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Pharmaceutical Statistics
Lecture 14
Hypothesis Testing:
The Difference Between Two Population
Means
Hypothesis Testing: The Difference Between Two
Population Means
•
•
Hypothesis testing involving the difference between two population means is most
frequently employed to determine whether or not it is reasonable to conclude
that the population means are not equal. Using the same methodology, it is
possible to test the hypothesis that the difference is equal to, greater than or
equal to, or less than or equal to some value other than zero.
In such cases, one of the following hypotheses may be formulated:
H0: μ1 – μ2 = 0, HA: μ1 – μ2 ≠ 0
H0: μ1 – μ2 ≥ 0, HA: μ1 – μ2  0
H0: μ1 – μ2 ≤ 0, HA: μ1 – μ2  0
A) When we withdraw two samples from two populations that
are normally distributed with known σ1 and σ2
• Test statistic for the null hypothesis of equal population
means (H0: μ1 – μ2 = 0, HA: μ1 – μ2 ≠ 0) is:

z

x1 x 2


(x1  x 2 )  (1  2 ) H
12  22
n1

n2
• The subscript H indicates that the difference is a hypothesized
parameter (in this case (μ1-μ2)H=0).
Example
Researchers are interested to know if there is any difference in the mean uric acid
levels between normal individuals and individuals with Down’s syndrome. They
collected samples from each population and uric acid level were determined in
each samples. Analysis results are summarized below:
Normal individuals
With Down’s Syndrome
Number of volunteers
15
12
Uric acid sample mean
(mg/100 mL)
4.5
3.4
Assume that samples were drawn from normally distributed population with
variance equal to 1 for the Down’s syndrome population and 1.5 for the normal
population.
A) Construct the proper hypothesis to test if there is significant difference between
the two population means;
B) Test your constructed hypothesis using the experimental data.
C) Decide on your hypothesis and
D) Conclude on the difference in mean uric acid levels between the two studied
populations
E) Calculate the P-value
Example
• We are interested in any significant difference, so we consider
the null hypothesis with equal means
H0: μ1 – μ2 = 0, HA: μ1 – μ2 ≠ 0
An alternative way of stating the hypothesis is as follows
H0: μ1 = μ2, HA: μ1 ≠ μ2

• The test statistic is:
z

x1 x 2


(x1  x 2 )  (1  2 ) H
12  22
n1
• Calculation of the test statistic:

z X N X D 

n2

(x N  x D )  (N  D ) H
 2N
nN

 D2
nD

(4.5  3.4)  0
1.1  2.57

1 1.5
0.4282

12 15
Example
• Decision rule: We use the z-stat. Let α=0.05, the critical values
of z are 1.96 and -1.96. Reject H0 unless -1.96  z  1.96
• Statistical decision to reject H0 since +2.57  +1.96
• Conclude that on the basis of the data, there is an indication
that the two population means are not equal.
• P-value: 0.0102 (the area to the right of 2.57 and the left of 2.57)
B) When we withdraw two samples from two populations that
are NOT normally distributed (σ1 and σ2 are unknown) and the
sample size is large
If we have no idea about the normality of the parent distributions and their
parameters, we use the CLT to justify our use of the z-table to find the reliability
coefficient. We use the samples standard deviations to calculate the standard error
of the difference. This is only true for large samples
• Test statistic for the null hypothesis of equal population
means (H0: μ1 – μ2 = 0, HA: μ1 – μ2 ≠ 0) is:

z

(x1  x 2 )  (1  2 ) H
s12 s22

n1 n 2

X 1 X
2
s
X 1 X
 [(s12/ n 1)  (s22 / n2)]
2
• The subscript H indicates that the difference is a hypothesized
parameter (in this case (μ1-μ2)H=0).
Example
• A study was designed to test the effect of disability on the beneficiary
effects of health promotion. The researchers developed a scale for
testing this effect (BHADP), the scale was administered to a sample of
132 disabled (D) and 137 nondisabled (ND) subjects with the following
results:
Sample
Mean Score
Standard
Deviation
D
31.83
7.93
ND
25.07
4.80
• The authors wish to know if they may conclude on the basis of these
results that, in general, disabled persons, on the average, score higher
on the BHADP scale. Construct the proper hypothesis and test it.
Note: use a significance level of 1%.
Example
• The statistics were computed from two independent samples that
behave as simple random samples from a normally distributed
population of disabled persons and a population of nondisabled
persons.
• Since the population variances are unknown, we will use sample
variances in the calculation of the test statistic.
• Since we have large samples, the central limit theorem allows us to
use z as a test statistic.


