T test and ANOVA

FF2613

Inferential Statistics, T Test,
ANOVA & Proportionate Test

Assoc. Prof . Dr Azmi Mohd Tamil
Dept of Community Health
Universiti Kebangsaan Malaysia

©drtamil@gmail.com 2012

FF2613

Inferential Statistics

Basic Hypothesis Testing


Inferential Statistic

4 When we conduct a study, we want to
make an inference from the data
collected. For example;

“drug A is better than drug B in treating
disease D"


Drug A Better Than Drug B?

4 Drug A has a higher rate of cure than
drug B. (Cured/Not Cured)
4 If for controlling BP, the mean of BP
drop for drug A is larger than drug B.
(continuous data – mm Hg)


Null Hypothesis

4 Null Hyphotesis;

“no difference of effectiveness between
drug A and drug B in treating disease D"


Null Hypothesis

4 H0is assumed TRUE unless data indicate
otherwise:
• The experiment is trying to reject the null
hypothesis
• Can reject, but cannot prove, a hypothesis
– e.g. “all swans are white”
» One black swan suffices to reject
» H0 “Not all swans are white”
» No number of white swans can prove the hypothesis –
since the next swan could still be black.


Can reindeer fly?
4 You believe reindeer can fly
4 Null hypothesis: “reindeer cannot fly”
4 Experimental design: to throw reindeer off the
roof
4 Implementation: they all go splat on the ground
4 Evaluation: null hypothesis not rejected
• This does not prove reindeer cannot fly: what you have
shown is that
– “from this roof, on this day, under these weather conditions,
these particular reindeer either could not, or chose not to,
fly”
4 It is possible, in principle, to reject the null
hypothesis
• By exhibiting a flying reindeer!

Significance
4 Inferential statistics determine whether a significant
difference of effectiveness exist between drug A
and drug B.
4 If there is a significant difference (p<0.05), then the
null hypothesis would be rejected.
4 Otherwise, if no significant difference (p>0.05), then
the null hypothesis would not be rejected.
4 The usual level of significance utilised to reject or
not reject the null hypothesis are either 0.05 or 0.01.
In the above example, it was set at 0.05.


Confidence interval

4 Confidence interval = 1 - level of
significance.
4 If the level of significance is 0.05, then
the confidence interval is 95%.
4CI = 1 – 0.05 = 0.95 = 95%
4If CI = 99%, then level of
significance is 0.01.

What is level of
significance? Chance?

Reject H0 Reject H0

.025 .025

-2.0639 0 2.0639
-1.96 1.96 t

Fisher’s Use of p-Values

4 R.A. Fisher referred to the probability to declare
significance as “p-value”.
4 “It is a common practice to judge a result significant, if
it is of such magnitude that it would be produced by
chance not more frequently than once in 20 trials.”
4 1/20=0.05. If p-value less than 0.05, then the
probability of the effect detected were due to chance
is less than 5%.
4 We would be 95% confident that the effect detected is
due to real effect, not due to chance.


Error

4 Although we have determined the level
of significance and confidence interval,
there is still a chance of error.
4 There are 2 types;
• Type I Error
• Type II Error


Error
REALITY

Treatments are Treatments are
DECISION not different different

Conclude Correct Decision Type II error
treatments are β error
not different
(Cell a) (Cell b)

Conclude Type I error Correct Decision
treatments are α error
different
(Cell c) (Cell d)

Error
Incorrect Null
Test of Correct Null Hypothesis Hypothesis
Significance (Ho not rejected) (Ho rejected)
Null Hypothesis
Not Rejected Correct Conclusion Type II Error
Null Hypothesis
Rejected Type I Error Correct Conclusion


Type I Error

• Type I Error – rejecting the null hypothesis
although the null hypothesis is correct
e.g.
• when we compare the mean/proportion of
the 2 groups, the difference is small but the
difference is found to be significant.
Therefore the null hypothesis is rejected.
• It may occur due to inappropriate choice of
alpha (level of significance).


