Transcript PRCACalculus

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Chapter 11
Comparing Two Populations or
Treatments
Suppose we have a population of adult men
with a mean height of 71 inches and
standard deviation of 2.5 inches. We also have a
population of adult women with a mean height of
65 inches and standard deviation of 2.3 inches.
Assume heights are normally distributed.
Suppose we take a random sample of 30 men and
a random sample of 25 women from their
respective populations and calculate the
difference in their heights (man’s height –
woman’s height).
If we did this many times, what would the
distribution of differences be like?
Male Heights
Female Heights
Randomly take
one of the sample
means for the
71 sM = 2.5
65 sF = 2.3
males
and one of
Suppose we took repeated
Suppose
we took repeated
thesamples
sample
means
samples of size n = 25
from
the
of size n = 30 from the
for the females
population of female heights
population
and of male heights and
and find the
calculated the sampledifference
calculated
means. in the sample means.
We would have the sampling
We heights.
would have the sampling
mean
distribution of xF
71
s xM 
Doing this
repeatedly, we will
create the sampling
distribution
of (xM – xF)
2.5
30
distribution of xM.
65
xM - xF
s xF 
2. 3
25
 2.5 
2
 2.3 
s x M xF  
  

 30   25 
6
2
Heights Continued . . .
•
Describe the sampling distribution of the
difference in mean heights between men and
women.
The sampling distribution is normally
2.5 2 2.3 2
distributed with
s


x M xF  71  65  6
•
x M x F
30
25
What is the probability that the difference in
mean heights of a random sample of 30 men
and a random sample of 25 women is less
than 5 inches?
P ((xM  xF )  5)  .0614
6
Properties of the Sampling
Distribution of x1 – x2
If the random samples on which x1 and x2 are based are
selected independently of one another, then
1. x1 x2 
x1  x2  1  2
2
2
1
2
distribution
s
s
2
2
s
s
1
2
of x1s– x2 is 
always

and
x1 x 2
2
2
2
The
sampling
s

s

s

2. x1 x 2
x1
x2 
Meancentered
value of
nof
n2
at thenvalue
1
2 1 – 2, so x1 – x2 isnan
1
x1 –unbiased
x2
statistic for estimating 1 – 2.
3. InThe
n1 and
n2 areofboth
large or theispopulation distributions
variance
the differences
are (at least
approximately)
normal, x1 and x2 each have (at
the sum
of the variances.
least approximately) normal distributions. This implies that
the sampling distribution of x1 – x2 is also (approximately)
normal.
The properties for the sampling distribution of x1
– x2 implies that x1 – x2 can be standardized to
obtain a variable with a sampling distribution
that is approximately the standard normal (z)
distribution.
When two random samples are independently selected
and n1 and n2 are both large or the population We must
s1 and
s2 iss1
distributions are (at least approximately)If
normal,
the
know
distribution of
unknown
we
and
s
in
2
x1  x2  ( 1  2 )
must
useto
t use
z 
order
2
2
s1 s2
distributions.
this

n1
n2
procedure.
is described (at least approximately) by the standard
normal (z) distribution.
Two-Sample t Test for Comparing
Two Populations
Null Hypothesis: H0: 1 – 2 = hypothesized value
Test Statistic:
t 
x1  x2  hypothesiz ed value
2
2
s
s
1
2
A conservative
of the P estimate
The hypothesized
is tn1 found
n2 by value
value
can be
using the
often
0, but
areoftimes
curve with
thethere
number
degrees
The appropriate df for the two-sample t test is
of
freedom
equal
to the smaller
of
when
we
are
interested
in

