HYPOTHESIS TESTING.ppt
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Oct 09, 2022
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About This Presentation
HYPOTHESIS TESTING
Slide Content
Introduction to
Hypothesis Testing
Hypothesis Testing
Hypothesis Tests
A hypothesis test is a process that uses sample statistics
to test a claim about the value of a population parameter.
A verbal statement, or claim, about a population parameter
is called a statistical hypothesis.
If a manufacturer of rechargeable batteries claims
that the batteries they produce are good for an
average of at least 1,000 charges, a sample would
be taken to test this claim.
To test the average of 1000 hours, a pair of
hypotheses are stated –one that represents the
claim and the other, its complement. When one of
these hypotheses is false, the other must be true.
A hypothesis test is a process that uses sample statistics
to test a claim about the value of a population parameter.
A verbal statement, or claim, about a population parameter
is called a statistical hypothesis.
If a manufacturer of rechargeable batteries claims
that the batteries they produce are good for an
average of at least 1,000 charges, a sample would
be taken to test this claim.
To test the average of 1000 hours, a pair of
hypotheses are stated –one that represents the
claim and the other, its complement. When one of
these hypotheses is false, the other must be true.
Stating a Hypothesis
A null hypothesis H
0is a statistical hypothesis that
contains a statement of equality such as , =, or .
To write the null and alternative hypotheses, translate
the claim made about the population parameter from a
verbal statement to a mathematical statement.
A alternative hypothesis H
ais the complement of the null
hypothesis. It is a statement that must be true if H
0is false
and contains a statement of inequality such as >, , or <.
“H subzero” or “Hnaught”
“H sub-a”
A null hypothesis H
0is a statistical hypothesis that
contains a statement of equality such as , =, or .
To write the null and alternative hypotheses, translate
the claim made about the population parameter from a
verbal statement to a mathematical statement.
A alternative hypothesis H
ais the complement of the null
hypothesis. It is a statement that must be true if H
0is false
and contains a statement of inequality such as >, , or <.
“H subzero” or “Hnaught”
“H sub-a”
Stating a Hypothesis
Example:
Write the claim as a mathematical sentence. State the
null and alternative hypotheses and identify which
represents the claim.
A manufacturer claims that its rechargeable batteries
have an average life of at least 1,000 charges.
H
0:
H
a:
1000
< 1000
Condition of
equality
1000 (Claim)
Complement of
the null
hypothesis
Example:
Write the claim as a mathematical sentence. State the
null and alternative hypotheses and identify which
represents the claim.
A manufacturer claims that its rechargeable batteries
have an average life of at least 1,000 charges.
H
0:
H
a:
1000
< 1000
Condition of
equality
1000 (Claim)
Complement of
the null
hypothesis
Stating a Hypothesis
Example:
Write the claim as a mathematical sentence. State the
null and alternative hypotheses and identify which
represents the claim.
Statesville college claims that 94% of their graduates
find employment within six months of graduation.
H
0:
H
a:
p= 0.94
p0.94
Condition of
equality
p= 0.94 (Claim)
Complement of
the null
hypothesis
Example:
Write the claim as a mathematical sentence. State the
null and alternative hypotheses and identify which
represents the claim.
Statesville college claims that 94% of their graduates
find employment within six months of graduation.
H
0:
H
a:
p= 0.94
p0.94
Condition of
equality
p= 0.94 (Claim)
Complement of
the null
hypothesis
Types of Errors
No matter which hypothesis represents the claim, always
begin the hypothesis test assuming that the null
hypothesis is true.
At the end of the test, one of two decisions will be made:
1. reject the null hypothesis, or
2. fail to reject the null hypothesis.
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.
No matter which hypothesis represents the claim, always
begin the hypothesis test assuming that the null
hypothesis is true.
At the end of the test, one of two decisions will be made:
1. reject the null hypothesis, or
2. fail to reject the null hypothesis.
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.
Types of Errors
Actual Truth of H
0
H
0 is true H
0 is false
Do not reject H
0
Reject H
0
Correct Decisio
n
Correct Decisio
n
Type II Error
Type I Error
Decision
Actual Truth of H
0
H
0 is true H
0 is false
Do not reject H
0
Reject H
0
Correct Decisio
n
Correct Decisio
n
Type II Error
Type I Error
Decision
Types of Errors
Example:
Statesville college claims that 94% of their graduates find
employment within six months of graduation. What will a
type I or type II error be?
H
0:
H
a:
p0.94
p= 0.94 (Claim)
A type I error is rejecting the null when it is true.
The population proportion is actually 0.94, but is rejected.
(We believe it is not 0.94.)
A type II error is failing to reject the null when it is false.
The population proportion is not 0.94, but is not rejected.
(We believe it is 0.94.)
Example:
Statesville college claims that 94% of their graduates find
employment within six months of graduation. What will a
type I or type II error be?
H
0:
H
a:
p0.94
p= 0.94 (Claim)
A type I error is rejecting the null when it is true.
The population proportion is actually 0.94, but is rejected.
(We believe it is not 0.94.)
A type II error is failing to reject the null when it is false.
The population proportion is not 0.94, but is not rejected.
(We believe it is 0.94.)
Level of Significance
In a hypothesis test, the level of significance is your
maximum allowable probability of making a type I error.
It is denoted by , the lowercase Greek letter alpha.
The probability of making a type II error is denoted by ,
the lowercase Greek letter beta.
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
Hypothesis tests
are based on .
In a hypothesis test, the level of significance is your
maximum allowable probability of making a type I error.
It is denoted by , the lowercase Greek letter alpha.
The probability of making a type II error is denoted by ,
the lowercase Greek letter beta.
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
Hypothesis tests
are based on .
Statistical Tests
The statistic that is compared with the parameter in
the null hypothesis is called the test statistic.
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.
X
2
s
2
2
zp
t(n< 30)
z (n30)μ
Standardized test
statistic
Test
statistic
Population
parameterˆp x
The statistic that is compared with the parameter in
the null hypothesis is called the test statistic.
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.
X
2
s
2
2
zp
t(n< 30)
z (n30)μ
Standardized test
statistic
Test
statistic
Population
parameterˆp x
P-values
If the null hypothesis is true, a P-value (or probability
value) of a hypothesis test is the probability of obtaining
a sample statistic with a value as extreme or more
extreme than the one determined from the sample data.
The P-value of a hypothesis test depends on the nature of
the test.
There are three types of hypothesis tests –a left-, right-,
or two-tailed test. The type of test depends on the
region of the sampling distribution that favors a rejection
of H
0. This region is indicated by the alternative
hypothesis.
If the null hypothesis is true, a P-value (or probability
value) of a hypothesis test is the probability of obtaining
a sample statistic with a value as extreme or more
extreme than the one determined from the sample data.
The P-value of a hypothesis test depends on the nature of
the test.
There are three types of hypothesis tests –a left-, right-,
or two-tailed test. The type of test depends on the
region of the sampling distribution that favors a rejection
of H
0. This region is indicated by the alternative
hypothesis.
Left-tailed Test
1. If the alternative hypothesis contains the less-than
inequality symbol (<), the hypothesis test is a left-tailed
test.
z
0 1 2 3-3-2-1
Test
statistic
H
0: μk
H
a: μ<k
Pis the area to
the left of the
test statistic.
1. If the alternative hypothesis contains the less-than
inequality symbol (<), the hypothesis test is a left-tailed
test.
z
0 1 2 3-3-2-1
Test
statistic
H
0: μk
H
a: μ<k
Pis the area to
the left of the
test statistic.
Right-tailed Test
2. If the alternative hypothesis contains the greater-than
symbol (>), the hypothesis test is a right-tailed test.
z
0 1 2 3-3-2-1
Test
statistic
H
0: μk
H
a: μ>k
P is the area to
the right of the
test statistic.
2. If the alternative hypothesis contains the greater-than
symbol (>), the hypothesis test is a right-tailed test.
z
0 1 2 3-3-2-1
Test
statistic
H
0: μk
H
a: μ>k
P is the area to
the right of the
test statistic.
Two-tailed Test2
1
3. If the alternative hypothesis contains the not-equal-to
symbol (), the hypothesis test is a two-tailed test. In a
two-tailed test, each tail has an area of P.
z
0 1 2 3-3-2-1
Test
statistic
Test
statistic
H
0: μ= k
H
a: μk
Pis twice the
area to the left
of the negative
test statistic.
P is twice the
area to the right
of the positive
test statistic.
1
3. If the alternative hypothesis contains the not-equal-to
symbol (), the hypothesis test is a two-tailed test. In a
two-tailed test, each tail has an area of P.
z
0 1 2 3-3-2-1
Test
statistic
Test
statistic
H
0: μ= k
H
a: μk
Pis twice the
area to the left
of the negative
test statistic.
P is twice the
area to the right
of the positive
test statistic.