Type II Error

• Type II Error – not rejecting the null
hypothesis although the null hypothesis is
wrong
• e.g. when we compare the mean/proportion
of the 2 groups, the difference is big but the
difference is not significant. Therefore the
null hypothesis is not rejected.
• It may occur when the sample size is
too small.

Example of Type II Error
Data of a clinical trial on 30 patients on comparison of pain control between
two modes of treatment.
Type of treatment * Pain (2 hrs post-op) Crosstabulation

Pain (2 hrs post-op)
No pain In pain Total
Type of treatment Pethidine Count 8 7 15
% within Type
53.3% 46.7% 100.0%
of treatment
Cocktail Count 4 11 15
% within Type
26.7% 73.3% 100.0%
of treatment
Total Count 12 18 30
% within Type
40.0% 60.0% 100.0%
of treatment

Chi-square =2.222, p=0.136

p = 0.136. p bigger than 0.05. No significant difference and the null hypothesis was not
rejected.

There was a large difference between the rates but were not
significant. Type II Error? ©drtamil@gmail.com 2012

Not significant since power of
the study is less than 80%.

Power is only
32%!


Check for the errors

4 You can check for type II errors of your
own data analysis by checking for the
power of the respective analysis
4 This can easily be done by utilising
software such as Power & Sample Size
(PS2) from the website of the Vanderbilt
University


Determining the
appropriate statistical test


Data Analysis

4 Descriptive – summarising data
4 Test of Association
4 Multivariate – controlling for confounders


Test of Association

4 To study the relationship between one
or more risk variable(s) (independent)
with outcome variable (dependent)
4 For example; does ethnicity affects the
suicidal/para-suicidal tendencies of
psychiatric patients.


Problem Flow Chart

Independent Variables

Ethnicity Marital Status

Suicidal Tendencies

Dependent Variable

Multivariat

4 Studies the association between
multiple causative factors/variables
(independent variables) with the
outcome (dependent).
4 For example; risk factors such as
parental care, practise of religion,
education level of parents & disciplinary
problems of their child (outcome).


Hypothesis Testing

4 Distinguish parametric & non-parametric
procedures
4 Test two or more populations using
parametric & non-parametric procedures
• Means
• Medians
• Variances


Hypothesis Testing
Procedures


Parametric Test
Procedures

4 Involve population parameters
• Example: Population mean
4 Require interval scale or ratio scale
• Whole numbers or fractions
• Example: Height in inches: 72, 60.5, 54.7
4 Have stringent assumptions
• Example: Normal distribution
4 Examples: Z test, t test

Nonparametric Test
Procedures

4 Statistic does not depend on population
distribution
4 Data may be nominally or ordinally
scaled
• Example: Male-female
4 May involve population parameters such
as median
4 Example: Wilcoxon rank sum test


Parametric Analysis –
Quantitative

Qualitative Quantitative Normally distributed data Student's t Test
Dichotomus
Qualitative Quantitative Normally distributed data ANOVA
Polinomial
Quantitative Quantitative Repeated measurement of the Paired t Test
same individual & item (e.g.
Hb level before & after
treatment). Normally
distributed data
Quantitative - Quantitative - Normally distributed data Pearson Correlation
continous continous & Linear
Regresssion


non-parametric tests
Variable 1 Variable 2 Criteria Type of Test
Qualitative Qualitative Sample size < 20 or (< 40 but Fisher Test
Dichotomus Dichotomus with at least one expected
value < 5)
Qualitative Quantitativ Data not normally distributed Wilcoxon Rank Sum
e
Dichotomus Test or U Mann-
Whitney Test
Qualitative Quantitativ Data not normally distributed Kruskal-Wallis One
e
Polinomial Way ANOVA Test
Quantitative Quantitativ Repeated measurement of the Wilcoxon Rank Sign
e
same individual & item Test
Quantitative - Quantitativ - Data not normally distributed Spearman/Kendall
e
continous continous Rank Correlation


