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Transcript
```Chapter
7
Hypothesis Testing
with One Sample
1 of 101
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
2 of 101
Section 7.1
Introduction to Hypothesis Testing
3 of 101
Section 7.1 Objectives
• State a null hypothesis and an alternative hypothesis
• Identify type I and type II 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
4 of 101
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
5 of 101
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.
6 of 101
Stating a Hypothesis
Null hypothesis
• A statistical hypothesis
that contains a statement
of equality such as ≤, =,
or ≥.
sub-zero” or “H naught.”
Alternative hypothesis
• A statement of strict
inequality such as >, ≠,
or <.
• Must be true if H0 is
false.
sub-a.”
complementary
statements
7 of 101
Stating a Hypothesis
• To write the null and alternative hypotheses, translate
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.
8 of 101
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 school publicizes that the proportion of its students
who are involved in at least one extracurricular activity is
61%.
Solution:
H0: p = 0.61
Equality condition (Claim)
Ha: p ≠ 0.61
Complement of H0
9 of 101
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 car dealership announces that the mean time for an
oil change is less than 15 minutes.
Solution:
H0: μ ≥ 15 minutes
Complement of Ha
Ha: μ < 15 minutes
Inequality condition (Claim)
10 of 101
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 company advertises that the mean life of its furnaces
is more than 18 years
Solution:
H0: μ ≤ 18 years
Complement of Ha
Ha: μ > 18 years
Inequality condition (Claim)
11 of 101
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
 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.
12 of 101
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.
13 of 101
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)
14 of 101
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
0.18
0.20
0.22
0.24
15 of 101
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.
16 of 101
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.
17 of 101
Level of Significance
Level of significance
• Your maximum allowable probability of making a
type I error.
 Denoted by α, 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:
 α = 0.10
α = 0.05
α = 0.01
• P(type II error) = β (beta)
18 of 101
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
s2
Standardized test
statistic
z (Section 7.2 n ≥ 30)
t (Section 7.3 n < 30)
z (Section 7.4)
χ2 (Section 7.5)
19 of 101
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.
20 of 101
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.
21 of 101
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
standardized
test statistic.
z
–3
–2
–1
0
1
2
3
Test
statistic
22 of 101
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
standardized
test statistic.
z
–3
–2
–1
0
1
2
3
Test
statistic
23 of 101
Two-tailed Test
• The alternative hypothesis Ha contains the not-equalto 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
standardized test
statistic.
P is twice the area to
the left of the
negative standardized
test statistic.
z
–3
–2
–1
Test
statistic
0
1
2
Test
statistic
3
24 of 101
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 school publicizes that the proportion of its students
who are involved in at least one extracurricular activity
is 61%.
Solution:
H0: p = 0.61
Ha: p ≠ 0.61
Two-tailed test
25 of 101
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 car dealership announces that the mean time for an
oil change is less than 15 minutes.
Solution:
H0: μ ≥ 15 min
Ha: μ < 15 min
Left-tailed test
P-value
area
-z
0
z
26 of 101
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 company advertises that the mean life of its
furnaces is more than 18 years.
Solution:
H0: μ ≤ 18 yr
Ha: μ > 18 yr
Right-tailed test
P-value
area
z
z
0
27 of 101
Making a Decision
Decision Rule Based on P-value
• Compare the P-value with α.
 If P ≤ α , then reject H0.
 If P > α, then fail to reject H0.
Claim
Decision
Claim is H0
Claim is Ha
Reject H0
There is enough evidence to
reject the claim
There is enough evidence to
support the claim
Fail to reject H0
There is not enough evidence
to reject the claim
There is not enough evidence
to support the claim
28 of 101
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 school publicizes that the proportion
of its students who are involved in at least one
extracurricular activity is 61%.
Solution:
• The claim is represented by H0.
29 of 101
Solution: Interpreting a Decision
• If you reject H0, then you should conclude “there is
enough evidence to reject the school’s claim that the
proportion of students who are involved in at least
one extracurricular activity is 61%.”
• If you fail to reject H0, then you should conclude
“there is not enough evidence to reject the school’s
claim that proportion of students who are involved in
at least one extracurricular activity is 61%.”
30 of 101
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): A car dealership announces that the
mean time for an oil change is less than 15 minutes.
Solution:
• The claim is represented by Ha.
• H0 is “the mean time for an oil change is greater than
or equal to 15 minutes.”