V1 V2 2
(n1 a
– 1)
or (n2 – 1).that
2
2
testing
for
difference
is
df 
s
s
1
2
V

where V1  not
and
V1 2
V22
2

n1 0.
n2
n1  1 n2  1
The computed number of df should be truncated to an integer.
Two-Sample t Test for Comparing
Two Populations Continued . . .
Null Hypothesis: H0: 1 – 2 = hypothesized value
Alternative Hypothesis:
P-value:
Ha: 1 – 2 > hypothesized value
Area under the appropriate t
curve to the right of the
computed t
Ha: 1 – 2
Area under the appropriate t
< hypothesized value curve to the left of the
computed t
Ha: 1 – 2 ≠ hypothesized
value
2(area to right of computed t)
if +t or
2(area to left of computed t) if
-t
Another Way to Write Hypothesis
Statements:
H0: 1 =
- 22= 0
Ha: 1 <
- 22< 0
Ha: 1 >
- 22> 0
Ha: 1 -≠22≠ 0
When the
hypothesized
value is 0, we
Be sure to
can rewrite
define
BOTH
these
1 and 2!
hypothesis
statements:
Two-Sample t Test for Comparing
Two Populations Continued . . .
Assumptions:
1) The two samples are independently selected
random samples from the populations of interest
2) The sample sizes are large (generally 30 or larger)
or the population distributions are (at least
approximately) normal.
When comparing two treatment groups, use the
following assumptions:
1) Individuals or objects are randomly assigned to
treatments (or vice versa)
2) The sample sizes are large (generally 30 or larger)
or the treatment response distributions are
approximately normal.
Are women still paid less than men for comparable
work? A study was carried out in which salary data
was collected from a random sample of men and from a
random sample of women who worked as purchasing
managers and who were subscribers to Purchasing
magazine. Annual salaries (in thousands of dollars) appear
below (the actual sample sizes were much larger). Use a =
.05 to determine
there isconvincing
evidence
that the
If we hadifdefined
as
the
mean
salary
1 purchasing managers is
mean annual salary for male
for female
purchasing
and 2 purchasing
greater than
the mean
annual managers
salary for female
as the mean salary for male purchasing
managers.
managers, then the correct alternative
hypothesis would be the difference in the
Men
81 69 81 76 76 74 69 76 79 65
means is less than 0.
Women 78 60 67 61 62 73 71 58 68 48
H0: 1 – 2 = 0
Ha: 1 – 2 > 0
Where 1 = mean annual salary for male
State the hypotheses:
purchasing managers and 2 = mean annual
salary for female purchasing managers
Salary War Continued . . .
Men
81 69
Wome 78 60
nH :  –  = 0
0
1
2
Ha: 1 – 2 > 0
81
67
76
61
76
62
74
73
69
71
76
58
79
68
65
48
Where 1 = mean annual salary for male
purchasing managers and 2 = mean annual
salary for female purchasing managers
Assumptions:
1)Given two independently selected random samples of male
and female purchasing managers.
Men
2) Since
the
sample
sizes
are
small,
we
must
Even thoughVerify
these the
are assumptions
samples from subscribers of
determine if it is plausible that the sampling Women
Purchasingformagazine,
of the study believed
distributions
each of thethe
twoauthors
populations
are approximately
normal. Since
the boxplots
it was reasonable
to view
the samples as 60
are reasonably
symmetrical
with
no outliers, of
it interest.
representative
of
the
populations
is plausible that the sampling distributions are
approximately normal.
80
Salary War Continued . . .
Men
81
69
81
76
76
74
69
76
79
65
Women 78
60
67
61
62
73
71
58
68
48
Where 1 = mean annual salary for male
H0: 1 – 2 = 0
purchasing managers and 2 = mean annual
Ha: 1 –What
2 > 0potential
typeforerror
salary
female purchasing managers
could we have made with
74.6  64.6  0  3.11
this conclusion?
t

Test Statistic:
2
(round down) this
8.62
Type I 5.4  Truncate
value.
10
10
P-value =.004
a = .05
Now find the area to the 2.916  7.3962
Since theright
P-value
< 3.11
a, weinreject
is convincing
dft-H0. There
 15 .14  15
of t =
the
2
2
.916the
7test
.396statistic
evidence that the mean salary
for2male
purchasing
Compute
 for
with
df =
15.mean
To find
the
P-value,
first
managers iscurve
higher
than
the
salary
female
9
9
and P-value
purchasing managers.
find the appropriate df.
The Two-Sample t Confidence Interval for the
Difference Between Two Population or
Treatment Means
The general formula for a confidence interval for 1 – 2
when
1) The two samples are independently selected random samples from
the populations of interest
2) The sample sizes are large (generally 30 or larger) or the population
distributions are (at least approximately) normal.
s12 s22
isFor a comparison
x1  x2  of(ttwo
critical
value) use the
 following
treatments,
n1 n2
assumptions:
The
t critical value
is based
onrandomly assigned to
1) Individuals
or objects
are