Statistical Tests - Qualitative

Variable 1 Variable 2 Criteria Type of Test
Qualitative Qualitative Sample size > 20 dan no Chi Square Test (X2)
expected value < 5
Qualitative Qualitative Sample size > 30 Proportionate Test
Dichotomus Dichotomus
Qualitative Qualitative Sample size > 40 but with at X2 Test with Yates
Dichotomus Dichotomus least one expected value < 5 Correction
Qualitative Quantitative
Qualitative Normallysize < 20 or data but Fisher Test Test
Sample distributed (< 40 Student's t
Dichotomus Dichotomus with at least one expected
value < 5)
Qualitative Quantitative Data not normally distributed Wilcoxon Rank Sum


Data Analysis

4Using SPSS;
http://161.142.92.104/spss/
4Using Excel;
http://161.142.92.104/excel/


FF2613

T Test, ANOVA &
Proportionate Test

Assoc. Prof . Dr Azmi Mohd Tamil
Dept of Community Health
Universiti Kebangsaan Malaysia


T - Test

Independent T-Test
Student’s T-Test

Paired T-Test

ANOVA

© d rta m il@ g m a il. o m
c 2012

Student’s T-test

William Sealy Gosset @
“Student”, 1908. The Probable
Error of Mean. Biometrika.


Student’s T-Test
4 To compare the means of two independent
groups. For example; comparing the mean
Hb between cases and controls. 2 variables
are involved here, one quantitative (i.e. Hb)
and the other a dichotomous qualitative
variable (i.e. case/control).

4 t=

Examples: Student’s t-
test

4 Comparing the level of blood cholestrol
(mg/dL) between the hypertensive and
normotensive.
4 Comparing the HAMD score of two
groups of psychiatric patients treated
with two different types of drugs (i.e.
Fluoxetine & Sertraline


Example

Group Statistics

DRUG N Mean Std. Deviation
DHAMAWK6 F 35 4.2571 3.12808
S 32 3.8125 4.39529

Independent Samples Test

t-test for Equality of Means
Sig. Mean
t df (2-tailed) Difference
DHAMAWK6 Equal variances
.48 65 .633 .4446
assumed


Assumptions of T test

4 Observations are normally distributed in
each population. (Explore)
4 The population variances are equal.
( L e v e n e ’s T e s t)

The 2 groups are independent of each
other. (Design of study)


Manual Calculation

4 Sample size > 30 4 Small sample size,
equal variance
X1 − X 2
t=
X1 − X 2 1 1
t= s0 +
n1 n2
2 2
s s
+ 1 2
n1 n2 (n1 − 1) s12 + (n2 − 1) s2
2
s0 =
2

(n1 − 1) + (n2 − 1)


Example – compare
cholesterol level
4 Hypertensive : 4 Normal :
Mean : 214.92 Mean : 182.19
s.d. : 39.22 s.d. : 37.26
n : 64 n : 36
• Comparing the cholesterol level between
hypertensive and normal patients.
• The difference is (214.92 – 182.19) = 32.73 mg%.
• H0 : There is no difference of cholesterol level
between hypertensive and normal patients.
• n > 30, (64+36=100), therefore use the first formula.

Calculation
X1 − X 2
t=
2 2
s s
1
+ 2
n1 n2
4t = (214.92- 182.19)________
((39.222/64)+(37.262/36))0.5
4 t = 4.137
4 df = n1+n2-2 = 64+36-2 = 98
4 Refer to t table; with t = 4.137, p < 0.001

If df>100, can refer Table A1.
We don’t have 4.137 so we
use 3.99 instead. If t = 3.99,
then p=0.00003x2=0.00006
Therefore if t=4.137,
p<0.00006.

Or can refer to Table A3.
We don’t have df=98,
so we use df=60 instead.
t = 4.137 > 3.46 (p=0.001)

Therefore if t=4.137, p<0.001.

