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```Chapter 10
Hypothesis Testing Using
a Single Sample
Sharing prescription drugs with others can
be dangerous. Is this a common
occurrence among teens?
OR
like these using sample data?
The National Association of Colleges and
In Chapter
9, wethat
usedthe
sample
datastarting
to
Employers
stated
average
In this
chapter,
we
will
use
sample
data
estimate
the value
of an unknown
salary
for
students
with
a bachelor’s
to test some
claim
or
hypothesis
population characteristic.
degree
in 2010characteristic
is \$48,351. Istothis
the
population
seetrue
if for
To do this, we use a test of hypotheses
or test procedure.
What is a test of hypotheses?
A test of hypotheses is a method that
Is it one of the
uses sample data to decide
between
two
values of
the
competing claims (hypotheses)
sample statistic
population characteristic.
that are likely to
occur?
Is theIsvalue
of that
the sample
statistic . . .
it one
isn’t likely
to
– a random
occurrence
due to natural
occur?
variation?
OR
– a value that would
be considered
surprising?
Hypothesis statements:
YouHare
usually
The null hypothesis, denoted by
0, is a
trying to
determine if
that is initially assumed to be true.
this claim is
The hypothesis statements are ALWAYS
The alternative hypothesis, denoted by
Ha, is the competing claim.
To determine what the alternative
hypothesis should be, you need to keep the
research objectives in mind.
Let’s consider a murder trial . . .
To determine which hypothesis is
What
is the
null
hypothesis?
correct,
the
jury
will listen to the You are trying to
is what you
if the
So we
will make
one This
ofdetermine
two
evidence. Only
if there
is “evidence
is true
Hbeyond
is innocent
evidence
reasonable decisions:
doubt”
would assume
before
you begin.
the null hypothesis be rejected in
supports
this
favor of the alternative hypothesis.
claim.
• Reject the null hypothesis
What is
the
alternative
hypothesis?
• Fail
to
reject
null hypothesis
If there
is not the
convincing
evidence, then
we would “fail to reject” the null
Ha: the hypothesis.
defendantRemember
is guilty that the actually
verdict that is returned is “GUILTY” or
“NOT GUILTY”. We never end up
determining the null hypothesis is true –
only that there is not enough evidence to
say it’s not true.
The Form of Hypotheses:
Null hypothesis
H0: population characteristic = hypothesized value
This one is considered a
two-tailed test because
you are interested
in both
Alternative
hypothesis
The null
hypothesis always
direction.
includes the equal
case.
Ha: population
characteristic
> hypothesized
value
This hypothesized
valuevalue
is
Ha: population characteristic
< hypothesized
a specific number
Ha: population characteristic
≠ hypothesized
This
signby
is the value
determined
Notice
that the alternative
These are
considered
onedetermined
by
the
context
of
the
problem
hypothesis
usesyou
the same
tailed tests
because
context
of the
population
characteristic
and
the
Let’s
practice
writing
are only interested in one
problem.
same hypothesized
value
as
the
hypothesis statements.
direction.
null hypothesis.
Sharing prescription drugs with others can be
dangerous. A survey of a representative sample
of 592 U.S. teens age 12 to 17 reported that 118
of those surveyed admitted to having shared a
prescription drug with a friend. Is this sufficient
evidence that more than 10% of teens have
shared prescription medication with friends?
State the hypotheses :
What is the
What words indicate
What
the
is the
population
hypothesized
direction of theThe
characteristic
ofp of
value?
true
proportion
H
:
p
=
.1
0
alternative hypothesis?
teensinterest?
who have shared
Ha: p > .1
prescription medication
with friends
Compact florescent (cfl) lightbulbs are
much more energy efficient than regular
incandescent lightbulbs. Ecobulb brand
60-watt cfl lightbulbs state on the package
“Average life 8000 hours”. People who purchase
this brand would be unhappy if the bulbs lasted
less than 8000 hours. A sample of these bulbs
will be selected and tested.
What is the
StateWhat
the words
hypotheses
: theis the
indicate
What
population
hypothesized
direction of thecharacteristic
of
value?
H : m alternative
= 8000 hypothesis?interest?
0
Ha: m < 8000
The true mean (m) life of
the cfl lightbulbs
Because in variation of the manufacturing
process, tennis balls produced by a particular
machine do not have the same diameters.
Suppose the machine was initially calibrated to
achieve the specification of m = 3 inches.
However, the manager is now concerned that the
diameters no longer conform to this specification.
If the mean diameter is not 3 inches, production
will have to be halted.
State the hypotheses :
What words indicateWhat
the is the population
direction
of
The true mean m
H0: m =of3 the characteristic
alternative hypothesis? diameter
interest?
of tennis
Ha: m ≠ 3
balls
For each pair of hypotheses, indicate which are
Must useand
a population
characteristic - x
not legitimate
explain why
is a statistics (sample)
a) H0 : m  15 ; Ha : m  15
greater
than!
b) H0 : xMust
 4be
; Honly
:
x

