anatomy

1. Jimma University
Institute of Health Sciences
Public Health Faculty, Epidemiology Department
Advanced Biostatistics
Novemeber, 2022
1

2. By the end of this course you will be able to:
 Know how to analyze continuous data.
 Know analyzing data using t-test, ANOVA, simple
linear regression, multiple linear regression,
correlation and others.
 Know how to analyze data using non-parametric test.
 Know how to analyze categorical data.
 Know analyzing data using chi-square test, logistic
regression.
 Data analysis will be performed by SPSS software.
2

3. Basic Concepts in Analysis ofVariance
 Analysis of variance (ANOVA), as the name implies,
it is a statistical technique that is intended to analyze
variability in data in order to infer the inequality
among population means.
 The Analysis of Variance (ANOVA) is a hypothesis
test to compare the means of more than two
populations.
 It is used to analyze "variation" among and between
groups, developed by Ronald Fisher.
3

4. 4
How ANOVA is related with t-test?
• When we have only two samples we can use the t-
test to compare the means of the samples.
• We use t-test in either of the following three ways:
A. One sample t-test:
 It is used to compare the estimate of a sample with a
hypothesized population mean to see if the sample is
significantly different.

5. 5
Example: The distance covered by marathon runners until
a physiological stress develops and whether they used drug
or not.
No.Distance in miles Drug use No. Distance in miles Drug use
1 14.5 No 10 18.4 Yes
2 13.4 No 11 16.9 Yes
3 14.8 Yes 12 12.6 No
4 19.5 Yes 13 13.4 No
5 14.5 No 14 16.3 Yes
6 18.2 Yes 15 17.1 Yes
7 16.3 No 16 11.8 No
8 14.8 No 17 13.3 Yes
9 20.3 Yes 18 14.5 No

6. 6
Test whether the mean distance covered before feeling physiological
stress is 15 miles.
• Hypothesis: Ho : µ=15 vs. HA: µ≠15
• Level of significance: α= 0.05
• Test statistics: 𝑡𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =
ҧ
𝑥−µo
ൗ
𝑠
√𝑛
=
15.59−15
ൗ
2.43
√18
= 1.03 ,
and P-value = 0.318.
• Tabulated value: 𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 = 𝑡 Τ
𝛼
2
, 𝑛 − 1 = 𝑡0.975, 17 = 2.11
• Decision/conclusion: Decision/conclusion can be in two ways.
 We reject Ho if the calculated value is greater than the
tabulated.
 We reject Ho if the p-value less than or equals to α-value.

7. 7
Interpretation:
• In the above two scenarios we do not reject Ho and
conclude that the population mean distance covered before
feeling physiological stress is 15 miles.
Exercise:
1. Test whether the mean total cholesterol level in the
Framingham offspring study was different from the
national mean value of 203. The following statistics on
total cholesterol levels of participants in the Framingham
offspring study were available n=3310, ҧ
𝑥 = 200.3 ,
s=36.8. Use α=0.05

8. 8
2. The data on expenditure on health care and prescription drugs.
Test whether there was significance evidence of a reduction in
expenditure from the reported value of $3302 per year. A
sample of 100 individuals were selected and their expenditures
on health care and prescription drugs in 2010 were
summarized as follows: ҧ
𝑥 = $3190 and s= $890. Use α=0.05

9. 9
B. Paired t-test
 Tests means of two related populations
• Paired or matched samples
• Repeated measures (before/after)
• Use difference between paired values
• Measurement might be:
"pre/post”, "before/after", “right/left, “parent/child”, etc.

10. 10
Example : The blood pressure (BP) of 10 mothers were measured
before and after taking a new drug.
No. BP before BP after Difference(di)
1 130 110 -20
2 125 130 5
3 140 120 -20
4 150 130 -20
5 120 110 -10
6 130 130 0
7 120 115 -5
8 135 130 -5
9 140 130 -10
10 130 120 -10

11. 11
Test whether there is no difference between the two measurements.
• Hypothesis: Ho : 𝜇ത
𝑑 = 0 vs. 𝜇ത
𝑑≠0
• Level of significance: α= 0.05
• Test statistics: 𝑡𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =
ത
𝑑−µo
ൗ
𝑠ഥ
𝑑
√𝑛
=
−9.5
ൗ
8.64
√10
= −3.48 ,
and P-value = 0.007.
• Tabulated value: 𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 = 𝑡 Τ
𝛼
2
, 𝑛 − 1 = 𝑡0.975, 9 = 2.26
• Conclusion: we reject Ho and conclude that there is a difference
between the two measurements or the new drug has showed the
difference on BP of mothers.
NB: The extension of paired t-test is repeated measure ANOVA.