Conclusion
• Therefore p < 0.05, null hypothesis rejected.
• There is a significant difference of
cholesterol level between hypertensive and
normal patients.
• Hypertensive patients have a significantly
higher cholesterol level compared to
normotensive patients.

Exercise (try it)
• Comparing the mini test 1 (2012) results between
UKM and ACMS students.
• The difference is 11.255
• H0 : There is no difference of marks between UKM
and ACMS students.
• n > 30, therefore use the first formula.


Exercise (answer)

4 Nullhypothesis rejected
4 There is a difference of marks between
UKM and ACMS students. UKM marks
higher than AUCMS


T-Test In SPSS

4 For this exercise, we will
be using the data from
the CD, under Chapter
7, sga-bab7.sav
4 This data came from a
case-control study on
factors affecting SGA in
Kelantan.
4 Open the data & select -
>Analyse
>Compare Means
>Ind-Samp T
Test…

T-Test in SPSS

4 We want to see whether
there is any association
between the mothers’ weight
and SGA. So select the risk
factor (weight2) into ‘Test
Variable’ & the outcome
(SGA) into ‘Grouping
Variable’.
4 Now click on the ‘Define
Groups’ button. Enter
• 0 (Control) for Group 1 and
• 1 (Case) for Group 2.
4 Click the ‘Continue’ button &
then click the ‘OK’ button.


T-Test Results
Group Statistics

Std. Error
SGA N Mean Std. Deviation Mean
Weight at first ANC Normal 108 58.666 11.2302 1.0806
SGA 109 51.037 9.3574 .8963

4 Compare the mean+sd of both groups.
• Normal 58.7+11.2 kg
• SGA 51.0+ 9.4 kg
4 Apparently there is a difference of
weight between the two groups.

Results & Homogeneity of
Variances
Independent Samples Test

Levene's Test for
Equality of Variances t-test for Equality of Means
95% Confidence
Interval of the
Mean Std. Error Difference
F Sig. t df Sig. (2-tailed) Difference Difference Lower Upper
Weight at first ANC Equal variances
1.862 .174 5.439 215 .000 7.629 1.4028 4.8641 10.3940
assumed
Equal variances
5.434 207.543 .000 7.629 1.4039 4.8612 10.3969
not assumed

4 Look at the p value of Levene’s Test. If p is not
significant then equal variances is assumed (use top
row).
4 If it is significant then equal variances is not assumed
(use bottom row).
4 So the t value here is 5.439 and p < 0.0005. The
difference is significant. Therefore there is an
association between the mothers weight and SGA.

How to present the
result?

Group N Mean test p

Normal 108 58.7+11.2 kg
T test
<0.0005
t = 5.439
SGA 109 51.0+ 9.4


Paired t-test

“Repeated measurement on the
same individual”


Paired T-Test

4 “Repeated measurement on the same
individual”
4 t=


Formula

d −0
t=
sd
n

(∑ d )
2

∑d i
2
−
n
sd =
n −1

df = n p − 1

Examples of paired t-test

4 Comparing the HAMD score between
week 0 and week 6 of treatment with
Sertraline for a group of psychiatric
patients.
4 Comparing the haemoglobin level
amongst anaemic pregnant women after
6 weeks of treatment with haematinics.


Example

Paired Samples Statistics

Mean N Std. Deviation
Pair DHAMAWK0 13.9688 32 6.48315
1 DHAMAWK6 3.8125 32 4.39529

Paired Samples Test

Paired Differences
Std. Sig.
Mean Deviation t df (2-tailed)
Pair DHAMAWK0 -
10.1563 6.75903 8.500 31 .000
1 DHAMAWK6


M a n u a l C a l c u l a t i o n

The measurement of the systolic and diastolic
blood pressures was done two consecutive
times with an interval of 10 minutes. You want
to d e te r m in e w h e th e r th e r e w a s a n y

difference between those two measurements.
4 H0:There is no difference of the systolic blood
pressure during the first (time 0) and second
measurement (time 10 minutes).


