Ch 12 SIGNIFICANT TESTrr.doc

12
Significance test
If a man will begin with certainties, he shall end in doubts; but if he
will be content to begin with doubts, he shall end in certainties ...
Francis Bacon
What is significant test?
Once sample data has been gathered through an
observational study or experiment, statistical inference allows
analysts to assess evidence in favor or some claim about the
population from which the sample has been drawn. The
methods of inference used to support or reject claims based on
sample data are known as tests of significance. Every test of
significance begins with a null hypothesis (H0).
Null hypothesis
[Q: Write shorts on: Null hypothesis, (BSMMU, MD
Radiology, January, 2010)]
Null hypothesis is the hypothesis of no difference .It states
that there is no difference between the experimental value
and the control value.

Biostatistics-91
Alternative hypothesis (H1) is the hypothesis which states
that there is difference between the experimental value and the
control value.
Null derives from the Latin word `nullus` meaning `none`.
When Null hypothesis is rejected, the alternative hypothesis is
accepted .This means, statistical hypothesis testing usually
focuses on the null hypothesis, which is a hypothesis of no
association between two variables.
When a statistical hypothesis is treated there are 4 possible
ways of interpreting the result.
1. The hypothesis Ho is true and our test accepts it.
2. The hypothesis Ho is false and our test rejects.
These two interpretations are a routine but the other two
interpretations may leads to errors
3. Hypothesis Ho is true still it is rejected.(Type 1 error)
4. The hypothesis Ho is false but it is accepted (Type II
Error)
Inference Accept it Reject it
Hypothesis is true Correct decision Type I error
Hypothesis is false Type II error Correct decision
Type I Error & Type II Error
[Q: Write shorts notes on: Errors of hypothesis
testing. (BSMMU, MD Radiology, January 2010, January
2011 January 2009),
Type I Error
Type I error is the false rejection of the null hypothesis.
Type I error is an ‘illusion’ i.e. it states that there is something
when there is really not.
Type II Error
Type II error is the false acceptance of the null hypothesis.

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Which one is dangerous
Say, in a diabetic patient, test done for DM, by significant test you
say that his diabetic test result is not significant that is he is not
diabetic (type II error). This (type II error) is more dangerous than
when you do mistake by saying a normal patient diabetic (type I
error).
Notes about Type I error:
· is the incorrect rejection of the null hypothesis
· is not affected by sample size as it is set in advance
· increases with the number of tests or end points (i.e. do 20
tests and 1 is likely to be wrongly significant)
Notes about Type II error:
· is the incorrect acceptance of the null hypothesis
· Beta depends upon sample size and alpha
· Beta gets smaller as the sample size gets larger
· Beta gets smaller as the number of tests or end points
increases
Power
[Q: Write short notes on: Power of test (BSMMU, MD
Radiology, January 2009, July 2010, January 2012)]
Power is the probability of correctly rejecting a false null
hypothesis. Power is therefore defined as:
1 - β, where β is the Type II error probability.
In other words, the power of a hypothesis test is the
probability of not committing a type II error.
If the power of an experiment is low, then there is a good
chance that the experiment will be inconclusive.

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The maximum power a test can have is 1, the minimum is 0.
Ideally we want a test to have high power, close to 1.
The following table shows the relationship between power
and error in hypothesis testing:
TRUTH DECISION
Accept H0 Reject H0
H0 is true correct decision P =
1-alpha
type I error P =
alpha (significance)
H0 is false type II error P = Beta correct decision P =
1-beta (power)
H0 = null hypothesis
P = probability
Test of significance
These are the test by means of which new hypothesis is either
rejected or accepted.
Tests done to measure significance
[Q: Enumerate the methods of inferential statistics.
(BSMMU, MD Radiology, January, 2009)]
1. t` test
a. Unpaired `t` test
b. Paired `t` test
2. X2
test (Chi-squared test)
3. Z test
4. `r` test ( correlation coefficient test) - will be described in
the next chapter
5. `F` test (or Fisher `s test or analysis of variance/ANOVA test)
[Q:

