1. Tests
of
significance
Dr P Raghavendra,
PG in Dept of Community Medicine,
SMC, Vijayawada

2. Overview
• Types of data
• Basic biostatistics – central tendency, standard
deviation, standard normal curve, Z score
• Basic terms – Sampling Variation, Null hypothesis,
P value, Standard error
• Steps in hypothesis testing
• Types of tests of significance
• SEDP
• Chi Square test
• Special Situations in chi square testing

3. Types of
Data

4. Qualitative Data
• Qualitative are those which can be answered
as YES or NO, Male or Female, etc.
• We can only measure their numbers,
eg: Number of males, Number of MDR-TB
cases among TB patients, etc.
• A set of qualitative data can be expressed as
proportions. Eg: Prevalence, success rate.

5. Quantitative Data
• Quantitative are those which can be measured
in numbers, like Blood pressure, Age, etc.
• A set of quantitative date can be expressed in
mean and its standard deviation.
• Mean is the average of all variables in the
data.
• Standard deviation is a measure of the
distribution of variables around the mean.

6. Basic Bio-
Statistics

7. Basic Bio-Statistics
• Mean, Median, Mode
These are the measures of central tendency.
• Standard Deviation
It is the measure of distribution of the data
around the central value.
• Biological Variability

8. Simple Normal Curve
Mean = Median = Mode

9. • Z value
– Gives the position of the variable with respect to
central value in terms of standard deviation
Z value = Observed Value – mean
standard deviation

10. Numbers!!
• Numbers are confusing.
• Consider an example:
– In a hypothetical population, there were 70
deaths among men and 90 deaths among women.
So, where is the death rate high, among
men? Or women??

11. Sampling Variation
• Research done on samples and not on
populations.
• Variability of observations occur among
different samples.
• This complicates whether the observed
difference is due to biological or sampling
variation from true variation.
• To conclude actual difference, we use tests of
significance.

12. Null Hypothesis
• 1st step in testing any hypothesis.
• Set up such that it conveys a meaning that
there exists no difference between the
different samples.
• Eg: Null Hypothesis – The mean pulse rate
among the two groups are same
(or)
there is no significant difference between
their pulse rates.

13. • By using various tests of significance we either:
–Reject the Null Hypothesis
(or)
–Accept the Null Hypothesis
• Rejecting null hypothesis  difference is significant.
• Accepting null hypothesis  difference is not
significant.

14. Level of Significance – “P” Value
• P is the probablity of the measured significance
in difference attributable to chance.
• It is the probability of null hypothesis being true.
• We accept or reject the null hypothesis based on
P value.
• If P value is small, it implies, the probability of
attributing chance to cause the significance in
difference is small and we reject null hypothesis.
• In any test we use, we find out P value or Z score
to attribute significance.

15. How much P is significant
• Depends on the type of study.
• Balance between Type – I and Type – II errors.
• Practically, P < 0.05 (5%) is considered
significant.
• P = 0.05 implies,
– We may go wrong 5 out of 100 times by rejecting
null hypothesis.
– Or, We can attribute significance with 95%
confidence.

16. Standard Error
• To put it simply, it is the standard deviation of
means or proportion.
• Gives idea about the dispersion of mean or
proportions obtained from multiple or
repeated samples of same population.

17. Steps in
Testing a
Hypothesis

18. General procedure in testing a
hypothesis
1. Set up a null hypothesis.
2. Define alternative hypothesis.
3. Calculate the test statistic (t, 2, Z, etc).
4. Determine degrees of freedom.
5. Find out the corresponding probability level (P
Value) for the calculated test statistic and its
degree of freedom. This can be read from
relevant tables.
6. Accept or reject the Null hypothesis depending
on P value

19. Classification of tests of significance
• For Qualitative data:-
1. Standard error of difference between 2 proportions
(SEp -p )
1 2
2. Chi-square test or 2
• For Quantitative data:-
1. Unpaired (student) ‘t’ test
2. The Z test
3. Paired ‘t’ test

20. Standard error of
difference between
2 proportions
(SEp -p )
1 2

21. Standard error of difference between
proportions (SEp1-p2)
• For comparing qualitative data between 2
groups.
• Usable for large samples only. (> 30 in each
group).
• We calculate SEp1-p2 by using the formula:

22. • And then we calculate the test statistic ‘Z Score’
by using the formula:
Z = P1 – P2
SEP1-P2
• If Z > 1.96 P < 0.05 SIGNIFICANT
• If Z < 1.96 P > 0.05 INSIGNIFICANT

23. Example – 1
• Consider a hypothetical study where cure rate
of Typhoid fever after treatment with
Ciprofloxacin and Ceftriaxone were recorded
to be 90% and 80% among 100 patients
treated with each of the drug. How can we
determine whether cure rate of Ciprofloxacin
is better than Ceftriaxone?

