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Two-samples tests, X
Dr. Mona Hassan Ahmed
Prof. of Biostatistics
HIPH, Alexandria University
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Z-test
(two independent proportions)
Z
P1  P2
P1 (1  P1 ) P2 (1  P2 )

n1
n2
P1= proportion in the first group
P2= proportion in the second group
n1= first sample size
n2= second sample size
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Critical z =
• 1.96 at 5% level of significance
• 2.58 at 1% level of significance
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Example
Researchers wished to know if urban and rural adult
residents of a developing country differ with respect to
prevalence of a certain eye disease. A survey revealed the
following information
Eye disease
Residence
Total
Yes
No
Rural
24
276
300
Urban
15
485
500
Test at 5% level of significance, the difference in the
prevalence of eye disease in the 2 groups
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Answer
P1 = 24/300 = 0.08
p2 = 15/500 = 0.03
0.08  0.03
Z
 2.87
0.08(1  0.08) 0.03(1  0.03)

300
500
2.87 > Z*
The difference is statistically significant
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t-Test (two independent means)
t 
x1  x 2
2
2
S p
S p

n1
n2
X1 = mean of the first group
X2
= mean of the second group
S2p = pooled variance
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(n 1  1)S 1  (n 2  1)S
S P
n1  n 2  2
2
2
2
2
Critical t from table is detected
at degree of freedom = n1+ n2 - 2
level of significance 1% or 5%
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Example
Sample of size 25 was selected from healthy
population, their mean SBP =125 mm Hg with
SD of 10 mm Hg . Another sample of size 17 was
selected from the population of diabetics, their
mean SBP was 132 mmHg, with SD of 12 mm
Hg .
Test whether there is a significant difference in
mean SBP of diabetics and healthy individual at
1% level of significance
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Answer
n 1  25
X 1  125 S1 = 12
n 2  17
X 2  132 S2 =11
H0 : 1 = 2
H1 : 1  2
α = 0.01
State H0
State H1
Choose α
(25  1)10  (17  1)12
SP
 117 .6
25  17  2
2
2
2
10
Answer
t
125  132
 2.503
117.6 117.6

25
17
Critical t at df = 40 & 1% level of significance = 2.58
Decision:
Since the computed t is smaller than critical t so
there is no significant difference between mean
SBP of healthy and diabetic samples at 1 %.
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Degrees of
freedom
1
5
10
17
20
24
25

Probability (p value)
0.10
0.05
0.01
6.314
12.706 63.657
2.015
2.571
4.032
1.813
2.228
3.169
1.740
2.110
2.898
1.725
2.086
2.845
1.711
2.064
2.797
1.708
2.060
2.787
1.645
1.960
2.576
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Paired t- test
(t- difference)
Uses:
To compare the means of two paired samples.
Example, mean SBP before and after intake
of drug.
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d
t 
Sd
n
di

d
 mean difference
n
Sd 
2
di
 
( di) 2
n 1
n
di = difference (after-before)
Sd = standard deviation of difference
n = sample size
Critical t from table at df = n-1
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Example
The following data represents the
reading of SBP before and after
administration of certain drug. Test
whether the drug has an effect on
SBP at 1% level of significance.
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Serial No.
1
2
3
4
5
6
SBP
(Before)
200
160
190
185
210
175
SBP
(After)
180
165
175
185
170
160
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Answer
Serial
No.
BP
Before
BP
After
di
After-Before
di2
1
200
180
-20
400
2
160
165
5
25
3
190
175
-15
225
4
185
185
0
0
5
210
170
-40
1600
6
175
160
-15
225
-85
2475
∑di
∑ di2
Total
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Answer
di  85

d

 14.17
n
Sd 
6
( 85) 2
2475 
6
 15.942
5
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Answer
 14.17
Computed t 
 2.17
15.942
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Critical t at df = 6-1 = 5 and 1% level of significance
= 4.032
Decision:
Since t is < critical t so there is no significant
difference between mean SBP before and after
administration of drug at 1% Level.
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Degrees of
freedom
1
5
10
17
20
24
25