V1 V2 2(or vice versa)
treatments
s12
s22
df 
V2 
where V1  n
and
V1 2
V22
n2
1
2) The sample
sizes are large (generally
30 or larger)
or

n1  1 n2  1
the treatment response distributions are approximately
df should be truncated to an integer.
normal.
In a study on food intake after sleep deprivation,
men were randomly assigned to one of two treatment
groups. The experimental group were required to sleep
only 4 hours on each of two nights, while the control group
were required to sleep 8 hours on each of two nights. The
amount of food intake (Kcal) on the day following the two
nights of sleep was measured. Compute a 95% confidence
interval for the true difference in the mean food intake for
the two sleeping conditions.
4-hour
sleep
3585
4470
3068
5338
2221
4791
4435
3187
3901
3868
3869
4878
3632
4518
8-hour
sleep
4965
3918
1987
4993
5220
3653
3510
4100
5792
4547
3319
3336
4304
4057
3099
3338
the mean xand
standard deviation for
x4 = 3924 s4 =Find
829.67
8 = 4069.27 s8 = 952.90
each treatment.
Food Intake Study Continued . . .
4-hour
sleep
3585
4470
3068
5338
2221
4791
4435
3187
3901
3868
3869
4878
3632
4518
8-hour
sleep
4965
3918
1987
4993
5220
3653
3510
4100
5792
4547
3319
3336
4304
4057
x4 = 3924 s4 = 829.67
3099
3338
x8 = 4069.27 s8 = 952.90
Assumptions:
1) Men were randomly assigned to two treatment groups
Verify the assumptions.
2) The assumption of normal response 4-hour
distributions is plausible because
8-hour
both boxplots are approximately
4000
symmetrical with no outliers.
Food Intake Study Continued . . .
4-hour
sleep
3585
4470
3068
5338
2221
4791
4435
3187
3901
3868
3869
4878
3632
4518
8-hour
sleep
4965
3918
1987
4993
5220
3653
3510
4100
5792
4547
3319
3336
4304
4057
3099
3338
x4 = 3924
s4 =upon
829.67
x8 = is
4069.27
Based
this interval,
there a s8 = 952.90
significant difference in the mean food
No,
since
is intwo
thesleeping
confidence
interval,
829.67 2conditions?
952
.902 there is not
intake
for0 the
(3924  4069.27)  2.052

