Transcript chapter12 - mistergallagher

Chapter 12
Asking and Answering
Questions About A
Population Mean
Created by Kathy Fritz
Let's review some statistical notation.
n
the sample size
𝒙
the mean of a sample
s
the standard deviation of a sample
m
the mean of the entire population
s
the standard deviation of the entire
population
When the purpose of a statistical study is to
learn about a population mean m, the sample mean
𝑥 can be used as an estimate of m.
To understand statistical inference procedures
based on 𝑥, you must first study how sampling
variability causes 𝑥 to vary in value from one
sample to another.
The sample size n and characteristics of the
population (its shape, mean value m, and standard
deviation s) are important in determining the
sampling distribution of 𝑥.
The Sampling Distribution of
the Sample Mean
Properties of the Sampling
Distribution of 𝑥
Central Limit Theorem
The paper “Mean Platelet Volume in Patients with Metabolic
Syndrome
and use
Its aRelationship
with Coronary
Artery
We can
statistical software
package
to select
Disease”500
(Thrombosis
Research
includes
data that
random samples
of n, =2007)
5 from
the population.
suggests that the distribution of
The sample mean platelet volume 𝑥 was computed for
= platelet
volume
each sample,xand
these 500
values were used to
What values
fornot
thehave
sample
meanhistogram.
would
be expected
for patients
who construct
do
metabolic
syndrome
is
a density
if you were
to take
a random
of size
5 from
approximately
normal
with
mean m =sample
8.25 and
standard
this population distribution?
deviation s = 0.75.
Platelet Volume Continued . . .
To
investigate
the effect
sample
size
Notice
that a sample
meanof
may
be 7.3
or on
9.1.the
A sample
sizeof
of
from
the
population
of patients
behavior
selected
500
samples
of sizewon’t
10,
What
if𝑥5,a we
larger
sample
is selected?
always
precise
information
about
500 provide
samplesvery
of size
20, and
500 samples
ofthe
sizemean
30.
volume
in the
population.
Densityplatelet
histograms
of the
resulting
𝑥 are created.
What do
do you
you notice
notice
What
about the
the means
standard
shape of
about
of
deviation
of these
these
distributions?
distributions?
The paper “Is the Overtime Period in an NHL Game Long
Enough?” (American
Statistician
, 2008)
gave data
on thewe
time
Using
a statistical
software
package,
(in minutes) from selected
the start 500
of the
game toofthe
first
goal
samples
each
sample
scored for the 281 regular
sizes n season
= 5, n =games
10, nfrom
= 20,the
n =200530.
2006 season that went into overtime. The density histogram
for the data is shown below.
We then constructed a histogram of the
500 𝑥 values for each of the four sample
sizes. Let’s consider
these 281 values as
a population. The
distribution is
strongly positively
skewed with mean
m = 13 minutes and
with a median of 10
minutes.
What do you notice
What
dodistributions
you
notice
Are
these
about
the
standard
about
the of
shape
of
centered
at
deviations
these
these
distributions?
approximately
m = 13?
distributions?
General Properties of Sampling
Distributions of 𝑥
Rule 1: 𝜇𝑥 = 𝜇
Rule 2: 𝜎𝑥 =
𝜎
𝑛
This rule is exact if the population is infinite, and is
approximately correct if the population is finite and
no more than 10% of the population is included in
the sample.
General Properties Continued . .
Rule 3:
When the population distribution is
normal, the sampling distribution of 𝑥 is
also normal for any sample size n.
n = 16
n=4
Population
General Properties Continued . . .
Rule 4: Central Limit Theorem
When n is sufficiently large, the sampling
distribution of 𝑥 is well approximated by
a normal curve,
even
when the population
How large
is “sufficiently
distribution is large”
not itself
normal.
anyway?
n = 16
n=4
Population
CLT can
safely be
applied if
n ≥ 30.
In a study of the courtship of mating scorpion flies, one
variable of interest was x = courtship time, which was
defined as the time from the beginning of a female-male
interaction until mating. Data suggest that it is
reasonable to think that the population mean and standard
deviation of 𝑥 are m = 117.1 minutes and s = 109.1 minutes.
The
distribution
of 𝑥 = of
mean
courtship time,
time for
Dosampling
you think
the distribution
x, courtship
is
a random sample
of 20 scorpion
fly mating
pairs would
approximately
normal?
Explain.
have mean
𝜇𝑥 = 𝜇 = 117.1 minutes
The standard deviation of 𝑥 is
𝜎
109.1
𝜎𝑥 =
=
= 24.40
𝑛
20
It is not reasonable to assume that
the shape of the sampling distribution
of 𝑥 is normal.
