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Chapter
10
Hypothesis Test
of a Claim
Regarding a
Population
Parameter
(p,μ)
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Chap 2
2
Section
10.1
The Language of
Hypothesis
Testing
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
An hypothesis is a statement regarding a
characteristic of one or more populations.
From the Greek, an hypothesis is a statement, or
claim, yet to be proved.
In statistics, hypothesis testing is the use of
sampling and probability theory to evaluate a
claim that someone has made regarding a
characteristic of the population.
10-4
CLAIMS REGARDING A CHARACTERISTIC OF A
POPULATION
1. In 2008, 62% of American adults regularly volunteered their time
for charity work. A researcher claims that this percentage is different
today.
2. According to a study published in 2006 the mean length of a phone
call on a cellular telephone was 3.25 minutes. A researcher claims
that call length has increased since then.
3. Using an old manufacturing process, the standard deviation of the
amount of wine put in a bottle was 0.23 oz. With new equipment,
the quality control manager claims the standard deviation has
decreased.
10-5
We test these types of hypotheses (claims)
using sample data because it is usually
impossible/impractical to gain access to the
entire population database.
If we have access to population data, obviously
we have no need for inferential statistics to
test these claims because we can just look at
the facts.
10-6
Steps in Hypothesis Testing
1.
Note a claim someone has made regarding the
population. Generate two mutually exclusive
math hypotheses including the claim.
2.
Collect evidence (sample data) pertinent to the
claim. Make a decision as to which hypothesis
is more likely.
3.
Analyze the decision to assess the plausibility
of the claim.
10-7
The null hypothesis, denoted H0, is a
statement to be tested.
The null hypothesis is a statement which is
assumed true until/unless evidence
(sample) indicates otherwise.
(Innocent until proven Guilty)
10-8
The alternative hypothesis, denoted H1, is
a statement that contradicts the H0 .
We look at sample data (evidence) and
decide to support/reject one of these
mutually exclusive hypotheses.
We use that decision to make a comment
about the plausibility of the claim we have
been given to evaluate.
10-9
“In Other Words”
The null hypothesis is a statement of status
quo or no difference and always contains a
statement of equality.
The null hypothesis is initially assumed to be
true until/unless we have evidence to the
contrary.
We look to see if there is evidence that
rejects the null and therefore supports the
alternative hypothesis.
10-10
“On Trial”
null hypothesis: man = innocent and is
assumed true until/unless we find evidence to
the contrary.
alternative hypothesis: man ≠ innocent
Evaluate evidence (sample data) and decide to
reject or fail-to-reject the null.
The evidence will cause us to support or reject the
initial claim of innocence.
10-11
in the courtroom…

We are trying Karl for impersonating a Math
Professor. He looks guilty, but we must assume he
is innocent anyway….

We ask 100 of his students whether Karl knows
anything about math. For a response. we use a
scale of 0 to 10, with 10 meaning he’s a math
genius, and 0 meaning he’s a math idiot (not a total
idiot).

The mean of our sample turns out to be 1.0445
on a scale of 0 to 10
in the courtroom…

We conclude that it would be such a rare event
(exceptionally small probability) for Karl to
actually be a Math Professor and get such a
pathetically low sample score of 1.0445, that our
initial assumption of innocence was probably
wrong, so we reject it…

