This document provides an overview of various statistical tests used to analyze data, including parametric and non-parametric tests. It discusses tests for comparing one or more groups, such as t-tests for two independent groups and ANOVA for more than two groups. Specific tests are described, including their assumptions and calculations. Examples are provided to illustrate how to apply and interpret various statistical tests.

1. TESTS OF SIGNIFICANCE
Moderator: Dr. S.K.Bhasin
Presenters: Migom
Parnava
Kartikey
1

2. Probability distribution
1. Normal distribution
2. Binomial distribution
3. Poissons distribution
2

3. Normal curve
• Gaussian Distribution
• Continuous
• Bell shaped
• Symmetrical
• Mean, Mode and Median coincide
3

4. Confidence level and Confidence limits
4

5. Z distribution
• Z transformation: Z= observation – mean = X - 
SD 
5

6. • P value:
• The probability of observing a result as extreme as or more extreme than the one
actually observed from chance alone (i.e., if the null hypothesis is true)
• Power of a test:
• Probability that a study or a trial will be able to detect a specified difference
• Power= 1- β
6

7. Scales of measurement
• Nominal scales
• E.g. outcome of surgical procedure
• Ordinal scales
• E.g. APGAR score, tumour staging
• Numerical scale
• E.g. Age, weight, number of fractures
7

8. Measures of central tendency
• Mean
• Median
• Mode
Some important concepts:
• Standard deviation
• Variance
• Standard error of mean
8

9. TESTS OF STATISTICAL SIGNIFICANCE
• Procedure for comparing observed data with a claim (hypothesis) whose
truth we want to assess
• Appropriate test of significance:
• Data
• Sample
• Purpose
9

10. Parametric tests Non parametric tests
Assumed distribution Normal Any
Typical data Numerical Ordinal or nominal
Usual central measure Mean Median , mode
Advantages Can draw more conclusions Simplicity ; less affected by outliers
Describe one group Mean , SD Median, interquartile range
Proportion
Independent measures,2 groups Unpaired t test Chi-square test, Fisher’s test
Mann-Whitney U test
Independent measures ,>2 groups ANOVA Kruskal – Wallis test
Chi-square test
Repeated measures,2 conditions Paired t test Wilcoxon sign rank, Mc Nemar’s
Chi-square test
Repeated measures,>2 conditions ANOVA Friedman’s test
Chi-square test
Regression Simple linear regression or
Non-linear regression
Non parametric regression
10

11. n > 30
t
Normally
Distributed
t
Transform for t or Sign test
No
Yes
No
Yes
Difference in means or medians (ordinal or numerical measures)
1 Group
11

12. Number of
Groups
Independent
Groups
n > 30
t
Normally
Distributed
Equal
Variances
t
Equal n’s
t
Transform
for t or
Wilcoxon
rank-sum
Transform
for t or
Wilcoxon
rank-sum
2 Groups
Yes
Yes
No
No
Yes
Yes
No
No
Yes
Difference in means or medians (ordinal or numerical measures)
12

13. Independent
Groups
n > 30
t ( Paired t)
Normally
Distributed
t ( Paired t)
Transform for t
Or
Wilcoxon signed
ranks
Yes
No
No
Yes
No
13

14. Independent
Groups
Normally
Distributed
Number of
Factors
One-Way or
Other ANOVA
Two-Way or
Other ANOVA
Kruskal-Wallis for
1 Factor
Normally
Distributed
Repeated
Measures ANOVA
Friedman
Yes
Yes
Yes
No
No
No
1 factor
2 or
More
Factors3 or More
Groups
Difference in means or medians (ordinal or numerical measures, three or more groups)
14

15. Number of Groups
np and
n(1-p) > 5
z Approximation
Independent
Groups
Small Expected
frequencies
Fisher's Exact
Test
X2 or Z
Approximation
Mcnemar or K
Independent
Groups
Small Expected
Frequencies
Collapse
Categories for
X2
X2
Cochran's Q
1 Group
2 Groups
3 or More
Group
Yes
Yes
Yes
Yes
No
No
No
No
Difference in proportions (nominal measure) 15

16. PARAMETRIC TEST
16

17. t distribution and t test
• symmetric
• mean of 0, but SD > 1, the SD is related to degrees of freedom (df)
• t test:
• Corresponds to Z test
• Assumptions:
• Observations are normally distributed
• Samples must be random
• Sample size is fewer than 30
t=
𝑋 − 
𝑆𝐷
√𝑛
=
𝑋 − 
𝑆𝐸
17
𝑋=Sample Mean
=Population Mean
SD= Standard Deviation
n= sample size
SE= Standard Error