Identifying Types of Tests
Example:
For each claim, state H
0 and H
a. Then determine whether
the hypothesis test is a left-tailed, right-tailed, or two-tailed
test.
a.) A cigarette manufacturer claims that less than one-
eighth of the US adult population smokes cigarettes.
b.) A local telephone company claims that the average
length of a phone call is 8 minutes.
H
a:p< 0.125 (Claim)
H
0:p0.125
Left-tailed test
H
a:μ8
H
0:μ= 8 (Claim)
Two-tailed test
Example:
For each claim, state H
0 and H
a. Then determine whether
the hypothesis test is a left-tailed, right-tailed, or two-tailed
test.
a.) A cigarette manufacturer claims that less than one-
eighth of the US adult population smokes cigarettes.
b.) A local telephone company claims that the average
length of a phone call is 8 minutes.
H
a:p< 0.125 (Claim)
H
0:p0.125
Left-tailed test
H
a:μ8
H
0:μ= 8 (Claim)
Two-tailed test
Making 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 H
0.
2.If P> , then fail to reject H
0.
Claim
Claim is H
0 Claim is H
a
Do not reject H
0
Reject H
0
There is enough evidence to
reject the claim.
Decision
There is not enough evidenc
e to reject the claim.
There is enough evidence to
support the claim.
There is not enough evidenc
e to support the claim.
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 H
0.
2.If P> , then fail to reject H
0.
Claim
Claim is H
0 Claim is H
a
Do not reject H
0
Reject H
0
There is enough evidence to
reject the claim.
Decision
There is not enough evidenc
e to reject the claim.
There is enough evidence to
support the claim.
There is not enough evidenc
e to support the claim.
Interpreting a Decision
Example:
You perform a hypothesis test for the following claim. How
should you interpret your decision if you reject H
0? If you
fail to reject H
0?
H
0:(Claim) A cigarette manufacturer claims that less
than one-eighth of the US adult population smokes
cigarettes.
If H
0is rejected, you should conclude “there is sufficient
evidence to indicate that the manufacturer’s claim is
false.”
If you fail to reject H
0, you should conclude “there is not
sufficient evidence to indicate that the manufacturer’s
claim is false.”
Example:
You perform a hypothesis test for the following claim. How
should you interpret your decision if you reject H
0? If you
fail to reject H
0?
H
0:(Claim) A cigarette manufacturer claims that less
than one-eighth of the US adult population smokes
cigarettes.
If H
0is rejected, you should conclude “there is sufficient
evidence to indicate that the manufacturer’s claim is
false.”
If you fail to reject H
0, you should conclude “there is not
sufficient evidence to indicate that the manufacturer’s
claim is false.”
Steps for Hypothesis Testing
1.State the claim mathematically and verbally. Identify the
null and alternative hypotheses.
2.Specify the level of significance.
3.Determine the standardized
sampling distribution and
draw its graph.
H
0: ?H
a: ?
= ?
4.Calculate the test statistic
and its standardized value.
Add it to your sketch.
z
0
Test statistic
This sampling distribution
is based on the
assumption that H
0is true.
z
0
Continued.
1.State the claim mathematically and verbally. Identify the
null and alternative hypotheses.
2.Specify the level of significance.
3.Determine the standardized
sampling distribution and
draw its graph.
H
0: ?H
a: ?
= ?
4.Calculate the test statistic
and its standardized value.
Add it to your sketch.
z
0
Test statistic
This sampling distribution
is based on the
assumption that H
0is true.
z
0
Continued.
Steps for Hypothesis Testing
5.Find the P-value.
6.Use the following decision rule.
7.Write a statement to interpret the decision in the context of
the original claim.
Is the P-value less than
or equal to the level of
significance?
Fail to reject H
0.
Yes
Reject H
0.
No
These steps apply to left-tailed, right-tailed, and
two-tailed tests.
5.Find the P-value.
6.Use the following decision rule.
7.Write a statement to interpret the decision in the context of
the original claim.
Is the P-value less than
or equal to the level of
significance?
Fail to reject H
0.
Yes
Reject H
0.
No
These steps apply to left-tailed, right-tailed, and
two-tailed tests.
Hypothesis Testing for
the Mean(Large Samples)
the Mean(Large Samples)
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 H
0.
2.If P> , then fail to reject H
0.
Recall that when the sample size is at least 30, the
sampling distribution for the sample mean is normal.
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 H
0.
2.If P> , then fail to reject H
0.
Recall that when the sample size is at least 30, the
sampling distribution for the sample mean is normal.
Using P-values to Make a Decision
Example:
The P-value for a hypothesis test is P= 0.0256. What
is your decision if the level of significance is
a.) 0.05,
b.) 0.01?
a.) Because 0.0256 is < 0.05, you should reject the null
hypothesis.
b.) Because 0.0256 is > 0.01, you should fail to reject
the null hypothesis.
Example:
The P-value for a hypothesis test is P= 0.0256. What
is your decision if the level of significance is
a.) 0.05,
b.) 0.01?
a.) Because 0.0256 is < 0.05, you should reject the null
hypothesis.
b.) Because 0.0256 is > 0.01, you should fail to reject
the null hypothesis.
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).
Example:
The test statistic for a right-tailed test is z= 1.56. Find the P-value.
z
01.56
The area to the right of z = 1.
56 is 1 –.9406 = 0.0594.
P-value = 0.0594
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).
Example:
The test statistic for a right-tailed test is z= 1.56. Find the P-value.
z
01.56
The area to the right of z = 1.
56 is 1 –.9406 = 0.0594.
P-value = 0.0594
Finding the P-value
Example:
The test statistic for a two-tailed test is z= 2.63.
Find the P-value.
The area to the left of z= 2.63 is 0.0043.
The P-value is 2(0.0043) = 0.0086
z
02.63
0.0043
Example:
The test statistic for a two-tailed test is z= 2.63.
Find the P-value.
The area to the left of z= 2.63 is 0.0043.
The P-value is 2(0.0043) = 0.0086
z
02.63
0.0043
The z-test for the meanis a statistical test for a
population mean. The z-test can be used when the
population is normal and is known, or for any
population when the sample size nis at least 30.
Using P-values for a z-Testxμ
z
σn
standard error
x
σ
σ
n
When n30, the sample standard deviation scan be
substituted for .
The teststatisticis the sample mean and the
standardizedteststatisticis z.x
population mean. The z-test can be used when the
population is normal and is known, or for any
population when the sample size nis at least 30.
Using P-values for a z-Testxμ
z
σn
standard error
x
σ
σ
n
When n30, the sample standard deviation scan be
substituted for .
The teststatisticis the sample mean and the
standardizedteststatisticis z.x
Using P-values for a z-Testxμ
z
σn
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Determine the standardized test
statistic.
4.Find the area that corresponds
to z.
Continued.
Using P-values for a z-Test for a Mean μ
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
z
σn
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Determine the standardized test
statistic.
4.Find the area that corresponds
to z.
Continued.
Using P-values for a z-Test for a Mean μ
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
Using P-values for a z-Test
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).
Using P-values for a z-Test for a Mean μ
In Words In Symbols
Reject H
0if P-value
is less than or
equal to .
Otherwise, fail to
reject H
0.
5.Find the P-value.
6.Make a decision to reject or fail to
reject the null hypothesis.
7.Interpret the decision in the
context of the original claim.
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).
Using P-values for a z-Test for a Mean μ
In Words In Symbols
Reject H
0if P-value
is less than or
equal to .
Otherwise, fail to
reject H
0.
5.Find the P-value.
6.Make a decision to reject or fail to
reject the null hypothesis.
7.Interpret the decision in the
context of the original claim.
Hypothesis Testing with P-values
Example:
A manufacturer claims that its rechargeable batteries
are good for an average of more than 1,000 charges. A
random sample of 100 batteries has a mean life of
1002 charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?
H
a: > 1000 (Claim)H
0:1000
The level of significance is = 0.01.
The standardized test statistic is1002 1000
14 100
1.43 xμ
z
σn
Continued.
Example:
A manufacturer claims that its rechargeable batteries
are good for an average of more than 1,000 charges. A
random sample of 100 batteries has a mean life of
1002 charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?
H
a: > 1000 (Claim)H
0:1000
The level of significance is = 0.01.
The standardized test statistic is1002 1000
14 100
1.43 xμ
z
σn
Continued.
Hypothesis Testing with P-values
Example continued:
A manufacturer claims that its rechargeable batteries
are good for an average of more than 1,000 charges. A
random sample of 100 batteries has a mean life of
1002 charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?1.43z
z
01.43
The area to the right of
z= 1.43 is P= 0.0764.
At the 1% level of significance, there is not enough evidence to
support the claim that the rechargeable battery has an average
life of at least 1000 charges.
P-value is greater than
= 0.01, fail to reject
H
0.
H
a: > 1000 (Claim)H
0:1000
Example continued:
A manufacturer claims that its rechargeable batteries
are good for an average of more than 1,000 charges. A
random sample of 100 batteries has a mean life of
1002 charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?1.43z
z
01.43
The area to the right of
z= 1.43 is P= 0.0764.