12. 12
Exercise:
Fifteen patients had a pre-treatment or baseline total cholesterol
level measured, and then after taking the drug for 6 weeks, each
patient's total cholesterol level was measured again. Test the
difference between the two measurements. Use α=0.05.
Data from cholesterol study:
No. Baseline 6 weeks
1 215 205
2 190 156
3 230 190
4 220 180
5 214 201
6 240 227
7 210 197
8 193 173
9 210 204
10 230 217
11 180 142
12 260 262
13 210 207
14 190 184
15 200 193

13. 13
C. Two independent samples t-test
• Used to compare two unrelated or independent groups.
Assumptions:
• The variance of the dependent variable in the two populations
are equal.
• The dependent variable is normally distributed within each
population.
• The data are independent (scores of one participant are not
related systematically to the scores of the others).

14. 14
Example:
Do the marathon runners grouped by their drug intake status differ in
their average distance coverage before they feel any physiological
stress?
• Hypothesis: Ho : 𝜇𝑑 = 𝜇𝑛𝑑 vs. 𝜇𝑑 ≠ 𝜇𝑛𝑑
• Level of significance: α= 0.05
• Test statistics: 𝑡𝑐𝑎𝑙𝑐𝑢𝑙𝑡𝑒𝑑 =
ҧ
𝑥𝑑− ҧ
𝑥𝑛𝑑
ൗ
𝑠𝑝
√𝑛
,
Where, 𝑠𝑝 =
(𝑛𝑑−1)𝑠𝑑
2+(𝑛𝑛𝑑−1)𝑠𝑛𝑑
2
𝑛𝑑+𝑛𝑛𝑑−2

15. 15
Given that ҧ
𝑥𝑑=13.98, 𝑠𝑑 = 1.33 and ҧ
𝑥𝑛𝑑 = 17.2, 𝑠𝑛𝑑=2.21
then, 𝑡𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 = −3.741, 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.002
• Tabulated value: 𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 = 𝑡 Τ
𝛼
2
, (𝑛1+𝑛2) − 2 = 𝑡0.975, 16 =
2.21
• Conclusion: we reject 𝐻0 and conclude that there is a
difference between drug user and non drug user in their
average distance coverage before they feel any
physiological stress.
NB: The extension of two independent sample t-test is
ANOVA.

16. 16
Exercise
A clinical trail is run to compare an experimental drug to a placebo
for its effectiveness in lowering systolic blood pressure . A total of
18 participants are enrolled in the study and randomly assigned to
receive either the experimental drug or placebo. After 6 weeks on the
assigned treatment, each patients systolic blood pressure is measured
and the data are shown here:
Experimental Drug: 125 130 135 121 140 137 129 145 115
Placebo: 145 140 132 129 145 150 160 140 120
Is there statistical evidence of a difference in mean systolic blood
pressures between treatment at 𝛼 = 0.05?

17. 17
Variables in any ANOVA
Your data should include at least two variables (represented in columns)
that will be used in the analysis.
 The independent variable should be categorical (nominal or
ordinal) and include at least two groups, and
 The dependent variable should be continuous (i.e., interval or
ratio). Each row of the dataset should represent a unique subject or
experimental unit.

18. Analysis of variance is used:
• To estimate and test hypotheses about
population means.
Assumptions:
 Each of the populations from which the
sample comes is normally
distributed.
 Each of the populations has the same
variance.
18

19. The ANOVA Procedure
• For the analysis of variance for the
different designs, we follow the most
common procedures.
 Description of data.
 Assumptions.
 Hypothesis.
 Test statistics.
 Read the tabulated value.
 Decision rule.
19

20. One way ANOVA
 Compare the population mean using t-test
become unreliable in case of more than two
samples.
 One-way analysis of variance to test the null
hypothesis that more than two population means
are equal.
 One factor (independent variable) is analyzed, also
called the “grouping” variable.
 Dependent variable should be interval or ratio. 20

21. participants Treatment
1 2 3 . . . k
𝑦11 𝑦12 𝑦13 . . . 𝑦1𝑘
𝑦21 𝑦22 𝑦23 . . . 𝑦2𝑘
. . . . .
. . . . .
. . . . .
𝑦𝑛11 𝑦𝑛22 𝑦𝑛33 . . . 𝑦𝑛𝑘𝑘
Mean ത
𝑦.1 ത
𝑦.2 ത
𝑦.3 ത
𝑦.𝑘
ത
𝑌. .
Data layout for One Way ANOVA
21