15
a
Must use same
c) H0number
: p  .1as; H
:
p
a
in H0!
 .1
d) H0 :Hm0 MUST
 2.3 ;be
Ha“=“
: m!  3.2
e) H0 : p  .5 ; Ha : p  .5
When you perform a hypothesis
test you make a decision:
reject H0 or fail to reject H0
When
one of
Each could possibly
beyoua make
wrong
these decisions, there is
decision; therefore,
there are
a possibility
thattwo
you
types of errors. could be wrong!
Type I error
• The error of rejecting H0 when H0 is
true
• The probability of a Type I error is
denoted by a.
a is called the significance level of the
This is the lower-case
test
Greek letter “alpha”.
Type II error
• The error of failing to reject H0
when H0 is false
• The probability of a Type II error
is denoted by b
This is the lower-case
Greek letter “beta”.
Here is another way to look at the
types of errors:
Suppose
H0H
is is
true
Suppose
Suppose
H
is
true
is
Suppose
0H
00
and
wereject
fail
to
and
false
weand
wewefail
it,
false
and
reject
it,
what
toreject
what
rejecttype
it,what
what
of
it,
type
of decision
decision
type
decision
of
decision
was
type
of
was
was
was
H0 is
true
H0 is
false
Reject
H0
Type I
error
Correct
Fail to
reject
H0
Type II
Correct
error
The U.S. Bureau of Transportation Statistics
reports that for 2009 72% of all domestic
passenger flights arrived on time (meaning within
15 minutes of its scheduled arrival time). Suppose
that an airline with a poor on-time record decides
to offer its employees a bonus if, in an upcoming
month, the airline’s proportion of on-time flights
exceeds the overall 2009 industry rate of .72.
State the hypotheses. Type I error – the airline
reward the
H0: p = .72
State a decides
Type Ito
error
employees when the
in
context.
H : p > .72
a
proportion of on-time
flights doesn’t exceeds .72
State a Type II error
Type II error – the airline
in context.
bonus when they deserve it.
In 2004, Vertex Pharmaceuticals, a biotechnology
company, issued a press release announcing that it
had filed an application with the FDA to begin clinical
trials on an experimental drug VX-680 that had been
found to reduce State
the growth
rate
of pancreatic
a Type
I error
in the and
colon cancer tumors
in animal
context
of studies.
this problem.
A potential consequence of making a Type I
What
is
athe
potential
error
would
be
that the
company
would
Data resulting
from
planned
clinical
trials
can be
tothis
devote
resources to the
error?
used toconsequence
test: continueof
development of the drug when it really is
Let m = the true mean growth
rate of tumors for patients
not effective.
taking the experimental drug
H0: m = mean growth rate of tumors for patients not
taking the experimental drug
Ha: m < mean growth rate of tumors for patients not
taking the experimental drug
A Type I error would be to incorrectly conclude
that the experimental drug is effective in
slowing the growth rate of tumors
In 2004, Vertex Pharmaceuticals, a biotechnology
company, issued a press release announcing that it
had filed an application with the FDA to begin
clinical trials on an experimental drug VX-680
Statetoa reduce
Type IIthe
error
in the
growth
rate of
context
this
problem.
pancreaticAand
colon
canceroftumors
in animal
potential
consequence
of making
a Type
What
a potential
erroriswould
be that the company might
studies. II
abandon development
of a drug that was
consequence
of this error?
Data resulting from theeffective.
planned clinical trials can
be used to test:
H0: m = mean growth rate of tumors for patients not
taking the experimental drug
Ha: m < mean growth rate of tumors for patients not
taking the experimental drug
A Type II error would be to conclude that the
drug is ineffective when in fact the mean
growth rate of tumors is reduced
The relationship between a and b
The ideal test procedure would result in both
Selecting a of
significance
level aand
= .05
a = 0 (probability
a Type I error)
b=0
results inof
a test
procedure
that, used over
(probability
a Type
II error).
and over with different samples, rejects a
This isSo
impossible
to
achieve
since
we
must
why
choose
ainsmall
a base
truenot
H0always
5
times
100.
a = .05data.
or a = .01)?
our decision(like
on sample
Standard test procedures allow us to select a,