At the 1% level of significance, there is not enough evidence to
support the claim that the rechargeable battery has an average
life of at least 1000 charges.
P-value is greater than
= 0.01, fail to reject
H
0.
H
a: > 1000 (Claim)H
0:1000
A rejection region(or critical region) of the sampling
distribution is 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 z
0
separates the rejection region from the nonrejection
region.
Rejection Regions and Critical Values
Example:
Find the critical value and rejection region for a right tailed test
with = 0.01.
z
02.575
The rejection region is to t
he right of z
0= 2.575.
= 0.01
distribution is 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 z
0
separates the rejection region from the nonrejection
region.
Rejection Regions and Critical Values
Example:
Find the critical value and rejection region for a right tailed test
with = 0.01.
z
02.575
The rejection region is to t
he right of z
0= 2.575.
= 0.01
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) z
0. 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 –,
4.Sketch the standard normal distribution. Draw a
vertical line at each critical value and shade the
rejection region(s).2
1
c.two-tailed, find the z-score that corresponds to
and 1 –.2
1
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) z
0. 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 –,
4.Sketch the standard normal distribution. Draw a
vertical line at each critical value and shade the
rejection region(s).2
1
c.two-tailed, find the z-score that corresponds to
and 1 –.2
1
Rejection Regions for a z-Test
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 H
0.
2.is notin the rejection region, then fail to reject H
0.
z
0z
0
Fail to reject H
o.
Reject H
o.
Left-Tailed Test
z < z
0
z
0z
0
Reject H
o.
Fail to reject H
o.
z > z
0
Right-Tailed Test
z
0z
0
Two-Tailed Test
z
0z < z
0
z > z
0
Reject H
o.
Fail to reject H
o.
Reject H
o.
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 H
0.
2.is notin the rejection region, then fail to reject H
0.
z
0z
0
Fail to reject H
o.
Reject H
o.
Left-Tailed Test
z < z
0
z
0z
0
Reject H
o.
Fail to reject H
o.
z > z
0
Right-Tailed Test
z
0z
0
Two-Tailed Test
z
0z < z
0
z > z
0
Reject H
o.
Fail to reject H
o.
Reject H
o.
Rejection Regions for a z-Test
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Sketch the sampling distribution.
4.Determine the critical value(s).
5.Determine the rejection
regions(s).
Continued.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Sketch the sampling distribution.
4.Determine the critical value(s).
5.Determine the rejection
regions(s).
Continued.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
Rejection Regions for a z-Test
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.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols or if 30
use
xμ
zn
σ
n
σ s.
If zis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
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.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols or if 30
use
xμ
zn
σ
n
σ s.
If zis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
Testing with Rejection Regions
Example:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 58 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05?
H
a: 8H
0:= 8 (Claim)
The level of significance is =
0.05.
Continued.
z
0 = 1.96
z
0
z
0 = 1.96
0.025 0.025
Example:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 58 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05?
H
a: 8H
0:= 8 (Claim)
The level of significance is =
0.05.
Continued.
z
0 = 1.96
z
0
z
0 = 1.96
0.025 0.025
Testing with Rejection Regions
Example continued:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 58 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05? H
a: 8H
0:= 8 (Claim)
The standardized test statistic is
z
0 = 1.96
z
0
z
0 = 1.967.8 8
0.5 58
xμ
z
σn
3.05.
The test statistic falls
in the rejection region,
so H
0is rejected.
At the 5% level of significance, there is enough evidence to
reject the claim that the average length of a phone call is 8
minutes.
Example continued:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 58 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05? H
a: 8H
0:= 8 (Claim)
The standardized test statistic is
z
0 = 1.96
z
0
z
0 = 1.967.8 8
0.5 58
xμ
z
σn
3.05.
The test statistic falls
in the rejection region,
so H
0is rejected.
At the 5% level of significance, there is enough evidence to
reject the claim that the average length of a phone call is 8
minutes.
Hypothesis Testing for
the Mean(Small Samples)
the Mean(Small Samples)
Critical Values in a t-Distribution
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.
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.
Finding Critical Values for t
Example:
Find the critical value t
0for a right-tailed test given = 0.01
and n= 24.
The degrees of freedom are d.f. = n–1 = 24 –1 = 23.
To find the critical value, use Table 5 with d.f. = 23 and 0.01
in the “One Tail, “ column. Because the test is a right-tail
test, the critical value is positive.
t
0= 2.500
Example:
Find the critical value t
0for a right-tailed test given = 0.01
and n= 24.
The degrees of freedom are d.f. = n–1 = 24 –1 = 23.
To find the critical value, use Table 5 with d.f. = 23 and 0.01
in the “One Tail, “ column. Because the test is a right-tail
test, the critical value is positive.
t
0= 2.500
Finding Critical Values for t
Example:
Find the critical values t
0and t
0for a two-tailed test
given = 0.10 and n= 12.
The degrees of freedom are d.f. = n–1 = 12 –1 = 11.
To find the critical value, use Table 5 with d.f. = 11 and 0.10
in the “Two Tail, “ column. Because the test is a two-tail
test, one critical value is negative and one is positive.
t
0= 1.796 and t
0= 1.796
Example:
Find the critical values t
0and t
0for a two-tailed test
given = 0.10 and n= 12.
The degrees of freedom are d.f. = n–1 = 12 –1 = 11.
To find the critical value, use Table 5 with d.f. = 11 and 0.10
in the “Two Tail, “ column. Because the test is a two-tail
test, one critical value is negative and one is positive.
t
0= 1.796 and t
0= 1.796
t-Test for a Mean μ(n< 30, Unknown)
The t-test for the meanis 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. xμ
t
sn
The degrees of freedom are d.f. = n–1 .
The teststatisticis the sample mean and the
standardizedteststatisticis t.x
The t-test for the meanis 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. xμ
t
sn
The degrees of freedom are d.f. = n–1 .
The teststatisticis the sample mean and the
standardizedteststatisticis t.x
t-Test for a Mean μ(n< 30, Unknown)
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Identify the degrees of freedom
and sketch the sampling
distribution.
4.Determine any critical values.
5.Determine any rejection region(s).
Continued.
Using the t-Test for a Mean μ(Small Sample)
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 5 in
Appendix B.
d.f. = n–1.
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Identify the degrees of freedom
and sketch the sampling
distribution.
4.Determine any critical values.
5.Determine any rejection region(s).
Continued.
Using the t-Test for a Mean μ(Small Sample)
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 5 in
Appendix B.
d.f. = n–1.
t-Test for a Mean μ(n< 30, Unknown)
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.
Using the t-Test for a Mean μ(Small Sample)
In Words In Symbols
xμ
t
s
n
If tis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
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.
Using the t-Test for a Mean μ(Small Sample)
In Words In Symbols
xμ
t
s
n
If tis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
Testing μUsing Critical Values
Example:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 18 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05?
H
a: 8H
0:= 8 (Claim)
The level of significance is =
0.05.
Continued.
The test is a two-tailed test.
Degrees of freedom are d.f. = 18 –1 = 17.
The critical values are t
0= 2.110 and t
0= 2.110
Example:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 18 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05?
H
a: 8H
0:= 8 (Claim)
The level of significance is =
0.05.
Continued.
The test is a two-tailed test.
Degrees of freedom are d.f. = 18 –1 = 17.
The critical values are t
0= 2.110 and t
0= 2.110
Testing μUsing Critical Values
Example continued:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 18 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05? H
a: 8H
0:= 8 (Claim)
The standardized test statistic is
z
0 = 2.110
z
0
z
0 = 2.1107.8 8
0.5 18
xμ
t
sn
1.70.
The test statistic falls in
the nonrejection region,
so H
0is not rejected.
At the 5% level of significance, there is not enough evidence to reject
the claim that the average length of a phone call is 8 minutes.
Example continued:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random
sample of 18 phone calls, the sample mean was 7.8
minutes and the standard deviation was 0.5 minutes.
Is there enough evidence to support this claim at =
0.05? H
a: 8H
0:= 8 (Claim)
The standardized test statistic is
z
0 = 2.110
z
0
z
0 = 2.1107.8 8
0.5 18
xμ
t
sn
1.70.
The test statistic falls in
the nonrejection region,
so H
0is not rejected.
At the 5% level of significance, there is not enough evidence to reject
the claim that the average length of a phone call is 8 minutes.
Testing μUsing P-values
Example:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?
H
a: > 1000 (Claim)H
0:1000
The level of significance is =
0.01.
The standardized test statistic is1002 1000
14 10
0.45 xμ
t
sn
Continued.
The degrees of freedom are d.f. = n–1 = 10 –1 = 9.
Example:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?
H
a: > 1000 (Claim)H
0:1000
The level of significance is =
0.01.
The standardized test statistic is1002 1000
14 10
0.45 xμ
t
sn
Continued.
The degrees of freedom are d.f. = n–1 = 10 –1 = 9.