z
•
Hypotheses:
(x1  x 2 )  (1  2 ) H
s12 s22

n1 n 2
H0: μD – μND ≤ 0
HA: μD – μND > 0
or alternatively:
H0: μD ≤ μND
HA: μD > μND
Disabled scores are higher!!
Disabled scores are higher!!
• Decision rule: Let α=0.01. this is a one-tailed (right) test with critical
value of z equal to 2.33 (from SND tables, Z0.99).
• Reject H0 if zcomputed≥2.33
• Calculation of the test statistic:
z
(31.83 25.07)  0
2
(7.93)
(4.80)

132
137
 8.42
2
• Reject H0 since zcomputed = 8.42 > 2.33
• The data indicate that, on average, disabled persons score higher on
the BHADP scale than do nondisabled person.
• For this test, p  0.001 (very highly significant)
C) When we withdraw two samples from two populations that are
normally distributed (σ1 and σ2 are unknown) and the sample size is
small
• Parent pops: Normally distributed
• Variance for pops: Unknown
• Sample size: Small
We can not use the z-distribution in this case. We need to use the t-table
to find the critical values and we need to use standard deviations of the
two samples to find the standard error in the t-score calculation.
Important note in this case:
The calculation of the standard error from s1,n1 and s2, n2 depend on the equality of
the parent populations variances
C.1) If they are
equal: we use tdistribution
C.2) If they are
not equal: we use
the t’-distribution
C.1) When we withdraw two samples from two populations that are normally
distributed (σ1 and σ2 are unknown and equal) and the sample size is small
•
•
In this case, we need to consider the samples variances + to use t-table.
Since the sample variance is dependent on the sample size, we need to take this
into consideration for samples with different sizes (n) to calculate the pooled
estimate of the common variance
[This formula pools both sample
2
2
s 
2
p
•
(n 1)s  (n 1)s
1
1
2
n1  n 2 2
variances with their corresponding
weight that based on the sample
size]
2
NOTE: If the sample size for both independent samples is equal, we take directly
the arithmetic mean of the two samples variances (simple average) .
The standard error of the estimate will be:
s 2p s 2p

X 1 X
•
s
2
X 1 X 2

n1

n2
The test statistic for testing H0: μ1 = μ2 is given by:

t

(x1  x 2 )  (1  2 ) H
s2p s2p

n1 n 2
For critical values: We use the t-table
with D.F= n1+n2-2
Example
In a study to investigate the lung destruction in cigarette smokers, a lung
destructive index was measured in a sample of lifelong nonsmokers and
smokers. A larger score indicates greater lung damage. The data is
summarized below:
Non-smokers
smokers
Number of volunteers
9
16
Average score
12.4
17.5
Standard deviation
4.8492
4.4711
We wish to know if we may conclude that smokers, in general, have
greater lung damage measured by this destructive index than do
nonsmokers? Assuming that the lung destructive index scores in both
populations are approximately normally distributed with equal variances,
construct the proper hypothesis and test it.
Example
• Hypotheses:
H0: μS ≤ μNS, HA: μS μNS
• Test statistic (t-test with pooled variance):
nS  9,x S  17.5, SS  4.4711............. nNS  16, x NS  12.4, S NS 4.8492
2
2
2
2
(n
1)s
(n
1)s
8(4.8492)
15(4.4711)
S
S
NS
NS
s 2p 

 21.2165
nS  nNS  2
23

t 



(x S  xNS )  (S  NS )0
s 2p
nS

s 2p
n NS

(17.5 12.4)  0
 2.6573
21.2165 21.2165

9
16
Example
•
Critical values and Decision rule:
let α=0.05 (right-tailed test). the critical values of t is +1.714 (from table with
D.F=23, cumm. prop=95%). Reject if tcaclulated > 1.714
Confidence=95%
AUC-∞t0.95
α=5% (0.05)
t0.95=1.714 for DF:23
Example
• Statistical decision: we reject H0 because 2.6573>1.714 (falls in the
rejection zone).
• Conclusion: we conclude that smokers may have greater lung
damage than nonsmokers.
α=5% (0.05)
tcalc=2.6573
t0.95=1.714 for DF:23
• p value: 0.01>P>0.005, since 2.500 2.65732.8073
tcalc=2.6573
C.2) When we withdraw two samples from two populations that are
normally distributed (σ1 and σ2 are unknown and NOT equal) and the
sample size is small
• We can not use the t-table to find the critical values for D.F= n1+n2-2.
• Solution: Instead of finding the critical values from the tables, we need to
compute it taking into consideration the critical values for each sampling
distributions and the weight of each sample.
How to compute the critical values
for two-tailed test?
'