the significance level of the test, but we have no
direct control over b.
The relationship between a and b
Suppose
thisisnormal
curve
If
the
null
hypothesis
false
and
the
Let’s consider the represents the sampling
alternative hypothesis is true, then the
tail distribution
would
represent
the the
following This
hypotheses:
pb,when
This
is thefor
part
of
true proportion
is believed
to
be the
greater
probabilitycurve
of failing
to
reject
a a
null
hypothesis
is
true.
that
represents
than .5 – so the curve should really be
false
H0Type
.
or
the
I error.
shifted to the right.
H0: p = .5
Ha: p > .5
Let a = .05
.5
The relationship between a and b
If the null hypothesis is false and the
Let’s consider
the hypothesis is true, then the
alternative
This
tailthat
would
b, the
following
hypotheses:
true
proportion
to
be bgreater
Notice
asisrepresent
abelieved
gets smaller,
probability
of
failing
to
reject
a
than .5 – sogets
the
curve
should
really
be
larger!
false to
H0the
.
shifted
right.
H0: p = .5
Ha: p > .5
Let a = .01
How does one decide what a
level to use?
After assessing the consequences of
type I and type II errors, identify the
largest a that is tolerable for the
problem. Then employ a test procedure
that uses this maximum acceptable value
–rather than anything smaller – as the
level of significance.
Remember, using a smaller a increases b.
Copper Rule, which defines drinking water as unsafe if the
concentration of lead is 15 parts per billion (ppb) or
greater or if the concentration of copper is 1.3 ppb or
greater.
aa Type
III
error
inmight
Which
type of
error
has
a more
serious
The manager
of aState
community
water
system
State
Type
error
incontext.
context.
level measurements
from aconsequence?
sample
water specimens
Since
most
people
wouldof
consider
theI? to
What
is
a
consequence
of
a
Type
What is
a consequence
ofIaerror
Type II?
test the following
hypotheses:
consequence
of the Type
H0: m = 15we
versus
Hawant
: m < 15to keep
more serious,
would
A Type I error
theselect
conclusion
that a water source
smallto
– so
a smaller
meets EPA standards
when the
water
significance
level
of ais=really
.01. unsafe.
There are possible health risks to the community
A Type II error leads to the conclusion that a
water source does NOT meet EPA standards
when the water is really safe.
The community might lose a good water source.
Large-Sample Hypothesis Test
for a Population Proportion
The fundamental idea behind hypothesis
testing is:
We reject H0 if the observed sample is very
unlikely to occur if H0 is true.
Recall the General Properties for
Sampling Distributions of p
These three properties imply that the standardized
variable
pˆ  p
z 
1. ˆ
p 1  p 
p
n
As
long normal
as the distribution
sample size is
has an approximately
p (1  p )standard
less than 10% of the population
 pˆn is
 large.
when
2.
m p
n
3. When n is large, the sampling
distribution of p is approximately
normal.
In June 2006, an Associated Press survey
was conducted to investigate how people
use the nutritional information provided
on food packages. Interviews were
conducted with 1003 randomly selected
adult Americans, and each participant was
asked a series of questions, including the following two:
Based on these data, is it reasonable to
Question
1: When
packaged
food,Americans
how often do
conclude
that
a majority
you
check thecheck
nutritional
labeling on
the package?
frequently
nutritional
labels
when
packaged
Question 2: How
often do you
purchasefoods?
for you, even after you’ve checked the nutrition
labels?
It was reported that 582 responded “frequently” to the
question about checking labels and 441 responded “very
often” or “somewhat often” to the question about
Nutritional Labels Continued . . .
H0: p = .5
Ha:We
p > .5
will create a test statistic using:
p = true proportion of adult
pˆ  p Americans who
z  nutritional labels
frequently check
p 1  p 
n
We use p > .5 to
test for a majority
582
pˆ 
who
.58 frequently
For this sample:
1003
check nutritional
labels.
A test statistic indicates how many standard
This
observed
is is
greater
than
deviations
thesample
sampleproportion
statistic (p)
from the
.5..58
Isit.5plausible
a characteristic
sample proportion
of p = .58
population
(p).
z occurred

