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Chapter 10
Asking and Answering
Questions about a
Population Proportion
Created by Kathy Fritz
In the original study, 1385 women were sent
a 19 question survey. Of the 561 surveys
returned, 229 women said they would like to
choose the sex of a future child. Of these
229 women, 140 choose to have a girl.
The
article
“Boy or
Which Gender Baby Would
What
is the
population
of Girl:
interest?
You Pick?” (LiveScience, March 23, 2005,
Are the
561 womenawho
responded
to the
www.livescience.com)
summarized
study
that was
survey representative of the population?
In thisinchapter,
learn techniques
for
published
Fertilityyou
andwill
Sterility.
The LiveScience
Does the high
nonresponse
testing
this claim.statement:
article
makes
the following
rate pose
a problem?
“When
given the opportunity to choose the sex of
their baby, women are just as likely to choose pink
socks as blue socks, a new study shows.”
Or, is observing a sample proportion as large as 0.611
very unlikely if the population proportion is 0.50?
Hypotheses and Possible
Conclusions
Null Hypothesis
Alternative Hypothesis
Hypotheses
In its simplest form, a hypothesis is a claim or statement
about the valueHypotheses
of a single are
population
characteristic.
ALWAYS
statements
about the population characteristic –
The following are examples
of hypotheses
about
NEVER the
sample statistic.
population proportions:
Hypothesis
Population Proportion of
Interest
The Hypothesis says . . .
p < 0.25
Where p is the proportion of
e-mail messages that included an
attachment
Less than 25% of the e-mail
messages sent included an
attachment
p > 0.8
Where p is the proportion of
e-mail messages that were longer
than 500 characters
More than 80% of the e-mail
messages sent were longer
than 500 characters
p = 0.3
Where p is the proportion of
e-mail messages that were sent to
multiple recipients
30% of the e-mail messages
sent were sent to multiple
recipients
What is a hypothesis test?
A hypothesis test uses sample data to choose
between two competing hypotheses about a
population characteristic.
Suppose that a particular community college claims that the
majority of students completing an associate’s degree transfer
to a 4-year college. You would then want to determine if the
sample data provide convincing
evidence
support
of the of
Notice
thatinthe
hypothesis
hypothesis p > 0.5.
interest is one of the two
competing hypotheses.
To test a claim: set up competing hypotheses
p ≤ 0.5 and p > 0.5
Hypothesis statements:
The null hypothesis, denoted by H0, is a claim
about a population characteristic that is initially
assumed to be true.
If theout
sample
not provide
such
In carrying
a testdata
of Hdo
versus
H
,
the
null
0
a
evidence,
H0inisfavor
not
rejected.
The
alternative
hypothesis,
denoted
by
Ha, is the
hypothesis
H
is
rejected
of
the
alternative
0
Notice that the conclusions are made about the
competing
claim.
hypothesis
Ha ONLY
if the
sample
data provide
null hypothesis
NOT
about
the alternative!
convincing evidence that H0 is false.
Two possible conclusions in a hypothesis test are:
• Reject H0
• Fail to reject H0
The Form of Hypotheses:
This one is considered a twoNull hypothesis
tailed test because you are
H0: interested
populationincharacteristic
both directions. = hypothesized value
Alternative hypothesis
The null hypothesis always
Ha: population characteristic
> hypothesized
value
includes the equal
case.
This hypothesized value is a
Ha: population characteristic
< hypothesized value
specific number determined by
Ha: Notice
population
≠This
hypothesized
value
sign
is uses
the context
of the
problem
thatcharacteristic
the alternative
hypothesis
These are considered one- determined by the
the same population characteristic and the
tailed tests because you are context of the
hypothesized
valuewriting
as the null
Let’s
onlysame
interested
in onepractice
problem.
hypothesis.
hypothesis
statements.
direction.
Let’s consider a murder trial . . .
What is the null hypothesis?
This is what you
assume is true
before you begin.
If there is not convincing evidence, then we
would
reject” the hypothesis?
null hypothesis.