(and we send him to jail where he belongs, for
impersonating a Math Professor.)
Forming Hypotheses
For each of the following claims, determine the null and
alternative hypotheses. State whether the test is two-tailed, lefttailed or right-tailed.
a)
In 2008, 62% of American adults regularly volunteered their
time for charity work. A researcher claims that this
percentage is different today. H0: p=0.62
b)
According to a study published in 2006, the mean length of
a phone call on a cellular telephone was 3.25 minutes. A
researcher claims that the mean length of a call has
increased since then. H0: μ = 3.25.
c)
Using an old manufacturing process, the standard deviation
of the amount of wine put in a bottle was 0.23 ounces. With
new equipment, the quality control manager claims the
standard deviation has decreased.
d)
H1: σ < 0.23, a left-tailed test
10-14
Types of Errors
Actual Truth of H0
Decision
H0 is true
H0 is false
Fail to reject H0 Correct Decision
Type II Error
Reject H0
Correct Decision
Type I Error
Larson & Farber, Elementary Statistics: Picturing the World, 3e
15
α = P(Type I Error)
= P(rejecting H0 when H0 is true)
β = P(Type II Error)
= P(not rejecting H0 when H1 is true)
10-16
Type I and Type II Errors
For each of the following claims, explain what it would
mean to make a Type I error. What would it mean to
make a Type II error?
a)
In 2008, 62% of American adults regularly
volunteered their time for charity work. A
researcher claims that this percentage is different
today.
b)
According to a study published in March, 2006 the
mean length of a phone call on a cellular telephone
was 3.25 minutes. A researcher claims the mean
length of a call has increased since then.
10-17
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 Greek letter “Alpha”.
Hypothesis tests
are based on .
The probability of making a type II error is denoted by ,
the 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 (Type IError) to be
small.
Commonly used levels of significance:
 = 0.10  = 0.05  = 0.01
Larson & Farber, Elementary Statistics: Picturing the World, 3e
18
The probability of making a Type I error, α,
is chosen by the researcher before the
sample data is collected.
The level of significance (LOS), α, is the
probability of making a Type I error.
As the probability of a Type I error
increases, the probability of a Type II error
decreases, and vice-versa.
10-19
We assume the null hypothesis is true.
We look at the evidence (sample) and then
decide to reject H0 or fail to reject H0 ,the null
hypothesis.
This is just like the court system where we
assume the defendant is innocent, and then
either find him “not guilty” (FTR H0 ) or
“guilty” (Reject H0 ).
10-20
Stating the Conclusion
According to a study published in 2006, the mean
length of a phone call on a cell phone was 3.25
minutes. A researcher claims that the mean length of a
call has increased since then.
a)
Suppose the sample evidence indicates that the
null hypothesis should be rejected (Reject H0 ).
State the wording of the conclusion.
b)
Suppose the sample evidence indicates that the
null hypothesis should not be rejected (Fail to Reject
H0 ). State the wording of the conclusion.
10-21
Statistical Tests
After stating the null and alternative hypotheses and
specifying the level of significance, a random sample is
taken from the population and sample statistics are
calculated.
The statistic that is compared with the parameter in
the null hypothesis is called the test statistic.
Population
parameter
μ
p
2
Test
statistic
x
p̂
s2
Standardized test
statistic
z (n  30)
t (n < 30)
z
X2
Larson & Farber, Elementary Statistics: Picturing the World, 3e
22
Hypothesis Testing:
Quick review of two final steps:
6. Decision: Reject/FTR
H0
7. Interpret Decision:
Suf / Insuf evidence to Support Claim
(Claim is the H1 )
Suf / Insuf evidence to Reject Claim
(Claim is the H 0 )
Interpreting a Decision
Claim
Decision
Claim is H0
Claim is H1
Reject H0
There is enough evidence to
reject the claim.
There is enough evidence to
support the claim.
Do not reject H0
There is not enough evidence
to reject the claim.
There is not enough evidence
to support the claim.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
24
Section
10.2
Hypothesis Tests
for a Population
Proportion
(P)
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
A researcher assumes (based on her prior research) that
the population proportion of people who are in favor of the
banning cell phone use while driving is 0.5
She obtains a random sample of 1000 people and finds
that 534 are in favor of the banning cell phone use while
driving.
so
pˆ = 534/1000 = 0.534
Would it be unusual to obtain a sample proportion of
0.534 or higher from a population whose proportion is
0.5?
What is convincing, or statistically significant, evidence?
10-26
When observed results are unlikely
compared to null hypothesis (H0 ), we say
the result is statistically significant and
we reject the null hypothesis (which we
always assume to be true at the start).
10-27
To determine if a sample proportion of 0.534 is
statistically significant, we build a probability model.
1. npq = 100(0.5)(0.5) = 250 ≥ 10
2. Sample n = 1000 is < 5% population size
We can use the normal model to describe the distribution.
The mean of the distribution
 pö  0.5
and the standard deviation is
 pö 