 5a.08
as
result of chance variation, or is it
.5.5
unusual to observe a sample proportion this large
1003
when p = .5?
Nutritional Labels Continued . . .
H0: p = .5
Ha: p > .5
p = true proportion of adult Americans who
frequently
check
nutritional
labels
the
normal
NextIn
we
findstandard
the P-value
for curve,
this seeing
a value
5.08 or larger is unlikely.
testofstatistic.
It’s probability
582 is approximately 0.
pˆ 
 .58
For this sample:
1003
Since the P-value
is so small,
.58P-value
 .5
The
is the probability
of obtaining
a we
z 
 5.08
reject
H0. There is convincing
test.5statistic
at least
as inconsistent
with H0
.5
evidence
to suggest
that the
as
was
observed,
assuming
H
is
true.
0
1003
P-value ≈ 0
frequently check the nutritional
labels on packaged0foods.
Computing P-values
The calculation of the P-value depends on the form
of the inequality in the alternative hypothesis.
• Ha: p > hypothesize value
z curve
P-value = area in upper tail
Calculated z
Computing P-values
The calculation of the P-value depends on the form
of the inequality in the alternative hypothesis.
• Ha: p < hypothesize value
z curve
P-value = area in lower tail
Calculated z
Computing P-values
The calculation of the P-value depends on the form
of the inequality in the alternative hypothesis.
• Ha: p ≠ hypothesize value
P-value = sum of area in two tails
z curve
Calculated z and –z
Using P-values to make a decision:
To decide whether or not to reject
H0, we compare the P-value to the
significance level a
If the P-value > a, we “fail to reject”
the null hypothesis.
If the P-value < a, we “reject” the null
hypothesis.
Summary of the Large-Sample z
Test for p
Null hypothesis:
Test Statistic:
H0: p = hypothesized value
pˆ  p
z 
p (1  P )
n
Alternative Hypothesis:
P-value:
Ha: p > hypothesized value Area to the right of calculated z
Ha: p < hypothesized value Area to the left of calculated z
Ha: p ≠ hypothesized value 2(Area to the right of z) of +z
or 2(Area to the left of z) of -z
Summary of the Large-Sample z
Test for p Continued . . .
Assumptions:
1.
p is a sample proportion from a random sample
2. The sample size n is large. (np > 10 and
n(1 - p) > 10)
3. If sampling is without replacement, the sample
size is no more than 10% of the population size
A report states that nationwide, 61% of high
school graduates go on to attend a two-year or
four-year college the year after graduation.
Suppose a random sample of 1500 high school
graduates in 2009 from a particular state
estimated the proportion of high school graduates
that attend college the year after graduation to be
58%. Can we reasonably conclude that the
proportion of this state’s high school graduates in
2009 who attended college the year after
graduation is different from the national figure?
Use a = .01.
H0: p = .61
Ha: p ≠ .61
Where p is the proportion of all
State
hypotheses.
2009
highthe
school
this state who attended college
College Attendance Continued . . .
H0: p = .61
Ha: p ≠ .61
Where p is the proportion of all 2009 high
school graduates in this state who
attended college the year after graduation
Assumptions:
• Given a random sample of 1500 high school
• Since 1500(.61) > 10 and 1500(.39) > 10, sample
size is large enough.
• Population size is much larger than the sample
size.
College Attendance Continued . . .
H0: p = .61
Ha: p ≠ .61
Where p is the proportion of all 2009 high
school graduates in this state who attended
Test statistic:
z 
.58  .61
 2.38
.61(.39)
What potential error could you
1500
Type II
The= area
P-value = 2(.0087)
.0174to the left
Useof
a -2.38
= .01 is
approximately .0087
Since P-value > a, we fail to reject H0. The evidence does
not suggest that the proportion of 2009 high
school graduates in this state who attended
college the year after graduation differs from
the national value.
In December 2009, a county-wide water
conservation campaign was conducted in a
particular county. In 2010, a random
sample of 500 homes was selected and water
usage was recorded for each home in the sample.
Suppose the sample results were that 220
households had reduced water consumption. The
county supervisors wanted to know if their data
supported the claim that fewer than half the
households in the county reduced water
consumption.
H0: p = .5
Ha: p < .5
State
the hypotheses.
where p is the
proportion
of all households in
Calculate
the county with reduced
water p.
usage
pˆ 
220
 .44
500
Water Usage Continued . . .
H0: p = .5
Ha: p < .5
where p is the proportion of all households in
the county with reduced water usage
Verify assumptions
1. p is from a random sample of households
2. Sample size n is large because np = 250 >10 and n(1-p) =
250 > 10
3. It is reasonable that there are more than 5000 (10n)
households in the county.
Water Usage Continued . . .
H0: p = .5
Ha: p < .5
where p is the proportion of all households in
the county with reduced water usage
Calculate
the up
test
Look
this
value
in the
.44  .5
statistic
P-value
z 
 2.What
68
potential
error
could
you
table
of z and
curve
areas
.5(.5)
Type I
500
P-value = .0037
Use a = .01
Since P-value < a, we reject H0. There is convincing
evidence that the proportion of households with
reduced water usage is less than half.
Water Usage Continued . . .
H0: p = .5
Ha: p < .5
where p is the proportion of all households in
the county with reduced water usage
Since P-value < a, we reject H0.
Confidence intervals are twoUsed a = .01
Compute
98%
confidence
tailed,
soain
we
need
to put
in
With .01
each
tail,
that.01
puts
interval:
the
upper
tail
(since
the –curve is
.98
in
the
middle
Notice that the
Let’s create
a confidence
symmetrical).
are is
testing
Ha: p < .05, a
hypothesizedSince
valueweinterval
this
the
appropriate
with
this
data.