31 of 101
Solution: Interpreting a Decision
• If you reject H0, then you should conclude “there is
enough evidence to support the dealership’s claim
that the mean time for an oil change is less than 15
minutes.”
• If you fail to reject H0, then you should conclude
“there is not enough evidence to support the
dealership’s claim that the mean time for an oil
change is less than 15 minutes.”
32 of 101
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
sketch its graph.
z
0
4. Calculate the test statistic
and its corresponding
standardized test statistic.
z
0
Standardized test
statistic
33 of 101
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.
34 of 101
Section 7.1 Summary
• Stated a null hypothesis and an alternative hypothesis
• Identified type I and type II 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
35 of 101
Section 7.2
Hypothesis Testing for the Mean
(Large Samples)
36 of 101
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
37 of 101
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 α.
1. If P ≤ α, then reject H0.
2. If P > α, then fail to reject H0.
38 of 101
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. α = 0.05?
Solution:
Because 0.0237 < 0.05, you should reject the null
hypothesis.
2. α = 0.01?
Solution:
Because 0.0237 > 0.01, you should fail to reject the
null hypothesis.
39 of 101
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).
40 of 101
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.
41 of 101
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.
42 of 101
Z-Test for a Mean μ
• Can be used when the population is normal and σ 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 σ.
43 of 101
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 α.
3. Determine the standardized test
statistic.
z
4. Find the area that corresponds
to z.
x 
 n
Use Table 4 in
Appendix B.
44 of 101
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).
6. Make a decision to reject or
fail to reject the null hypothesis.
Reject H0 if P-value
is less than or equal
to α. Otherwise, fail
to reject H0.
7. Interpret the decision in the
context of the original claim.
45 of 101
Example: Hypothesis Testing Using Pvalues
In auto racing, a pit crew claims that its mean pit stop
time (for 4 new tires and fuel) is less than 13 seconds. A
random selection of 32 pit stop times has a sample mean
of 12.9 seconds and a standard deviation of 0.19 second.
Is there enough evidence to support the claim at
α = 0.01? Use a P-value.
46 of 101
Solution: Hypothesis Testing Using Pvalues
•
•
•
•
H0: μ ≥ 13 sec
Ha: μ < 13 sec (Claim)
 = 0.01
Test Statistic:
x 
z
 n
12.9  13

0.19 32
 2.98
• P-value
• Decision: 0.0014 < 0.01
Reject H0 .
At the 1% level of significance,
you have sufficient evidence to
support the claim that the mean pit
stop time is less than 13 seconds.
47 of 101
Example: Hypothesis Testing Using Pvalues
The National Institute of Diabetes and Digestive and
Kidney Diseases reports that the average cost of
bariatric (weight loss) surgery is \$22,500. You think this
information is incorrect. You randomly select 30
bariatric surgery patients and find that the average cost
for their surgeries is \$21,545 with a standard deviation
of \$3015. Is there enough evidence to support your
claim at α = 0.05? Use a P-value. (Adapted from National
Institute of Diabetes and Digestive and Kidney Diseases)
48 of 101
Solution: Hypothesis Testing Using Pvalues
• H0: μ = \$22,500
• Ha: μ ≠ 22,500 (Claim)
• α = 0.05
• Test Statistic:
x 
z
 n
21,545  22,500

3015 30
 1.73
• P-value
• Decision: 0.0836 > 0.05
Fail to reject H0 .
At the 5% level of significance,
there is not sufficient evidence to
support the claim that the mean
cost of bariatric surgery is
different from \$22,500.
49 of 101
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.
50 of 101
Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution
1. Specify the level of significance α.
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 α,
b. right-tailed, find the z-score that corresponds to an area
of 1 – α,
c. two-tailed, find the z-score that corresponds to ½α and
1 – ½α.
4. Sketch the standard normal distribution. Draw a vertical
line at each critical value and shade the rejection region(s).
51 of 101
Example: Finding Critical Values
Find the critical value and rejection region for a twotailed test with α = 0.05.
1 – α = 0.95
Solution:
½α = 0.025
½α = 0.025
0 z0 =z01.96
–z0 = z–1.96
0
z
The rejection regions are to the left of –z0 = –1.96
and to the right of z0 = 1.96.
52 of 101
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
53 of 101
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.