  0.016.
0.5 1 0.5
1000
10-28
Sampling distribution of the sample proportion
10-29
Recall that our simple random sample yielded a
sample proportion of 0.534, so
z
pö   pö
 pö

0.534  0.5


 2.15
0.5 1 0.5
1000
Our sample was 2.15 std deviations above the
hypothesized proportion of 0.5 which is an
“unlikely” result.
Therefore, we reject the null hypothesis (H0 ) that
the population proportion was 0.50.
10-30
Why does it make sense to reject the null hypothesis if
the sample proportion is more than 2 standard deviations
away from the hypothesized proportion?
The area under
the standard
normal curve to
the right of
z = 2 is 0.0228, or
only 2.3% of the
time would you
expect to get a
sample of 0.532
10-31
Hypothesis Testing
If the sample proportion Z-score is too many
standard deviations (generally 2 or more) from
the proportion stated in the null hypothesis H0
(assumed true), then we delete that assumption
based on evidence (our sample) and
“reject the H0”
10-32
Testing Hypotheses Regarding a
Population Proportion, p
The best point estimate of p, the proportion of
the population is given by
x
pˆ 
n
where x is the number of “success” individuals
in the sample and n is the sample size.

10-33
The sampling distribution of pˆ is approximately
normal, with mean  pˆ  p and
standard deviation
p(1 p)
 pˆ 

n

provided that the following requirements are
satisfied:
1. The sample is a simple random sample.
2.
np(1-p) ≥ 10 and n < 5% N
3. The sampled values are independent of each
other.
10-34
A. Critical Value Test or
Classical Approach
Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
10-35
A. Critical Value Test or
Classical Approach
Step 2: Select a level of significance, α, based
on the max allowable probability of making a
Type I error.
10-36
A. Critical Value Test or
Classical Approach
Step 3: Compute the standardized test statistic
pˆ  p0
z0 
p0 (1 p0 )
n
Note: p0 refers to the value of p in the H0
hypothesis

10-37
Use TI-84 or Table V to
determine the critical value.
Left-Tailed
(critical value)
10-38
Use TI-84 or Table V to
determine the critical value.
Right-Tailed
(critical value)
10-39
Use TI-84 or Table V to
determine the critical value.
Two-Tailed
(critical value)
10-40
A. Critical Value Test or
Classical Approach
Step 4: Compare the Critical Z Value (based on α)
with the “Standardized Test Statistic or STS”
(which is the Z-score of your sample).
If the STS is outside (more extreme than) the Critical
Z, then your sample falls in the Rejection Region and
your Decision is to Reject the null (H0 ).
If the STS is not in the Rejection Region, then your
Decision is to Fail to Reject the null (H0 ).
10-41
B. P-Value Approach
The P-Value Approach has the same objective as the
Critical Value Approach, which is to decide whether to
Reject or Fail-To-Reject (FTR) the null H0 based on the
LOS ( α )of the problem
The BIG advantage is that the TI-84 does all the work
of calculating the STS (below) and the std dev of the
sample.
z0 
pˆ  p0
p0 (1 p0 )
n
10-42
B. P-Value Approach
If the P-value < α, the Z of sample is in the
Rejection Region, so your Decision is :
Reject the null hypothesis H0.
If P ≥ α , then the Z of your sample is not in the
Rejection Region, so your Decision is:
Fail-To-Reject the H0
Now, finally “Interpret the Decision” you just made by
making a statement about the Claim…
10-43
Testing a Hypothesis about a
Population Proportion: Large Sample Size
In 1997, 46% of Americans said they did not trust the
media “when it comes to reporting the news fully,
accurately and fairly”.
In a 2007 poll of 1010 adults nationwide, 525 stated
they did not trust the media.
At the α = 0.05 level of significance (LOS), is there
evidence to support the claim that the percentage of
Americans that do not trust the media has increased
since 1997?
10-44
We want to evaluate the claim that p > 0.46.
First, we must verify that we can perform the
hypothesis test using z-scores. For the sampling
distribution of pˆ to be approximately normal, so we
require npq be at least 10.
1. This isa simple random sample.
2. np0q0 = 1010(0.46)(0.54) = 251 > 10
3. The sample size is less than 5% of the population
size, so the assumption of independence is met.
10-45
Step 1: H0: p ≤ 0.46
H1: p > 0.46 CLAIM
Step 2: LOS = α = 0.05
525
 0.52.
Step 3: The sample proportion is pˆ 
1010
The standardized test statistic (STS) is:

0.52  0.46
z0 
 3.83
0.46(0.54)
1010
10-46
A. Critical Value Approach
Step 4: This is a right-tailed test, and the critical
value at the α = 0.05 level is:
z0.05 = 1.645 (See Table V)
Step 5: The STS is z0 = 3.83, which is
greater than the critical value 1.645.
Therefore, the sample lies in the Rejection
Region, so we decide to:
Reject the null hypothesis.
10-47
B. P-Value Approach
Step 4: Since this is a right-tailed test, the
P- value is the area to the right of (more extreme)
the STS test statistic z0=3.83.
P-value = normcdf(3.83, 1E99) = 6.4E-5 ≈ 0.
Step 5: Since the P-value is less than α = 0.05,
we decide to: Reject the H0
10-48
“Interpret the Decision”
Step 6: Based on our sample, there is sufficient
evidence at the α = 0.05 level of significance to
support the claim that the percentage of
Americans that do not trust the media has
increased since 1997.
10-49
TI-84 (Proportion)
STAT:TESTS:5:1-PropZTest
 po = p used in Ho
 x = number of successes in the sample
(must be an Integer)
 n from your sample
 For prop line: use the Ha symbol


Draw: will show the shaded part of the curve
and also show z and P-value.
TI-84 (Proportion)
STAT:TESTS:5:1-PropZTest
 po = 0.46
 x = 525
(must be an Integer)
 n = 1010
 For prop line: > po (Right-tailed Test)


Draw: shows z = 3.8133 and p=1E-4

Note: The TI-84 does not know LOS, so it
does not know where the Reject Region is.
Section 10.3
Hypothesis Tests
for a Population
Mean
(μ)
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
To test hypotheses regarding the population
mean assuming the population standard
deviation is unknown, we use the t-distribution
rather than the Z-distribution. When we replace
σ with s,
x  0
t0 
s
n
follows Student’s t-distribution with n –1
degrees of freedom.
10-53
Properties of the t-Distribution: Table VI
1.
2.
3.
The t-distribution is different for different
degrees of freedom.
The t-distribution is centered at 0 and is
symmetric about 0.
The area in the tails of the t-distribution is a
little greater than the area in the tails of the
standard normal distribution because using s
as an estimate of σ introduces more variability
to the t-statistic.
10-54
Properties of the t-Distribution
4.
5.
As the sample size n increases, the density
curve of t gets closer to the standard normal z
density curve.
This result occurs because as the sample size
increases, the values of “s” get closer to the
values of “σ” by the Law of Large Numbers.
10-55
Testing Hypotheses Regarding a
Population Mean
To test hypotheses regarding the population
mean, we require that:
1.
3.
The sample is obtained using simple random
sampling.
2. The sample has no outliers, and the
population from which the sample is drawn is
normally distributed OR the sample size is
large (n ≥ 30).
The sampled values are independent of each
other.
10-56
Step 1: Determine the null and alternative
hypotheses. They can be structured in
one of three ways:
10-57
Classical Approach
Step 2: Compute the standardized test statistic
x  0
t0 
s
n
which follows the Student’s t-distribution with n – 1
degrees of freedom.
Step 3. Use Table VI to determine the critical t value.
10-58
P-Value Approach
Step 1: Compute the STS:
x  0
t0 
s
n
Use Table VI or the TI-84 to approximate the
P-value.
10-59
P-Value Approach
Left-Tailed
10-60
P-Value Approach
Right-Tailed
10-61
P-Value Approach
Two-Tailed
10-62
P-Value Approach
If the P-value < α, the “t” of sample is in the
Rejection Region, so your Decision is :
Reject the null hypothesis H0.
If P ≥ α , then the “t” of your sample is not in
the Rejection Region, so your Decision is:
Fail-To-Reject the H0
Now, finally “Interpret the Decision” you just made by
making a statement about the Claim…
10-63
The procedure is robust, which means that
minor departures from normality will not
adversely affect the results of the test.
However, for small samples, if the data have
outliers, or multiple modes, this procedure
should not be used because the distribution will
not be approximately normal.
10-64
Testing a Hypothesis about a
Population Mean, Large Sample
Assume the resting metabolic rate (RMR) of healthy
males in complete silence is 5710. Researchers
measured the RMR of 45 healthy males who were
listening to calm classical music and found their mean
RMR to be 5500 with a standard deviation of 992.
At the α = 0.05 level of significance, is there evidence to
support the researcher’s claim that the mean RMR of
males listening to calm classical music is not 5710?
10-65
Solution
Step 1: H0: μ = 5710
versus
H1: μ ≠ 5710 Claim
Step 2: The level of significance is α = 0.05.
Step 3: The sample mean is x = 5500 and the sample
standard deviation is s = 992. The STS is
  5710
5500
t0 
 1.4201
992 45
10-66
Solution: Classical Approach
Step 4: Since this is a two-tailed test, the critical values
at the α = 0.05 level with (n –1) = 44 to be
t0.025 = ± 2.021
Step 5: Since the test statistic, t0 = – 1.4201, is between
the critical values, (not in the Reject Region)
we fail to reject the null hypothesis.
10-67
TI-84 (n<30)