22. The total sum of squares
 It is the sum of the squares of the
deviations of individual observations
from the mean of all the
observations taken together. This
total sum of squares is defined as:
𝑺𝑺𝑻 = ෍
𝒋=𝟏
𝒌
෍
𝒊=𝟏
𝒏
(𝒚𝒊𝒋 − ഥ
𝒀. . )𝟐
22

23. The Within Groups Sum of Squares
 It is the computing of within each
group of the sum of the squared
deviations of the individual
observations from their mean.
 The expression for these calculations
is written as follows:
𝑺𝑺𝑾 = ෍
𝒋=𝟏
𝒌
෍
𝒊=𝟏
𝒏
(𝒚𝒊𝒋 − ഥ
𝒀.𝒋 )𝟐
23

24. The Between Sum of Squares
 It is the squared deviation of the
group mean from the grand mean and
multiply the result by the size of the
groups.The formula is given by:
𝑺𝑺𝑩 = ෍
𝒋=𝟏
𝒌
𝒏𝒋(ഥ
𝒚.𝒋−ഥ
𝒀..)𝟐
24

25. ANOVA table
 It is a convenient display of
calculations of between, within and
total sum of squares, the
associated degrees of freedom and
mean squares.
 It is composed of none
negative values.
 A general ANOVA table
follows:
25

26. Source of
variation
Sum of Square df Mean square Variance
ration
Between
𝑺𝑺𝑩 = ෍
𝒋=𝟏
𝒌
𝒏𝒋(ഥ
𝒚.𝒋−ഥ
𝒀..)𝟐
K-1
𝑀𝑆𝐵 =
𝑆𝑆𝐵
𝐾 − 1
𝐹 =
𝑀𝑆𝐵
𝑀𝑆𝑊
Within
𝑺𝑺𝑾 = ෍
𝒋=𝟏
𝒌
෍
𝒊=𝟏
𝒏
(𝒚𝒊𝒋 − ഥ
𝒀.𝒋 )𝟐
N-K
𝑀𝑆𝑊 =
𝑆𝑆𝑊
𝑁 − 𝑘
Total
𝑺𝑺𝑻 = ෍
𝒋=𝟏
𝒌
෍
𝒊=𝟏
𝒏
(𝒚𝒊𝒋 − ഥ
𝒀. . )𝟐
N-1
ANOVA Table
26

27.  The statistic used in ANOVA is called
an F statistic.
 As F gets smaller, the groups are less
distinct.
 As F gets larger, the groups are more
distinct.
 If calculated F > Ftabulated → reject
HO.
27

28.  The one-way analysis of variance
model may be written as follows:
𝒚𝒊𝒋 = 𝝁 + 𝝉𝒋 + 𝝐𝒊𝒋 ; 𝑖 = 1,2, . . . , 𝑛𝑗,
𝑗 = 1,2, . . . , 𝑘.
The terms in this model are defined as
follows:
1. 𝝁 represents the mean of all the k
population means and is called the
grand mean.
28

29. 2. 𝝉𝒋 represents the difference between
the mean of the jth population and
the grand mean and is called the
treatment effect.
3. 𝝐𝒊𝒋 represents the amount by which
an individual measurement differs
from the mean of the population
to which it belongs and is called the
error term.
29

30. One way ANOVA: Example
Height (in inches) of children with four different groups
Group 1 Group 2 Group 3 Group 4
60 50 48 47
67 52 49 67
42 43 50 54
67 67 55 67
56 67 56 68
62 59 61 65
64 67 61 65
59 64 60 56
72 63 59 60
71 65 64 65

31. Hypothesis
𝐻𝑜: 𝜇1 = 𝜇2 = 𝜇3 = 𝜇4 (On average
the four groups have the same
height).
𝐻𝐴: Not all 𝜇′
𝑠 are equal (At least one
group’s average height different
from the average height of at least
one other group).

