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Comparing Populations
Proportions and means
The sampling distribution of differences of Normal
Random Variables
If X and Y denote two independent normal random
variables, then :
D = X – Y is normal with
mean D   X  Y
standard deviation  D   X2   Y2
Comparing proportions
Situation
• We have two populations (1 and 2)
• Let p1 denote the probability (proportion) of
“success” in population 1.
• Let p2 denote the probability (proportion) of
“success” in population 2.
• Objective is to compare the two population
proportions
Consider the statistic:
x1 x2
D  pˆ1  pˆ 2 = n1 n2
This statistic has a normal distribution
D   pˆ   pˆ  p1  p2
1
z 
pˆ1  pˆ 2
 pˆ  pˆ
1
2

2
pˆ1  pˆ 2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n1
Consider the statistic:
x1 x2
D  pˆ1  pˆ 2 = n1 n2
This statistic has a normal distribution with
D   pˆ   pˆ  p1  p2
1
2
 D = pˆ  pˆ   p2ˆ   p2ˆ
1
2
1
2

p1 1  p1  p2 1  p2 

n1
n1

pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
Thus
z 
D  D
D

pˆ1  pˆ 2 -  p1  p2 
 pˆ  pˆ
1


2
pˆ1  pˆ 2 -  p1  p2 
p1 1  p1  p2 1  p2 

n1
n1
pˆ1  pˆ 2 -  p1  p2 
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n1
Has a standard normal distribution
We want to test either:
1. H 0 : p1  p2 vs H A : p1  p2
or
2. H 0 : p1  p2 vs H A : p1  p2
or
3. H 0 : p1  p2 vs H A : p1  p2
If p1 = p2 (p say) then the test statistic:
z 
D  D
D

pˆ1  pˆ 2 -  p1  p2 
 pˆ  pˆ
1


2
pˆ1  pˆ 2 -  p1  p2 
p1 1  p1  p2 1  p2 

n1
n2
pˆ1  pˆ 2
1 1
p 1  p    
 n1 n2 

pˆ1  pˆ 2
1 1
pˆ 1  pˆ    
 n1 n2 
has a standard normal distribution.
where
x1  x2
pˆ 
n1  n2
is an estimate of the common
value of p1 and p2.
Thus for comparing two binomial probabilities
p1 and p2
The test statistic
z
pˆ1  pˆ 2
1 1
pˆ 1  pˆ    
 n1 n2 
where
x1
x2
pˆ1 
, pˆ 2 
n1
n2
x1  x2
and pˆ 
n1  n2
The Critical Region
The Alternative
Hypothesis HA
The Critical Region
H A : p1  p2
z   z / 2 or z  z / 2
H A : p1  p2
z  z
H A : p1  p2
z   z
Example
• In a national study to determine if there was an
increase in mortality due to pipe smoking, a
random sample of n1 = 1067 male nonsmoking
pensioners were observed for a five-year period.
• In addition a sample of n2 = 402 male pensioners
who had smoked a pipe for more than six years
were observed for the same five-year period.
• At the end of the five-year period, x1 = 117 of the
nonsmoking pensioners had died while x2 = 54 of
the pipe-smoking pensioners had died.
• Is there a the mortality rate for pipe smokers
higher than that for non-smokers
We want to test:
H 0 : p1  p2 vs H A : p1  p2
The test statistic:
z
pˆ1  pˆ 2
 pˆ  pˆ
1
2

pˆ1  pˆ 2
1 1
pˆ 1  pˆ   
 n1 n2 
Note:
x1
117
pˆ1 

 0.1097
n1 1067
x2
54
pˆ 2 

 0.1343
n2 402
x1  x2
117  54
pˆ 

n1  n2 1067  402
171

 0.1164
1469
The test statistic:
z 

pˆ1  pˆ 2
1 1
pˆ 1  pˆ   
 n1 n2 
0.1097  .1343
1 
 1
0.11641  0.1164 


 1067 402 
 1.315
We reject H0 if:
z   z  -z0.05  1.645
Not true hence we accept H0.
Conclusion: There is not a significant ( =
0.05) increase in the mortality rate due to
pipe-smoking
Estimating a difference proportions using
confidence intervals
Situation
• We have two populations (1 and 2)
• Let p1 denote the probability (proportion) of
“success” in population 1.
• Let p2 denote the probability (proportion) of
“success” in population 2.
• Objective is to estimate the difference in the
two population proportions d = p1 – p2.
Confidence Interval for d
100P% = 100(1 – ) % :
= p1 – p2
pˆ1  pˆ 2  z / 2  pˆ1  pˆ 2
pˆ1  pˆ 2  z / 2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
Example
• Estimating the increase in the mortality rate
for pipe smokers higher over that for nonsmokers d = p2 – p1
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 
pˆ 2  pˆ1  z / 2