Testing μUsing P-values
Example continued:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?0.45t
z
00.45
Using the d.f. = 9 row from Table 5, you can
determine that Pis greater than = 0.25 and is
therefore also greater than the 0.01
significance level. H
0would fail to be rejected.
At the 1% level of significance, there is not enough evidence to
support the claim that the rechargeable battery has an average
life of at least 1000 charges.
H
a: > 1000 (Claim)H
0:1000
Example continued:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there
enough evidence to support this claim at = 0.01?0.45t
z
00.45
Using the d.f. = 9 row from Table 5, you can
determine that Pis greater than = 0.25 and is
therefore also greater than the 0.01
significance level. H
0would fail to be rejected.
At the 1% level of significance, there is not enough evidence to
support the claim that the rechargeable battery has an average
life of at least 1000 charges.
H
a: > 1000 (Claim)H
0:1000
Hypothesis Testing for
Proportions
Proportions
z-Test for a Population Proportion
The z-test for a populationis a statistical test for a
population proportion. The z-test can be used when a
binomial distribution is given such that np5 and nq5.ˆ
ˆ
ˆ ˆp
p
pμpp
z
σ pq n
The teststatisticis the sample proportion and the
standardizedteststatisticis z.ˆp
The z-test for a populationis a statistical test for a
population proportion. The z-test can be used when a
binomial distribution is given such that np5 and nq5.ˆ
ˆ
ˆ ˆp
p
pμpp
z
σ pq n
The teststatisticis the sample proportion and the
standardizedteststatisticis z.ˆp
Hypothesis Test for Proportions
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Sketch the sampling distribution.
4.Determine any critical values.
Continued.
Using a z-Test for a Proportion p
Verify that np5 and nq5.
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Sketch the sampling distribution.
4.Determine any critical values.
Continued.
Using a z-Test for a Proportion p
Verify that np5 and nq5.
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 4 in
Appendix B.
Hypothesis Test for Proportionsˆpp
z
pq n
Using a z-Test for a Proportion p
Verify that np5 and nq5.
In Words In Symbols
If zis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
5.Determine any rejection regions.
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.
z
pq n
Using a z-Test for a Proportion p
Verify that np5 and nq5.
In Words In Symbols
If zis in the rejection
region, reject H
0.
Otherwise, fail to
reject H
0.
5.Determine any rejection regions.
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.
Hypothesis Test for Proportions
Example:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
Verify that the products npand nqare at least 5.
np= (500)(0.94) = 470 and nq= (500)(0.06) = 30
H
a: p> 0.94 (Claim)H
0:p0.94
Continued.
Example:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
Verify that the products npand nqare at least 5.
np= (500)(0.94) = 470 and nq= (500)(0.06) = 30
H
a: p> 0.94 (Claim)H
0:p0.94
Continued.
Hypothesis Test for Proportionsˆpp
z
pq n
Example continued:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
H
a: p> 0.94 (Claim)H
0:p0.94
Continued.
z
02.33
Because the test is a right-tailed test and = 0.01, the
critical value is 2.33.0.95 0.94
(0.94)(0.06) 500
0.94
Test statistic
z
pq n
Example continued:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
H
a: p> 0.94 (Claim)H
0:p0.94
Continued.
z
02.33
Because the test is a right-tailed test and = 0.01, the
critical value is 2.33.0.95 0.94
(0.94)(0.06) 500
0.94
Test statistic
Hypothesis Test for Proportions
Example continued:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
H
a: p> 0.94 (Claim)H
0:p0.94
z
02.330.94z
The test statistic falls in
the nonrejection region,
so H
0is not rejected.
At the 1% level of significance, there is not enough evidence to
support the college’s claim.
Example continued:
Statesville college claims that more than 94% of their
graduates find employment within six months of
graduation. In a sample of 500 randomly selected
graduates, 475 of them were employed. Is there
enough evidence to support the college’s claim at a
1% level of significance?
H
a: p> 0.94 (Claim)H
0:p0.94
z
02.330.94z
The test statistic falls in
the nonrejection region,
so H
0is not rejected.
At the 1% level of significance, there is not enough evidence to
support the college’s claim.
Hypothesis Test for Proportions
Example:
A cigarette manufacturer claims that one-eighth of the
US adult population smokes cigarettes. In a random
sample of 100 adults, 5 are cigarette smokers. Test
the manufacturer's claim at = 0.05.
Verify that the products npand nqare at least 5.
np= (100)(0.125) = 12.5 and nq= (100)(0.875) = 87.5
H
a: p0.125H
0:p= 0.125 (Claim)
Continued.
Because the test is a two-tailed test and = 0.05, the
critical values are ±1.96.
Example:
A cigarette manufacturer claims that one-eighth of the
US adult population smokes cigarettes. In a random
sample of 100 adults, 5 are cigarette smokers. Test
the manufacturer's claim at = 0.05.
Verify that the products npand nqare at least 5.
np= (100)(0.125) = 12.5 and nq= (100)(0.875) = 87.5
H
a: p0.125H
0:p= 0.125 (Claim)
Continued.
Because the test is a two-tailed test and = 0.05, the
critical values are ±1.96.
Hypothesis Test for Proportions
Example continued:
A cigarette manufacturer claims that one-eighth of the
US adult population smokes cigarettes. In a random
sample of 100 adults, 5 are cigarettes smokers. Test the
manufacturer's claim at = 0.05.
H
a: p0.125H
0:p= 0.125 (Claim)
z
0 = 1.96
z
0
z
0 = 1.96ˆpp
z
pq n
0.05 0.125
(0.125)(0.875)100
2.27
The test statistic
is
At the 5% level of significance, there is enough evidence
to reject the claim that one-eighth of the population
smokes.
Reject H
0.
2.27
Example continued:
A cigarette manufacturer claims that one-eighth of the
US adult population smokes cigarettes. In a random
sample of 100 adults, 5 are cigarettes smokers. Test the
manufacturer's claim at = 0.05.
H
a: p0.125H
0:p= 0.125 (Claim)
z
0 = 1.96
z
0
z
0 = 1.96ˆpp
z
pq n
0.05 0.125
(0.125)(0.875)100
2.27
The test statistic
is
At the 5% level of significance, there is enough evidence
to reject the claim that one-eighth of the population
smokes.
Reject H
0.
2.27
Hypothesis Testing for
Variance and Standard
Deviation
Variance and Standard
Deviation
Critical Values for the χ
2
-Test
Finding Critical Values for the χ
2-
Distribution
1.Specify the level of significance .
2.Determine the degrees of freedom d.f. = n–1.
3.The critical values for the χ
2
-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 –.2
1
c.two-tailed test, use the values that corresponds to
d.f. and and d.f. and 1 –.2
1
2
-Test
Finding Critical Values for the χ
2-
Distribution
1.Specify the level of significance .
2.Determine the degrees of freedom d.f. = n–1.
3.The critical values for the χ
2
-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 –.2
1
c.two-tailed test, use the values that corresponds to
d.f. and and d.f. and 1 –.2
1
Finding Critical Values for the χ
2
Example:
Find the critical value for a left-tailed test when n= 19
and = 0.05.
There are 18 d.f. The area to the right of the critical
value is 1 –= 1 –0.05 = 0.95.
From Table 6, the critical value is χ
2
0= 9.390.
Example:
Find the critical value for a two-tailed test when n= 26
and = 0.01.
From Table 6, the critical values are χ
2
L= 10.520 and
χ
2
R= 46.928.
There are 25 d.f.The areas to the right of the critical
values are = 0.005 and 1 –= 0.995. 2
1 2
1
2
Example:
Find the critical value for a left-tailed test when n= 19
and = 0.05.
There are 18 d.f. The area to the right of the critical
value is 1 –= 1 –0.05 = 0.95.
From Table 6, the critical value is χ
2
0= 9.390.
Example:
Find the critical value for a two-tailed test when n= 26
and = 0.01.
From Table 6, the critical values are χ
2
L= 10.520 and
χ
2
R= 46.928.
There are 25 d.f.The areas to the right of the critical
values are = 0.005 and 1 –= 0.995. 2
1 2
1
The Chi-Square Test
The χ
2
-test for a variance or standard deviationis a
statistical test for a population variance or standard
deviation. The χ
2
-test can be used when the population is
normal.2
2
2
( 1)ns
χ
σ
The teststatisticis s
2
and the standardizedteststatistic
follows a chi-square distribution with degrees of
freedom d.f. = n–1.
The χ
2
-test for a variance or standard deviationis a
statistical test for a population variance or standard
deviation. The χ
2
-test can be used when the population is
normal.2
2
2
( 1)ns
χ
σ
The teststatisticis s
2
and the standardizedteststatistic
follows a chi-square distribution with degrees of
freedom d.f. = n–1.
The Chi-Square Test
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Determine the degrees of freedom
and sketch the sampling
distribution.
4.Determine any critical values.
Continued.
Using the χ
2
-Test for a Variance or Standard Deviation
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 6 in
Appendix B.
d.f. = n –1
1.State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2.Specify the level of significance.