A hot dog manufacturer claims that one of its brands of hot
dogs has a mean fat content of m = 18 grams per hot dog.
Consumers of this brand would probably not be disturbed if the
mean was less than 18 grams, but would be unhappy if it
exceeded 18 grams.
In this situation, the variable of interest is
x = fat content of a hot dog
For purposes of this example, suppose we know that s, the
standard deviation of the x distribution, is equal to 1 gram.
An independent testing organization is asked to analyze a
random sample of 36 hot dogs. The fat content for each of the
36 hot dogs is measured and the sample mean is calculated to be
𝑥 = 18.4 grams.
Does this result suggest that the manufacturer’s
claim that the population mean is 18 is incorrect?
Hot Dogs Continued . . .
Let’s look at the sampling distribution of 𝑥 :
• The sample size, n = 36, is large enough to say that the
sampling distribution of 𝑥 will be approximately normal.
• The standard deviation
of the
𝑥 distribution
is
Since the
sample
size is
greater𝜎 than130, the Central
𝜎𝑥 =
=
= 0.1667
Limit𝑛 Theorem
36 applies.
• If the manufacturer’s claim is correct, you also know
that
𝜇𝑥 = 𝜇 = 18
Hot Dogs Continued . . .
You know that even if m = 18, 𝑥 will not usually be exactly
18 due to sampling variability. But, is it likely that you
would see a sample mean at least as large as 18.4 when the
population mean is really 18?
Using the normal distribution, you can compute the
probability of observing a sample mean this large. If the
manufacturer’s claim is correct,
The value 𝑥 = 18.4 is enough greater
than 18 that
18.4 − 18
𝑃(be
𝑥 ≥skeptical
18.4) ≈ of
𝑃 the
𝑧 ≥manufacturer’s claim.
you should
0.1667
= 𝑃 𝑧 ≥ 2.4 = 0.0082
Values of 𝑥 as large as 18.4 will be observed
only about 0.82% of the time when a random
sample of size 36 is taken from a population
with m = 18 and s = 1.
A Confidence Interval for a
Population Mean
t-distributions
A One-Sample t Confidence
Interval for m
Confidence intervals for m when
s is known
The general formula for a confidence interval estimate is
standard error
statistic ± (critical value)
of the statistic
From Section 12.1, you know that:
1. The sampling distribution of 𝑥 is centered at m.
𝜎
2. The standard deviation
of
𝑥
is
𝜎
=
.
𝑥
z critical value
𝑛
because
the
3. As long as n is large
(n ≥ 30),
thesampling
sampling distribution of
𝑥statistic
is approximately
distribution of 𝑥 is
that normal.
approximately normal
provides
an
standard
This suggests that a confidence interval for a population
when n is large
estimate
of
m
error
mean when the sample size is large and s is known
is .of
. .𝑥
𝑥 ± 𝑧 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝜎
𝑛
Cosmic radiation levels rise with increasing
altitude, promoting researchers to consider how
pilots and flight crews might be affected by
increased exposure to cosmic radiation. A study reported
It is
almost
neverradiation
the casedose
thatof
you
would
know
a mean
annual
cosmic
219
mrem
forthe
a
value
the personnel
populationof
mean
(which
is why you
would
sample
of of
flight
Xinjiang
Airlines.
Suppose
be using
sample
to estimate
but would
this mean
is based
on data
a random
sample it)
of 100
flight not
crew
Based
on
this
sample,
plausible
values
of
m
,
the
mean
know and
the that
valuesof
themrems.
population
standard deviation.
members
=
35
Let:
annual cosmic radiation exposure for all Xinjiang
Airlines
flight
crewcosmic
members,
are between
212.14
mFor
= mean
annual
exposure
for
alland
this
reason,
you
willradiation
probably
never use
this
225.86
mrem.
Xinjiang
Airlines
flight
crew
members
confidence
interval
formula.
A 95% confidence interval for m is . . .
Let’s see what to do when the population standard
35
deviation𝜎s is unknown.
𝑥 ± 𝑧 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑛
= 219 ± (1.96)
100
= (212.15, 225.86)
Important Properties of t
Distributions
The t distribution
corresponding
to anynormal
particular
Just as there
are many different
number
of degreesthere
of freedom
bell different
shaped and
distributions,
are alsoismany
centered at
zero (just like
the standardare
normal (z)
t distributions.
t distributions
distribution).
distinguished by a positive whole number
called degrees
ofspread
freedom
(df).
2) Each t distribution
is more
out
than the
standard normal distribution.