What
is“fail
the to
alternative
H0: the defendant is innocent
end upisdetermining
the null
HaWe
: thenever
defendant
guilty
hypothesis is true – only that there is not
To determine
which hypothesis
enough evidence
to say it’s is
notcorrect,
true.
the jury will listen to the evidence. Only if
there is “evidence beyond a reasonable
doubt” would the null hypothesis be rejected
in favor of the alternative hypothesis.
In a study, researchers were interested in
determining if sample data support the claim that
more than one in four young adults live with their
parents.
Define the population characteristic:
p = the proportion
young
who as:
It is acceptable
to writeofthe
null adults
hypothesis
live with their parents
H0: p ≤ 0.25
State the hypotheses :
What is the
hypothesized
H0: p = 0.25
value?
What words indicate the
direction
alternative
Ha: pof> the
0.25
hypothesis?
A study included data from a survey of 1752 people ages
13 to 39. One of the survey questions asked participants
how satisfied they were with their current financial
situation. Suppose you want to determine if the survey
data provide convincing evidence that fewer than 10% of
adults 19 to 39 are very satisfied with their current
financial situation.
Define the population characteristic:
p = the proportion of adults ages 19 to 39 who are very
satisfied with their current financial situation
State the hypotheses :
H0: p = 0.10
Ha: p < 0.10
What words indicate the
direction of the alternative
hypothesis?
The manufacturer of M&Ms claims that 40% of plain
M&Ms are brown. A sample of M&Ms will be used to
determine if the proportion of brown M&Ms is different
from what the manufacturer claims.
Define the population characteristic:
p = the proportion of plain M&Ms that are brown
State the hypotheses :
What words indicate the
direction
H0: p
= 0.40of the alternative
hypothesis?
Ha: p ≠ 0.40
For each pair of hypotheses, indicate which are
not legitimate and explain why
Must use a population
characteristic!
a) H0 : p  0.15 ; Ha : p  0.15
ˆinclude
does
NOT
b) H0 : pˆHa 0
.4 ; H
a : p  0.4
Must
usestatement!
same
equality
asa :inpH0!0.1
c) H0 : p number
 0.1 ; H
H0 MUST include the
d) H0 : p
 0.23statement!
; Ha : p  0.32
equality
e) H0 : p  .5 ; Ha : p  .5
Potential Errors in
Hypothesis Testing
Type I Errors
Type II Errors
Significance Level
When you perform a hypothesis
test you make a decision:
reject H0 or fail to reject H0
Each could possibly be a wrong decision;
therefore, there are two types of
errors.
Type I error
A Type I error is the error of rejecting H0
when H0 is true.
The probability of a Type I error is denoted
by a.
This is the lower-case Greek
letter “alpha”.
In a hypothesis test, the probability of a
Type I error, a, is also called the significance
level.
Type II error
A Type II error is 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”.
The U.S. Bureau of Transportation Statistics reports that
for 2008, 65.3% 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 the airline’s
proportion of on-time flights exceeds the overall industry
rate of 0.653 in an upcoming month.
Let p = the actual proportion of the airline’s flights that
are on time during the month of interest.
The hypotheses are:
State
StateaaType
Type II
I error
errorin
in
AA Type
IIIerror
is concluding
concluding that
that
Type
is
not
context.
H0: p = 0.653 the airline on-time
rate
exceeds
the airline’s on-time
proportion
overallthan
industry
rating when in
Ha: p > 0.653 the
is better
the industry
fact
the airline
have a
proportion
whendoes
the not
airline
better
on-time
really did
have record.
a better on-time
record.
Boston Scientific developed a new heart stent used to treat
arteries blocked by heart disease. The new stent, called the
Liberte, is made of thinner metal than heart stents currently in
use, making it easier for doctors to direct the stent to a
blockage.
A consequence of making a Type I error
In order to obtain approval to sell the new Liberte stent, the
be that the
new
stent is
approved
Food and Drug would
Administration
(FDA)
required
Boston
Scientific
for sale.
patients will
experience
reto provide evidence
thatMore
the proportion
of patients
receiving
blockedaarteries.
the Liberte stent who experienced
re-blocked artery was less
than 0.1.