32. One way ANOVA: Example
Calculate the sum of
squares between groups:
Mean for group 1 = 62.0
Group 1 Group 2 Group 3 Group 4
60 50 48 47
Mean for group 2 = 59 7
Mean for group 3 = 56.3
Mean for group 4 = 61.4
Grand mean= 59.85
67 52 49 67
42 43 50 54
67 67 55 67
56 67 56 68
62 59 61 65
64 67 61 65
59 64 60 56
72 63 59 60
71 65 64 65
SSB = [(62-59.85)2 + (59.7-59.85)2 + (56.3-59.85)2 + (61.4-59.85)2 ]x10
= 19.65x10 = 196.5
32

33. One way ANOVA: Example
Calculate the sum of
squares within groups:
(60-62)2+(67-62)2+
(42-62)2+ (67-62)2+
(56-62)2+ (62-62)2+
(64-62)2+ (59-62)2+
(72-62)2+ (71-62)2+
(50-59.7)2+ (52-59.7)
2+ (43-59.7)2+67-59.7)
2+ (67-59.7)2+ (69-
59.7)2…+….(sum of
40 squared deviations)
= 2060.6
Group 1 Group 2
60 50
67 52
42 43
67 67
56 67
62 59
64 67
59 64
72 63
71 65
Group 3 Group 4
48 47
49 67
50 54
55 67
56 68
61 65
61 65
60 56
59 60
64 65

34. One way ANOVA: Example
Source of
variation
d.f. Sum of
squares
Mean Sum of F-statistic
Squares
Between 3 196.5 65.5 1.14
Within 36 2060.6 57.2
Total 39 2257.1

35. One way ANOVA: Example
Source of variation d.f. Sum of squares Mean Sum of F-statistic
Squares
Between 3 196.5 65.5 1.14
Within 36 2060.6 57.2
Total 39 2257.1
𝐹𝑡𝑎𝑏 = 𝐹 0.025,3,36 =3.4 , which implies Fcal is less than Ftab.
Do not reject Ho.

37. One way ANOVA: Example
SPSS output

38. Interpretation:
we reject Ho if P-value is less than level of
significance (Probability of rejecting the null
hypothesis). Therefore there is no significant
difference among the four group means,
implies that on average the height are the same
in each group for the population.

39. Exercise:
A clinical trial is run to compare weight loss programs and
participants are randomly assigned to one of the
comparison programs and are counseled on the details of
the assigned program. Participants follow the assigned
program for 8 weeks. The outcome of interest is weight
loss, defined as the difference in weight measured at the
start of the study (baseline) and weight measured at the end
of the study (8 weeks), measured in pounds.

40. Test whether there is a statistically significant difference
in the mean weight loss among the four diets. Use α =
0.05.
Data from clinical trail
Low-Calories Low-Fat Low-Carbohaydrate Control
8 2 3 2
9 4 5 2
6 3 4 -1
7 5 2 0
3 1 3 3

41. Multiple Comparison Procedures:
 A statistically significant ANOVA (F-test)
only tells you that at least two of the
groups differ, but not which ones differ.
 If the ANOVA F-test shows there is a
significant difference in means between the
groups we may want to perform multiple
comparisons between all pair-wise means
to determine how they differ.

42.  Determining which groups differ
(when it’s unclear) requires more
sophisticated analyses to correct for
the problem of
multiple comparisons.

43. • List of some of multiple comparison procedures
 Bonferroni procedure.
 Duncan’s new multiple-range test.
 Dunnett’s multiple comparison test.
 Newman-Keuls test.
 Scheffe’s test.
 Tukey’s test.
 Tukey’s HSD (Honestly Significant
Difference).
 Holm t-test
-

44. Choice of Multiple Comparisons of Means
 The choice of a multiple comparisons method
in ANOVA will depend on the type of
experimental design used and the comparisons
of interest to the analyst. For example, Tukey (
1949) developed his procedure specifically for
pairwise comparisons when the sample sizes of
the treatments are equal.

45.  Scheffe (1953) developed a more general
procedure for comparing all possible linear
combinations of treatment means ( called
contrasts).
 Suitable for pair-wise comparison between all
groups.
 Corrects for the increased risk of a Type I error.

46. Reading Assignment:
Read the advantage and disadvantage of:
1. Tukey
2. Scheffee
3. Bonferroni

47. Exercise
Ilich-Ernst et al. (A-4) investigated dietary intake
of calcium among a cross section of 113 healthy
women ages 20–88. The researchers formed four
age groupings as follows: Group A, 20.0–45.9
years; group B, 46.0–55.9 years; group C, 56.0–
65.9 years; and group D, over 66 years. Calcium
from food intake was measured in mg/day. The
data below are consistent with summary statistics
given in the following.