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 What are the measurements of association between
quantitative and quantitative variables? (BSMMU,
MD Radiology, July, 2010)
 How do we measure associations between variable?
(BSMMU, MD Radiology, July, 2010)]
Significance tests for equality of means with normal data
Test Description
Unpaired t-test Equality of means of two
independent groups of data
Paired t-test Equality of means of paired groups of
data.
ANOVA Equality of means of more
than two independent groups
of data
Parametric and non parametric test

Biostatistics-95
[Q:
 Enumerate parametric and non-parametric tests.
(BSMMU, MD Radiology, January, 2011)
 Write shorts notes on: Parametric and nonparametric
test (BSMMU, MD Radiology, January, 2009)]
Parametric test
Conventional statistical procedures are also called parametric
tests. In a parametric test, a sample statistic is obtained to
estimate the population parameter. Because this estimation
process involves a sample, a sampling distribution, and a
population, certain parametric assumptions are required to
ensure all components are compatible with each other.
When applied (assumption)
1. the population from which samples are taken should be
normally distributed
2. the variance of the samples are same (applies to t test)
Example
1. t test
2. Pearson’s correlation coefficient test
Non parametric test
As the name implies, non-parametric tests do not require
parametric assumptions because interval data are converted
to rank-ordered data.
When applied
[Q: When non-parametric tests are utilized. (BSMMU, MD
Radiology, January, 2011)
1. Make no assumptions about the underlying distribution of
the sample

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2. when data are qualitative, i.e. measured by ordinal scale,
or not normally distributed
Non-parametric tests do not require parametric assumptions
because interval data are converted to rank-ordered data.
Example
1. X2
test (Chi-squared test)
2. `F` test (or Fisher `s test or analysis of variance/ANOVA
test)
3. Mann –Whitney test
4. Wilcoxon rank sum test
5. Wilcoxon signed rank test
Some commonly used parametric and their
corresponding non parametric tests
normal theory
based tests
corresponding non
parametric tests
purpose of tests
t test for
independent
sample
Mann-Whitney U
test
Wilcoxon rank-sum
test
compare two
independent
sample
paired t test Wilcoxon matched
pairs signed-rank
test
examine a set of
difference
Pearson’s
correlation
coefficient
spearman's rank
correlation
coefficient
assesses the linear
association
between two
variables
one way analysis
of variance (F test)
Kruskal-Wallis
analysis of variance
by ranks
compare three or
more groups
two way analysis
of variance
Friedman two way
analysis of variance
Compare groups
classified by two
different factors

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t-Tests (Student's t-Tests)
"Student" (real name: W. S. Gossett [1876-1937]) developed
statistical methods to solve problems stemming from his
employment in a brewery.
`t` test is applied to find the significant difference between the
two means.
Criteria (assumption)
[Q: Write down the preconditions for t-test. (BSMMU, MD
Radiology, July, 2011)
Criteria for applying `t` test:
1. Quantitative data
2. Random sample
3. Sample size less than 30
4. Variable normally distributed
In case of more than 30 sample z test is done formula of which
is same as t test but unlike t test where p value is counted from
t table, p value of z test is counted as follows:
o Z<1.96, p value is >0.05
o Z>1.96, p value is <0.05
o Z>2.58, p value is <0.01
Types
`t` test are two types :
a. Unpaired `t` test.
b. Paired `t` test
Unpaired t-Tests

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When applied
Applied when the two group of samples are not dependent
with one another (different group).
Criteria
o sample size from the two groups may or may not be
equal
o in addition to the assumption that the data is from a
normal distribution, there is also the assumption that the
standard deviation (SD)s is approximately the same in
both groups
Formula
1 2
2 2
1 2
m
`t` test(unpaired) =
m
SE SE
-
+
Example
Problem: In a nutritional study, 13 children were given a
usual diet plus vitamins A and D tablets while the second
comparable group of 12 children was taking the usual diet.
After 12 months, the gain in weight in pounds was noted as
given below. Can we say that vitamins A and D were
responsible for this difference?
Gain of weight in children on vitamins (Group I): 5, 3, 4, 3, 2, 6,
3, 2, 3, 6, 7, 5, 3
Gain of weight in children on usual diet (Group II): 1, 3, 2, 4, 2,
1, 3, 4, 3, 2, 2, 3,
Calculation: We know,
1 2
2 2
1 2
m
`t` test (unpaired) =
m
SE SE
-
+