Let p = the proportion of patients receiving the Liberte
stent who experience a re-blocked artery
H0: p = 0.1
Ha: p < 0.1
A Type
beintothe
StateI aerror
Typewould
I error
conclude
thatofthe
context
thisre-block
problem.
proportion for the new stent is
less than 0.1 when it really is
0.1 (or greater).
Boston Scientific developed a new heart stent used to treat
arteries blocked by heart disease. The new stent, called the
Liberte, is made of thinner metal than heart stents currently in
use, making it easier for doctors to direct the stent to a
blockage.
In order to obtain approval to sell the new Liberte stent, the
A consequence of making a Type II error
Food and Drug Administration (FDA) required Boston Scientific
would be that the new stent is not
to provide evidence that the proportion of patients receiving
for sale.
Patients and
doctors
the Liberte stentapproved
who experienced
a re-blocked
artery
was less
will not benefit from the new design.
than 0.1.
Let p = the proportion of patients receiving the Liberte
stent who experience a re-blocked artery
H0: p = 0.1
Ha: p < 0.1
A Type II error would be that
State a Type II error in the
you are not convinced that the
context of this problem.
re-block proportion for the new
stent is less than 0.1 when it
really is less than 0.1.
The relationship between a and b
The ideal test procedure would result in both
a = 0 (probability of a Type I error) and b = 0
(probability of a Type II error).
Selecting a significance level a = 0.05
results in a test procedure that, used over
This isand
impossible
to achieve
since rejects
we must
over with different
samples,
a base
our decisiontrue
on sample
H0 aboutdata.
5 times in 100.
Standard test procedures allow us to select a,
the significance
of thechoose
test,abut
So whylevel
not always
smallwe
a have no
direct control over
(like ab=. 0.05 or a = 0.01)?
The relationship between a and b
If the null hypothesis is false and the
alternative hypothesis is true, then the true
Let’sproportion
consideristhe
believed
to berepresent
greater than
This
tail would
b, 0.5 –
Thisreally
is the
of the
following
hypotheses:
so the
curve
bepart
shifted
the
theshould
probability
of
failing
toto curve
that
represents
right.
reject
a false H0a
. or the Type
I error.
H0: p = 0.5
Ha: p > 0.5
Let a = 0.05
.5
The relationship between a and b
If the null hypothesis is false and the
alternative hypothesis is true, then the true
Notice that as a gets
Let’s consider
proportionthe
is believed to be greater than 0.5 –
smaller, b gets larger!
so hypotheses:
the curve should really be shifted to the
following
right.
This tail would
represent b, the
probability of failing to reject a false H0.
H0: p = 0.5
Ha: p > 0.5
Let a = 0.01
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.
Heart Stents Revisited . . .
Let p = the proportion of patients receiving the Liberte
stent who experience a re-blocked artery
H0: p = 0.1 versus Ha: p < 0.1
A consequence of making a Type I error would be that
the new stent is approved for sale. More patients will
experience re-blocked arteries.
A Type I error
has a amore
consequence.
A consequence
of making
Typeserious
II error
would be that
the new stent is not approved for sale. Patients and
Becausewill
thisnot
represents
an unnecessary
risk to patients
doctors
benefit from
the new design.
(given that other stents with lower re-block proportions
are available), a small
valuetype
for a
, such
0.01,
Which
of
errorashas
thewould
morebe
selected.
serious consequence?
The Logic of Hypothesis
Testing
An Informal Example
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 conclude that
Question
1: When
purchasing
packaged
food, howcheck
often do
a majority
of adult
Americans
frequently
you checklabels
the nutritional
labeling on
the package?
nutritional
when purchasing
packaged
foods?
Question 2: How often do you purchase food that is bad
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
purchasing bad foods even after checking the labels.
Nutritional Labels Continued . . .
