Download Chapter 11: Comparing Two Populations or Treatments

Chapter 11: Comparing Two
Populations or Treatments
Section 11.1: Inferences
Concerning the Difference
Between Two Population or
Treatment Means Using
Independent Samples
Notation
Mean
Value
Variance
Standard
Deviation
Population or Treatment 1
1
12
1
Population or Treatment 2
2
22
2
Sample
Size
Mean
Variance
Standard
Deviation
Population or Treatment 1
n1
x1
s12
s1
Population or Treatment 2
n2
x2
s 22
s2
Properties of the Sampling Distribut ion of x1  x2
If the random samples on which x1 and x 2 are based
are selected independen tly of one another, then
1. x1  x2  ( mean value of x1  x2 )   x1  x2  1  2
Thus, the sampling distributi on of x1  x2 is always
centered at the value of 1  2 , so x1  x2 is an unbiased
statistic for estimating 1  2 .
2. 
2
x1  x 2
 12  22
 ( variance of x1  x2 )     

n1 n2
2
x1
2
x2
and
 x x
1
2
 12  22
 (standard deviation of x1  x2 ) 

n1 n2
3. If n1 and n 2 are both large or if the population distributi ons are (at least approximat ely) normal,
then both x1 and x 2 have (at least approx) a normal distributi on. This implies that the sampling
distributi on of x1  x2 is also normal or approximat ely normal.
•
Properties 1 and 2 follow from the
following general results:
1. The mean value of a difference in means is
the difference of the two individual mean
values.
2. The variance of a difference of independent
quantities is the sum of the two individual
variances.
When n1 and n2 are both large or when the population
distributions are (at least approximately) normal, the
distribution of
z
x1  x2  ( 1  2 )
 12
n1

 22
n2
Is described (at least approximately) by the standard normal
(z) distribution.
When two random samples are independently selected and when
n1 and n2 are both large or if the population distributions are
normal, the standardized variable
t
x1  x2  ( 1  2 )
has approximately a t distribution with
(V1  V2 )
df 
2
2
V1
V2

n1  1 n2  1
s12 s22

n1 n2
where
s12
s22
V1  and V2 
n1
n2
The degrees of freedom should be rounded down to obtain an
integer value.
Example
• Do children diagnosed with attention
deficit/hyperactivity disorder (ADHD) have
smaller brains than children without this
condition? This question was the topic of a
research study described in a paper. Brain
scans were completed for 152 children
with ADHD and 139 children of similar age
without ADHD. Summary values for total
cerebral volume (in milliliters) are given in
the following table:
n
152
x
1059.4
s
117.5
Children without ADHD 139
1104.5
111.3
Children with ADHD
Does this data provide evidence that the mean brain volume
of children with ADHD is smaller than the mean for children
without ADHD? Let’s test the relevant hypotheses using a .05
level of significance.
We will use the 9 steps to test the hypothesis.
1. 1  true mean brain volume for children with ADHD
2  true mean brain volume for children without ADHD
1  2  difference in mean brain volume.
2. H 0 : 1  2  0
3. H a : 1  2  0
4. Significan ce Level :   .05
5. Test Statistic : t 
x1  x2  hypothesiz ed value
2
1
2
2
s
s

n1 n2

x1  x2  0
s12 s22

n1 n2
6. Assumptions: The paper states that the study controlled for age
and that the participants were “recruited from the local community.”
This is not equivalent to random sampling, but the authors of the
paper (five of whom were doctors at well-known medical institutions)
believed that it was reasonable to regard these samples as
representative of the two groups under study. Both sample sizes are
large, so it is reasonable to proceed with the two-sample t test.
7. Calculatio ns :
(1059.4  1104.5)  0
 45.10
 45.10
t


 3.36
2
2
13
.
415
90.831  89.120
(117.5) (111.3)

152
139
8. P - value : We first compute the number of degrees of freedom for the two - sample t test :
s12
s22
V1   90.831 V2 
 89.120
n1
n2
(V1  V2 )2
(90.831  89.120)2
32,382.362
df 


 288.636
V12
V22
(90.831)2 (89.120)2
112.191


n1  1 n2  1
151
138
We round down the degrees of freedom to 288. P-value is
approx. 0 because the t score is too large.
9. Conclusion: Because P-value ≈ 0 < .05, we reject H0. There is
convincing evidence that the mean brain volume for children with
ADHD is smaller than the mean for children without ADHD.
Two-Sample t Test for Comparing
Two Treatments
When
1. treatments are randomly assigned to
individuals or objects (or vice versa:
individuals or objects are randomly
assigned to treatments)
2. the sample sizes are large (in general 30
or larger) or the treatment response
distributions are approximately normal
The two-sample t test can be used to test H0: μ1 –
μ2 = hypothesized value, where μ1 and μ2
represent the mean response for Treatments 1
and 2, respectively.
In this case, these two conditions replace the
assumptions previously stated for comparing two
population means. The validity of the
assumption of normality of the treatment
response distributions can be assessed by
constructing a normal probability plot or a
boxplot of the response values in each sample.