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2
( )
= ( 1)
sum x x
SE n n
-
-
d.f = (n1-1) + (n2-1)
Here, in group I,
X = 5, 3, 4, 3, 2, 6, 3, 2, 3, 6, 7, 5, 3
m1 = mean of gr. I ( x )
SE1 = SE of gr. I
n1 = 13
In group II,
X = 1, 3, 2, 4, 2, 1, 3, 4, 3, 2, 2, 3,
m2 = mean of gr. II ( x )
SE2 = SE of gr. II
n2 = 12
For group I
x
x (m1) x x
- 2
( )
x x
- Sum
2
( )
x x
-
5 4 1 1 32
3 -1 1
4 0 0
3 -1 1
2 -2 4
6 2 4
3 -1 1
2 -2 4
3 -1 1
6 2 4
7 3 9
5 1 1
3 -1 1

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Sum x
= 52
2
1
( )
= ( 1)
sum x x
SE n n
-
-
1
32
= 13(13 1)
SE -
For group II
x
x (m2) x x
- 2
( )
x x
- Sum
2
( )
x x
-
1
2.5
-1.5 2.25
11
3 0.5 0.25
2 -0.5 0.25
4 1.5 2.25
2 -0.5 0.25
1 1.5 2.25
3 0.5 0.25
4 1.5 2.25
3 0.5 0.25
2 -0.5 0.25
2 -0.5 0.25
3 0.5 0.25
1
32
= 156
SE
2
1
32
= = 0.205
156
SE

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2
11
= 12(12 1)
SE -
2
11
= 132
SE
2
2
11
= = 0.083
132
SE
1 2
2 2
1 2
m
`t` test (unpaired) =
m
SE SE
-
+
4 2.5 1.5
`t` test (unpaired) = = =2.79
0.5366
0.205 0.083
-
+
d.f = (13-1) + (12-1) = 23
At 23 d.f the highest obtainable value of `t` at 0.05 level of
significance is 2 .069 as found on reference to `t` table. The `t`
value in this experiment is calculated at 2.79 which is much
higher than the highest 2.069 obtainable by chance. Thus the
probability of occurrence (p) of the value obtained (2.79) by
chance is much less than 0.05
So vitamins A and D were responsible for the difference in
increase of weight in two groups.
[Q:
 25 obese individuals have mean blood pressure 140
mm Hg with SD = 5. In another 16 non-obese
individuals, mean blood pressure is 120 mm Hg with
SD =4. Comment on the relation between obesity and
blood pressure. [BSMMU, Radiology, January, 2012)
 The following table shows the statistics by gender

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Gender Number
Mean
height
in CM
SD
Boys 169 168 14
Girls 54 153 8
Determine whether heights differ with gender. (BSMMU,
MD Radiology, July, 2009)]
Paired t test
When applied
Applied when the variables are dependent with one another
(same group).
Formula
t (paired) =
d d
SD SE
n

Here,
d =mean of difference before and after treatment.
n =no of observation in the single group.
SD = standard deviation
2
( )
=
1
sum d d
n


d= difference of observation before and after
treatment.
Example
Problem: The fasting glucose level of 7 subjects was 4, 6, 8, 5,
9, 3, 7 m mol/L. After 75 gm of oral glucose the blood sugar
level were 8,10,13,9,8,12,14 m mol/ L respectively. Use

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appropriate statistical test whether oral glucose had
significantly increase the blood sugar level or not.
Answer: Here paired `t` test to be applied
Blood sugar
Fasting
After 75 gm
glucose
d d (d-
d )2
Sum
(d-
d )2
4 8 4 4.85 .72 38.84
6 10 4 .72
8 13 5 .02
5 9 4 .72
9 8 1 14.82
3 12 9 17.22
7 14 7 4.62
=
d
t SD
n
Here,
2
( )
= 1
sum d d
SD n
-
-
=
38.84
= 2.43
16
t
4.85
=
2.43
7
4.85
=
2.43
2.64
4.85
= × 2.64
2.43