24. Solution
• Step – 1: Set up a null hypothesis
– H0: “There is no significant difference between
cure rates of Ciprofloxacin and Ceftriaxone.”
• Step – 2: Define alternative hypothesis
– Ha: “ Ciprofloxacin is 1.125 times better in curing
typhoid fever than Ceftriaxone”
• Step – 3: Calculate the test statistic – ‘Z Score’

25. Step – 3: Calculating Z score
Z= P1 – P2
SEP1-P2
– Here, P1 = 90, P2 = 80
– SEP1-P2 will be given by the formula:
So, Z = 10/5 = 2

26. • Step – 4: Find out the corresponding P Value
– Since Z = 2 i.e., > 1.96, hence, P < 0.05
• Step – 5: Accept or reject the Null hypothesis
– Since P < 0.05, So we reject the null hypothesis (H0)
There is no significant difference between cure rates of
Ciprofloxacin and Ceftriaxone
– And we accept the alternate hypothesis (Ha) that,
Ciprofloxacin is 1.125 times better in curing typhoid
fever than Ceftriaxone

27. Example – 2
• Consider another study, where 72 TB patients
were divided into 2 treatment groups A and B.
Group A were given DOTS and group B were
given a newly discovered regimen (Regimen-X)
found effective in Nigeria. The cure rates at
the end of treatment (assume 100%
compliance) among group A and group B were
90% and 80% respectively. Comment on
whether this difference is significant or not.

28. If you think the answer is same for example 1
and 2, remember that –
• Numbers what we see, can be misguiding.
• Mind can be biased.
• Hence always try and test the significance of
the difference observed.

29. Solution
• Step – 1: Set up a null hypothesis
– H0: “There is no significant difference between
cure rates of DOTS and Regimen-X.”
• Step – 2: Define alternative hypothesis
– Ha: “ DOTS is 1.125 times better in curing typhoid
fever than Regimen-X”
• Step – 3: Calculate the test statistic – ‘Z Score’

30. Step – 3: Calculating Z score
Z= P1 – P2
SEP1-P2
– Here, P1 = 90, P2 = 80
– SEP1-P2 will be given by the formula:
So, Z = 1.2

31. • Step – 4: Find out the corresponding P Value
– Since Z = 1.2 i.e., < 1.96, hence, P > 0.05
• Step – 5: Accept or reject the Null hypothesis
– Since P > 0.05, So we accept the null hypothesis
(H0)
There is no significant difference between cure
rates of DOTS and Regimen-X and the new
regimen-X is as good as DOTS.

32. Limitations of SEP1-P2
• Sample must be large.
– At least 30 in each sample.
• It can compare only 2 groups.
– Quite often we might be testing significance of
difference between many groups.
– In such cases we might not be able to do SEP1-P2

33. Chi-Square
Test

34. Chi square ( 2) test
• This test too is for testing qualitative data.
• Its advantages over SEDP are:
– Can be applied for smaller samples as well as for
large samples.
– Can be used if there are sub-groups or more than
2 groups

35. Chi square ( 2) test
Prerequisites for Chi square ( 2) test to be
applied:
– It must be quantitative data.
– The sample must be a random sample.
– None of the observed values must be zero.
– Yates correction must be applied if any
observation is less than 5 (but 0).

36. Steps in Calculating 2 value
1. Make a contingency table mentioning the
frequencies in all cells.
2. Determine the expected value (E) in each cell.
E= Row total x Column total rxt
Grand total T
3. Calculate the difference between observed and
expected values in each cell (O-E).
Contd…

37. 4. Calculate 2 value for each cell
2 of each cell = (O – E)2
E
5. Sum up 2 value of each cell to get 2 value of
the table.
2= (O – E)2
E

38. Example 3
• Consider a study done in a hospital where
cases of breast cancer were compared against
controls from normal population against
possession of a family history of Ca Breast.
100 in each group were studied for presence
of family history. 25 of cases and 15 among
controls had a positive family history.
Comment on the significance of family history
in breast cancer.