48. A B C D
1820 191 724 1652
2588 1098 613 1309
2670 644 918 1002
1022 136 949 966
1555 1605 877 788
222 1247 1368 472
1197 1529 1692 471
1249 1422 697 771
1520 445 849 869
489 990 1199 513
2575 489 429 731
1426 2408 798 1130
1846 1064 631 1034
1088 629 1016 1261

49. 912 - 1025 42
1383 - 948 767
1483 - 1020 775
1723 - 805 1393
727 - 631 533
1463 - 641 734
1777 - 760 485
1129 - 1085 449
- - 775 236
- - 1307 831
- - 344 698
- - 961 167

50. - - 239 824
- - 944 448
- - 1096 991
- - - 590
- - - 994
- - - 1781
- - - 937
- - - 1022
- - - 1073
- - - 948
- - - 222
- - - 721
- - - 375
- - - 1187

51. - - - 752
- - - 804
- - - 1182
- - - 1243
- - - 985
- - - 1295
- - - 1676
- - - 754
A.State both null and alternative hypothesis.
B.Set up ANOVA table and test whether there
is a significance difference between the
groups at 5% significance level.

52. Two way ANOVA

53.  When we are examining the effect of two
independent variables, this is called a Two-
Way ANOVA.
 In a Two-way ANOVA, the effects of two
factors can be investigated simultaneously.
 Two-way ANOVA permits the investigation of
the effects of either factor alone and also the
two factors together.

54. Three questions are answered by a Two-way
ANOVA with Interactions:
 Is there any effect of Factor A on the
outcome?
(Main Effect of A).
 Is there any effect of Factor B on the
outcome?
(Main Effect of B).

55.  Is there any effect of the interaction of Factor
A and Factor B on the outcome?
(Interactive Effect of AB)
This means that we will have three sets of
hypotheses, one set for each question.

56. • To analyse the two-way ANOVA design we
have to follow the same steps as the one-way
ANOVA design
- State the Null Hypotheses
- Partition the Variability
- Calculate the Sum of Squares
- Calculate the Mean Squares
- Calculate the F-Ratios

57. • Assume that there are two factors A and B
• The effect of a single variable is known as a
main effect
• The effect of two variables considered
together is known as an interaction
• We are going to assess then
- Main effect of variable A
- Main effect of variable B
- Interaction between A and B

58. • There are three sets of null hypotheses
1. The population means of the first factor are equal.
This is like the one-way ANOVA for the row
factor.
2. The population means of the second factor are
equal. This is like the one-way ANOVA for the
column factor.
3. There is no interaction between the two factors.
This is similar to performing a test for
independence with contingency tables.

59. Two way ANOVA
• The Main Effect involves the independent
variables one at a time.
• The Interaction Effect is the effect that one
factor has on the other factor.
• The Within Variation/error is the sum of
squares within each treatment group.

60. • Main Effect
- The main effect involves the independent
variables one at a time.
- The interaction is ignored for this part. Just the
rows or just the columns are used, not mixed.
- This is the part which is similar to the one-way
analysis of variance. Each of the variances
calculated to analyze the main effects are like
the between variances.

61. • Interaction Effect
- The interaction effect is the effect that one
factor has on the other factor.
- The degrees of freedom here is the product of
the two degrees of freedom for each factor.

62. • Interaction Effect
• If the interaction is significant, the model is
called an interaction model
• If the interaction is not significant, the model is
called an additive model
• Explanatory variables are often referred to as
factors

63. • Interaction or profile plots
- An interaction plot is a way to look at outcome
means for two factors simultaneously
- A plot with parallel lines suggests an additive
model
- A plot with non-parallel lines suggests an
interaction model

64. • Within Variation
- The Within variation is the sum of squares
within each treatment group. You have one less
than the sample size for each treatment group.
- The within variance is the within variation
divided by its degrees of freedom.
- The within group is also called the error.

65. Two way ANOVA: Model

66. Two way ANOVA: Model
• Error assumptions
- All errors are normal distributed.
- All errors have expected value zero.
- All errors have the same variance.
- All errors are independent distributed.

67. Two way ANOVA: Model
Null Hypotheses
i.e., the means of the different
levels of factor A are equal
i.e., the means of the different
levels of factor B are equal
i.e., there is no interaction

68. Two way ANOVA: Model
Alternative Hypotheses
HA1 : The means of the different
levels of factor A are not equal
HB1 : The means of the different
levels of factor B are not equal
H : There is interaction
AB1

71. SST =
SS(A)+SS(B)+SS(AB)+SSE

72. Two Way ANOVA with Interactions Table
Source of
Variation
Sum of
Squares df Mean Square F-ratio F-critical
Factor A SSA a-1 MSA= SSA / (a-1) F= MSA/MSE F [(a-1),ab(n-1)]
Factor B SSB b-1 MSB= SSB / (b-1) F= MSB/MSE F [(b-1),ab(n-1)]
Interaction SSAB (a-1)(b-1) MSI = SSAB / (a-1)(b-1) F= MSAB/MSE F [(a-1)(b-1),ab(n-1)]
Error SSE ab(n-1) MSE= SSE / ab(n-1)
TOTAL SST abn-1
a: Number of treatment levels (categories)
for Factor A.
b: Number of treatment levels (categories)
for Factor B.
n: number of observations per cell

73. Two way ANOVA :Example
• The data in the following table are from a two-
factor experiment in a health education teacher
training program.
• Three new textbooks (factor A) were tested with
two methods of instruction (factor B), and 36
health education trainees were randomly allocated
to the six treatment groups with six subjects per
group. The trainees were tested before and after
four weeks of instruction, and the increases in test
scores were recorded.

