Variance or
Standard Deviation
In Words
In Symbols
5. Determine any rejection
region(s).
(n  1)s 2
6. Find the standardized test
statistic.
 
7. Make a decision to reject or fail
to reject the null hypothesis.
If χ2 is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed.
2
2
94
Example: Hypothesis Test for the
Population Variance
A dairy processing company claims that the variance of
the amount of fat in the whole milk processed by the
company is no more than 0.25. You suspect this is
wrong and find that a random sample of 41 milk
containers has a variance of 0.27. At α = 0.05, is there
enough evidence to reject the company’s claim? Assume
the population is normally distributed.
Larson/Farber 4th ed.
95
Solution: Hypothesis Test for the
Population Variance
•
•
•
•
•
H0: σ2 ≤ 0.25
Ha: σ2 > 0.25
α = 0.05
df = 41 – 1 = 40
Rejection Region:
  0.05
χ2
55.758
43.2
Larson/Farber 4th ed.
• Test Statistic:
 
2
(n  1) s 2
2
(41  1)(0.27)

0.25
 43.2
• Decision: Fail to Reject H0
At the 5% level of significance,
there is not enough evidence to
reject the company’s claim that the
variance of the amount of fat in the
whole milk is no more than 0.25.
96
Example: Hypothesis Test for the
Standard Deviation
A restaurant claims that the standard deviation in the
length of serving times is less than 2.9 minutes. A
random sample of 23 serving times has a standard
deviation of 2.1 minutes. At α = 0.10, is there enough
evidence to support the restaurant’s claim? Assume the
population is normally distributed.
Larson/Farber 4th ed.
97
Solution: Hypothesis Test for the
Standard Deviation
•
•
•
•
•
H0: σ ≥ 2.9 min.
Ha: σ < 2.9 min.
α = 0.10
df = 23 – 1 = 22
Rejection Region:
  0.10
χ2
14.042
11.536
Larson/Farber 4th ed.
• Test Statistic:
2 
(n  1) s 2
2
(23  1)(2.1) 2

2.92
 11.536
• Decision: Reject H0
At the 10% level of significance,
there is enough evidence to
support the claim that the standard
deviation for the length of serving
times is less than 2.9 minutes.
98
Example: Hypothesis Test for the
Population Variance
A sporting goods manufacturer claims that the variance
of the strength in a certain fishing line is 15.9. A random
sample of 15 fishing line spools has a variance of 21.8.
At α = 0.05, is there enough evidence to reject the
manufacturer’s claim? Assume the population is
normally distributed.
Larson/Farber 4th ed.
99
Solution: Hypothesis Test for the
Population Variance
•
•
•
•
•
H0: σ2 = 15.9
Ha: σ2 ≠ 15.9
α = 0.05
df = 15 – 1 = 14
Rejection Region:
1
  0.025
2
χ2
5.629
26.119
19.194
Larson/Farber 4th ed.
• Test Statistic:
 
2
(n  1) s 2
2
(15  1)(21.8)

15.9
 19.194
• Decision: Fail to Reject H0
At the 5% level of significance,
there is not enough evidence to
reject the claim that the variance in
the strength of the fishing line is
15.9.
100
Section 7.5 Summary
• Found critical values for a χ2-test
• Used the χ2-test to test a variance or a standard
deviation
Larson/Farber 4th ed.
101