39. Solution
• From the numbers, it suggests that family
history is 1.66 (25/15) times more common in
Ca breast. So is it a risk factor in our
population?
• We need to test for the significance of this
difference.
• We shall apply 2 test.
• As in SEDP, we should follow the same steps.

40. Solution
• Step – 1: Set up a null hypothesis
– H0: “There is no significant difference between
incidence of family history among cases and controls.”
• Step – 2: Define alternative hypothesis
– Ha: “Family history is 1.66 times more common in Ca
breast”
• Step – 3: Calculate the test statistic – ‘ 2’

41. Step – 3: Calculating 2
1. Make a contingency table mentioning the
frequencies in all cells.
Risk factor (Family History)
Group Total
Present absent
Cases 25 ( a) 75 (b) 100
Controls 15 c
( ) 85 d
( ) 100
Total 40 160 200

42. Step – 3: Calculating 2
2. Determine the expected value (E) in each
cell.
E= Row total x Column total rxt
Grand total T
– For (a), Ea = 100 x 40 / 200 = 20
– For (b), Eb = 100 x 160 / 200 = 80
– For (c), Ec = 100 x 40 / 200 = 20
– For (d), Ed = 100 x 160 / 200 = 80

43. Step – 3: Calculating 2
(O – E)2
O E (O – E) (O – E)2
E
a 25 20 5 25 1.25
b 75 80 -5 25 0.3125
c 15 20 5 25 1.25
d 85 80 -5 25 0.3125
2 = 3.125

44. Solution
• Step – 4: Determine degrees of freedom.
DoF is given by the formula:
DoF = (r-1) x (c-1)
where r and c are the number of rows and
columns respectively
Here, r = c = 2.
Hence, DoF = (2-1) x (2-1) = 1

45. Solution
• Step – 5: Find out the corresponding P Value
– P values can be calculated by using the 2
distribution tables.
0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47
5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46

46. Solution
• Step – 6: Accept or reject the Null hypothesis
• In our given scenario,
2 = 3.125
• This is less than 3.84 (for P = 0.05 at dof =1)
• Hence Null hypothesis is Accepted, i.e.,
“There is no significant difference between
incidence of family history among cases and
controls”

47. Example - 4
• Calculate 2 for the following data and
comment on the significance of blood groups
in type of leprosy.
Non
Lepromatous
Blood group Non-leprosy Lepromatous Total
leprosy
leprosy
A 30 49 52 131
B 60 49 36 145
O 47 59 48 154
AB 13 12 16 41
Total 150 169 152 471

48. The expected frequencies will be
Lepromatous Non Lepromatous
Blood Group Non-leprosy
leprosy leprosy
(131/471) x 150 (131/471) x 169 (131/471) x 152
A
= 41.7 = 47.0 = 42.3
(145/471) x 150 (145/471)x 169 (145/471)x 152
B
= 46.2 =52.0 = 46.8
(154/471)x 150 (154/471)x 169 (154/471)x 152
O
= 49.0 = 55.3 = 49.7
(41/171)x 150 (41/171)x 169 (41/171)x 152
AB
= 13.1 = 14.7 = 13.2

49. On substituting 2 formula, 2 in each cell will be:
Non
Lepromatous
Blood group Non Leprosy Lepromatous Total
leprosy
leprosy
A 3.28 0.09 2.22 5.59
B 4.12 0.17 2.49 6.78
O 0.08 0.25 0.06 0.39
AB 0.01 0.50 0.58 1.09
2 13.85

50. Solution
• DoF = (r-1) x (c-1) = (3-1)(4-1) =6
• For DoF of 6, a 2 of 13.85 means P will be
between 0.01 and 0.05.
• Hence, We can conclude that the observations
are significant.
0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47
5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46

51. Special situations in 2 test
• In 2 x 2 tables only, a simplification of formula
for 2 can be applied:
2= (ad – bc)2 x N
(a+b)(c+d)(a+c)(b+d)
(N = a+b+c+d)
• Yates correction

52. Special situations in 2 test
• Yates correction cannot be done in
multinomial groups.
• This can be used only in 2x2 tables.
• In multinomial groups, such groups containing
values <5 can be merged to form a bigger
group to avoid the pitfall.

53. Thank you