74. Two way ANOVA :Example
• Data are classified by textbook, the row factor,
and method of instruction, the column factor. In
this experiment, the random allocation of subjects
was done simultaneously to all combinations of
the two sets of levels
• The data is shown in the following table.

75. Two way ANOVA :Example
• Increase in test scores after four weeks of instruction using three
textbooks and two teaching methods. Factor B
Text Lecture Discussion
book
1 30 43 12 18 22 16 36 34 15 18 40 45
2 21 26 10 14 17 16 33 31 28 15 29 26
3 42 30 18 10 21 18 41 46 19 23 38 48
Factor A

76. Two way ANOVA :Example
• Several means will be used in the analysis.
• The means here include
- the cell means ( y ),
ij•
- two sets of marginal means
• row ( y )and
i
• column (
••
y ) and
•j •
- the overall mean ( y ).
•••

77. Overall mean
Cell
means Two way ANOVA :Example
• Increase in test scores after four weeks of instruction using three
textbooks and two teaching methods. Factor B
Method of instruction
Text Lecture Discussion
book
1 30 43 12 18
2 21 26 10 14
3 42 30 18 10
22 16 36 34 15
17 16 33 31 28
21 18 41 46 19
Column marginal means
18 40 45
15 29 26
23 38 48
Row
mea
s

78. Two way ANOVA :Example
SPSS output

79. Two way ANOVA :Example
SPSS output

80. Tests of Between-Subjects Effects
Dependent Variable: Testscore
Source
Type III Sum
of Squares df
Mean
Square F Sig.
Corrected Model 1288.472a 5 257.694 2.494 .053
Intercept
25016.694 1 25016.694 242.110 .000
Text.Book 342.722 2 171.361 1.658 .207
Instructin.Methods 910.028 1 910.028 8.807 .006
Text.Book *
Instructin.Methods 35.722 2 17.861 .173 .842
Error 3099.833 30 103.328
Total
29405.000 36
Corrected Total 4388.306 35
a. R Squared = .294 (Adjusted R Squared = .176)

81. Two way ANOVA :Example
SPSS output.
• The F (F=.173) statistic and its associated
p-value (P=.842) for interaction indicate
that there is no statistically significant
interaction of the two independent
variables.
• There is a statistically significant effect for
the methods of instruction (P<0.05).
• But no significant effect associated with the
textbooks (P>0.05).

82. Two way ANOVA :Example
SPSS output
• The lack of an interaction effect is reflected in the
cell means that are plotted in Figure
Two way ANOVA :Example
SPSS output

83. Two way ANOVA :Example
SPSS output
• Interaction measures the degree of
similarity between the responses
to factor A at different levels of
factor B. The lines connecting the
three cell means for the discussion
method are roughly parallel with
the lines connecting cell means
for the lecture method, reflecting
the absence of interaction.
• If these two lines were not parallel
or crossed each other, then the
interaction effect would have been
statistically significant.

84. Apply all the procedures of two-way ANOVA on the following
data on the effects of a drug on both men and women with
different doses.

85. Repeated Measure ANOVA

86.  Is a special form of one way ANOVA when the
same subjects receive different treatments.
 For example, surveying same individual at
monthly intervals is a repeated measure
design.
 The same individual receive different
treatment.

87.  Because repeated measurements are not
independent, one-way ANOVA cannot be used
to test the hypothesis that three or more
population means are equal.
 Instead, repeated-measures analysis of variance,
a generalization of the paired t-test, should be
considered for testing this hypothesis.

88.  “within-subjects” design - subjects are
measured on Dependent Variable more than
once.
Advantages:
• With Repeated Measures ANOVA the stable
variance/systematic variance due to individual
differences is removed.
• Increased power - due to decreased “error” variance.