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12.80
=
2.43
=5.26
d.f = (n-1) = 6
P value of 5.26 at 6 d.f is < 0.01 which is the significant. So
null hypothesis rejected.
Problem for practice
1. Serum protein is lower in females than in males. Justify this conclusion by
applying appropriate statistical technique to the data given below:
Sex Number n
Mean serum protein
level in gm/per 100 ml
SD
Males 18 7.21
0.26
Females 7 6.90 1.28
Answer: t = .64, p > 0.10]
3. In an Investigation on neonatal blood pressure in relation to maturity
following results were obtained:
Babies 9 days
old
Number Mean systolic Standard
deviation
Normal 54 75 6
Neonatal
asphyxia
14 69 5
Is the difference in mean systolic BP between two groups statistically
significant?
[Answer: Yes, t = 3.44. p <0.001]
3. In order to determine the effect of certain oral contraceptive on weight
gain, nine ‘healthy females were weighed prior to the start of its use and
again at the end of a 3-months Period.
Subject No. Initial weigh in kgs weight after 3 months
1 48.0 49.2
2 56.4 57.2
3
4
52.0
60.0
56.0
58.0
5 54.0 56.0
6 56.0 57.2

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7 48.0 47.2
8 56.0 56.4
9 52.0 52.8
[Answer: Yes, t = 1.26. p>0.10]
Chi- square test (x2
)
[Q:
 Write shorts on: Chi-square test. (BSMMU, MD
 What are the condition for chi square test? Name the
steps of chi-square test. (BSMMU, MD, July, 2009)
 Write down the preconditions for chi-square test.
(BSMMU, MD Radiology, July, 2011)
 Write short notes on: Chi-squares test, (BSMMU, MD
 What are the condition for chi square test? Name the
steps of chi-square test. (BSMMU, MD Radiology, July,
2009)]
It is applied to show the association between observed value
and expect value.
Application of Chi- squared test
1. To asses the efficacy of the new vaccine or new drug
2. To compare the effectiveness between two or more drug.
Contingency table
A frequency table in which a sample is classified according to
two different attributes is called a contingency table.
A contingency table having two rows and two columns is
known as 2x2 contingency table. A contingency table having
`r' row and `c' columns is referred as r x c contingency table.

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2x2contingency Table
Disease Smokers Non-smokers Total
Cancer 6 4 10
No cancer 94 96 190
Total 100 100 200
Sample = 2 (smokers and non smokers). each sample divided
in two categories.
Importance of 2x2 tables:
They are used in-
X2
test
Relative risk
Odd ratio
Specificity
Sensitivity
Positive predictive value
Negative predictive value
Accuracy
Incidence
Formula
2
2
(O-E)
x =sum E
Here,
O= Observed value
E = Expected value
×
=
Row total Column total
Formula of E Grand total
d.f= (Row-1) X (Column-1)
Example

Biostatistics-107
Problem: Find the efficacy of drug from the data given in the table.
Group Result Total
Died Survived
Control, on
placebo
10 25 35
Experiment,
on drug
5 60 65
Total 15 85 100
Calculation:
 The expected values (E) of different groups are determined
by the formula.
×
=
Column total Row total
E Sample total
 Now find the X2 value contributed by each cell by the
formula
2
2
( )
=
O E
X E
-
and
 Sum up the x2
values of all cells. Total value has to be taken
into account to draw the conclusion about the efficacy of
the drug.
1.
 Expected number (E)of the died in control group Is
15
= ×35 = 5.25
100

2
X value of this cell
2 2
( ) (10 5.25) 22.5625
= = = = 4.2978
5.25 29.75
O E
E
- -
2.

Biostatistics-108
 Expected number (E) of the survived in control group
85
= ×35 = 29.75
100

2
X
2
(25 29.75) 22.5625
= = = 0.7584
29.75 29.75
-
3.
 E for the died in experiment group
15
= ×65 = 9.75
100

2
2
(5 9.72) 22.5625
= = = 2.3140
9.75 9.75
-
4.
 E for the survived in experiment group =Total –E
for died=65 - 9.75=55.25