.
44
(.
56
)
would
also
be
in
the
lower
tail.
 level

of .5 is NOT in the
.
44

2
.
326
confidence
What is the appropriate


500
.98
98% confidence
 to use? 
confidence level
interval and that we
(.
388
,
.
492
)
“rejected”
H0!
.5
College Attendance Revisited . . .
H0: p = .61
Ha: p ≠ .61
Where p is the proportion of all 2009 high
school graduates in this state who attended
Since P-value > a, we fail to reject H0.
Use a = .01
Let’s compute
a a
This is a two-tailed
test so
for
Notice that the gets splitconfidence
evenly intointerval
both tails,
this
problem.
.58
(.42) 
hypothesized value leaving 99%
in the
middle.
.58  2.576

of .5 IS
in
the
98%

.99
1500


confidence interval
and that we “failed
(.547, .613)
to reject” H0!
Hypothesis Tests
for a Population Mean
Let’s review the assumptions for a
confidence interval for a population mean
The assumptions are the same for a
1) x is the large-sample
sample mean from
a random
hypothesis
testsample,
for a
2) the sample size n population
is large (n mean.
> 30), and
3) , the population standard deviation, is known
or unknown
This
Thisisisthe
thetest
teststatistic
statistic
when
when isisunknown.
known.
x m
z 

n
P-value is area under
the z curve
x m
t 
s
n
P-value is area under
the t curve with df=n-1
The One-Sample t-test for a
Population Mean
Null hypothesis:
Test Statistic:
H0: m = hypothesized value
x m
t 
s
n
Alternative Hypothesis:
P-value:
Ha: m > hypothesized value Area to the right of calculated t
with df = n-1
Ha: m < hypothesized value Area to the left of calculated t
with df = n-1
Ha: m ≠ hypothesized value 2(Area to the right of t) of +t
or 2(Area to the left of t) of -t
The One-Sample t-test for a
Population Mean Continued . . .
Assumptions:
1.
x and s are the sample mean and sample
standard deviation from a random sample
2. The sample size n is large (n > 30) or the
population distribution is at least approximately
normal.
A study conducted by researchers at Pennsylvania
State University investigated whether time
perception, an indication of a person’s ability to
concentrate, is impaired during nicotine withdrawal.
After a 24-hour smoking abstinence, 20 smokers were
45-second period. Researchers wanted to see whether
smoking abstinence had a negative impact on time
perception, causing elapsed time to be overestimated.
Suppose the resulting data on perceived elapsed time
(in seconds) were as follows:
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
What is the mean and standard
x = 59.30 s = 9.84
n = 20
deviation
of the sample?
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
Where m is the true mean perceived elapsed
H0: m = 45
time for smokers who haveState
abstained
from
the
Ha: m > 45
smoking for 24-hours
Since the boxplot is approximately hypotheses.
Assumptions:
symmetrical, it is plausible that the
1) It is reasonable
todistribution
believe thatisthe sampleVerify
of smokers
population
To do this, we need to
graph the
assumptions.
is representative
of
all
smokers.
approximately normal.
data using a boxplot or normal
2) Since the sample size is probability
not
plot
at least 30, we must
determine if it is plausible
that the population
distribution is approximately
40 50 60 70
normal.
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
H0: m = 45
Ha: m > 45
Where m is the true mean perceived elapsed
time for smokers whoCompute
have abstained
from
the test
smoking for 24-hours
Test statistic:
P-value ≈ 0
statistic and P-value.
59.30  45
t 
 6.50
9.84
20
a = .05
Since P-value < a, we reject H0. There is convincing
evidence that the mean perceived elapsed time is
greater than the actual elapsed time of 45 seconds.
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
Compute
the appropriate
Where
m
is
the
true
mean
perceived elapsed
H0: m = 45
confidence
time for smokers
who have interval.
abstained from
Ha: m > 45
smoking for 24-hours
Since P-value < a, we reject H0.
Notice that the
a = .05
hypothesized value
 9.84
 