18. 18

19. • Dependent t test
• Paired sample t test, within subjects or repeated measures
• Compares the means between two related groups on the same, continuous variable
• Assumptions:
• Dependent variable should be continuous variable
• Independent variable should be two categorical, related groups or matched pairs
• No significant outliers
• Distribution of differences in the dependent variable between the two related
groups should be approximately normally distributed
19

20. Example:
• The measurement of the systolic and diastolic blood pressures was done two consecutive
times with an interval of 10 minutes. You want to determine whether there was any
difference between those two measurements
• Null hypothesis: H0:There is no difference of the systolic blood pressure during the first
(time 0) and second measurement (time 10 minutes).
SBP 1 SBP2 D D2
164 163 1.00 1.00
164 155 9.00 81.00
156 158 -2.00 4.00
147 131 16.00 256.00
186 178 8.00 64.00
170 160 10.00 100.00
20

21. • Σ D = 46, Σ D2 = 506, n = 6
• Mean D = 46/6 = 7.6
21
D 𝑫 D- 𝑫 (D- 𝑫) (D- 𝑫)2
1 7.6 1-7.6 -6.6 43.56
9 7.6 9-7.6 1.4 1.96
-2 7.6 -2-7.6 -9.6 92.16
16 7.6 16-7.6 8.4 70.56
8 7.6 8-7.6 0.4 0.16
10 7.6 10-7.6 2.4 5.76
214.16

22. • SD =  Σ (D- 𝑫)2 /(n-1)
• SD= 214.16/5 = 42.832= 6.54
• t=
𝐷 − 
𝑆𝐷
√𝑛
• t= 7.6/(6.54/  6) = 7.6/(6.54/2.44)= 7.6/2.68=2.835
• T=2.835
• df = n – 1 = 6 – 1 = 5
• Refer to t table:
t= 2.571 at α = 0.05, df=5
• So we will reject the null hypothesis
• Conclusion: There is a significant difference of the systolic blood pressure
between the first and second measurement. The mean average of first
reading is significantly higher compared to the second reading
22

23. 23

24. Independent t test
• Independent samples t test/ student’s t test/ two sample t test
• Compares the means between two unrelated groups on the same continuous,
dependent variable
• Determines whether there is a statistically significant difference between the
means of two unrelated groups
• Assumptions:
• Dependent variable should be measured on a continuous scale
• Independent variable should consist of two categorical, independent groups
• Independence of observations
• No significant outliers
• Dependent variable should be approximately normally distributed for each group of the
independent variable
• Homogeneity of variance (Levene’s test)
24

25. • t(n1+n2-2) =
𝑋1− 𝑋2
𝑆𝐷 𝑃
1
𝑛1+ 1
𝑛2
• SDP =
𝑛1 −1 𝑆𝐷1
2+ 𝑛2 −1 𝑆𝐷2
2
𝑛1+𝑛2 −2
• 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
X1 = mean of first group
X2 = mean of second group
SDP = pooled standard deviation
25

26. SDP =
𝑛1 −1 𝑆𝐷1
2+ 𝑛2 −1 𝑆𝐷2
2
𝑛1+𝑛2 −2
SD1 = 10
SD2 =12
State H0 H0 : 1 = 2
State H1 H1 : 1  2
Choose α α = 0.01
SDP =
25 −1 102
+ 17−1 122
25+17−2
= 10.84
t40 =
125−132
10.84 1
25+ 1
17
= 2.15
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 % level of significance
𝑋1 = 125
𝑋2= 132
𝑛1= 25
𝑛2= 17
26
t(n1+n2-2) =
𝑋1− 𝑋2
𝑆𝐷 𝑃
1
𝑛1+ 1
𝑛2

27. 27

28. F distribution
• Distribution of ratios
• F ratio =
𝑆 𝑥
2
𝑆 𝑦
2 =
𝑙𝑎𝑟𝑔𝑒𝑟 𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
(𝑛 𝑥 −1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚)
(𝑛 𝑦 −1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚)
28

29. ANOVA(ANalysis Of VAriance)
• Three or more groups
• OMNIBUS test
• Assumptions:
• Dependent variable should be continuous
• Dependent variable is normally distributed in each group
• Homogeneity of variance
• Independence of observations
• No significant outliers 29

30. 30

31. 31
SST
(total/overall) sum
of square
SSC
(column/between/ treatment)
Sum of squares
SSE
(within/ error)
Sum of squares
SSC = ( 𝑋 - 𝑋)2
Partitioning sum of squares:
SSE =  ( X - 𝑋 )2
Variance = average squared deviation of an observation from the distribution mean
Sample variance (S2) =
∑(𝜒− 𝜇)2
𝑛−1
Sum of Squares, SS =  ( - )2