3. Determine the critical value(s).
4. Determine the rejection region(s).
In Symbols
State H0 and Ha.
Identify α.
Use Table 4 in
Appendix B.
54 of 101
Using Rejection Regions for a z-Test for a
Mean μ
In Words
5. Find the standardized test
statistic.
6. Make a decision to reject or fail
to reject the null hypothesis.
In Symbols
x 
or if n  30
 n
use   s.
z
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
7. Interpret the decision in the
context of the original claim.
55 of 101
Example: Testing with Rejection Regions
Employees at a construction and mining company claim
that the mean salary of the company’s mechanical
engineers is less than that of the one of its competitors,
which is \$68,000. A random sample of 30 of the
company’s mechanical engineers has a mean salary of
\$66,900 with a standard deviation of \$5500. At
α = 0.05, test the employees’ claim.
56 of 101
Solution: Testing with Rejection Regions
• H0: μ ≥ \$68,000
• Ha: μ < \$68,000 (Claim)
• α = 0.05
• Rejection Region:
z  1.10
• Test Statistic
x   66,900  68,000
z

 n
5500 30
 1.10
• 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
\$68,000.
57 of 101
Example: Testing with Rejection Regions
The U.S. Department of Agriculture claims that the
mean cost of raising a child from birth to age 2 by
husband-wife families in the U.S. is \$13,120. A random
sample of 500 children (age 2) has a mean cost of
\$12,925 with a standard deviation of \$1745. At
α = 0.10, is there enough evidence to reject the claim?
(Adapted from U.S. Department of Agriculture Center for
Nutrition Policy and Promotion)
58 of 101
Solution: Testing with Rejection Regions
• H0: μ = \$13,120 (Claim) • Test Statistic
x   12,925  13,120
• Ha: μ ≠ \$13,120
z

 n
1745 500
• α = 0.10
 2.50
• Rejection Region:
• Decision: Reject H0 .
At the 10% level of significance,
you have enough evidence to
reject the claim that the mean
cost of raising a child from birth
to age 2 by husband-wife families
in the U.S. is \$13,120.
59 of 101
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
60 of 101
Section 7.3
Hypothesis Testing for the Mean
(Small Samples)
61 of 101
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
62 of 101
Finding Critical Values in a t-Distribution
1. Identify the level of significance α.
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, α ” column with a negative
sign,
b. right-tailed, use “One Tail, α ” column with a positive
sign,
c. two-tailed, use “Two Tails, α ” column with a
negative and a positive sign.
63 of 101
Example: Finding Critical Values for t
Find the critical value t0 for a left-tailed test given
α = 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, α” column.
• Because the test is lefttailed, the critical value is
negative.
0.05
t0 = –1.725 0
t
64 of 101
Example: Finding Critical Values for t
Find the critical values –t0 and t0 for a two-tailed test
given α = 0.10 and n = 26.
Solution:
• The degrees of freedom are
d.f. = n – 1 = 26 – 1 = 25.
• Look at α = 0.10 in the
“Two Tail, α” column.
• Because the test is twotailed, one critical value is
negative and one is positive.
65 of 101
t-Test for a Mean μ (n < 30, σ 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, σ 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.
66 of 101
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 α.
3. Identify the degrees of freedom.
d.f. = n – 1.
4. Determine the critical value(s).
Use Table 5 in
Appendix B.
5. Determine the rejection region(s).
67 of 101
Using the t-Test for a Mean μ
(Small Sample)
In Words
6. Find the standardized test
statistic and sketch the
sampling distribution
7. Make a decision to reject or
fail to reject the null
hypothesis.
In Symbols
x 
t
s n
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
68 of 101
Example: Testing μ with a Small Sample
A used car dealer says that the mean price of a 2008
Honda CR-V is at least \$20,500. You suspect this claim
is incorrect and find that a random sample of 14 similar
vehicles has a mean price of \$19,850 and a standard
deviation of \$1084. 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)
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Solution: Testing μ with a Small Sample
•
•
•
•
•
• Test Statistic:
H0: μ ≥ \$20,500 (Claim)
x   19,850  20,500
Ha: μ < \$20,500
t

 2.244
s n
1084 14
α = 0.05
• Decision: Reject H0 .
df = 14 – 1 = 13
Rejection Region:
At the 5% level of
t ≈ –2.244
significance, there is enough
evidence to reject the claim
that the mean price of a 2008
Honda CR-V is at least
\$20,500.