H0: p = 0.5
Ha: p > 0.5
p = true proportion of adult Americans who frequently
check nutritional labels
We use p > 0.5 to test for a majority of
582frequently check
For this sample:
adult Americans
who
pˆ 
 .58
nutritional
1003labels.
Thus, we have convincing evidence to suggest that
the null hypothesis is not true.
We would reject H0.
A Procedure for Carrying
Out a Hypothesis Test
Test Statistic
P-value
Test Statistic
A test statistic is computed using sample data.
The value of the test statistic is used to
determine the P-value associated with the test.
P-values
The P-value (also sometimes called the
observed significance level) is a measure of
inconsistency between the null hypothesis and
the observed sample.
It is the probability, assuming that H0 is true,
of obtaining a test statistic value at least as
inconsistent with H0 as what actually resulted.
You reject the null hypothesis when the P-value
is small.
Using P-values to make a decision:
A decision in a hypothesis test is based on
comparing the P-value to the chosen significance
level a.
H0 is rejected if the P-value ≤ a.
H0 is not rejected if the P-value > a.
What decision would be made if a = 0.01?
For example, suppose that P-value = 0.0352 and
a = 0.05. Then, because
0.0352 ≤ 0.05
H0 would be rejected.
Recall the 5 Steps for Performing
a Hypothesis Test
Step
This Step Includes . . .
H Hypotheses
1. Describe the population characteristic of interest.
2. Translate the research question or claim into null
and alternative hypotheses.
1. Identify the appropriate test and test statistic.
M Method
C Check
C Calculate
C Communicate
Results
2. Select a significance level for the test.
1. Verify that any conditions for the selected test
are met.
1. Find the values of any sample statistics needed to
calculate the value of the test statistic.
2. Calculate the value of the test statistic.
3. Determine the P-value for the test.
1. Compare the P-value to the selected significance
level and make a decision to either reject H0 or
fail to reject H0.
2. Provide a conclusion in words that is in context
and addresses the question of interest.
Large Sample Hypothesis Test
for a Population Proportion
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
A Large-Sample Test for a
Population Proportion
Appropriate when the following conditions are met:
1.
The sample is a random sample from the population of
interest or the sample is selected in a way that would
result in a representative sample.
2. The sample size n is large. This condition is met when
both np > 10 and n (1 - p) > 10.
When these conditions are met, the following test statistic
can be used:
pˆ  p0
z 
p0 (1  P0 )
n
Where p0 is the hypothesized
value from the null hypothesis
A Large-Sample Test for a
Population Proportion Continued . . .
Null hypothesis:
H0: p = p0
When the Alternative
Hypothesis Is . . .
The P-value Is . . .
Ha: p > p0
Area under the z curve to the right of
the calculated value of the test statistic
Ha: p < p0
Area under the z curve to the left of
the calculated value of the test statistic
Ha: p ≠ p0
2·(area to the right of z) if z is positive
Or
2·(area to the left of z) if z is negative
In a study, 2205 adolescents ages 12 to 19 took a
cardiovascular treadmill test. The researchers conducting
the study believed that the sample was representative of
adolescents nationwide. Of the 2205 adolescents tested,
750 had a poor level of cardiovascular fitness. Does this
sample provide convincing evidence that more than thirty
percent of adolescents have a poor level of cardiovascular
fitness?
Hypothesis:
Let p = proportion of American adolescents who have a
poor level of cardiovascular fitness
H0: p = 0.30
Ha: p > 0.30
Cardiovascular Fitness Continued . . .
H0: p = 0.30 versus Ha: p > 0.30
Method:
Because
the aanswers
to the level,
four key
are 1) the
To select
significance
youquestions
must consider
hypothesis
testing,
2) sample Type
data, 3)
one Type
categorical
variable,
consequences
of potential
I and
II errors.
and 4) one sample, consider a large sample hypothesis test for
a population proportion.
What are the Type I and Type II errors? Which has
the most serious consequence?
Significance level:
a = 0.05
Because neither type of error is much worse than
the other, you might choose a value of 0.05.
Cardiovascular Fitness Continued . . .