32. Year 1 scores Year 2 scores Year 3 scores
82 71 64
93 62 73
61 85 87
74 94 91
69 78 56
70 66 78
53 71 87
𝑋1 = 71.71 𝑋2 = 75.29 𝑋3 = 76.57
Overall mean 𝑋 = 74.52
32
Question: Is there a difference in mean scores of 3 groups of students

33. • SSC, dfcolumns = C – 1 MSC =
𝑆𝑆𝐶
𝑑𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠
=
88.67
2
= 44.33
• SSE, dferror = N – C MSE =
𝑆𝑆𝐶
𝑑𝑓𝑒𝑟𝑟𝑜𝑟
=
2812.57
18
= 156.25
• SST, dftotal = N-1 F =
𝑀𝑆𝐶
𝑀𝑆𝐸
= F =
44.33
156.25
= 0.28
MSC- Mean sum squares column
MSE- Mean sum squares error
N – Total number of subjects
C – Number of columns
33

34. F =
𝑑𝑓=2
𝑑𝑓=18
Fcritical = 3.55 (Fα, df c, df E = F.05, 2, 18 )
Null hypothesis for ANOVA:
H0: 1 = 2 =3 = 4…………
F-stat smaller than F-critical
FAIL TO REJECT null hypothesis
Source of variance Df SS MS F
Between (columns) 2 12.69 6.35 0.04
Within (error) 18 2812.57 156.25
Total 20 2825.26
34

35. 35

36. Failure of assumptions:
• Data is not randomly distributed
• Transform data
• Non parametric test (Kruskal-Wallis H test)
• Violation of homogeneity of variance
• Welch test
• Brown and Forsythe test
• Lack of independence of data
• Very little can be done!
36

37. Two way ANOVA
• Mean differences between groups that have been split on two independent variable
(factors)
• Tells us if there is an interaction between two independent variables on the dependent
variable
• Assumptions:
• Dependent variable is continuous
• Independent variable - two or more categorical, independent groups
• Independence of observations
• No significant outliers
• Dependent variable - normally distributed
• Homogeneity of variances
37

38. 38
Question: If a difference existed in insulin sensitivity depending on thyroid level or body
mass index (BMI)
4 Treatment combinations: overweight hyperthyroid subjects, overweight controls, normal
weight hyperthyroid subjects and normal weight controls
3 Questions:
A. Do differences exist between hyperthyroid subjects and controls?
B. Do differences exist between overweight and normal weight subjects?
C. Do differences exist due to neither thyroid status nor weight alone but to the
combination of factors?

39. 39
A. Difference between Patients and Controls
B. Difference between Overweight and Normal weight subjects
C. Difference owing only to Combination of Factors
Weight
Subjects Overweight Normal
Patients 1.00 1.00
Controls 0.50 0.50
Weight
Subjects Overweight Normal
Patients 0.50 1.00
Controls 0.50 1.00
Weight
Subjects Overweight Normal
Patients 0.50 1.00
Controls 1.00 0.50

40. INTERACTION
No interaction: effects are additive Significant interaction: effects are
multiplicative
40

41. NON PARAMETRIC TEST
41

42. Sign test
• Used to determine if there is a median difference between paired or matched
observations
• Assumptions:
• Dependent variable is continuous
• Independent variable consists of two categorical, related groups or matched pairs
• Paired observations for each participant is independent
42

43. Example:
• Data:
• H0; N=8
H1; N>8 (if any observation is equal to 8, eliminate it and decrease by one)
• Test statistic: S+ = number of sample observation greater than 8 = 7
we will reject H0 if S+ is ‘sufficiently large’
• P-value: the number of observations greater than 8 is a binomial random variable,X,
and S+ is the observed value of X
• n = 10 trials, Probality (p)= 0.5
• P value= p (X ≥ 7) = 1 – P (X ≤ 6) = 1 – 0.828 = 0.172
• P value (0.172) > α = 0.05 Fail to reject Null hypothesis
2.4 15.6 14.3 11.2 9.4 3.9 11.6 8.4 12.5 6.8
- + + + + - + + + -
43

44. 44

45. Wilcoxon signed rank test
• Non parametric equivalent to dependent t test
• Used when assumption of normality has been violated and use of dependent t
test is inappropriate
• To compare two sets of scores that come from the same participants
• Assumptions:
• Dependent variable is continuous or ordinal
• Independent variable should consist of two categorical, related groups or
matched pairs
• Distribution of the differences between two related groups is symmetrical
45