59
.
30

1
.
729
of 45 is NOT in the
20 

90%
confidence
Since
this is a one-tailed
test,
a )
(55.497
, 63.103
interval
wetail. .05 goes in
goes inand
thethat
upper
“rejected”
H0! leaving .90 in the
the lower tail,
middle.
A growing concern of employers is time spent in activities
like surfing the Internet and emailing friends during work
hours. The San Luis Obispo Tribune summarized the
findings of a large survey of workers in an article that
ran under the headline “Who Goofs Off More than 2
Hours a Day? Most Workers, Survey Says” (August 3,
2006). Suppose that the CEO of a large company wants
to determine whether the average amount of wasted time
during an 8-hour day for employees of her company is less
than the reported 120 minutes. Each person in a random
daily wasted time at work. The resulting data are the
following:
108 112 117 130 111
131 113 113 105 128
What is the mean and standard
x = 116.80 s = 9.45 n = 10
deviation of the sample?
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120
Ha: m < 120
Where m is the true mean daily wasted
time for employees of this company
The boxplot reveals some skewness,
State
the
Verify
the
but
there
is
no
outliers.
It
is
plausible
Assumptions:
hypotheses.
assumptions.
that
the
population
distribution
is
1) The given sample was a random sample of employees
approximately normal.
2) Since the sample size is not
at least 30, we must
determine if it is plausible
that the population
110
120
130
distribution is approximately
normal.
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120
Ha: m < 120
Where m is the true mean daily wasted
time for employees of this company
the
testwe
116potential
.80 Compute
120 error
What
could