1)
z curve
t curve for 2 df
0
Why is the z
curve taller
than the t curve
for 2 df?
Important Properties of t
Distributions Continued . . .
3) As the number of degrees of freedom increases,
the spread of the corresponding t distribution
decreases.
t curve for 8 df
t curve for 2 df
0
Important Properties of t
Distributions Continued . . .
4) As the number of degrees ofFor
freedom
increases,
what df
would the
the corresponding sequence of t tdistributions
distribution be
approaches the standard normalapproximately
distribution. the
same as a standard
normal distribution?
z curve
t curve for 2 df
t curve for 5 df
0
Finding t critical values
Table 3 t Critical Values
Central area captured / Confidence level
df
.80
80%
.90
90%
.95
95%
.98
98%
.99
99%
.998
99.8%
.999
99.9%
1
3.08
6.31
12.71
31.82
63.66
318.31
636.62
2
1.89
2.92
4.30
3
1.64
2.35
3.18
4
1.53
2.13
2.78
5
1.48
2.02
2.57
6
1.44
4.94
2.45
7
1.42
4.90
2.37
8
1.40
1.86
2.31
2.90
3.36
4.50
5.04
9
1.38
1.83
2.26
2.82
3.25
4.30
4.78
10
1.37
1.81
2.23
2.76
3.17
4.14
4.59
Suppose
you
wish 31.60
to
9.93
23.53
compute a 95%
4.54
5.84
10.21
12.92
confidence interval using
3.75
4.60
8.61
sample
data. 7.17
The sample
3.37
4.03
size is n5.89
= 10. 6.86
3.14
3.71= 10 –
5.21
df
1 = 9 5.96
3.00 t critical
3.50 value
4.79is 2.26
5.41
The
6.97
One-Sample t Confidence Interval
for a Population Mean m
Appropriate when the following conditions are met:
This isfrom
usedthe
when
the population
• The sample is a random sample
population
of
s results
is unknown.
interest or the sample is standard
selected indeviation
a way that
in a
sample that is representative of the population.
•
The sample size is large (n ≥ 30) or the population
distribution is normal. t critical values are found
in Table 3 or using
technologyinterval for
When these conditions are met, a confidence
the population mean is
𝑠
𝑥 ± (𝑡 critical value)
𝑛
The t critical value is based on df = n – 1 and the desired confidence
level.
One-Sample t Confidence Interval
for a Population Mean m Continued …
Interpretation of Confidence Interval
You can be confident that the actual value of the
population mean is included in the computed interval. In a
given problem, this statement should be worded in
context.
Interpretation of Confidence Level
The confidence level specifies the long-run proportion of
the time that this method is expected to be successful in
capturing the actual population mean.
During a flu outbreak, many people visit emergency rooms.
Before being treated, they often spend time in crowded
waiting rooms where other patients may be exposed. A study
was performed investigating a drive-through model where flu
patients are evaluated while they remain in their cars.
In the study, 38 people were each given a scenario for a flu
case that was selected at random from the set of all flu cases
actually seen in the emergency room. The scenarios provided
the “patient” with a medical history and a description of
were
interested
estimating
the
symptomsResearchers
that would allow
the
patient to in
respond
to questions
from themean
examining
physician.
processing
time for flu patients using the
drive-through
model.
The patients were processed
using a drive-through
procedure
that was implemented in the parking structure of Stanford
University Use
Hospital.
The time to process
each this
case mean.
from
95% confidence
to estimate
admission to discharge was recorded.
The following sample statistics were computed from the
data:
n = 38
𝒙 = 26 minutes
s = 1.57 minutes
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 1 (Estimate):
You want to estimate the value of m, the mean time to
process a flu case using the new drive-through model.
Step 2 (Method):
Because the answers to the four key questions are: 1)
estimation, 2) sample data, 3) one numerical variable, and
4) one sample, consider using a one-sample t
confidence interval for a population mean. A
confidence level of 95% was specified for this
example.
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 3 (Check):
• Because the sample size is large (38 > 30), you do not
need to worry about whether the population
distribution is approximately normal.
• The sample is a random sample because the 38 flu
cases were randomly selected from the population of
all flu cases seen at the emergency room
Step 4 (Calculate):
t critical value (from Table 3) = 2.02
𝑥 ± 𝑡 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑠
1.57
= 26 ± 2.02
𝑛
38
= (25.486, 26.514)
Drive-through Model Continued . . .
The following sample statistics were computed from the data:
n = 38
𝑥 = 26 minutes
s = 1.57 minutes
Step 5 (Communicate Results):
Confidence Interval
You can be 95% confident that the actual mean
processing time for emergency room flu cases using the
new drive-through model is between 25.486 minutes and
26.514 minutes
Confidence Level
The method used to construct this interval
estimate is successful in capturing the actual
value of the population mean about 95% of the
time.