3.Determine the degrees of freedom
and sketch the sampling
distribution.
4.Determine any critical values.
Continued.
Using the χ
2
-Test for a Variance or Standard Deviation
In Words In Symbols
State H
0and H
a.
Identify .
Use Table 6 in
Appendix B.
d.f. = n –1
The Chi-Square Test2
2
2
( 1)ns
χ
σ
Using the χ2-Test for a Variance or Standard Deviation
In Words In Symbols
If χ
2
is in the
rejection region,
reject H
0.
Otherwise, fail to
reject H
0.
5.Determine any rejection regions.
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.
2
2
( 1)ns
χ
σ
Using the χ2-Test for a Variance or Standard Deviation
In Words In Symbols
If χ
2
is in the
rejection region,
reject H
0.
Otherwise, fail to
reject H
0.
5.Determine any rejection regions.
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.
Hypothesis Test for Standard
Deviation
Example:
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard
deviation is found to be 28.8. Test this professor’s
claim at the = 0.01 level.
H
a: < 30 (Claim)H
0:30
Continued.
This is a left-tailed test with d.f.= 9 and = 0.01.
X
20.01
Deviation
Example:
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard
deviation is found to be 28.8. Test this professor’s
claim at the = 0.01 level.
H
a: < 30 (Claim)H
0:30
Continued.
This is a left-tailed test with d.f.= 9 and = 0.01.
X
20.01
Hypothesis Test for Standard
Deviation2
2
2
( 1)ns
χ
σ
Example continued:
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard
deviation is found to be 28.8. Test this professor’s
claim at the = 0.01 level.
H
a: < 30 (Claim)H
0:30
X
20.01 2
2
(10 1)(28.8)
30
8.29
X
2
0= 2.088
χ
2
0= 2.088
Fail to reject H
0.
At the 1% level of significance, there is not enough
evidence to support the professor’s claim.
Deviation2
2
2
( 1)ns
χ
σ
Example continued:
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard
deviation is found to be 28.8. Test this professor’s
claim at the = 0.01 level.
H
a: < 30 (Claim)H
0:30
X
20.01 2
2
(10 1)(28.8)
30
8.29
X
2
0= 2.088
χ
2
0= 2.088
Fail to reject H
0.
At the 1% level of significance, there is not enough
evidence to support the professor’s claim.
Hypothesis Test for Variance
Example:
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
Continued.
This is a two-tailed test with d.f.= 22 and = 0.05.
X
2
X
2
RX
2
L1
0.025
2
1
0.025
2
Example:
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
Continued.
This is a two-tailed test with d.f.= 22 and = 0.05.
X
2
X
2
RX
2
L1
0.025
2
1
0.025
2
Hypothesis Test for Variance
Example continued:
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
Continued.
The critical values are χ
2
L= 10.982 and χ
2
R= 36.781.
X
21
0.025
2
1
0.025
2
36.78110.982
Example continued:
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
Continued.
The critical values are χ
2
L= 10.982 and χ
2
R= 36.781.
X
21
0.025
2
1
0.025
2
36.78110.982
Hypothesis Test for Variance
Example continued:
A local balloon company claims that the variance for the
time one of its helium balloons will stay afloat is 5 hours.
A disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
X
2
36.78110.9822
2
2
( 1)ns
χ
σ
(23 1)(4.5)
5
19.8
Fail to reject H
0.
At = 0.05,there is not enough
evidence to reject the claim that the
variance of the float time is 5 hours.
19.8
Example continued:
A local balloon company claims that the variance for the
time one of its helium balloons will stay afloat is 5 hours.
A disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At = 0.05, does she have
enough evidence to reject the company’s claim?
H
a:
2
5 H
0:
2
= 5 (Claim)
X
2
36.78110.9822
2
2
( 1)ns
χ
σ
(23 1)(4.5)
5
19.8
Fail to reject H
0.
At = 0.05,there is not enough
evidence to reject the claim that the
variance of the float time is 5 hours.
19.8
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Comprehensive
Hypothesis Test -cont
A confidence interval estimate of a population
parameter contains the likely values of that
parameter. We should therefore reject a claim
that the population parameter has a value that
is not included in the confidence interval.
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Comprehensive
Hypothesis Test -cont
A confidence interval estimate of a population
parameter contains the likely values of that
parameter. We should therefore reject a claim
that the population parameter has a value that
is not included in the confidence interval.
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Caution: In some cases, a conclusion based
on a confidence interval may be different from
a conclusion based on a hypothesis test. See
the comments in the individual sections.
Comprehensive
Hypothesis Test -cont
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Caution: In some cases, a conclusion based
on a confidence interval may be different from
a conclusion based on a hypothesis test. See
the comments in the individual sections.
Comprehensive
Hypothesis Test -cont
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Recap
In this section we have discussed:
Null and alternative hypotheses.
Test statistics.
Significance levels.
P-values.
Decision criteria.
Type I and II errors.
Power of a hypothesis test.
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Recap
In this section we have discussed:
Null and alternative hypotheses.
Test statistics.
Significance levels.
P-values.
Decision criteria.
Type I and II errors.
Power of a hypothesis test.
Slide71
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Section 8-3
Testing a Claim About a
Proportion
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
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Section 8-3
Testing a Claim About a
Proportion
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
Slide72
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Key Concept
This section presents complete procedures
for testing a hypothesis (or claim) made about
a population proportion. This section uses
the components introduced in the previous
section for the P-value method, the traditional
method or the use of confidence intervals.
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Key Concept
This section presents complete procedures
for testing a hypothesis (or claim) made about
a population proportion. This section uses
the components introduced in the previous
section for the P-value method, the traditional
method or the use of confidence intervals.
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1) The sample observations are a simple random
sample.
2) The conditions for a binomial distributionare
satisfied (Section 5-3).
3)The conditions np5 and nq5 are satisfied, so
the binomial distribution of sample proportions
can be approximated by a normal distribution with
µ= npand = npq.
Requirements for Testing Claims
About a Population Proportion p
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1) The sample observations are a simple random
sample.
2) The conditions for a binomial distributionare
satisfied (Section 5-3).
3)The conditions np5 and nq5 are satisfied, so
the binomial distribution of sample proportions
can be approximated by a normal distribution with
µ= npand = npq.
Requirements for Testing Claims
About a Population Proportion p
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Notation
p= population proportion (used in the
null hypothesis)
q= 1 –p
n= number of trials
p= x(sampleproportion)
n
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Notation
p= population proportion (used in the
null hypothesis)
q= 1 –p
n= number of trials
p= x(sampleproportion)
n
Slide75
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p–p
pq
n
z=
Test Statistic for Testing
a Claim About a Proportion
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p–p
pq
n
z=
Test Statistic for Testing
a Claim About a Proportion
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P-Value Method
Use the same method as described
in Section 8-2 and in Figure 8-8.
Use the standard normal
distribution (Table A-2).
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P-Value Method
Use the same method as described
in Section 8-2 and in Figure 8-8.
Use the standard normal
distribution (Table A-2).
Slide77
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Traditional Method
Use the same method as described
in Section 8-2 and in Figure 8-9.
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Traditional Method
Use the same method as described
in Section 8-2 and in Figure 8-9.
Slide78
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Confidence Interval Method
Use the same method as described
in Section 8-2 and in Table 8-2.
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Confidence Interval Method
Use the same method as described
in Section 8-2 and in Table 8-2.
Slide79
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Example:An article distributed by the Associated
Press included these results from a nationwide
survey: Of 880 randomly selected drivers, 56%
admitted that they run red lights. The claim is that
the majority of all Americans run red lights. That is, p
> 0.5. The sample data are n= 880, and p= 0.56.
np= (880)(0.5) = 440 5
nq= (880)(0.5) = 440 5
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Example:An article distributed by the Associated
Press included these results from a nationwide
survey: Of 880 randomly selected drivers, 56%
admitted that they run red lights. The claim is that
the majority of all Americans run red lights. That is, p
> 0.5. The sample data are n= 880, and p= 0.56.
np= (880)(0.5) = 440 5
nq= (880)(0.5) = 440 5
Slide80
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
Referring to Table A-2, we see that for values of z= 3.50
and higher, we use 0.9999 for the cumulative area to the left
of the test statistic. The P-value is 1 –0.9999 = 0.0001.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
Referring to Table A-2, we see that for values of z= 3.50
and higher, we use 0.9999 for the cumulative area to the left
of the test statistic. The P-value is 1 –0.9999 = 0.0001.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
Slide81
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
Since the P-value of 0.0001 is less than the significance
level of = 0.05, we reject the null hypothesis.
There is sufficient evidence to support the claim.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
Since the P-value of 0.0001 is less than the significance
level of = 0.05, we reject the null hypothesis.
There is sufficient evidence to support the claim.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
Slide82
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
z= 3.56
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the P-value Method.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
z= 3.56
Slide83
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Example:An article distributed by the Associated
Press included these results from a nationwide
survey: Of 880 randomly selected drivers, 56%
admitted that they run red lights. The claim is that
the majority of all Americans run red lights. That is,
p> 0.5. The sample data are n= 880, and p= 0.56.