89.  Can use smaller sample sizes.
 Allows for study over time (also can assess
interaction between time and factor).
Disadvantages
 carryover effects - early treatments affect later
ones.
 practice effects - subjects’ experience with test
can influence their score on the Dependent
Variable.
 Fatigue-as a result of over exposure or
excessive activity the result will be influenced.

90. How many time points are being compared? >2
 For before and after studies, a paired t-test
will suffice.
 For more than two time periods, you need
repeated-measures ANOVA.
 Serial paired t-tests is incorrect, because this
strategy will increase your type I error.

91. Answers the following questions, taking into
account the fact the correlation within
subjects:
 Are there significant differences across time
periods?
 Are there significant differences between
groups in their changes over time?

92. Examples: Ergometer data. Within Subjects
factor with 5 levels (3, 6, 9, 12, 15
min).
subject minute 3 mintue 6 mintue 9 minute 12 minute 15
1 7 7 23 36 70
2 12 22 26 26 20
3 11 6 9 31 30
4 10 18 16 40 25
5 6 12 9 28 37
6 13 21 30 55 65
7 5 0 2 10 11
8 15 18 22 37 42
9 0 2 0 16 11
10 6 8 27 32 54

93. Sphericity Condition
Sphericity: refers to the equality of variances of
the differences between treatment
levels.
 If we were to take each pair of treatment
levels and calculate the differences between
each pair of scores, then it is necessary that
these differences have equal variances.

94. Mauchly’s test Statistic
 If significant, the variances are significantly
different from equal, and a correction must be
applied to produce a valid F-ratio:
Corrections applied to degrees of freedom to
produce a valid F-ratio:
 When G-G Sphericity Epsilon estimates < .75,
use Greenhouse-Geisser estimate
 When G-G sphericity Epsilon estimates> .75,
use Huynh-Feldt estimate

95. SPSS Output
General Linear Model
Within-Subjects Factors
Measure: MEASURE_1
time Dependent Variable
1 Minute3
2 mintue6
3 mintue9
4 minute12
5 minute15
Descriptive Statistics
Mean Std. Deviation N
Minute 3
8.50 4.503 10
mintue 6
11.40 7.961 10
mintue 9
16.40 10.803 10
minute 12
31.10 12.556 10
minute 15
36.50 21.131 10

96. Mauchly's Test of Sphericitya
Measure: MEASURE_1
Within Subjects
Effect
Mauchly's
W
Approx.
Chi-Square df Sig.
Epsilonb
Greenhouse
-Geisser
Huynh-
Feldt
Lower-
bound
treatment
.024 27.594 9 .001 .371 .428 .250
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed
dependent variables is proportional to an identity matrix.
a. Design: Intercept
Within Subjects Design: treatment
b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests
are displayed in the Tests of Within-Subjects Effects table.

97.  Mauchly’s Test of Sphericity indicated that
sphericity was violated.
 Since Sphericity is violated, we must use
either the G-G or H-F adjusted ANOVAs.

98. Tests of Within-Subjects Effects
Measure: MEASURE_1
Source
Type III
Sum of
Squares df
Mean
Square F Sig.
Partial Eta
Squared
time Sphericity
Assumed 6115.880 4 1528.970 18.359 .000 .671
Greenhouse-
Geisser 6115.880 1.485 4117.754 18.359 .000 .671
Huynh-Feldt 6115.880 1.710 3575.916 18.359 .000 .671
Lower-bound 6115.880 1.000 6115.880 18.359 .002 .671
Error(time) Sphericity
Assumed 2998.120 36 83.281
Greenhouse-
Geisser 2998.120 13.367 224.289
Huynh-Feldt 2998.120 15.393 194.776
Lower-bound 2998.120 9.000 333.124

99. Tests of Between-Subjects Effects
Measure: MEASURE_1
Transformed Variable: Average
Source
Type III
Sum of
Squares df
Mean
Square F Sig.
Partial Eta
Squared
Intercept 21590.420 1 21590.420 45.801 .000 .836
Error 4242.580 9 471.398

100. Pairwise Comparisons
Measure: MEASURE_1
(I)
treatment
(J)
treatment
Mean
Difference
(I-J)
Std.
Error Sig.b
95% Confidence
Interval for Differenceb
Lower
Bound
Upper
Bound
1 2 -2.900 1.656 1.000 -9.011 3.211
3 -7.900 2.718 .174 -17.929 2.129
4 -22.600* 3.194 .001 -34.386 -10.814
5 -28.000* 6.354 .017 -51.445 -4.555
2 1 2.900 1.656 1.000 -3.211 9.011
3 -5.000 2.380 .651 -13.783 3.783
4 -19.700* 2.848 .001 -30.209 -9.191
5 -25.100* 6.457 .037 -48.926 -1.274
3 1 7.900 2.718 .174 -2.129 17.929
2 5.000 2.380 .651 -3.783 13.783
4 -14.700* 2.633 .003 -24.416 -4.984
5 -20.100* 4.792 .023 -37.782 -2.418