 ( 814.1, 523.6)
convincing evidence that15the
mean
food
intake
for the
15the interval.
Calculate
two sleep conditions are different.
We are 95% confident that the true difference in the mean
Interpret the interval in context.
food intake for the two sleeping conditions is between 814.1 Kcal and 523.6 Kcal.
Pooled t Test
• Used when the variances of the two
populations are equal (s1 = s2)
• CombinesP-values
information
from
both
computed
using
thesamples
pooled t to
create a “pooled”
thethe
common
procedureestimate
can be farof
from
actual
variance which
in place variances
of the two
P-valueisif used
the population
are not equal.
sample standard deviations
When the population variances are equal,
• Is not the
widely
used
due to is
itsbetter
sensitivity
to any
pooled
t procedure
at detecting
departure
from the
variance
assumption
deviations
fromequal
H0 than
the two-sample
t
test.
Suppose that an investigator wants to determine
if regular aerobic exercise improves blood
pressure. A random sample of people who jog
regularly and a second random sample of people
who do not exercise regularly are selected
independently of one another.
Can we conclude that the difference in mean blood
pressure is attributed to jogging?
What about other factors like weight?
One way to avoid these difficulties
would be to pair subjects by weight
then assign one of the pair to jogging
and the other to no exercise.
Summary of the Paired t test for Comparing
Two Population or Treatment Means
Null Hypothesis: H0: d = hypothesized value
xd  hypothesiz ed value
Where d tisTthe
he mean
hypothesized
of the value is
Test Statistic:
sd
differences
usually
in the
0 –paired
meaning
that there
n
is no and
difference.
Where n is the number observations
of sample differences
xd and sd are the mean and
standard deviation of the sample differences. This test is based on df = n – 1.
Alternative Hypothesis:
Ha: d > hypothesized value
Ha: d < hypothesized value
Ha: d ≠ hypothesized value
P-value:
Area to the right of calculated t
Area to the left of calculated t
2(area to the right of t) if +t
or 2(area to the left of t) if -t
Summary of the Paired t test for Comparing
Two Population or Treatment Means
Continued . . .
Assumptions:
1. The samples are paired.
2. The n sample differences can be viewed as a
random sample from a population of differences.
3. The number of sample differences is large
(generally at least 30) or the population
distribution of differences is (at least
approximately) normal.
Is this an example of paired samples?
An engineering association wants to see if
there is a difference in the mean annual
salary for electrical engineers and chemical
engineers. A random sample of electrical
engineers is surveyed about their annual
income. Another random sample of chemical
engineers is surveyed about their annual
income.
No, there is no pairing of
individuals, you have two
independent samples
Is this an example of paired samples?
A pharmaceutical company wants to test its
new weight-loss drug. Before giving the drug
to volunteers, company researchers weigh
each person. After a month of using the drug,
each person’s weight is measured again.
Yes, you have two observations on
each individual, resulting in paired
data.
Can playing chess improve your memory? In a study,
students who had not previously played chess participated
in a program in which they took chess lessons and played
chess daily for 9 months. Each student took a memory test
before starting the chess program and again at the end of
the 9-month period.
If we had subtracted Post-test minus
Pre-test,
then
Student
1
2
3
4
5
6
7
8the alternative
9
10
11
12
hypothesis
be the
Pre-test
510 610 640 675 600
550 610would
625 450
720mean
575 675
difference
greater
0. 680
Post-test
850 790 850 775 700
775 700 is850
690 than
775 540
Difference
-340
-180
-210
-100
-100
-225
-90
-225
-240
-55
35
H0: d = 0
First, find the differences
H a: 
d < 0 the hypotheses.
State
pre-test minus post-test.
Where d is the mean memory score difference
between students with no chess training and
students who have completed chess training
-5
Playing Chess Continued . . .
Student
1
2
3
4
5
6
7
8
9
10
11
12
Pre-test
510
610
640
675
600
550
610
625
450
720
575
675
Post-test
850
790
850
775
700
775
700
850
690
775
540
680
Difference
-340
-180
-210
-100
-100
-225
-90
-225
-240
-55
35
-5
H 0 : d = 0
H a : d < 0
Where d is the mean memory score difference between
students with no chess training and students who have
completed chess training
Assumptions:
1) Although the sample of studentsVerify
is not a assumptions
random sample, the
investigator believed that it was reasonable to view
the 12 sample differences as representative of all
such differences.
2) A boxplot of the differences is approximately
symmetrical with no outliers so the
assumption
of normality is plausible.
Playing Chess Continued . . .
Student
1
2
3
4
5
6
7
8
9
10
11
12
Pre-test
510
610
640
675
600
550
610
625
450
720
575
675
Post-test
850
790
850
775
700
775
700
850
690
775
540
680
Difference
-340
-180
-210
-100
-100
-225
-90
-225
-240
-55
35
-5
H 0 : d = 0
H a : d < 0
Where d is the mean memory score difference between
students with no chess training and students who have
State the conclusion in
completed chess training
Test Statistic:
t 
Compute
144.6  0 the
context.
test statistic
 4.56
109.74
P-value.
12
P-value ≈ 0
df = 11
and
a = .05
Since the P-value < a, we reject H0. There is
convincing evidence to suggest that the mean
memory score after chess training is higher than
the mean memory score before training.
Paired t Confidence Interval for d
When
1.
2.
3.
The samples are paired.
The n sample differences can be viewed as a random sample
from a population of differences.
The number of sample differences is large (generally at least 30)
or the population distribution of differences is (at least
approximately) normal.
the paired t interval for d is
 sd 
xd  (t critical value)