46. Candidate 1 2 3 4 5 6 7 8 9 10 11 12 13
Mock 40 65 53 70 87 42 80 63 51 82 27 71 29
Final exam 45 68 47 75 88 60 77 69 60 88 30 73 35
Find the difference
Difference 5 3 -6 -4 1 18 -3 6 9 6 3 2 6
Rearrange 1 2 3 3 3 4 5 6 6 6 6 9 18
Rank 1 2 4 4 4 6 7 9 9 9 9 12 13
Rank 1 2 3 4 5 6 7 8 9 10 11 12 13
Rank 7 4 9 6 1 13 4 9 12 9 4 2 9
T+ = 70, T- = 19
46

47. • Wilcoxon table
1 tail, 5%, n=13
The critical value is 21
Test statistic is 19
Test statistic < Critical value
REJECT NULL HYPOTHESIS
47

48. Mann-Whitney U test/ Wilcoxon Rank Sum test
• Compares median
• Assumption:
• Dependent variable should be ordinal or continuous
• Independent variable should have two categorical or continuous variable
• Independence of observation
• Two distributions of the independent variables have the same shape
48

49. Treatment A Treatment B
3 9
4 7
2 5
6 10
2 6
5 8
• H0 : No difference between the ranks of each treatment
• H1 : There is a difference between the ranks of each treatment
• α = 0.05
Rank Original
value
New Rank
1 2 1.5
2 2 1.5
3 3 3
4 4 4
5 5 5.5
6 5 5.5
7 6 7.5
8 6 7.5
9 7 9
10 8 10
11 9 11
12 10 12
stat = Rank sum – n(n+1)/2
A = 23 – 6(6+1)/2 = 2
B = 55 – 6(6+1)/2 = 34
stat = 2, critical = 5
stat < critical Reject null hypothesis
Treatment A Treatment B
3 11
4 9
1.5 5.5
7.5 12
1.5 7.5
5.5 10
(23) (55)
49

50. 50

51. Kruskal-Wallis H test
• Rank based non parametric alternative of One way ANOVA
• OMNIBUS test
• Assumptions:
• Dependent variable is ordinal or continuous and not normally distributed
• Independent variable consist of two or more categorical, independent groups
• Independence of variance
• Distributions in each groups has the same shape
51

52. A Priori, or Planned Comparisons
• Multiple groups of scores, but specific comparisons have been planned prior to data
collection
• E.g. If we have 3 groups and it has been decided before data collection that comparisons
will be made between groups 1 and 2, groups 1 and 3 or groups 2 and 3
• Orthogonal t-test
• Modification:
t = 𝑋1−𝑋2
2 𝑀𝑆𝐸 𝑛
Independent
52

53. The Bonferroni correction
• Error rate correction formula for overcomparison
• Multiple comparison correction when several dependent or independent
statistical tests are being performed
• Alpha value is lowered to account for the number of comparisons performed
• Sets the alpha value for the entire set of ‘C’ comparisons
α = αFW  C
(New alpha to (Family wise (Number of comparisons)
correct for Type I error) error alpha)
53

54. e.g.
• 5 treatment groups
• Compare between following sets of groups:
• 1 to 2, 2 to 3, 3 to 4, 4 to 5, 1 to 3, 2 to 4, 3 to 5, 4 to 6, (1 & 2) to (3 & 4) and (4 to 5)
10 comparisons
• Bonferroni correction
0.05  10 = 0.005 (new critical alpha)
54

55. A Posteriori, or Post Hoc Comparisons
• Tukey's test
• The Scheffé post hoc test
• The Newman–Kuels post hoc test
• Dunnett's procedure
55

56. Tukey’s HSD Procedure
• Done if null hypothesis is rejected
• Calculates a value based upon the mean squared error, the sample size and a value from a
studentized Q range distribution
• If difference between two sample means is greater than HSD, then they are statistically
significant
Honestly significant difference
HSD = Qα, C, N-C MSE/𝑛
α = level of significance
C = number of columns/groups
N = total number of observations
n = number of observations in each group
56

57. Example: 5 groups with 9 observations in each
1 2 3 4 5
Mean 14.5 13.8 13.3 14.3 13.1
• HO : means are equal among all the samples
• One way ANOVA:
• F = 37.84 > Fcritical = 2.61, Reject Null hypothesis
df SS MS F
Column 4 13.32 3.33 37.84
Error 40 3.53 0.088
Total 44 16.85
57