n1
n2
0.10971  0.1097 0.13431  0.1343
0.1343  0.1097  1.960

1067
0.0247  0.0382
 0.0136 to 0.0629
 1.36% to 6.29%
402
Comparing Means
Situation
• We have two normal populations (1 and 2)
• Let 1 and 1 denote the mean and standard
deviation of population 1.
• Let 2 and 2 denote the mean and standard
deviation of population 1.
• Let x1, x2, x3 , … , xn denote a sample from a
normal population 1.
• Let y1, y2, y3 , … , ym denote a sample from a
normal population 2.
• Objective is to compare the two population means
We want to test either:
1. H 0 : 1  2 vs H A : 1  2
or
2. H 0 : 1  2 vs H A : 1  2
or
3. H 0 : 1  2 vs H A : 1  2
Consider the test statistic:
z
xy

 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
H 0 : 1  2 is true
If:
z
xy

2
1
n


2
2
m

xy
2
x
2
y
s
s

n m
• will have a standard Normal distribution
• This will also be true for the approximation
(obtained by replacing 1 by sx and 2 by sy) if
the sample sizes n and m are large (greater than
30)
Note:
n
n
x
x
i 1
i
n
sx 
y
i 1
m
i
i 1
i
n 1
n
n
y
 x  x 
2
sy 
2


y

y
 i
i 1
m 1
The Alternative
Hypothesis HA
The Critical Region
H A : 1  2
z   z / 2 or z  z / 2
H A : 1  2
z  z
H A : 1  2
z   z
Example
• A study was interested in determining if an
exercise program had some effect on reduction of
Blood Pressure in subjects with abnormally high
blood pressure.
• For this purpose a sample of n = 500 patients with
abnormally high blood pressure were required to
adhere to the exercise regime.
• A second sample m = 400 of patients with
abnormally high blood pressure were not required
to adhere to the exercise regime.
• After a period of one year the reduction in blood
pressure was measured for each patient in the
study.
We want to test:
H 0 : 1  2
The exercise group did not have a higher
average reduction in blood pressure
vs
H A : 1  2
The exercise group did have a higher
average reduction in blood pressure
The test statistic:
z
xy

 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
Suppose the data has been collected and:
n
n
x
x
i 1
i
n
 10.67
sx 
 x  x 
y
i 1
m
i
i 1
n 1
n
n
 yi
2
 7.83
sy 
y
i 1
i
 3.895
 y
m 1
2
 4.224
The test statistic:
z
xy
2
x
2
y
s
s

n m

10.67  7.83
3.895
2
500

4.224 

2.84

 10.4
0.273765
2
400
We reject H0 if:
z  z  z0.05  1.645
True hence we reject H0.
Conclusion: There is a significant ( = 0.05)
effect due to the exercise regime on the
reduction in Blood pressure
Estimating a difference means using
confidence intervals
Situation
• We have two populations (1 and 2)
• Let 1 denote the mean of population 1.
• Let 2 denote the mean of population 2.
• Objective is to estimate the difference in the
two population proportions d = 1 – 2.
Confidence Interval for
d = 1 – 2
ˆ1  ˆ 2  z / 2  ˆ ˆ
1
x  y  z / 2
2
x
2
2
y
s
s

n m
Example
• Estimating the increase in the average
reduction in Blood pressure due to the
excercize regime d = 1 – 2
x  y  z / 2
2
x
2
y
s
s

n m

3.895
10.67  7.83  1.960
2
500
2.84  1.96(.273765)
2.84  0.537
2.303 to 3.337

4.224 

2
400
Comparing Means – small samples
Situation
• We have two normal populations (1 and 2)
• Let 1 and 1 denote the mean and standard
deviation of population 1.
• Let 2 and 2 denote the mean and standard
deviation of population 1.
• Let x1, x2, x3 , … , xn denote a sample from a
normal population 1.
• Let y1, y2, y3 , … , ym denote a sample from a
normal population 2.
• Objective is to compare the two population means
We want to test either:
1. H 0 : 1  2 vs H A : 1  2
or
2. H 0 : 1  2 vs H A : 1  2
or
3. H 0 : 1  2 vs H A : 1  2
Consider the test statistic:
z