 n
Where df = n - 1
Playing Chess Revisited . . .
Student
1
2
3
4
5
6
7
8
9
10
11
12
Pre-test
510
610
640
675
600
550
610
625
450
720
575
675
Post-test
850
790
850
775
700
775
700
850
690
775
540
680
Difference
-340
-180
-210
-100
-100
-225
-90
-225
-240
-55
35
-5
 109.74 
 144.6  1.796
  ( 201.5,  87.69)
 12 
Compute a 90% confidence interval for the
Wedifference
are 90% confident
the before
true mean
mean
in memorythat
scores
in memory
scores
before
chess
chessdifference
training and
the memory
scores
after
training and
the
memory scores after
chess
training.
chess training is between -201.5 and 87.69.
Large-Sample Inferences
Concerning the Difference
Between Two Population or
Treatment Proportions
Some people seem to think that duct tape
can fix anything . . . even remove warts!
Investigators at Madigan Army Medical Center
tested using duct tape to remove warts versus
the more traditional freezing treatment.
Suppose that the duct tape treatment will
successfully remove 50% of warts and that the
traditional freezing treatment will successfully
remove 60% of warts.
Let’s investigate the sampling
distribution of pfreeze - ptape
pfreeze = the true proportion of
ptape = the true proportion of
warts that are
warts that are
successfully removed
successfully removed
by freezing
by using duct tape
Randomly take
pfreeze = .6
ptape = .5
one of the sample
Suppose we repeatedly treated
Supposeforwe repeatedly treated
proportions
100 warts using the duct
tape
100
warts using the traditional
the
freezing
method and calculated
the freezing
treatment
and
treatment and
one ofare
the
sample the proportion of
proportion of warts that
calculated
proportions
for
successfully removed. We
would
warts
that are successfully
the duct tape
have the
.6 sampling
.6(.4) distribution
removed.
have the
treatment
and We would .5
s pˆ

.5(.5)
s

100.
of ptape
sampling
distribution ofpˆ pfreeze100
find
the
difference.
freeze
tape
Doing this
repeatedly, we will
create the sampling
distribution of
(pfreeze – ptape)
pfreeze - ptape
s pˆ
ˆ
freeze  ptape
.1

.6(.4) .5(.5)

100
100
Properties of the Sampling
Distribution of p1 – p2
If two random samples are selected independently
of one
When performing
a
another, the following properties hold:
hypothesis
test,forwe
Since
the value
p1
will
null
and
p2 use
are the
unknown,
1.  pˆ1  pˆ2  p1  p2
hypothesis
that p11
we
will combine
Use:
This says that the sampling distribution of p1 – p2 is centered at p1
and
We
andp2pare
to–pˆequal.
estimate
ˆ
2pn

n
p
– p2 so p1 – p2 is an unbiased statistic for estimating
p
.
11 1 2
2 2
ˆ
p

will
not
know
theof
the
common
value
c
p1 (1  p1 ) p2 (1  p2 )
n1  n2for p

common
value
2. s pˆ1  pˆ2 
p1 and p2
1
n1
n2
and p2.
3. If both n1 and n2 are large (that is, if n1p1 > 10,
n1(1
– p1) > 10, n2p2 > 10, and n2(1 – p2) > 10), then p1 and p2
each have a sampling distribution that is approximately
normal, and their difference p1 – p2 also has a sampling
distribution that is approximately normal.
Summary of Large-Sample z Test
for p1 – p2 = 0
Null Hypothesis: H0: p1 – p2 = 0
Test Statistic:
Use:
z 
n1 pˆ1  n2 pˆ2
pˆc 
n1  n2
Alternative Hypothesis:
H a: p 1 – p 2 > 0
H a: p1 – p 2 < 0
H a: p 1 – p 2 ≠ 0
pˆ1  pˆ2  ( p1  p2 )
pˆc (1  pˆc ) pˆc (1  pˆc )