2

Biostatistics-109
2 2
(60 55.25) (4.75) 22.5625
= = = = 0.4083
55.25 55.25 55.25
-
Total
2
X
value=4.2976+0.7584+2.3140+0.4083=7.7783
On referring to table x2
, as 1 degree of freedom, the value of
x2
under probability O.05 is 3.841 and under 0.02 is 5.412.
Calculated values of x2
7.7783, which is higher than 5.41. The
difference in mortality is significant at 2% level hence the
drug is efficacious.
Yates Correction:
If x2
test is applied in a fourfold table (also called four cell
table or 2 by 2 table or 2 X 2 contingency table) will not give
reliable one degree of freedom if the observed ‘value in any
cell is less than 5. In such cases to apply x2
test, Yates’
correction is necessary. Yates’ correction means subtraction
1/2 (or 0.5) from the absolute difference between the
observed value and expected value and then squares it to
divide by expected value. Hence the formula for the corrected
Chi-squared statistic for a 2 by 2 table is
2
[( ) 1/ 2]
O E
sum E
- -
If X2
test is applied in a table larger than 2 by 2 table Yates
Correction can not be possible.

Biostatistics-110
Even after Yates correction if X2
value lies much below the
table below. In such cases some other appropriate test may
be applied.
Problem for practice
1. From the table given below, can you conclude that there is
an association between the socioeconomic status of women
and the period of breast feeding?
Socioeconomic
status
No of women as per duration of
breast feeding in months
Total
1—6 7—12 12+
I, lI 1 16 2 19
III I 18 34 53
IV.V 2 11 25 38
Total 4 45 61 110
[Answer: Yes, x2 = 19.89 p <0.001]
2.
Site of
Infarction
No. of patients with
bradycardia
No. of patients with
bradycardia
Total
Posterior 31 35 66
Anterior 6 28 34
Total 37 63 100
Test whether the Incidence of bradycardia has any
predilection for the site of Infarction. (JIMA Vol. 57. Dec 1,
1971. p. 426).
[Answer: with Yate’s correction, x2
= 7.07 p <0.05]
[Q:

Biostatistics-111
 In a cross sectional study, 1000 & 400 were non
smokers & smokers respectively. Among them
120 and 30 people were found hypertensive in
non-smokers and smokers respectively. Is
smoking associated with hypertension?
(BSMMU, MD Radiology, July, 2011)
 Among 50 radiologists land 150 pathologists, 20
and 15 individuals were found to be sterile
respectively. Comment on the findings with
statistical basis. (BSMMU, MD Radiology,
January, 2011)
 100 patients treated by a new drug, of which 20
patients died. Another 100 patients treated by old
drug, of which 30 patients died. Comment on the
efficacy of the new drug. (BSMMU, Radiology,
January, 2012)
 Melamine containing powdered milk (MCPM) is
though to be associated with neprolethiasis. A
research team observed that 20 of 190 subjects
who were regularly taking MCPM showed
echogenic shadow in the KUB region as against
15 of 300 subjects showed similar echogenic
shadows who never consumed MCPM. Did
MCPM contribute to echogenicity? (BSMMU,
MD Radiology, July, 2010)
 A random sample 200 females show 120 as using
oral contraceptive pill (OCP) and test as non-
users. Among the users 4 & among the non users
12 were found to suffer from cancer cervix. Do
you consider that cancer cervix prevalence is
related with OCP? (BSMMU, MD Radiology,
July, 2009)
 Among the 50 subjects with retinopathy and 650-
control subjects, 40 and 160 subjects respectively
found diabetic. Is there any association between

Biostatistics-112
DM and retinopathy? (BSMMU, MD Radiology,
January, 2009)
 Among 50 radiologists and 150 histopathologists,
20 and 15 individuals respectively found to be
sterile. Comment on the findings with statistical
basis. (BSMMU, MD Radiology, January, 2010)
 Out of 300 cases of neonatal jaundice, 200 treated
by phototherapy where 10 patients did not
improved as against 20 who received no
phototherapy. How effective the phototherapy is?
(BSMMU, MD Radiology, January, 2010)
 Out of 300 volunteers 170 were vaccinated by
hepatitis B vaccine. Later on 5 people of
vaccinated group and 40 people of non-
vaccinated group and 40 people of non-
vaccinated group developed viral hepatitis.
Comment on the efficiency of the vaccine.
(BSMMU, MD Radiology, July, 2011)]
Proportion Test (Z test)
It is used to see the significance of difference between two
proportions.
Formula
1 2
1 1 2 2
1 2
=
p p
Z
p q p q
n n