xy
 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
If the sample sizes (m and n) are large the
statistic
t
xy
2
x
2
y
s
s

n m
will have approximately a standard
normal distribution
This will not be the case if sample
sizes (m and n) are small
The t test – for comparing means –
small samples (equal variances)
Situation
• We have two normal populations (1 and 2)
• Let 1 and  denote the mean and standard
deviation of population 1.
• Let 2 and  denote the mean and standard
deviation of population 1.
• Note: we assume that the standard deviation
for each population is the same.
1 = 2 = 
Let
n
n
x
x
i 1
i
n
sx 
y
i 1
m
i
i 1
i
n 1
n
n
y
 x  x 
2
sy 
2


y

y
 i
i 1
m 1
The pooled estimate of .
Note: both sx and sy are estimators of .
These can be combined to form a single
estimator of , sPooled.
sPooled 
n  1sx2  m  1s 2y
nm2
The test statistic:
xy
t
s
2
Pooled
n

s
2
Pooled
m
xy

1 1
sPooled

n m
If 1 = 2 this statistic has a t distribution
with n + m –2 degrees of freedom
The Alternative
Hypothesis HA
The Critical Region
H A : 1  2
t  t / 2 or t  t / 2
H A : 1  2
t  t
H A : 1  2
t  t
t / 2 and t
are critical points under the t distribution with
degrees of freedom n + m –2.
Example
• A study was interested in determining if
administration of a drug reduces cancerous
tumor size.
• For this purpose n +m = 9 test animals are
implanted with a cancerous tumor.
• n = 3 are selected at random and
administered the drug.
• The remaining m = 6 are left untreated.
• Final tumour sizes are measured at the end
of the test period
We want to test:
H 0 : 1  2
The treated group did not have a lower
average final tumour size.
vs
H A : 1  2
The exercise group did have a lower
average final tumour size.
The test statistic:
xy
t
1 1
sPooled

n m
Suppose the data has been collected and:
drug treated
untreated
1.89
2.08
1.79
1.28
1.29
1.75
n
x
 xi
n
 1.657
i 1
n
sx 
n
y
y
i 1
m
1.90
i
2.32
 x  x 
i 1
2.16
2
i
n 1
 0.3215
n
 1.915
sy 
2


y

y
 i
i 1
m 1
 0.3693
The test statistic:
sPooled 
n  1sx2  m  1s 2y
nm2
20.3215  50.3693
 0.3563
7
2

2
1.657  1.915
 .258
t

 1.025
.252
1 1
0.3563 
3 6
We reject H0 if:
t  t   t0.05  1.895
with d.f. = n + m – 2 = 7
Hence we accept H0.
Conclusion: The drug treatment does not
result in a significant ( = 0.05) smaller final
tumour size,
Summary of Tests
One Sample Tests
Situation
Test Statistic
Sample form the Normal
distribution with unknown
mean and known variance
(Testing )
z
Sample form the Normal
distribution with unknown
mean and unknown variance
(Testing )
Testing of a binomial
probability 
Sample form the Normal
distribution with unknown
mean and unknown variance
(Testing )
t
z
n x  0 
H0
  
0
n x   0 
s
pˆ  p0
p0 (1  p0 )
n

n  1s 2
U
 02
  
 
p = p0
 0
HA
  
  
  
  
  
  
p ≠p0 
p >p0
 p0
p <
  0
Critical Region
z < -z/2 or z > z/2
z > z
z <-z
t < -t/2 or t > t/2
t > t
t < -t
z < -z/2 or z > z/2
z > z
z < -z
U   12 / 2 n  1 or
  0
U   2 n  1
  0
U   12 n  1
U   2 / 2 n  1
Two Sample Tests
Situation
Two independent samples
from the Normal distribution
with unknown means and
known variances
(Testing 1 - 2)
Test Statistic
 x1  x2 
z
 12
n1

H0
HA
Critical Region
1   2 1   2 z < -z/2 or z > z/2
 22
1   2 z > z
n2
1   2 z < -z
Two independent samples
from the Normal distribution
with unknown means and
unknown but equal
variances.
(Testing 1 - 2)
t
x1  x2 
sp
1   2 1   2 t < -t/2 or t > t/2
1 1

n1 n2
1   2 t > t
1   2 t < -t
Estimation of a the
difference between two
binomial probabilities, p1-p2
z
ˆ1  ˆ2
1 1
 
n
 1 n2 
ˆ (1  ˆ ) 
I am using  instead of p.
1   2
1   2 z < -z/2 or z > z/2
 1   2 z > z
1   2
z < -z