101. 4 1
22.600* 3.194 .001 10.814 34.386
2 19.700* 2.848 .001 9.191 30.209
3 14.700* 2.633 .003 4.984 24.416
5 -5.400 4.525 1.000 -22.094 11.294
5 1
28.000* 6.354 .017 4.555 51.445
2 25.100* 6.457 .037 1.274 48.926
3 20.100* 4.792 .023 2.418 37.782
4 5.400 4.525 1.000 -11.294 22.094
Based on estimated marginal means
*. The mean difference is significant at the .05 level.
b. Adjustment for multiple comparisons: Bonferroni.

102. Interpretations:
 Participants' balance errors were measured
after 3, 6, 9, 12 and 15 minutes of exercise on
an ergometer.
 The results of a One-Way Repeated Measures
ANOVA show that the number of balance
errors was significantly affected by fatigue
p<.001.
 Since Mauchley’stest of sphericity was
violated, the Greenhouse-Geissercorrection
was used.

103.  Eta2 effect size 0.671 indicated that the
effect of fatigue on balance errors was
substantial.
 Bonferroni post-hoc tests comparing
adjacent fatigue conditions revealed a
significant difference in the number of
balance errors between 12 and 15 minutes of
exercise. No other comparisons were
significant.

104. Exercise 2 :
Depression score at baseline and four time
points for two groups (treatment and control).
Analyze the data by performing One way
ANOVA repeated measure technique.

105. Subject id time 1 time 2 time 3 time 4 treatment baeline age
1 104 5 3 3 3 0 11 31
2 106 3 5 3 3 0 11 30
3 107 2 4 0 5 0 6 25
4 114 4 4 1 4 0 8 36
5 116 7 18 9 21 0 66 22
6 118 5 2 8 7 0 27 29
7 123 6 4 0 2 0 12 31
8 126 40 20 23 12 0 52 42
9 130 5 6 6 5 0 23 37
10 135 14 13 6 0 0 10 28
11 141 26 12 6 22 0 52 36

106. 12 145 12 6 8 4 0 33 24
13 201 4 4 6 2 0 18 23
14 202 7 9 12 14 0 42 36
15 205 16 24 10 9 0 87 26
16 206 11 0 0 5 0 50 26
17 210 0 0 3 3 0 18 28
18 213 37 29 28 29 0 111 31
19 215 3 5 2 5 0 18 32
20 217 3 0 6 7 0 20 21
21 219 3 4 3 4 0 12 29
22 220 3 4 3 4 0 9 21
23 222 2 3 3 5 0 17 32
24 226 8 12 2 8 0 28 25
25 227 18 24 76 25 0 55 30
26 230 2 1 2 1 0 9 40
27 234 3 1 4 2 0 10 19
28 238 13 15 13 12 0 47 22
29 101 11 14 9 8 1 76 18
30 102 8 7 9 4 1 38 32
Continued…

107. 31 103 0 4 3 0 1 19 20
32 108 3 6 1 3 1 10 30
33 110 2 6 7 4 1 19 18
34 111 4 3 1 3 1 24 24
35 112 22 17 19 16 1 31 30
36 113 5 4 7 4 1 14 35
37 117 2 4 0 4 1 11 27
38 121 3 7 7 7 1 67 20
39 122 4 18 2 5 1 41 22
40 124 2 1 1 0 1 7 28
41 128 0 2 4 0 1 22 23
42 129 5 4 0 3 1 13 40
43 137 11 14 25 15 1 46 33
44 139 10 5 3 8 1 36 21
45 143 19 7 6 7 1 38 35
Continued…

108. 46 147 1 1 2 3 1 7 25
47 203 6 10 8 8 1 36 26
48 204 2 1 0 0 1 11 25
49 207 102 65 72 63 1 151 22
50 208 4 3 2 4 1 22 32
51 209 8 6 5 7 1 41 25
52 211 1 3 1 5 1 32 35
53 214 18 11 28 13 1 56 21
54 218 6 3 4 0 1 24 41
55 221 3 5 4 3 1 16 32
56 225 1 23 19 8 1 22 26
57 228 2 3 0 1 1 25 21
58 232 0 0 0 0 1 13 36
59 236 1 4 3 2 1 12 37
Continued…

109. END!