70 of 101
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.
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Solution: Testing μ with a Small Sample
•
•
•
•
•
H0: μ = 6.8 (Claim)
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 5% level of
significance, there is not
enough evidence to reject
the claim that the mean pH
is 6.8.
–1.816
72 of 101
Example: Using P-values with t-Tests
A Department of Motor Vehicles office claims that the
mean wait time is less than 14 minutes. A random
sample of 10 people has a mean wait time of 13 minutes
with a standard deviation of 3.5 minutes. At α = 0.10,
test the office’s claim. Assume the population is
normally distributed.
73 of 101
Solution: Using P-values with t-Tests
• H0: μ ≥ 14 min
• Ha: μ < 14 min (Claim)
TI-83/84 setup:
Calculate:
Draw:
P ≈ 0.1949
• Decision: Since 0.1949 > 0.10, fail to reject H0.
At the 10% level of significance, there is not enough
evidence to support the office’s claim that the mean
wait time is less than 14 minutes.
74 of 101
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
75 of 101
Section 7.4
Hypothesis Testing for Proportions
76 of 101
Section 7.4 Objectives
• Use the z-test to test a population proportion p
77 of 101
z-Test for a Population Proportion
z-Test for a Population Proportion p
• A statistical test for a population proportion p.
• Can be used when a binomial distribution is given
such that np ≥ 5 and nq ≥ 5.
• The test statistic is the sample proportionp̂ .
• The standardized test statistic is z.
z
pˆ   pˆ
 pˆ
pˆ  p

pq n
78 of 101
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.
3. Determine the critical value(s).
In Symbols
State H0 and Ha.
Identify α.
Use Table 4 in
Appendix B.
4. Determine the rejection region(s).
79 of 101
Using a z-Test for a Proportion p
In Words
5. Find the standardized test
statistic and sketch the
sampling distribution.
6. Make a decision to reject or
fail to reject the null
hypothesis.
In Symbols
z
p̂  p
pq n
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
7. Interpret the decision in the
context of the original claim.
80 of 101
Example: Hypothesis Test for
Proportions
A research center claims that less than 50% of U.S.
adults have accessed the Internet over a wireless
network with a laptop computer. In a random sample of
100 adults, 39% say they have accessed the Internet over
a wireless network with a laptop computer. At α = 0.01,
is there enough evidence to support the researcher’s
claim? (Adopted from Pew Research Center)
Solution:
• Verify that np ≥ 5 and nq ≥ 5.
np = 100(0.50) = 50 and nq = 100(0.50) = 50
81 of 101
Solution: Hypothesis Test for Proportions
•
•
•
•
H0: p ≥ 0.5
Ha: p ≠ 0.45
α = 0.01
Rejection Region:
z = –2.2
• Test Statistic
pˆ  p
0.39  0.5
z

pq n
(0.5)(0.5) 100
 2.2
• Decision: Fail to reject H0 .
At the 1% level of significance,
there is not enough evidence to
support the claim that less than
50% of U.S. adults have accessed
the Internet over a wireless
network with a laptop computer.
82 of 101
Example: Hypothesis Test for
Proportions
A research center claims that 25% of college graduates
think a college degree is not worth the cost. You decide
to test this claim and ask a random sample of 200
college graduates whether they think a college degree is
not worth the cost. Of those surveyed, 21% reply yes. At
α = 0.10 is there enough evidence to reject the claim?
Solution:
• Verify that np ≥ 5 and nq ≥ 5.
np = 200(0.25) = 50 and nq = 200 (0.75) = 150
83 of 101
Solution: Hypothesis Test for Proportions
•
•
•
•
H0: p = 0.25 (Claim)
Ha: p ≠ 0.25
α = 0.10
Rejection Region:
z ≈ –1.31
• Test Statistic
pˆ  p
0.21  0.25
z

pq n
(0.25)(0.75) 200
 1.31
• Decision: Fail to reject H0 .
At the 10% level of
significance, there is enough
evidence to reject the claim that
a college degree is not worth
the cost.