In a study, seven chimpanzees learned to use
an apparatus that dispensed food when
either of two ropes was pulled.
When one of the ropes was pulled, only the chimp
controlling the apparatus received food. When the other
rope was pulled, food was dispensed both to the chimp
controlling the apparatus and also to a chimp in the
adjoining cage.
The accompanying data represent the number of times out
of 36 trials that each of seven chimps chose the option
that would provide food to both chimps.
23 22 21 24 19 20 20
Construct a 99% confidence interval for the mean number
of times out of 36 trials chimpanzees would choose the
option that would provide food to both chimps.
Chimp Problem Continued . . .
Step 1 (Estimate):
The mean number of times out of 36 chimps choose the
charitable response, m, will be estimated.
Step 2 (Method):
Because the answers to the four key questions are
estimation, sample data, one numerical variable, and one
sample, a one-sample t confidence interval for a population
mean will be considered. A confidence level of 99% was
specified.
Chimp Problem Continued . . .
Step 3 (Check):
1. It was stated that it is reasonable to regard the
sample as representative of the population
2. The normal probability
plot is reasonably
straight, so it seems
plausible that the
Because the
sample size is small (n = 7), you need
population
distribution
consider whether it is reasonable to think that
isto
approximately
the distribution of the number of charitable
normal.
responses out of 36 for the population of all
chimps is at least approximately normal. This is
typically done by plotting the data.
Chimp Problem Continued . . .
Step 4 (Calculation):
𝑥 = 21.29 𝑠 = 1.80
𝑥 ± 𝑡 critical value
𝑠
= 21.29 ± 3.71
𝑛
= (18.77, 23.81)
1.80
7
Step 5 (Communicate Results):
Based on this sample, you can be 99% confident that the
population mean number of charitable responses (out of 36
trials) is between 18.76 and 23.81.
Choosing a Sample Size
The sample size required to estimate a population
mean m with a specified margin of error M is
𝜎
𝑀 =Recall,
1.96 the margin of error is
𝑛 likelySolving
the maximum
estimation
for n, you
error expected whenget
the. . .
2
statistic
is used as an estimator.
1.96𝜎
𝑛=
𝑀
If the value of s is unknown, it may be estimated based
on previous information or, for a population that is not
range
skewed, by using
.
4
A college financial advisor wants to estimate the
mean cost of textbooks per quarter for students
at the college. For the estimate to be useful, it
should have a margin of error of $20 or less.
How large a sample should be used to be confident
of achieving this level of accuracy?
Suppose the financial advisor thinks that the
amount spent on books varies widely, but that
most values are between $150 to $550.
A reasonable estimate
of s isup,
: a sample size of 97 or
Rounding
𝑟𝑎𝑛𝑔𝑒 550 − 150
=larger is recommended.
= 100
4
4
Using this estimate of the population standard deviation,
the required sample size is:
2
2
1.96𝜎
1.96(100)
𝑛=
=
= 96.04
𝑀
4
Testing Hypotheses About a
Population Mean
Hypotheses:
When testing hypotheses about a population
mean, the null hypothesis will have the form:
H0: m = m0
where m0 is a particular hypothesized value.
The alternative hypothesis has one of the
following three forms, depending on the research
question being addressed:
Ha: m > m0
Ha: m < m0
Ha: m ≠ m0
Test Statistic
If n is large (n ≥ 30) or if the population
distribution is approximately normal, the
appropriate test statistic is
𝑥−𝜇
𝑡= 𝑠
𝑛
If the null hypothesis is true, this test statistic has a
t distribution with df = n – 1.
This means that the P-value for a hypothesis test about a
population mean will be based on a t distribution and not
the standard normal distribution.
Computing P-values
1. Upper-tailed test:
Ha: m > m0
t curve
2. Lower-tailed test:
Ha: m < m0
P-value = area
P-value = area
in upper tail
in lower tail
Calculated t
3. Two-tailed test:
Ha: m ≠ m0
t curve
Calculated -t
t curve
P-value = sum
of area in
two tails
Calculated –t and t
The One-Sample t-test for a
Population Mean
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 results in a
sample that is representative of the population.
2. The sample size is large (n ≥ 30) or the population distribution
is normal.
When these conditions are met, the following test statistic
can be used:
𝑥 − 𝜇0
𝑡= 𝑠
𝑛
Where m0 is the hypothesized value from the null hypothesis.