t1
 /2
w1 
w1t1  w 2 t 2
w1  w 2
2
1
s
n1
How to compute the critical values
for one-tailed test?
'

t1

w1 
w1t1  w 2 t 2
w1  w 2
2
1
s
n1
s22
w2 
n2
s22
w2 
n2
t1  t1 / 2 ....... for: n1 1
t1  t1 ....... for: n1 1
t2  t1 / 2 ....... for: n2 1
t2  t1 ....... for: n2 1
And we use S1
and S2 in the
formula of t-test
_
_
(x  x )  (1   2) 0
t 
s12 s22

n1 n 2
1
2
'

t1
 /2
For a two sided test, reject H0 if the
computed value of t is either greater
than or equal to the critical values t`(1-α/2)
or less than or equal to the negative of
that value.
w1t1  w 2 t 2
w1  w 2
s12
w1 
n1
s22
w2 
n2
α/2=2.5% (0.025)
t1  t1 / 2 ....... for: n1 1
t2  t1 / 2 ....... for: n2 1
t1-α/2=t97.5%=t0.975
t' 1 
w1t1  w 2 t 2
w1  w 2
α=5% (0.05)
s12
w1 
n1
s2
w2 
2
n2
t1-α=t95%=t0.95
For a one-sided test with the rejection region in the right tail of the sampling
distribution, reject H0 if the computed t is equal to or greater than the critical t`1-α.
t1  t1 ....... for: n1 1
For a one-sided test with the rejection region in the left tail of the sampling distribution,
t2  t1 ....... for: n2 1 reject H if the computed tis equal to or smaller than the negative of the critical t`
0
1-α
computed.
Example
Researchers wish to know if two populations differ with respect to the mean
value of the total serum complement activity (CH50). The data consist of CH50
determinations on apparently normal subjects (n2=20 ) and subjects with
disease (n1=10 ). The sample means and standard deviations are:
_
x1  62.6 s1  33.8
_
x 2  47.2 s 2 101.1
The research team want to know if they can conclude that the two population
means are different. Assuming that both populations are approximately
normally distributed, construct the proper hypothesis and test it?.
The populations are normally distributed with unknown variances that are
unequal. With this in mind we will use t’-stat.
Example
• Hypothesis:
H0: μ1 – μ2 = 0
HA: μ1 – μ2 ≠ 0
• Test statistic:
_
t
_
(x 1  x 2 )  (1  2 )0
s12 s22

n1 n2
• We obtain the critical value by the equation:
w1t1  w2t2
t`  
(1 )
w1  w2
2
Two-tailed test!!
Example
t1'  /2 
w1t1  w 2 t 2 114.244(2.2622)  5.1005(2.0930)

 2.255
w1  w
114.244  5.1005
2
s12 33.8 2
w1  
114.244
n1
10
s22 10.12
w2 

 5.1005
n2
20
t1  t1 / 2 ....... for : n1 1...........  2.2622
t2  t1 / 2 ....... for : n 2 1..........  2.0930
Using t-table with DF=9 and cumm prop=0.975
(let α=0.05)
Using t-table with DF=19 and cumm prop=0.975
(let α=0.05)
• Since we found the t`(1-α/2) to be equal to 2.255, our critical values will be
±2.255 (two-tailed test)
• Our decision rule is reject H0 if the computed t is either ≥2.255 or ≤2.255.
• Calculation of the test statistic:
t
(62.6  47.2)  0