We will use the Traditional Method.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
This is a right-tailed test, so the critical region is an area of
0.05. We find that z= 1.645 is the critical value of the
critical region. We reject the null hypothesis.
There is sufficient evidence to support the claim.
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Example:An article distributed by the Associated
Press included these results from a nationwide
survey: Of 880 randomly selected drivers, 56%
admitted that they run red lights. The claim is that
the majority of all Americans run red lights. That is,
p> 0.5. The sample data are n= 880, and p= 0.56.
We will use the Traditional Method.
H
0: p= 0.5
H
1: p> 0.5
= 0.05
pq
n
p –p
z=
0.56 –0.5
(0.5)(0.5)
880
= = 3.56
This is a right-tailed test, so the critical region is an area of
0.05. We find that z= 1.645 is the critical value of the
critical region. We reject the null hypothesis.
There is sufficient evidence to support the claim.
Slide84
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the confidence interval method.
For a one-tailed hypothesis test with significance level ,
we will construct a confidence interval with a confidence
level of 1 –2. We construct a 90% confidence interval.
We obtain 0.533 < p< 0.588. We are 90% confident that
the true value of pis contained within the limits of 0.533
and 0.588. Thus we support the claim that p> 0.5.
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Example:An article distributed by the Associated Press
included these results from a nationwide survey: Of 880
randomly selected drivers, 56% admitted that they run red
lights. The claim is that the majority of all Americans run red
lights. That is, p> 0.5. The sample data are n= 880, and p=
0.56. We will use the confidence interval method.
For a one-tailed hypothesis test with significance level ,
we will construct a confidence interval with a confidence
level of 1 –2. We construct a 90% confidence interval.
We obtain 0.533 < p< 0.588. We are 90% confident that
the true value of pis contained within the limits of 0.533
and 0.588. Thus we support the claim that p> 0.5.
Slide85
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CAUTION
When testing claims about a population proportion,
the traditional method and the P-value method are
equivalent and will yield the same result since they
use the same standard deviation based on the claimed
proportion p. However, the confidence interval uses
an estimated standard deviation based upon the sample
proportion p. Consequently, it is possible that the
traditional and P-value methods may yield a different
conclusion than the confidence interval method.
A good strategy is to use a confidence interval to
estimate a population proportion, but use the P-value
or traditional method for testing a hypothesis.
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CAUTION
When testing claims about a population proportion,
the traditional method and the P-value method are
equivalent and will yield the same result since they
use the same standard deviation based on the claimed
proportion p. However, the confidence interval uses
an estimated standard deviation based upon the sample
proportion p. Consequently, it is possible that the
traditional and P-value methods may yield a different
conclusion than the confidence interval method.
A good strategy is to use a confidence interval to
estimate a population proportion, but use the P-value
or traditional method for testing a hypothesis.
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(determining the sample proportion of households with cable TV)
p= = = 0.64
x
n
96
(96+54)
and 54 do not” is calculated using
psometimes must be calculated
“96 surveyed households have cable TV
psometimes is given directly
“10% of the observed sports cars are red”
is expressed as
p= 0.10
Obtaining P
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(determining the sample proportion of households with cable TV)
p= = = 0.64
x
n
96
(96+54)
and 54 do not” is calculated using
psometimes must be calculated
“96 surveyed households have cable TV
psometimes is given directly
“10% of the observed sports cars are red”
is expressed as
p= 0.10
Obtaining P
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods
and 152 peas with yellow pods. According to Mendel’s
theory, 1/4 of the offspring peas should have yellow pods.
Use a 0.05 significance level with the P-value method to test
the claim that the proportion of peas with yellow pods is
equal to 1/4.
We note that n= 428 + 152 = 580,
so p= 0.262, and p= 0.25.
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods
and 152 peas with yellow pods. According to Mendel’s
theory, 1/4 of the offspring peas should have yellow pods.
Use a 0.05 significance level with the P-value method to test
the claim that the proportion of peas with yellow pods is
equal to 1/4.
We note that n= 428 + 152 = 580,
so p= 0.262, and p= 0.25.
Slide88
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods and
152 peas with yellow pods. According to Mendel’s theory, 1/4 of
the offspring peas should have yellow pods. Use a 0.05
significance level with the P-value method to test the claim that
the proportion of peas with yellow pods is equal to 1/4.
H
0: p= 0.25
H
1: p0.25
n= 580
= 0.05
p= 0.262
0.262 –0.25
(0.25)(0.75)
580
= = 0.67z=
p –p
pq
n
Since this is a two-tailed test, the P-value is twice the area
to the right of the test statistic. Using Table A-2,
z= 0.67 is 1 –0.7486 = 0.2514.
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods and
152 peas with yellow pods. According to Mendel’s theory, 1/4 of
the offspring peas should have yellow pods. Use a 0.05
significance level with the P-value method to test the claim that
the proportion of peas with yellow pods is equal to 1/4.
H
0: p= 0.25
H
1: p0.25
n= 580
= 0.05
p= 0.262
0.262 –0.25
(0.25)(0.75)
580
= = 0.67z=
p –p
pq
n
Since this is a two-tailed test, the P-value is twice the area
to the right of the test statistic. Using Table A-2,
z= 0.67 is 1 –0.7486 = 0.2514.
Slide89
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods and
152 peas with yellow pods. According to Mendel’s theory, 1/4 of
the offspring peas should have yellow pods. Use a 0.05
significance level with the P-value method to test the claim that
the proportion of peas with yellow pods is equal to 1/4.
The P-value is 2(0.2514) = 0.5028. We fail to reject the null
hypothesis. There is not sufficient evidence to warrant rejection
of the claim that 1/4 of the peas have yellow pods.
H
0: p= 0.25
H
1: p0.25
n= 580
= 0.05
p= 0.262
0.262 –0.25
(0.25)(0.75)
580
= = 0.67z=
p –p
pq
n
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Example:When Gregory Mendel conducted his famous
hybridization experiments with peas, one such experiment
resulted in offspring consisting of 428 peas with green pods and
152 peas with yellow pods. According to Mendel’s theory, 1/4 of
the offspring peas should have yellow pods. Use a 0.05
significance level with the P-value method to test the claim that
the proportion of peas with yellow pods is equal to 1/4.
The P-value is 2(0.2514) = 0.5028. We fail to reject the null
hypothesis. There is not sufficient evidence to warrant rejection
of the claim that 1/4 of the peas have yellow pods.
H
0: p= 0.25
H
1: p0.25
n= 580
= 0.05
p= 0.262
0.262 –0.25
(0.25)(0.75)
580
= = 0.67z=
p –p
pq
n
Slide90
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Recap
In this section we have discussed:
Test statistics for claims about a proportion.
P-value method.
Confidence interval method.
Obtaining p.
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Recap
In this section we have discussed:
Test statistics for claims about a proportion.
P-value method.
Confidence interval method.
Obtaining p.
Slide91
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Section 8-4
Testing a Claim About a
Mean: Known
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
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Edition, Tom Wegleitner, Centreville, VA
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Section 8-4
Testing a Claim About a
Mean: Known
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
Slide92
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Key Concept
This section presents methods for testing a
claim about a population mean, given that the
population standard deviation is a known
value. This section uses the normal
distribution with the same components of
hypothesis tests that were introduced in
Section 8-2.
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Key Concept
This section presents methods for testing a
claim about a population mean, given that the
population standard deviation is a known
value. This section uses the normal
distribution with the same components of
hypothesis tests that were introduced in
Section 8-2.
Slide93
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Requirements for Testing Claims About a
Population Mean (with Known)
1) The sample is a simple random
sample.
2) The value of the population standard
deviation is known.
3) Either or both of these conditions is
satisfied: The population is normally
distributed orn> 30.
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Requirements for Testing Claims About a
Population Mean (with Known)
1) The sample is a simple random
sample.
2) The value of the population standard
deviation is known.
3) Either or both of these conditions is
satisfied: The population is normally
distributed orn> 30.
Slide94
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Test Statistic for Testing a Claim
About a Mean (with Known)
n
x–µ
x
z=
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Test Statistic for Testing a Claim
About a Mean (with Known)
n
x–µ
x
z=
Slide95
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=z=
x–µ
x
n
98.2 –98.6
= −6.64
0.62
106
Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common
belief that the mean body temperature of healthy adults is equal
to 98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
This is a two-tailed test and the test statistic is to the left of the
center, so the P-value is twice the area to the left of z= –6.64. We
refer to Table A-2 to find the area to the left of z= –6.64 is 0.0001,
so the P-value is 2(0.0001) = 0.0002.
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=z=
x–µ
x
n
98.2 –98.6
= −6.64
0.62
106
Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common
belief that the mean body temperature of healthy adults is equal
to 98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
This is a two-tailed test and the test statistic is to the left of the
center, so the P-value is twice the area to the left of z= –6.64. We
refer to Table A-2 to find the area to the left of z= –6.64 is 0.0001,
so the P-value is 2(0.0001) = 0.0002.