Probability (p value)
0.10
0.05
0.01
6.314
12.706 63.657
2.015
2.571
4.032
1.813
2.228
3.169
1.740
2.110
2.898
1.725
2.086
2.845
1.711
2.064
2.797
1.708
2.060
2.787
1.645
1.960
2.576
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Chi-Square test
It tests the association between variables...
The data is qualitative .
It is performed mainly on frequencies.
It determines whether the observed
frequencies differ significantly from
expected frequencies.
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(O i  E i )
ComputedΧ  
Ei
2
2
Where E = expected frequency
O = observed frequency
Raw total  Column total
E
Grand total
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Critical X2 at df = (R-1) ( C -1)
Where R = raw C = column
I f 2 x 2 table
X2* = 3.84 at 5 % level of significance
X2* = 6.63 at 1 % level of significance
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Example
In a study to determine the effect of heredity in a certain
disease, a sample of cases and controls was taken:
Family
history
Positive
Negative
Total
Disease
Cases
Control
80
120
140
160
220
280
Total
200
300
500
Using 5% level of significance,
test whether family history has an effect on disease
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Answer
Family
history
positive
O
E
Negative
O
E
Total
Disease
Cases
Control
Total
80
88
120
112
200
140
132
220
160
168
280
300
500
X2 = (80-88)2/88 + (120-112)2/112 + (140-132)2/132 + (160-168)2/168
= 2.165 < 3.84
Association between the disease and family history is not
significant
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• The odds ratio was developed to quantify
exposure – disease relations using casecontrol data
• Once you have selected cases and controls 
ascertain exposure
• Then, cross-tabulate data to form a 2-by-2
table of counts
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2-by-2 Crosstab Notation
Exposed +
Exposed Total
• Disease status
Disease +
Disease -
Total
A
C
A+C
B
D
B+D
A+B
C+D
A+B+C+D
A+C = no. of cases
B+D = no. of non-cases
• Exposure status A+B = no. of exposed individuals
C+D = no. of non-exposed individuals
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Exposed +
Exposed -
Disease +
Disease -
A
C
B
D
exposure odds, cases  o1 
A
C
B
exposure odds, controls  o0 
D
o1 A / C AD
odds ratio  OR  

o0 B / D BC
Cross-product
ratio
AD
OR 
BC
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Example
• Exposure variable = Smoking
• Disease variable = Hypertension
D+
DAD (30)( 22)
E+
30
71
OR 

 9.3
BC
(71)(1)
E1
22
Total
31
93
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Interpretation of the Odds Ratio
• Odds ratios are relative risk
estimates
• Relative risk are risk multipliers
• The odds ratio of 9.3 implies 9.3×
risk with exposure
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OR > 1
OR = 1
OR < 1
Positive association
Higher risk
No association
Negative association
Lower risk (Protective)
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• In the previous example
OR = 9.3
• 95% CI is 1.20 – 72.14
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Multiple Levels of Exposure
Smoking level
Cases
Controls
Heavy smokers
Moderate smokers
213
61
274
147
Light smokers
Non-smokers
Total
14
8
296
82
115
618
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Multiple Levels of Exposure
• k levels of exposure  break up data into (k –
1) 22 tables
• Compare each exposure level to non-exposed
• e.g., heavy smokers vs. non-smokers
Cases
Heavy smokers
Non-smokers
Controls
213
274
8
115
(213)(115)
OR 
 11.2
(274)(8)
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Multiple Levels of Exposure
Smoking level
Cases
Controls
213
274
OR3 =(213)(115)/(274)(8)=11.2
Moderate smokers
61
147
OR2 =(61)(115)/(147)(8) = 6.0
Light smokers
14
82
Non-smokers
8
115
605
115
Heavy smokers
Total
OR1 =(14)(115)/(82)(8)
= 2.5
Notice the trend in OR
(dose-response relationship)
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Small Sample Size Formula For the Odds Ratio
It is recommend to add ½ to each cell before
calculating the odds ratio when some cells are
zeros
OR Small Sample
OR Small Sample =
(A+0.5)(D+0.5)
=
(B+0.5)(C+0.5)
(31+0.5)(22+0.5)
(71+0.5)(0+0.5)
=19.8
D+
D-
E+
31
71
E-
0
22
Total
31
93
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