Here,
p1 = proportion of one group
p2 = proportion of other group
q1 = 100-P1
q2 = 100-P2
n1 = no of independent sample in one group
n2 = no of independent sample in other group
p value of z test is counted as follows:

Biostatistics-113
o Z < 1.96, p value is >0.05
o Z > 1.96, p value is <0.05
o Z > 2.58, p value is <0.01
o Z > 3 p value is <0.001
Example
In Dhaka and Rajshahi universities there are 80% and 20%
female students. Independent samples of 100 and 200
students are drawn in these two universities. Do the data
reveal a real difference between two universities?
1 2
1 1 2 2
1 2
=
p p
Z
p q p q
n n


Here,
p1 =80, q1 = 100-80 = 20
p2= 20, q2=100-20=80
n1=100, n2=200
So,
Z =
80 20
80×20 20×80
100 200
-
+
60
=
16 8
+
60
=
24
=12.24
If Z =3 or more p < 0.001
So p< 0.001. Result is significant. Null hypothesis is rejected.
Problem for Practice:
Find out the significance of difference between the following
two proportions
proportions
-I
proposition -
II

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78% 75%
N=150 N=100
[Answer: Z = 12.83 p <0.001, highly significant]
Here,
p1 =
12 100
250
X
= 4.8%
p2 =
18 100
150
X
=12%
q1 = (100-4.8) = 95.2
q2 = (100-12) = 88
n1 = 250
n2 = 150
1 2
1 1 2 2
1 2
=
p p
Z
p q p q
n n


4.8 12
=
4.8 95.2 12 88
250 150
X X


=2.4
Z >1.96, so p< 0.05
Interpretation: Null hypothesis rejected. Test is significant.
There is difference in birth of low in rural area than urban
area.
[Q: Number of female of the random samples of sizes
50 and 60 drawn from two primary school students was

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20 and 30 respectively. Test whether the pro proportion
of female students of the two schools differ
significantly. Let a = 0.05. (BSMMU, MD Radiology,
January, 2011)]
One sample z test
Formula for One sample z test for mean
( )
x
Z
SE x



Here,
x = mean of sample
x = mean of population (given value)
SE=
SD
n
Formula for one sample z test for proportion

Biostatistics-116
p P
Z
pq
n


Here,
p = Proportion of sample
P = Proportion of population
n = Sample size
q = (1-p) or (100-p)
[Q: In a community the prevalence of under weight
baby is 20%. A survey of 64 newborn babies shows
under weight baby to be 10%. Is there any difference?
Is the sample taken from this community?]
Answer:
Here,
P = 10%
P = 20%
q = (100-10) = 90
n = 64
p P
Z
pq
n


10
900
64


10
3.75


=2.6
Level of significance: Calculate value of Z> 1.96, so

Biostatistics-117
p< 0.05
Interpretation: Null hypothesis is rejected. There is
difference b/w prevalence rate of sample & population
so; sample is not taken from that population.
Analysis of variance (ANOVA) or F test
[Write down the preconditions for chi-square test. ANOVA
& t-test. (BSMMU, MD Radiology, July, 2011)]
When the group more than two this test is used two see
whether any variation between the groups or within the
groups.
Formula
( )
= ( )
MS Mean square between group
F MS Mean square within group
=
SOS
MS df SOS= Sum of variance
Total d.f = N-1
d.f between group = Group no. – 1
d.f within groups = Total d.f - d.f between groups
To find out F` value following factors to be calculated
1.
sum x2
2.
2
GT
sum N (where N=total no .observation in all group,
GT is grand total)
3.
Total variance
2
2
=
GT
sum x N
- (or total sum of square)
4.
2
( )
.
cell total
sum cell total no

Biostatistics-118
5. Variance between group
2 2
CT GT
sum N N
-
6. Variance within group= Total variance- Variance between
group (or sum of square within group)
7. Total d.f and group d.f.
Now tabulate the above factor / results as below
Sources of
Variation
d.f Sum of
squares
(SOS)
mean
squares
(SOS
/d.f)
=
MS between group
F MS within group
P
value
Total
Between group
Within group
Example
Problem: Find out the P value of following groups
Group-1 Group-II Group-III
2 12 25
4 14 28
6 16 30
8 18 35
Answer:
Group - III
X X2
25 625
28 784
30 900