Slide96
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
Slide97
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
Because the P-value of 0.0002 is less than the significance level
of = 0.05, we reject the null hypothesis. There is sufficient
evidence to conclude that the mean body temperature of healthy
adults differs from 98.6°F.
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the P-valuemethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
Because the P-value of 0.0002 is less than the significance level
of = 0.05, we reject the null hypothesis. There is sufficient
evidence to conclude that the mean body temperature of healthy
adults differs from 98.6°F.
Slide98
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the traditionalmethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
We now find the critical values to be z= –1.96
and z= 1.96. We would reject the null
hypothesis, since the test statistic of z= –6.64
would fall in the critical region.
There is sufficient evidence to conclude that the mean body
temperature of healthy adults differs from 98.6°F.
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the traditionalmethod.
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
z= –6.64
We now find the critical values to be z= –1.96
and z= 1.96. We would reject the null
hypothesis, since the test statistic of z= –6.64
would fall in the critical region.
There is sufficient evidence to conclude that the mean body
temperature of healthy adults differs from 98.6°F.
Slide99
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the confidence intervalmethod.
For a two-tailed hypothesis test with a 0.05
significance level, we construct a 95%
confidence interval. Use the methods of Section
7-2 to construct a 95% confidence interval:
98.08 < < 98.32
We are 95% confident that the limits of 98.08 and 98.32
contain the true value of , so it appears that 98.6 cannot be
the true value of .
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
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Example:We have a sample of 106 body temperatures having a
mean of 98.20°F. Assume that the sample is a simple random
sample and that the population standard deviation is known to
be 0.62°F. Use a 0.05 significance level to test the common belief
that the mean body temperature of healthy adults is equal to
98.6°F. Use the confidence intervalmethod.
For a two-tailed hypothesis test with a 0.05
significance level, we construct a 95%
confidence interval. Use the methods of Section
7-2 to construct a 95% confidence interval:
98.08 < < 98.32
We are 95% confident that the limits of 98.08 and 98.32
contain the true value of , so it appears that 98.6 cannot be
the true value of .
H
0: = 98.6
H
1: 98.6
= 0.05
x= 98.2
= 0.62
Slide100
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Underlying Rationale of Hypothesis
Testing
If, under a given assumption, there is an extremely
small probability of getting sample results at least
as extreme as the results that were obtained, we
conclude that the assumption is probably not
correct.
When testing a claim, we make an assumption
(null hypothesis) of equality. We then compare the
assumption and the sample results and we form
one of the following conclusions:
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Underlying Rationale of Hypothesis
Testing
If, under a given assumption, there is an extremely
small probability of getting sample results at least
as extreme as the results that were obtained, we
conclude that the assumption is probably not
correct.
When testing a claim, we make an assumption
(null hypothesis) of equality. We then compare the
assumption and the sample results and we form
one of the following conclusions:
Slide101
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If the sample results (or more extreme results) can easily
occur when the assumption (null hypothesis) is true, we
attribute the relatively small discrepancy between the
assumption and the sample results to chance.
If the sample results cannot easily occur when that
assumption (null hypothesis) is true, we explain the
relatively large discrepancy between the assumption and
the sample results by concluding that the assumption is
not true, so we reject the assumption.
Underlying Rationale of
Hypotheses Testing -cont
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If the sample results (or more extreme results) can easily
occur when the assumption (null hypothesis) is true, we
attribute the relatively small discrepancy between the
assumption and the sample results to chance.
If the sample results cannot easily occur when that
assumption (null hypothesis) is true, we explain the
relatively large discrepancy between the assumption and
the sample results by concluding that the assumption is
not true, so we reject the assumption.
Underlying Rationale of
Hypotheses Testing -cont
Slide102
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Recap
In this section we have discussed:
Requirements for testing claims about population
means, σknown.
P-value method.
Traditional method.
Confidence interval method.
Rationale for hypothesis testing.
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Recap
In this section we have discussed:
Requirements for testing claims about population
means, σknown.
P-value method.
Traditional method.
Confidence interval method.
Rationale for hypothesis testing.
Slide103
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Section 8-5
Testing a Claim About a
Mean: Not Known
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
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Section 8-5
Testing a Claim About a
Mean: Not Known
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
Slide104
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Key Concept
This section presents methods for testing a
claim about a population mean when we do
not know the value of σ. The methods of this
section use the Student tdistribution
introduced earlier.
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Key Concept
This section presents methods for testing a
claim about a population mean when we do
not know the value of σ. The methods of this
section use the Student tdistribution
introduced earlier.
Slide105
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Requirements for Testing Claims
About a Population
Mean (with Not Known)
1) The sample is a simple random sample.
2) The value of the population standard
deviation is notknown.
3) Either or both of these conditions is
satisfied: The population is normally
distributed or n> 30.
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Requirements for Testing Claims
About a Population
Mean (with Not Known)
1) The sample is a simple random sample.
2) The value of the population standard
deviation is notknown.
3) Either or both of these conditions is
satisfied: The population is normally
distributed or n> 30.
Slide106
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Test Statistic for Testing a
Claim About a Mean
(with Not Known)
P-values and Critical Values
Found in Table A-3
Degrees of freedom (df) = n–1
x–µ
x
t=
s
n
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Test Statistic for Testing a
Claim About a Mean
(with Not Known)
P-values and Critical Values
Found in Table A-3
Degrees of freedom (df) = n–1
x–µ
x
t=
s
n
Slide107
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Important Properties of the
Student tDistribution
1.The Student tdistribution is different for different sample sizes
(see Figure 7-5 in Section 7-4).
2.The Student tdistribution has the same general bell shape as
the normal distribution; its wider shape reflects the greater
variability that is expected when sis used to estimate .
3.The Student tdistribution has a mean of t= 0 (just as the
standard normal distribution has a mean of z= 0).
4.The standard deviation of the Student tdistribution varies with
the sample size and is greater than 1 (unlike the standard
normal distribution, which has = 1).
5.As the sample size ngets larger, the Studenttdistribution gets
closer to the standard normal distribution.
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Important Properties of the
Student tDistribution
1.The Student tdistribution is different for different sample sizes
(see Figure 7-5 in Section 7-4).
2.The Student tdistribution has the same general bell shape as
the normal distribution; its wider shape reflects the greater
variability that is expected when sis used to estimate .
3.The Student tdistribution has a mean of t= 0 (just as the
standard normal distribution has a mean of z= 0).
4.The standard deviation of the Student tdistribution varies with
the sample size and is greater than 1 (unlike the standard
normal distribution, which has = 1).
5.As the sample size ngets larger, the Studenttdistribution gets
closer to the standard normal distribution.
Slide108
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Choosing between the Normal and Student t
Distributions when Testing a Claim about a
Population Mean µ
Use the Student tdistribution when is not
known and either or both of these conditions is
satisfied:The population is normally
distributed or n> 30.
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Choosing between the Normal and Student t
Distributions when Testing a Claim about a
Population Mean µ
Use the Student tdistribution when is not
known and either or both of these conditions is
satisfied:The population is normally
distributed or n> 30.
Slide109
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The sample is a simple random sample and we are
not using a known value of σ. The sample size is
n= 13 and a normal quartile plot suggests the
weights are normally distributed.
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
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The sample is a simple random sample and we are
not using a known value of σ. The sample size is
n= 13 and a normal quartile plot suggests the
weights are normally distributed.
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
Slide110
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H
0: = 0.8535
H
1: > 0.8535
= 0.05
x= 0.8635
s= 0.0576
n= 13
= 0.626
t=
x–µ
x
s
n
=
0.8635 –0.8535
0.0576
13
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
The critical value, from Table A-3, is t= 1.782
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H
0: = 0.8535
H
1: > 0.8535
= 0.05
x= 0.8635
s= 0.0576
n= 13
= 0.626
t=
x–µ
x
s
n
=
0.8635 –0.8535
0.0576
13
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
The critical value, from Table A-3, is t= 1.782
Slide111
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t = 0.626
Because the test statistic of t= 0.626 does not
fall in the critical region, we fail to reject H
0.
There is not sufficient evidence to support the
claim that the mean weight of the M&Ms is
greater than 0.8535 g.
H
0: = 0.8535
H
1: > 0.8535
= 0.05
x= 0.8635
s= 0.0576
n= 13
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
Critical Valuet= 1.782
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t = 0.626
Because the test statistic of t= 0.626 does not
fall in the critical region, we fail to reject H
0.
There is not sufficient evidence to support the
claim that the mean weight of the M&Ms is
greater than 0.8535 g.