STAT:TESTS:2:T-Test
o = 5710 ( Ho )
 X-bar = 5500 n = 45
 Sx = 992 from sample
 For  line: ≠ o (H1 symbol)


Draw: t = - 1.4201
p = 0.1626
Solution: P-Value Approach
Step 5: Since the P-value is greater than the LOS
(0.1626 > 0. 05), our Decision is to Fail to Reject H0
Step 6: There is insufficient evidence at the α = 0.05
level of significance to support the claim that
the mean RMR of males listening to calm
classical music is not 5710.
10-69
Testing a Hypothesis about a Population Mean,
Small Sample
According to the US Mint, standard quarters weigh 5.67 grams
(28g/oz). A researcher suspects the new “US State” quarters have
a weight that is heavier than the normal 5.67 grams. He randomly
selects 18 “state” quarters, weighs them and obtains the following
data.
5.70
5.67
5.73
5.61
5.70
5.67
5.65
5.62
5.73
5.65
5.79
5.73
5.77
5.71
5.70
5.76
5.73
5.72
At the α = 0.05 level of significance, is there evidence to support
the claim that “State” quarters weigh more than 5.67 grams?
10-70
Solution
Step 1: H0: μ ≤ 5.67
versus
H1: μ > 5.67 Claim
Step 2: The level of significance is α = 0.05.
Step 3: From the data, the sample mean is calculated to
be 5.7022 and the sample standard deviation is
s = 0.0497. The STS statistic is
5.7022  5.67
t0 
 2.75
.0497 18
10-71
TI-84 (n<30)

STAT:TESTS:2:T-Test
o = 5.67 ( Ho )
 X-bar = 5.7022 n = 18
 Sx = 0.0497 from sample
 For  line: > o (H1 symbol)


Draw: t = 2.7488
p = 0.0069
Solution: P-Value Approach
Step 5: Since the P-value is less than the level of
significance (0.0069 < 0.05), our Decision is to
Reject the H0 .
Step 6: There is sufficient evidence at the α = 0.05 level
of significance to support the claim that the
mean weight of the “State” quarters is greater
than 5.67 grams.
Note: t–crit = 1.740 and STS = 2.7488, so sample is
in the Reject Region (Critical Value Method).
10-73
Section 10.4
Hypothesis Tests
for a Population
Standard
Deviation
(σ)
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Chi-Square Distribution
If a random sample of size “n” is obtained from a
normally distributed population with mean “μ”
and standard deviation “σ”, then
2 
(n 1)s 2
2
where s2 is a sample variance has a chi-square
distribution with n – 1 degrees of freedom.