Biostatistics-119
Group -1
X X2
2 4
4 16
6 36
8 64
Sum
x =
20
Sum x2
=
120
2 2 2 2
Sum X =Sum X of Group -1+Sum X of Group - II +Sum X of Group - III
=120 + 920 + 3534
= 4574
GT = CT of Group-1+ CT of Group-II + CT of Group-III
= 20+120+118
=198
2 2
( ) (198) 39204
= = = 3267
12 12
GT
N
CT (Cell total)
1. CT In group I = 2+4+6+8=20
2
( ) 400
= =100
4
CT
Cell total no
2. In group II CT =12+14+16+18=60
2
( ) 3600
= = 900
4
CT
Cell total no
35 1225
Sum
x =
118
Sum x2
=
3534
Group - II
X X2
1 2 144
14 196
16 256
18 324
Sum x
= 60
Sum x2
= 920

Biostatistics-120
3. In group III CT = 25+28+30+35=118
2
( )
= 4481
CT
Sum Cell total no
Total variance
2
2
= - =1307
GT
Sum X N
Variance between groups
2
( ) 2
= Cell total no
= 4481 3267 =1214
CT GT
Sum N
-
-
Variance within group
= Total variance - Variance between groups
=1307 -1214
=93
Tabulation:
The following table shows the summery of ANOVA
Sources of
variations
SOS (sum of
variance)
d.f Mean
square
SOS/d.f
F P
Between
groups
1214 2 607 60.7 <0.01
Within groups 93 9 10
Total variance 1307 11
Total d.f = N (n1 + n2 +n3) -1 = 12-1=11
d.f between groups = Group no. - 1 = 3 - 1=2
d.f within groups = Total d.f - d.f between groups = 11 – 2 = 9

Biostatistics-121
Mean square (MS)
=
SOS
df
 Between groups =60.7
 Within groups = 10
=
MS between group
F
MS within group
607
= = 60.7
10
For 2/9 d.f i.e. 2 d.f across or horizontally 9 d.f vertically the
table value of F = 8.02 at P value 0.01or 1%.
Here the calculated value is 60.7 which is much greater than
table value. Hence the result is highly significant. Null
hypothesis rejected.
Non-Parametric Tests
Non-Parametric tests are often used in place of their parametric counterparts
when certain assumptions about the underlying population are questionable.
For example, when comparing two independent samples, the Wilcoxon
Mann-Whitney test does not assume that the difference between the
samples is normally distributed whereas its parametric counterpart, the two
sample t-test does. Non-Parametric tests may be, and often are, more
powerful in detecting population differences when certain assumptions are
not satisfied.
All tests involving ranked data, i.e. data that can be put in order, are non-
parametric.
Wilcoxon Mann-Whitney Test
The Wilcoxon Mann-Whitney Test is one of the most powerful of the non-
parametric tests for comparing two populations. It is used to test the null
hypothesis that two populations have identical distribution functions against
the alternative hypothesis that the two distribution functions differ only with
respect to location (median), if at all.

Biostatistics-122
The Wilcoxon Mann-Whitney test does not require the assumption that the
differences between the two samples are normally distributed.
In many applications, the Wilcoxon Mann-Whitney Test is used in place of the
two sample t-test when the normality assumption is questionable.
This test can also be applied when the observations in a sample of data are
ranks, that is, ordinal data rather than direct measurements.
One and two tailed tests
o Two – tailed (-sided) test is concerned with differences
between observations in either direction, (i.e. checks both
the upper and lower tails of the normal distribution).
Example: if two alternative treatment A and B are
compared then whether either A or B may be better.
o One – tailed ( - sided ) test is only concerned with
differences between observations in one direction ( i.e.
only one tail of the normal distribution curve) e.g.
whether drug A is better than B drug or a placebo.
Exercise
[Here are the questions on biostatics on measuring different significance
test of MD/MS/M. Phil examination of different discipline to practice for
exam.]
Significance of difference
1. Find out the significance of difference between the observation in
following groups & find out the level of significance:
Group-I Group-II
30,35,38,40,42 30,38,40,45,37
Ans. : T value-0.1837691675
P value 0.8587670160
Degrees of Freedom 8
2. Find out the significance of difference between 2 proportions:
proportions -I proposition -II
20% 50%
N=250 N=300