H
0: = 0.8535
H
1: > 0.8535
= 0.05
x= 0.8635
s= 0.0576
n= 13
Example:Data Set 13 in Appendix B of the text includes weights
of 13 red M&M candies randomly selected from a bag containing
465 M&Ms. The weights (in grams) have a mean x= 0.8635 and a
standard deviation s= 0.0576 g. The bag states that the net weight
of the contents is 396.9 g, so the M&Ms must have a mean weight
that is 396.9/465 = 0.8535 g in order to provide the amount claimed.
Use the sample data with a 0.05 significance level to test the claim
of a production manager that the M&Ms have a mean that is
actually greater than 0.8535 g. Use the traditional method.
Critical Valuet= 1.782
Slide112
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The critical value in the preceding example was t=
1.782, but if the normal distribution were being used,
the critical value would have been z= 1.645.
The Student tcritical value is larger (farther to the
right), showing that with the Student tdistribution, the
sample evidence must be more extremebefore we can
consider it to be significant.
Normal Distribution Versus
Student tDistribution
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The critical value in the preceding example was t=
1.782, but if the normal distribution were being used,
the critical value would have been z= 1.645.
The Student tcritical value is larger (farther to the
right), showing that with the Student tdistribution, the
sample evidence must be more extremebefore we can
consider it to be significant.
Normal Distribution Versus
Student tDistribution
Slide113
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P-Value Method
Use software or a TI-83/84 Plus
calculator.
If technology is not available, use Table
A-3 to identify a range of P-values.
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P-Value Method
Use software or a TI-83/84 Plus
calculator.
If technology is not available, use Table
A-3 to identify a range of P-values.
Slide114
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a) In a left-tailed hypothesis test, the sample size
is n= 12, and the test statistic is t= –2.007.
b) In a right-tailed hypothesis test, the sample size
is n= 12, and the test statistic is t = 1.222.
c) In a two-tailed hypothesis test, the sample size
is n = 12, and the test statistic is t= –3.456.
Example:Assuming that neither software nor
a TI-83 Plus calculator is available, use Table A-
3 to find a range of values for the P-value
corresponding to the given results.
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a) In a left-tailed hypothesis test, the sample size
is n= 12, and the test statistic is t= –2.007.
b) In a right-tailed hypothesis test, the sample size
is n= 12, and the test statistic is t = 1.222.
c) In a two-tailed hypothesis test, the sample size
is n = 12, and the test statistic is t= –3.456.
Example:Assuming that neither software nor
a TI-83 Plus calculator is available, use Table A-
3 to find a range of values for the P-value
corresponding to the given results.
Slide115
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
Slide116
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
a) The test is a left-tailed test with test
statistic t= –2.007, so the P-value is the
area to the left of –2.007. Because of the
symmetry of thetdistribution, that is the
same as the area to the right of +2.007.
Any test statistic between 2.201 and 1.796
has a right-tailed P-value that is between
0.025 and 0.05. We conclude that
0.025 < P-value < 0.05.
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
a) The test is a left-tailed test with test
statistic t= –2.007, so the P-value is the
area to the left of –2.007. Because of the
symmetry of thetdistribution, that is the
same as the area to the right of +2.007.
Any test statistic between 2.201 and 1.796
has a right-tailed P-value that is between
0.025 and 0.05. We conclude that
0.025 < P-value < 0.05.
Slide117
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a
range of values for the P-value corresponding to the
given results.
b) The test is a right-tailed test with test
statistic t= 1.222, so the P-value is the
area to the right of 1.222. Any test statistic
less than 1.363 has a right-tailed P-value
that is greater than 0.10. We
conclude that P-value > 0.10.
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Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a
range of values for the P-value corresponding to the
given results.
b) The test is a right-tailed test with test
statistic t= 1.222, so the P-value is the
area to the right of 1.222. Any test statistic
less than 1.363 has a right-tailed P-value
that is greater than 0.10. We
conclude that P-value > 0.10.
Slide118
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c) The test is a two-tailed test with test statistic
t= –3.456. The P-value is twice the area to
the right of –3.456. Any test statistic
greater than 3.106 has a two-tailed P-value
that is less than 0.01. We conclude that
P-value < 0.01.
Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
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c) The test is a two-tailed test with test statistic
t= –3.456. The P-value is twice the area to
the right of –3.456. Any test statistic
greater than 3.106 has a two-tailed P-value
that is less than 0.01. We conclude that
P-value < 0.01.
Example:Assuming that neither software nor a TI-83
Plus calculator is available, use Table A-3 to find a range
of values for the P-value corresponding to the given
results.
Slide119
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Recap
In this section we have discussed:
Assumptions for testing claims about
population means, σunknown.
Student tdistribution.
P-value method.
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Recap
In this section we have discussed:
Assumptions for testing claims about
population means, σunknown.
Student tdistribution.
P-value method.
Slide120
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Section 8-6
Testing a Claim About a
Standard Deviation or
Variance
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
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Section 8-6
Testing a Claim About a
Standard Deviation or
Variance
Created by Erin Hodgess, Houston, Texas
Revised to accompany 10
th
Edition, Tom Wegleitner, Centreville, VA
Slide121
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Key Concept
This section introduces methods for testing a
claim made about a population standard
deviation σor population variance σ
2
. The
methods of this section use the chi-square
distribution that was first introduced in
Section 7-5.
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Key Concept
This section introduces methods for testing a
claim made about a population standard
deviation σor population variance σ
2
. The
methods of this section use the chi-square
distribution that was first introduced in
Section 7-5.
Slide122
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Requirements for Testing
Claims About or
2
1. The sample is a simple random
sample.
2. The population has a normal
distribution. (This is a much stricter
requirement than the requirement of a
normal distribution when testing
claims about means.)
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Requirements for Testing
Claims About or
2
1. The sample is a simple random
sample.
2. The population has a normal
distribution. (This is a much stricter
requirement than the requirement of a
normal distribution when testing
claims about means.)
Slide123
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n= sample size
s
2= sample variance
2= population variance
(given in null hypothesis)
Chi-Square Distribution
Test Statistic
2
=
(n–1) s
2
2
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n= sample size
s
2= sample variance
2= population variance
(given in null hypothesis)
Chi-Square Distribution
Test Statistic
2
=
(n–1) s
2
2
Slide124
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P-Values and Critical Values for
Chi-Square Distribution
Use Table A-4.
The degrees of freedom = n–1.
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P-Values and Critical Values for
Chi-Square Distribution
Use Table A-4.
The degrees of freedom = n–1.
Slide125
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Properties of Chi-Square
Distribution
All values of
2
are nonnegative, and the
distribution is not symmetric
(see Figure 8-13, following).
There is a different distribution for each
number of degrees of freedom
(see Figure 8-14, following).
The critical values are found in Table A-4
using n –1 degrees of freedom.
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Properties of Chi-Square
Distribution
All values of
2
are nonnegative, and the
distribution is not symmetric
(see Figure 8-13, following).
There is a different distribution for each
number of degrees of freedom
(see Figure 8-14, following).
The critical values are found in Table A-4
using n –1 degrees of freedom.
Slide126
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Properties of Chi-Square
Distribution -cont
Figure 8-13
Properties of the Chi-Square
Distribution
There is a different distribution for each
number of degrees of freedom.
Chi-Square Distribution for 10
and 20 Degrees of Freedom
Figure 8-14
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Properties of Chi-Square
Distribution -cont
Figure 8-13
Properties of the Chi-Square
Distribution
There is a different distribution for each
number of degrees of freedom.
Chi-Square Distribution for 10
and 20 Degrees of Freedom
Figure 8-14
Slide127
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Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
= 2.765
2
=
(n–1)s
2
2
(13 –1)(7.2)
2
15
2
=
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Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
= 2.765
2
=
(n–1)s
2
2
(13 –1)(7.2)
2
15
2
=
Slide128
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H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
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H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
Slide129
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The critical values of 4.404 and 23.337 are
found in Table A-4, in the 12th row (degrees
of freedom = n–1) in the column
corresponding to 0.975 and 0.025.
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
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The critical values of 4.404 and 23.337 are
found in Table A-4, in the 12th row (degrees
of freedom = n–1) in the column
corresponding to 0.975 and 0.025.
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
Slide130
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Because the test statistic is in the critical
region, we reject the null hypothesis. There
is sufficient evidence to warrant rejection of
the claim that the standard deviation is equal
to 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
Copyright
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as
Because the test statistic is in the critical
region, we reject the null hypothesis. There
is sufficient evidence to warrant rejection of
the claim that the standard deviation is equal
to 15.
H
0: = 15
H
1: 15
= 0.05
n= 13
s= 7.2
2
= 2.765
Example:For a simple random sample of adults, IQ scores are
normally distributed with a mean of 100 and a standard deviation
of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s= 7.2. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance
level to test the claim that = 15.
Slide131
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Recap
In this section we have discussed:
Tests for claims about standard deviation
and variance.
Test statistic.
Chi-square distribution.
Critical values.
Copyright
© 2007
Pearson
Education,
Inc
Publishing
as
Recap
In this section we have discussed:
Tests for claims about standard deviation
and variance.
Test statistic.
Chi-square distribution.
Critical values.
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