84 of 101
Section 7.4 Summary
• Used the z-test to test a population proportion p
85 of 101
Section 7.5
Hypothesis Testing for Variance and
Standard Deviation
86 of 101
Section 7.5 Objectives
• Find critical values for a χ2-test
• Use the χ2-test to test a variance or a standard
deviation
87 of 101
Finding Critical Values for the χ2-Test
1. Specify the level of significance α.
2. Determine the degrees of freedom d.f. = n – 1.
3. The critical values for the χ2-distribution are found in Table 6
in Appendix B. To find the critical value(s) for a
a. right-tailed test, use the value that corresponds to d.f. and
α.
b. left-tailed test, use the value that corresponds to d.f. and
1 – α.
c. two-tailed test, use the values that corresponds to d.f. and
½α, and d.f. and 1 – ½α.
88 of 101
Finding Critical Values for the χ2-Test
Right-tailed
Left-tailed

1–

 02
1
Two-tailed
1

2
1–
 L2
 02
1

2
 R2
89 of 101
Example: Finding Critical Values for χ2
Find the critical χ2-value for a left-tailed test when
n = 11 and α = 0.01.
Solution:
• Degrees of freedom: n – 1 = 11 – 1 = 10 d.f.
• The area to the right of the critical value is
1 – α = 1 – 0.01 = 0.99.
  0.01
χ0 = 2.558
From Table 6, the critical value is  02  2.558.
90 of 101
Example: Finding Critical Values for χ2
Find the critical χ2-value for a two-tailed test when
n = 9 and α = 0.05.
Solution:
• Degrees of freedom: n – 1 = 9 – 1 = 8 d.f.
• The areas to the right of the critical values are
1
  0.025
2
1
1    0.975.
2
From Table 6, the critical values are  L2  2.180 and
 R2  17.535 .
91 of 101
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.
92 of 101
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 α.
3. Determine the degrees of
freedom.
d.f. = n – 1
4. Determine the critical value(s).
Use Table 6 in
Appendix B.
93 of 101
Using the χ2-Test for a Variance or
Standard Deviation
In Words
In Symbols
5. Determine the rejection region(s).
(n  1)s 2
6. Find the standardized test statistic
and sketch the sampling
distribution.
 
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.
2
2
94 of 101
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.
95 of 101
Solution: Hypothesis Test for the
Population Variance
•
•
•
•
•
H0: σ2 ≤ 0.25 (Claim) • Test Statistic:
2
Ha: σ2 > 0.25
(
n

1)
s
(41  1)(0.27)
2
 

2
α = 0.05

0.25
df = 41 – 1 = 40
 43.2
Rejection Region:
• Decision: Fail to Reject H0 .
  0.05
 2  43.2  02  55.758
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 of 101
Example: Hypothesis Test for the
Standard Deviation
A company claims that the standard deviation of the
lengths of time it takes an incoming telephone call to be
transferred to the correct office is less than 1.4 minutes.
A random sample of 25 incoming telephone calls has a
standard deviation of 1.1 minutes. At α = 0.10, is there
enough evidence to support the company’s claim?
Assume the population is normally distributed.
97 of 101
Solution: Hypothesis Test for the
Standard Deviation
•
•
•
•
•
H0: σ ≥ 1.4 min.
• Test Statistic:
Ha: σ < 1.4 min. (Claim)
2
2
(
n

1)
s
(25

1)(1.1)
α = 0.10
2 

2

1.42
df = 25 – 1 = 24
 14.816
Rejection Region:
• Decision: Reject H0 .
At the 10% level of significance, there
is enough evidence to support the
claim that the standard deviation of the
lengths of time it takes an incoming
telephone call to be transferred to the
correct office is less than 1.4 minutes.
98 of 101
Example: Hypothesis Test for the
Population Variance
A sporting goods manufacturer claims that the variance
of the strengths of 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.
99 of 101
Solution: Hypothesis Test for the
Population Variance
•
•
•
•
•
H0: σ2 = 15.9 (Claim) • Test Statistic:
2
Ha: σ2 ≠ 15.9
(n

1)s
(15  1)(21.8)
2
 

2
α = 0.05
15.9

df = 15 – 1 = 14
 19.195
Rejection Region:
• Decision: Fail to Reject H0
1
  0.025
2
1
  0.025
2
2
 L2  5.629   19.195  R  26.119
At the 5% level of significance,
there is not enough evidence to
reject the claim that the variance in
the strengths of the fishing line is
15.9.
100 of 101
Section 7.5 Summary
• Found critical values for a χ2-test
• Used the χ2-test to test a variance or a standard
deviation