Biostatistics-123
3. Find out the significance of difference between the following 2 groups
using unpaired `t` test :
Group-I Group-II
12,14,18,20,30 50,55,52,60,65
Ans.: T value -9.0399242925
P value 0.0000179377
4. Find out the significance of difference between the following two
proportions. Jan-94
80% 48%
N=75 N=100
proportions
28% 48%
N=75 N=100
proportions.
35% 82%
N=.50 N=55
ANOVA
1. Find out the analysis of variance between the following groups:
Group-I Group-II Group-III
5,8,7,6 15,18, 20,22 30,32,35,40
2. Calculate the ANOVA and find out the level of significance between
following 3 groups:
10,8,12,6,7 10,12,14,16,13 15,10,12,14,09
3. Calculate the ANOVA and find out the level of significance between
following 3 groups:

Biostatistics-124
2,4,6,8,10 6,3,5,9 ,12 7,11,8,10,09
`t` value
4. Find out the value of a paired `t` from the following observation:
Before Rx After Rx
70,65,72,75,78 52,50,55,60,65
Ans.: T value 17.8944272419
P value 0.0000573181
5. Calculate the unpaired `t` test and find out the level of significance
between the observation of the following groups:
Group-I Group-II
20,22,30,35,40 50,55,48,52,60
Ans.: T value -7.9357539324
P value 0.0013651102
6. Calculate the paired `t` test from the following observation:
Before Rx After Rx
8,15,20,12,25 4,6,8,3,9
Ans.: T value 5.0636968354
P value 0.0071619599
7. Calculate the significance of different using unpaired `t` test in two
groups:
Group-1 Group-II
2,4,6,8,10 15,18,20,25,30
Ans.: T value -5.1827528546
P value 0.0008399206
Efficacy

Biostatistics-125
1. 150 pts suffering from bone TB were given clinical trial on new anti-TB
drug A and the old drug `B`. Of these, 98 were given À` and rest `B`.
Those who received À` only did not respond as compared to 20 of
the patients who received `B`. Find out the efficacy of two drugs. [Ja-
2000]
2. 200 pts suffering from bone TB were selected to test the efficacy of
INH and Rifampicin. Of these 150 received INH & 100 Rifampicin.
Those who received INH, 85 in no. were improved as compared to 78
of the pts who received rifampicin. Find out which is more efficacious.
[Ja-95]
3. 200 pregnant women were selected to test the efficacy of `TT`. of
these, 120 were given `TT` and 12 were infected as compared to 15
unvaccinated group. How do you test whether the TT has any
protective role against tetanus? [Ja-95]
4. 170 pts selected for comparison of the effect of Diclofen & Ketoralac
against post operative pains. 100 were given diclofen and the rest
Ketoralac. of these 55 were improved who received diclofen & 62
improved who received Ketoralac. Use appropriate statistical test to
find out which drug is superior in releasing of pain? Ja-95
5. In a clinical trial of TT to 300 pregnant women, 200 were vaccinated
and 100 were remained unvaccinated. Of the unvaccinated group, 4
developed tetanus as compared to 20 of the unvaccinated group. How
do you prove the efficacy of the vaccine? July-93
6. 200 pts were operated for cholecystectomy by two methods À`&`B`
with method À` 100 and rest by method `B`. With method À` 80 pts
were successfully operated as compared to 60 pts with `B`. How do
you prove the superiority of the method? [Ja-94]
7. 150 patients suffering from RA were selected for a clinical trial of
diclofen and of these, 100 given the drug and rest 50 placebo. 70
patients who received drug were improved as compared to 10 of the
placebo group. How do you test that diclofen providing at significant
effect.
One sample z test
1. Prevalence of HTN pt, in a society was found to be 5%, 15 yrs back. A
recent survey of 100 adult showed / revealed HTN percent to be 10%.
Is there any increasing prevalence of hypertension?
[Z=1.66 Z< 1.96 P> 0.05]
2. In EPI Project Director (EPI) said that the total vaccination coverage is
80%. In a sample of 300 the vaccination coverage is actually 70%. Is
there any difference?
[Z=3.8 Z> 3 P> 0.001]

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