15.4
 1.41
10.92
( 3 3.8) 2 (1 0.1) 2

10
20
• Statistical decision: since -2.255  1.41  2.255, we can not reject the H0.
• On the basis of these results we can not conclude that the two population
means are different.
• The p value of this test  0.05
Paired Comparisons
• Previously we discussed the difference between two
population means assuming that the samples were
independent.
• Some times we may want to assess the effectiveness of a
treatment or experimental procedure making use of
observations resulting from nonindependent samples.
• A hypothesis test based on this type of data is called a
paired comparison test.
Why do we need paired test??
The objective in paired comparison tests is to eliminate maximum number of
sources of extraneous variations by making the pairs similar with respect to as
many variables as possible.
Example
Not Paired
laser
Paired
No laser
No laser
To study the effect of gold nanoparticles in treating tumors, we induce tumor and then
target it with gold nanoparticles which serve as “nanoheaters” upon radiation with laser.
In “Not Paired” case, we use two groups that one receives gold nanoparticles and the
other does not (control). The difference between treated and control may be due simply a
difference in external characteristics between the mice in both groups. To eliminate this
difference, we can use one group of mice and induce two identical tumors in the same
mouse as you see in the picture to the right. Or we can use the same mice with
before/after strategy.
laser
Why do we need paired test??
• Related or paired observations may be obtained in a
number of ways:
– The same subjects may be measured before and after
receiving some treatment.
– Same subjects with part of their bodies
(treatment/control, previous slide)
– In comparing two methods of analysis, the material to be
analyzed may be divided equally so that one half is
analyzed by one method and one half is analyzed by
another.
Paired Test
• Instead of performing the analysis with individual observations, we use
di, the difference between pairs of observations as the variable of
interest.
• When the n sample differences computed from the n pairs of
measurements constitute a simple random sample from a normally
distributed population of differences, the test statistic for testing
hypothesis about the population mean difference μd is:
d  d 0
t 
Sd / n
where:
d is the mean for sample differences
d is the hypothesized population mean differe
0
Sd is the standard deviation of the sample differe
n is the number of sample differences
Paired Test
• The t statistic is distributed as Student’s t with n-1 degrees
of freedom (this is the general case but not always!!).
• We do not have to worry about the equality of variances
in paired comparisons, since our variable is the difference
in the reading of the same subject or object.
• If the population variance of the difference is known, we
can use z-stat (this is not applicable always)
• If the assumption of normality for the distribution of the
differences can not be made, we can use large n and thus
use the CLT to justify our use of the z-stat (we approximate
σ by the use of sample S)
Example
• In a study to evaluate the effect of very low calorie diet
(VLCD) on the weight of 9 subjects, the following data was
collected before (B) and after (A) treatment:
B
(Kg)
117.3 111.4
98.6
104.3 105.4 100.4
81.7
89.5
78.2
A
(Kg)
83.3
75.8
82.9
62.7
69
63.9
85.9
82.3
77.7
• The researchers wish to know if these data provide
sufficient evidence to allow them to conclude that the
treatment is effective in causing weight reduction in those
individuals. Assuming that differences between A&B are
approximately normally distributed, construct the proper
hypothesis and test it (you need to know that this is
paired t-test!!!!!)?.
Example
• We may obtain the differences in one of two ways: by
subtracting the before weights from the after weights (A –
B) or by subtracting the after weights from the before
weights (B – A).
• If we choose (di=A – B), the differences are:
-34, -25.5, -22.8, -21.4, -23.1, -22.7, -19, -20.5, -14.3
• Assumptions: the observed differences constitute a simple
random sample from a normally distributed population of
differences with unknown variance (t-test).
• Hypotheses:
H0: μd ≥0 (left-tailed test)
HA: μd  0 (indicating weight reduction)
Example
• Hypotheses (for (di=A – B)):
H0: μd ≥ 0 (left-tailed test)
HA: μd  0 (indicating weight reduction)
NOTE:
If we had obtained the differences by subtracting the after weights
from the before weights (B – A) our hypotheses would have
been:
H0: μd ≤ 0 (right-tailed test)
HA: μd > 0 (indicating weight reduction)
• If the question had been such that a two-tailed test was indicated
(any difference/change), the hypotheses would have been:
H0: μd = 0
HA: μd ≠ 0
Example
• The test statistic:
t
d  d
0
Sd / n
• Decision rule: Let α=0.05, the critical value of t is
-1.860, reject H0 if the computed t is less than or equal to the
critical value.
di 2 03.3
H0: μd ≥ 0
d

 22.5889
n
9
(di  d) 2
2
α=5% (0.05)
sd 
 28.2961
n 1
22.588 9 0
t
 12.74
28.2961
-1.860=-t1-α=-t0.95 (n-1=8)
t'calc=-12.74
9
Reject H0, since -12.7395 is in the rejection reg
We may conclude that the diet program is effect
P-value<0.001 since 12.74<-5.041 see the table