H0: p = 0.30 versus Ha: p > 0.30
Check:
1. The researchers believed the sample to be representative
of adolescents nationwide.
2. The sample size is large enough because
np0 = 2205(0.3) = 661.5 ≥ 10 and
n (1 - p0) = 2205(0.7) = 1543.5 ≥ 10
Cardiovascular Fitness Continued . . .
H0: p = 0.30 versus Ha: p > 0.30
z = 4.00
P-value ≈ 0
Communicate Results:
Decision: 0 < 0.05, Reject H0
Conclusion: The sample provides convincing evidence that
more than 30% of adolescents have a poor
fitness level.
Notice that the conclusion answers the
question that was posed in the problem.
A Few Final Things to Consider
1. What about Small Samples?
In np ≥ 10 and n (1 – p) ≥ 10, the standard normal
distribution is a reasonable approximation to the
distribution of the z test statistic when the null
hypothesis is true.
If the sample size is not large enough to satisfy the large
sample conditions, the distribution of the test statistic
may be quite different from the standard normal
distribution.
Thus, you can’t use the standard normal distribution to
calculate P-values.
A Few Final Things to Consider
2. Choosing a Potential Method
Take a look back at Table 7.1 (on page 420 and also on the
inside back cover of the text).
As you get into the habit of answering the four key
questions for each new situation that you encounter, it
will become easier to use this table to select an
appropriate method in a given situation.
Power and Probability of
Type II Error
Suppose that the manager of a grocery
store is thinking about expanding the
store’s selection of organically grown
produce.
Because organically grown produce costs more than produce
that is not organically grown, the manager thinks this
expansion will be profitable only if the proportion of store
customers who would pay more for organic produce is
greater than 0.30.
What
is a Type
I error
What
a Type sample
II error
The
manager
plans
to ask each person
in aisrandom
for this context?
for to
this
context?
of customers
if he or she would be willing
pay
more for
organic produce. He will then use the resulting data to test
H0: p = 0.30 versus Ha: p > 0.30
using a significance level of a = 0.05.
Grocery Store Problem Continued . . .
H0: p = 0.30 versus Ha: p > 0.30
How likely is it that the null hypothesis is rejected?
Grocery Store Problem Continued . . .
H0: p = 0.30 versus Ha: p > 0.30
How likely is it that the null hypothesis is rejected?
Let’s investigate this further.
A package delivery service advertises that at least 90% of
all packages brought to its office by 9 a.m. for delivery in
the same city are delivered by noon that day.
Let p denote the proportion of all such packages actually
delivered by noon.
H0: p = 0.90 versus Ha: p < 0.90
Suppose the value for the proportion of all packages
delivered by noon is The
actually
p = 0.80.
Using a states that
alternative
hypothesis
significance level of 0.01the
andcompany’s
a sample of
n = is
225
claim
untrue.
packages, what is the probability that the departure
from H0 represented by this alternative value (0.80) will
be detected (H0 rejected)?
The calculations for the probabilities of
Type II errors and of power are NOT in
the AP Statistics course description.
Package Delivery Continued . . .
Let p denote the proportion of all such packages actually
delivered by noon.
H0: p = 0.90 versus Ha: p < 0.90
a = 0.01
This area of the alternative
p = 0.90
versus
pa = from
0.80
This area
of
the alternative
The
probability
departure
distribution
is0that
whenthe
H0 is
rejected.
distribution
whenhas
H0these
has not
H0sampling
represented
by
this
alternative
0.90
The
distribution
of
𝑝 whenofpvalue
=is0.80
It
represents
the
probability
been rejected.
(0.80)
will
be
detected
(H
rejected)
properties:
0
correctly rejecting a false H0, called is
It represents
the
probability
reasonable
large.
0.90
𝜇𝑝𝑝 = 0.80
Since
n
is
large,
the
distribution
the power of the test.
of a Type II error
0.02
𝜎𝑝𝑝 == 0.027
is approximately normal.