10-75
Characteristics of the Chi-Square
Distribution
1.
2.
3.
4.
It is not symmetric.
The shape of the chi-square distribution
depends on the degrees of freedom, just
as with Student’s t-distribution.
As the number of degrees of freedom
increases, the chi-square distribution
becomes more nearly symmetric.
The values of χ2 are always nonnegative
(greater than or equal to 0).
10-76
10-77
Testing Hypotheses about a Population
Variance or Standard Deviation
To test hypotheses about the population
variance or standard deviation, we can use the
following steps, provided that:


The sample is obtained using simple
random sampling.
The population is normally distributed.
10-78
Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
10-79
Compute the standardized test statistic
 02 
(n  1)s 2

2
0
Use Table VII to determine the critical value
using n – 1 degrees of freedom.
10-80
Classical Approach
Left -Tailed
10-81
Classical Approach
Right-Tailed
10-82
Classical Approach
Two-Tailed
10-83
Testing a Hypothesis about a
Population Standard Deviation
A can of soda-pop states that the can contains 355 ml
(11.5 oz) of soda. The can-filling machine has a
specification of ± 3.2 ml max standard deviation, but a
quality control engineer suspects the machine is not
calibrated correctly. She wants to verify the machine is not
under- or over-filling the cans, so she randomly selects 9
cans of the soda and measures the contents. She obtains the
following volume data:
351
360
358
356
359 358
355
361
352
Test her claim that the soda standard deviation, σ, is greater than
3.2 ml at the α = 0.05 level of significance.
10-84
Solution
Step 1: H0: σ = 3.2
H1: σ > 3.2 Claim
This is a right-tailed test.
Step 2: The level of significance is α = 0.05.
Step 3: From the data, the sample standard
deviation is computed to be s = 3.464.
The standardized test statistic is
2
(9
1)(3.464)
2
0 
 9.374
2
3.2
10-85
Solution: Classical Approach
Step 4: Since this is a right-tailed test, we
determine the critical value at the α = 0.05 with 8
degrees of freedom to be χ20.05= 15.507.
Step 5: Since STS  02  9.374 is less than the
critical value, 15.507, so we Fail to Reject H0
Step 6: There is insufficient evidence at α = 0.05
 the claim that the standard deviation
LOS to support
of the soda can contents is greater than 3.2 ml.
10-86
Section 10.5
Which Method
Do I Use?
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
10-88
Section 10.6
The Probability
of a Type II Error
and
the Power of the
Test
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
The Probability of a Type II Error
Step 1: Determine the sample mean that
separates the rejection region
from the non-rejection region.
10-90
Computing the Probability of a
Type II Error
Earlier, we tested the hypothesis that the mean
trade volume of Apple stock was greater than 35.14
million shares, H0: μ = 35.14 versus H1: μ > 35.14,
based upon a random sample of size n = 40 with
the population standard deviation, σ, assumed to be
15.07 million shares at the α = 0.1 level of
significance.
Compute the probability of a type II error given that
the population mean is μ = 40.62.
10-91
Solution
Step 1: Since z0.9=1.28, we let z = 1.28,
μ0 = 35.14, σ = 15.07, and n = 40
find the sample mean that separates the
rejection region from the nonrejection
region:
x  35.14
1.28 
 x  38.19
15.07
40
For any sample mean less than 38.19, we do
not reject the null hypothesis.

10-92
Solution
10-93
Solution
10-94
Solution
Step 3: P(Type II error) = β
= P(do not reject H0 given H1 is true)
= P( x< 38.19 given that μ =40.62)

=



38.19  40.62 
PZ 
 P(Z  1.02)
15.07




40
β = 0.1539
10-95
Power of the Test
The probability of rejecting the null hypothesis
when the alternative hypothesis is true is (1 –
β) which is referred to as the power of the test.
The higher the power of the test, the more
likely the test will reject the null when the
alternative hypothesis is true.
10-96
Recall that β = 0.1539.
The power of the test is
1 – β = 1 – 0.1539 = 0.8461.
There is a 84.61% chance of rejecting the null
hypothesis when the true population mean is
40.62.
10-97
Chap 2
98