The One-Sample t-test for a
Population Mean Continued . . .
Null hypothesis:
H0: m = m0
When the conditions are met and the null hypothesis is true, the t test
statistic has a t distribution with df = n – 1.
When the Alternative
Hypothesis Is . . .
The P-value Is . . .
Ha: m > m0
Area under the t curve to the right of
the calculated value of the test statistic
Ha: m < m0
Area under the t curve to the left of
the calculated value of the test statistic
Ha: m ≠ m0
2·(area to the right of t) if t is positive
Or
2·(area to the left of t) if t is negative
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 asked to estimate how much
time had elapsed during a 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
deviation of the sample?
n = 20
Smoking Abstinence Continued . . .
Step 1 (Hypotheses):
The population mean is
Youperceived
should choose
a significance
level based
m = mean
elapsed
time for smokers
who have
on a consideration
of
abstained
from smoking of
forthe
24consequences
hours
Type I and Type II errors.
Null hypothesis:
H0:neither
m = 45 type of
In this situation, because
errorhypothesis:
is much more serious
the other, a
Alternative
Ha:than
m > 45
value for a of 0.05 is a reasonable choice.
Step 2 (Method):
Because the answers to the four key questions are:
1) hypothesis test, 2) sample data, 3) one numerical
variable, and 4) one sample, consider a one-sample t test
for a population mean. When the null hypothesis is true,
this statistic will have a t distribution with df = 20 – 1 = 19.
Significance level: a = 0.05
Smoking Abstinence Continued . . .
Step 3 (Check):
• The researchers conducting the study indicated that
they believed that the sample was selected in a way that
would result in a sample that was representative of all
smokers in general.
• Because n is only 20 in this example, you need to verify
that the normality condition is reasonable.
A boxplot of the sample data is
shown here. Although the boxplot
is not perfectly symmetric, it is
not too skewed and there are no
outliers. It is reasonable to think that the population
distribution is at least approximately normal.
Smoking Abstinence Continued . . .
Step 4 (Calculate):
n = 20
Test statistic:
s = 9.84
𝑥 = 59.30
𝑡=
59.30−45
9.84
20
= 6.50
This is an upper-tailed test, so the P-value is the area under the t curve
with df = 19 to the right side of the computed t value.
Associated P-value:
P-value = area under t curve to the right of 6.50
= P(t > 6.50) ≈ 0
Smoking Abstinence Continued . . .
Step 5 (Communicate Results):
Decision:
0 < 0.05, Reject H0
Conclusion: There is convincing evidence that the mean
perceived time elapsed is greater than the actual time
elapsed of 45 seconds. It would be very unlikely to see a
sample mean this extreme just by chance when H0 is true.
Statistical Versus Practical
Significance
Carrying out a hypothesis test amounts to deciding
whether the value obtained for the test statistic could
plausibly have resulted when H0 is true.
When the value of the test statistic leads to rejection of
H0, it is customary to say that the result is statistically
significant a the chosen significance level a.
However, statistical significance does NOT mean that the
true situation differs from what the null states in any
practical sense.
See the following example.
Let m denote the actual mean score on a standardized test
for children in a large school district. The mean score for
all children in the United States is known to be 100.
District administrators are interested in testing
H0: m = 100 versus Ha: m >100 using a significance level of
a = 0.001. Data from a random sample of 2500 children in
the district resulted in 𝑥 = 101.0 and s = 15.0. Minitab
output from a one-sample t test is shown here:
One-Sample T
Test of mu = 100 vs > 100
n
Mean
2500
101.000
From a practical point95%
of view, a 1-point
Lower
difference
here
may
not be important.
StDev
SEMean
Bound
T
15.000
0.300
100.506
3.33
P
0.000
From the Minitab output, P-value ≈ 0, so H0 is rejected.
But, there is only a difference of 1 between the sample
mean of 101 and the population mean of 100.
Avoid These Common
Mistakes
Avoid These Common Mistakes
1. You will need to think about the distinction
between proportions and means when choosing
an appropriate method.
The best way to distinguish which method to use is to
focus on the type of data -
Think
Numerical
Think
Categorical
Proportions
Means
Avoid These Common Mistakes
2. Be sure to keep in mind that conditions are
important. Use of the one-sample t confidence
interval and hypothesis test REQUIRE that
conditions are met.
Be sure to check these conditions are met before
using these methods.
Avoid These Common Mistakes
3. Remember that the results of a hypothesis
test can never provide strong support for the
null hypothesis. Make sure that you don’t
confuse “I am not convinced that the null is
false” with the statement “I am convinced that
the null hypothesis is true”. These are not the
same!