Reject
Fail to
reject
Notice:
Power + b = 1
a
0.8
b
0.9
0.9
Package Delivery Continued
. . what happens if the
Let’s. see
population
proportion
Let p denote the proportionactual
of all such
packages
actuallyis
p = 0.85.
delivered by noon.
H0: p = 0.90 versus Ha: p < 0.90
a = 0.01
Notice that the discrepancy between the
p0 = 0.90 versus pa = 0.85
hypothesized
value and the actual value of the
The sampling
distribution
of 𝑝 when
has these
population
characteristic
( |p0 –pp=a|0.85
) is smaller,
but
properties:
the probability of making a Type II error has
𝜇𝑝 = 0.85 increased. Thus,
Since power
n is large,
the distribution
is smaller.
𝜎𝑝 = 0.024
is approximately normal.
Reject
Fail to
reject
a
b
Effects of Various Factors on the
Power of a Test
• The larger the size of the differences
between the hypothesized value and the
actual value of the population
characteristic, the higher the power.
Package Delivery Continued
. . . we use a significance
Suppose
level
of 0.05.actually
Let p denote the proportion of all such
packages
delivered by noon.
H0: p = 0.90 versus Ha: p < 0.90
a = 0.05
p0 = 0.90 versus pa = 0.85
The sampling distribution of 𝑝 when p = 0.85 has these
properties:
𝜇𝑝 = 0.85
Since n is large, the distribution
Notice that the larger
𝜎𝑝 = 0.024
is approximately normal.
the significance level
Fail to
a, the smaller Reject
reject
probability of making a
Type II error. Thus,
power is larger.
a
0.85
b
0.9
Effects of Various Factors on the
Power of a Test
• The larger the size of the differences
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.
Package Delivery Continued
. . what happens if the
Let’s. see
was n =actually
500
Let p denote the proportion of sample
all suchsize
packages
instead of n = 225.
delivered by noon.
H0: psample
= 0.90 versus Ha: p < 0.90
The larger the
size, the smaller 𝜎𝑝 a = 0.05
p0 = 0.90 versus pa = 0.85
The sampling distribution of 𝑝 when p = 0.9 has a standard deviation of
Notice that the probability of a
𝜎𝑝 = 0.0134.
Type II error (b) is much smaller,
The sampling distribution of 𝑝 when p = 0.85 has a standard deviation of
the power of the test larger.
𝜎making
𝑝 = 0.0159.
The shorter,
taller curves
dotted
are
the curves
sampling
are the sampling
distributions
distributions
when
n = 500.
when n = 225.
Reject
Fail to
reject
a
0.85
b
0.9
Effects of Various Factors on the
Power of a Test
• The larger the size of the differences
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
Avoid These Common
Mistakes
Be sure to include all the relevant
information:
1. Hypothesis. Whether specified in symbols or
described in words, it is important that both the null
and the alternative hypothesis be clearly stated. If
using symbols, be sure to define them in the context
of the problem.
2. Test procedure. You should be clear about what test
procedure was used and why you think it was
reasonable to use this procedure.
3. Test statistic. Be sure to include the value of the
test statistic and the associated P-value.
4. Conclusion in context. Always provide a conclusion
that is in the context of the problem and that
answers the question posed.
Avoid These Common Mistakes
1. A hypothesis test can never show strong
support for the null hypothesis. Make sure
that you don’t confuse “There is no reason to
believe the null is not true” with the statement
“There is convincing evidence that the null
hypothesis is true”. These are very different
statements!
This is like saying the defendant
is “innocent” instead of “not
guilty”.
Avoid These Common Mistakes
2. If you have complete information for the
population (census), don’t carry out a
hypothesis test!
Avoid These Common Mistakes
3. Don’t confuse statistical significance with
practical significance. When a null hypothesis
has been rejected, be sure to step back and
evaluate the result in the light of its practical
importance.
For example, you may be convinced that the proportion
who respond favorably to a proposed new medical
treatment is greater than p0 = 0.4. But if your
estimate for the proposed new treatment is 0.405, it
may not be of any practical use.