58. • Q α, C, N-C = Q o.5, 5, 40 = 4.04
• HSD = Qα, C, N-C MSE/𝑛
= 4.04 0.088/9= 0.4
• Arrange means in increasing order:
• Pairs not underscored by the same line are significantly different from one another
• 1 and 4 are not significantly different from one another. 1 and 4 are significantly different
from 2, 3, 5
• 3 and 5 are not significantly different from one another
• 2 is significantly different from 3 and 5
5 3 2 4 1
Mean 13.1 13.3 13.8 14.3 14.5
58

59. 59

60. CHI SQUARE TEST
• Test of significance of association between two or more qualitative variables
• Chi-square test is a very versatile statistical procedure used to test for differences in
proportions as well as an association between two variable
The assumptions made for applying chi square test are:
Random sample data
Sufficiently large sample
Normal distribution of deviation and not of data
Nominal/ordinal/interval data may be used (qualitative)
Logical/empirical basis for classifying data into nominal group
60

61. Chi-square is used for:
Test of proportion
Test of association
Test of goodness of fit
Chi-square test is a test of significance of association between two variables
2 =  (O – E)2
E
Where, O= observed frequencies
And E= expected frequencies
Expected frequencies are calculated as:
𝐑𝐨𝐰 𝐭𝐨𝐭𝐚𝐥 𝐗 𝐂𝐨𝐥𝐮𝐦𝐧 𝐭𝐨𝐭𝐚𝐥
𝐆𝐫𝐚𝐧𝐝 𝐭𝐨𝐭𝐚𝐥
Degree of freedom for critical 2 = (Row - 1) (Column - 1)
61

62. Example:
• In a study to determine the effect of heredity in a certain disease, a sample of
cases and controls was taken:
• Using 5% level of significance, test whether family history has an effect on
disease
Family history Disease
Total
Cases Controls
Positive 80 120 200
Negative 140 160 300
Total 220 280 500
62

63. 2= (80-88)2/88 + (120-112)2/112 + (140-132)2/132 + (160-168)2/168
= 2.165
2 < 3.84
No significant association
Family history Disease
Total
Cases Controls
Positive
O
E
80
88
120
112
200
Negative
O
E
140
132
160
168
300
Total 220 280 500
63

64. Fisher’s exact test
 An exact test for 2 × 2 contingency tables. It is used when the sample size is too small to
use the chi-square test.
 Preferred to chi-square when two characteristics are being compared, each at two levels
because it provides the exact probability
 If any expected frequency is less than 2 or if more than 20% of the expected frequencies
are less than 5
64

65. 65
sex Pain medication A Pain medication B Total
Male 2(2.5) 7(6.5) 9
Female 4(3.5) 9(9.5) 13
Total 6 16 22
1.Make sure no assumptions are violated.
2.State null hypothesis and alternative hypothesis
H0= there is no association between patient’s gender and pain medication used
H1=there is an association between gender and pain medication
3.State α (0.05)
4.Calculate df (NA)
5. Find critical chi square value(NA)
6.Calculate p value
Fisher’s directly calculate p value
Here p value=1 (>0.05)
Thus , there is not a statistically significant association between gender and pain
medication.

66. Continuity Correction (Yates' correction)
• It involves subtracting ½ from the difference between observed and expected
frequencies in the numerator of χ2 before squaring; it has the effect of making the
value for χ2 smaller
• A smaller value for χ2 means that the null hypothesis will not be rejected as often as
it is with the larger, uncorrected chi-square; that is, it is more conservative
• Thus, the risk of a type I error (rejecting the null hypothesis when it is true) is smaller
• However, the risk of a type II error (not rejecting the null hypothesis when it is false
and should be rejected) then increases
66

67. McNemar test
•In studies in which the outcome is a binary (yes/no) variable, researchers
may want to know whether the proportion of subjects with (or without)
the characteristic of interest changes after an intervention or the passage
of time.
•In these types of studies, we need a statistical test that is similar to the
paired t test and appropriate with nominal data.
67

68. Example:
• The researchers wanted to know whether changes occurred in the bowel
function of patients following cholecystectomy. They collected information
on the number of patients who had one or fewer versus more than one stool
per day :
Before
cholecystectomy
1 month after
cholecystectomy Total
≤1 >1
≤1 25 15 40
>1 0 11 11
Total 25 26 51
68

69. •Because 15 is larger than 3.84( at α = 0.05, critical value is 3.84), we can
reject the null hypothesis and conclude that there is a difference in the
proportion of patients having more than one stool per day before and after
cholecystectomy.
•As with the z statistic, it is possible to use a continuity correction with the
McNemar test. The correction involves subtracting 1 from the absolute value
in the numerator before squaring it.
69

70. 70
Thank You