 and
1.07
Test Statistic: thave
statistic
TypeP-value.
II
.45
P-value =.150
10
a = .05
Since p-value > a, we fail to reject H0. There is not
sufficient evidence to conclude that the mean daily
wasted time for employees of this company is less than
120 minutes.
Power and the Probability
of Type II Error
Recall Type I and Type II Errors:
Suppose
H0H
is is
true
Suppose
Suppose
H
is
true
is
Suppose
0H
00
and
wereject
fail
to
and
false
weand
wewefail
it,
false
and
reject
it,
what
toreject
what
rejecttype
it,what
what
of
it,
type
of decision
decision
type
decision
of
decision
was
type
of
was
was
was
Reject
H0
H0 is
true
H0 is
false
Type I
error
Correct
Power
Fail to
Type II
Correct
reject that
The probability
we correctly reject H0 is
error
called the power of the test.
H
0
Suppose that the student body president at a
university is interested in studying the amount of
money that students spend on textbooks each
semester. The director of financial aid services
Thethat
power
of a test
depends
true value of the
believes
average
amount
spentononthe
textbooks
mean!
Becauseand
the uses
actual
value
m is unknown,
is \$500
each
semester,
this
to of
However,
if
the
true
mean
was
\$525,
it
is
less
likely
If
the
true
mean
is
greater
than
\$500,
then
we of
we
cannot
know
the
power
for
the
actual
value
determine
the
amount
of
financial
aid
for
which
a
that
the reject
sampleHwould
be mistaken
for
a sample
from
should
true
mean
is ONLY
a
m
.
0. BUT, if the
student
isgreater,
eligible.if
The
student
body
president
the
population
the
meanthen
were
\$500.
So,mean
it is might
more
little
say
\$505,
the
sample
BUT,
wethat
can we
gain
insight
to the
power
m by
plans to
each
student
in
acorrectly
random
sample
how
likely
will
reject
H0of
.\$500.
look
like
we
expect
if
the
true
mean
were
investigating
somethis
what
if scenarios
...
muchThus
he or
she
spent
on
books
semester
and
we wouldn’t have convincing evidence to reject
use the data to test (using a H
=0.05)
the following
.
hypotheses:
H0: m = 500 versus Ha: m > 500
Let’s consider a one-sided, upper tail test.
Fail to Reject H0
Reject H0
Power = 1 - b
b
m0
a
ma
If the null hypothesis is false, then
m > hypothesized value
Textbooks Continued . . .
H0: m = 500
Ha: m > 500
Suppose that  = \$85 and
n = 100. (Since n is large, the
sampling distribution of x is
approximately normal.)
What is the probability of committing a Type I
error?
a = .05
This is the z value with
area
to its of
left.
= 500 is true, for .95
what
values
the sample
If m
mean would you reject the null hypothesis?
Rejection Region
x  500
Use:
1.645 
What is the value of this x?
.95
We would reject H0
for x > 513.98.
85
100
Textbooks Continued . . .
Suppose that  = \$85 and
n = 100.
Ha: m > 500 We would reject H
0
for x > 513.98.
If the null hypothesis
is false, then m > 500.
What is theThis
probability
of a Type
II
error (isb)?
area (to the left
of
x
=
513.98)
What if m =520?b.
H0: m = 500
513.98  520
z 
 .708
85
100
Rejection Region
b = .239Look this up in the
table of areas for z
curves
Textbooks Continued . . .
H0: m = 500
Ha: m > 500
Suppose that  = \$85 and
n = 100.
We would reject H0
for x > 513.98.
What is the power of the test if m = 520?
b = .239
Power
is
the
probability
of
power is in the
Power = Notice
1 -correctly
.239that
= .761
rejecting H .
SAME curve as b 0
Rejection Region
Power = 1 – b
Textbooks Continued . . .
Suppose that  = \$85 and
n = 100.
Ha: m > 500 We would reject H
0
for x > 513.98.
If we reject H0, then m
Notice that, as the distance
between
> 500.
Find b and
power.
the
null hypothesized
valueiffor
and
What
m =m530?
our alternative value for m increases, b
decreases AND power increases.
b = .03
power = .97
H0: m = 500
Rejection Region
Textbooks Continued . . .
Suppose that  = \$85 and
n = 100.
Ha: m > 500 We would reject H
0
for x > 513.98.
If the null hypothesis
Notice that, as theisdistance
between
false, then
m > 500.
Find b and
power.
the
null hypothesizedWhat
valueiffor
and
m =m 510?
our alternative value for m decreases, b
increases AND power decreases.
b = .68
power = .32
H0: m = 500
Rejection Region
Textbooks Continued . . .
Suppose that  = \$85 and
n = 100.
H0: m = 500
Ha: m > 500
b will increase and power will
decrease.
What happens if we use a = .01?
Rejection Region
b
Rejection Region
Power
b
Power
What happens to a, b, & power when
the sample size is increased?
Fail to Reject H0
a
The standard
b decreases
and
The
significance
deviation
will
power
increases!
Reject
levelH(a)
remains
0
decrease
making
thecurve
same taller
– so the
the
valueskinnier.
where the
and
rejection region
begins must move.
m0
b
Power
ma
Effects of Various Factors on
the Power of a Test
• The larger the size of the discrepancy
between the hypothesized value and the
actual value of the population
characteristic, the higher the power.
• The larger the significance level a, the
higher the power of a test.
• The larger the sample size, the higher
the power of a test
b and Power for a t Test
When using a t-test, the population standard
deviation  is unknown. b not only depends on a,
n, and the actual value of m, but b also depends 
so we must have an estimate of .
The b curves (on the next slide) can be used to
estimate b and the power of a test based upon
the value of d.
d 
alternativ e value - hypothesiz ed value
σ
b curves
Consider testing
H0: m = 100 versus Ha: m > 100
and
focus
onpopulation
the alternative
value mis= normal,
110.
When
the
distribution
Suppose
= 10,
n = 7,hypotheses
the t that
test for
testing
has smaller b than does any other test
110

100
d procedure that
 1has the same significance
level a.
10
Calculate d.
Use the df = 6
Power ≈ .4 curve to
estimate b
b ≈ .6
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