n1
n2
P-value:
area to the right of calculated z
area to the left of calculated z
2(area to the right of z) if +z or
2(area to the left of z) if -z
Another Way to Write Hypothesis
statements:
H
H00:: pp11 -=pp2 2= 0
Haa:: pp11 ->pp22> 0
H
Haa:: pp11 -<pp22< 0
H
Haa:: pp11 -≠pp22≠ 0
H
Be sure to
define both
p1 & p2!
Summary of Large-Sample z Test
for p1 – p2 = 0 Continued . . .
Assumption:
1) The samples are independently chosen
Since
p1 andor
p2 treatments
are unknown were
we must use
random
samples
and
p2 to verify
that the samples
are
assignedp1at
random
to individuals
or objects
large enough.
2) Both sample sizes are large
n1p1 > 10, n1(1 – p1) > 10, n2p2 > 10, n2(1 – p2)
> 10
Investigators at Madigan Army Medical Center tested
using duct tape to remove warts. Patients with warts
were randomly assigned to either the duct tape treatment
or to the more traditional freezing treatment. Those in
the duct tape group wore duct tape over the wart for 6
days, then removed the tape, soaked the area in water,
and used an emery board to scrape the area. This process
was repeated for a maximum of 2 months or until the wart
was gone. The data follows:
n
Number with wart
successfully removed
Liquid nitrogen freezing
100
60
Duct tape
104
88
Treatment
Do these data suggest that freezing is less
successful than duct tape in removing warts?
Duct Tape Continued . . .
Treatment
n
Number with wart successfully removed
Liquid nitrogen freezing
100
60
Duct tape
104
88
H 0: p 1 – p 2 = 0
H a: p 1 – p 2 < 0
Where p1 is the true proportion of warts that
would be successfully removed by freezing and p2
is the true proportion of warts that would be
successfully removed by duct tape
Assumptions:
1) Subjects were randomly assigned to the two treatments.
2) The sample sizes are large enough because:
n1p1 = 100(.6) = 60 > 10
n1(1 – p1) = 100(.4) = 40 > 10
n2p2 = 100(.85) = 85 > 10 n2(1 – p2) = 100(.15) = 15 > 10
Duct Tape Continued . . .
Treatment
n
Number with wart successfully removed
Liquid nitrogen freezing
100
60
Duct tape
104
88
H 0 : p1 – p2 = 0
H a : p1 – p2 < 0
z 
.6  .85  0
.73(.27) .73(.27)

100
104
pˆc 
 4.03
60  88
 .73
100  104
P-value ≈ 0
a = .01
Since the P-value < a, we reject H0. There is
convincing evidence to suggest the proportion of
warts successfully removed is lower for freezing
than for the duct tape treatment.
A Large-Sample Confidence
Interval for p1 – p2
When
1)The samples are independently chosen random samples or
treatments were assigned at random to individuals or
objects
2) Both sample sizes are large
n1p1 > 10, n1(1 – p1) > 10, n2p2 > 10, n2(1 – p2) > 10
a large-sample confidence interval for p1 – p2 is
pˆ  pˆ   z critical value
1
2
pˆ1 (1  pˆ1 ) pˆ2 (1  pˆ2 )

n1
n2
The article “Freedom of What?” (Associated Press,
February 1, 2005) described a study in which high school
students and high school teachers were asked whether
they agreed with the following statement: “Students
should be allowed to report controversial issues in their
student newspapers without the approval of school
authorities.” It was reported that 58% of students
surveyed and 39% of teachers surveyed agreed with the
statement. The two samples – 10,000 high school
students and 8000 high school teachers – were selected
from schools across the country.
Compute a 90% confidence interval for the
difference
in proportion of students who
agreed with the
statement and the proportion
of teachers who
agreed with the statement.
Newspaper Problem Continued . . .
p1 = .58
p2 = .39
Based
this confidence
interval,
does there
1) Assume that
it ison
reasonable
to regard these
two samples
as being
independently
selected
representative
of the populations
of
appear
to be and
a significant
difference
in proportion
interest.of students who agreed with the statement and the
2) Both sample
sizes are large
enough who agreed with the
proportion
of teachers
n1p1 = 10000(.58) > 10, n1(1 – p1) = 10000(.42) > 10,
statement? Explain.
n2p2 = 8000(.39) > 10, n2(1 – p2) = 8000(.61) > 10
.58(.42) .39(.61)
(.58  .39)  1.645

 (.178, .202)
10000
8000
We are 90% confident that the difference in
proportion of students who agreed with the
statement and the proportion of teachers who
agreed with the statement is between .178 and .202.