Calculation

4 Calculate the difference between first &
second measurement and square it.
Total up the difference and the square.


Calculation

4∑ d = 112 ∑ d2 = 1842 n = 36
4 Mean d = 112/36 = 3.11
4 sd = ((1842-1122/36)/35)0.5 d −0
t=
sd
sd = 6.53 n
4 t = 3.11/(6.53/6)
t = 2.858 (∑ d )
2

4 df = np – 1 = 36 – 1 = 35. ∑ d i2 −
n
sd =
n −1
4 Refer to t table;
df = n p − 1

Refer to Table A3.
We don’t have df=35,
so we use df=30 instead.
t = 2.858, larger than 2.75
(p=0.01) but smaller than 3.03
(p=0.005). 3.03>t>2.75
Therefore if t=2.858,
0.005<p<0.01.

Conclusion
with t = 2.858, 0.005<p<0.01
Therefore p < 0.01.
Therefore p < 0.05, null hypothesis
rejected.
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.

Paired T-Test In SPSS

7, sgapair.sav
controlled trial on
haematinic effect on Hb.
>Analyse
>Compare Means
>Paired-Samples T

Test…

Paired T-Test In SPSS

between the prescription
on haematinic to
anaemic pregnant
mothers and Hb.
4 We are comparing the
Hb before & after
treatment. So pair the
two measurements (Hb2
& Hb3) together.
4 Click the ‘OK’ button.


Paired T-Test Results
Paired Samples Statistics

Std. Error
Mean N Std. Deviation Mean
Pair HB2 10.247 70 .3566 .0426
1 HB3 10.594 70 .9706 .1160

4 Thisshows the mean & standard
deviation of the two groups.


Paired T-Test Results
Paired Samples Test

Paired Differences
95% Confidence
Interval of the
Std. Error Difference
Mean Std. Deviation Mean Lower Upper t df Sig. (2-tailed)
Pair 1 HB2 - HB3 -.347 .9623 .1150 -.577 -.118 -3.018 69 .004

4 This shows the mean difference of Hb
before & after treatment is only 0.347
g%.
4 Yet the t=3.018 & p=0.004 show the
difference is statistically significant.

How to present the
result?

Mean D
Group N Test p
(Diff.)

Before
treatment
Paired T-
(HB2) vs
70 0.35 + 0.96 test 0.004
After
t = 3.018
treatment
(HB3)


ANOVA


ANOVA –
Analysis of Variance

4 Extension of independent-samples t test
4 Comparesthe means of groups of
independent observations
• Don’t be fooled by the name. ANOVA does
not compare variances.
4 Can compare more than two groups


One-Way ANOVA
F-Test
4 Tests the equality of 2 or more population means
4 Variables
• One nominal scaled independent variable
– 2 or more treatment levels or classifications
(i.e. Race; Malay, Chinese, Indian & Others)
• One interval or ratio scaled dependent variable
(i.e. weight, height, age)
4 Used to analyse completely randomized
experimental designs


Examples

4 Comparing the blood cholesterol levels
between the bus drivers, bus conductors
and taxi drivers.
4 Comparing the mean systolic pressure
between Malays, Chinese, Indian &
Others.


One-Way ANOVA
F-Test Assumptions

4 Randomness & independence of errors
• Independent random samples are drawn
4 Normality
• Populations are normally distributed
4 Homogeneity of variance
• Populations have equal variances


Example
Descriptives

Birth weight
N Mean Std. Deviation Minimum Maximum
Housewife 151 2.7801 .52623 1.90 4.72
Office work 23 2.7643 .60319 1.60 3.96
Field work 44 2.8430 .55001 1.90 3.79
Total 218 2.7911 .53754 1.60 4.72

ANOVA

Birth weight
Sum of
Squares df Mean Square F Sig.
Between Groups .153 2 .077 .263 .769
Within Groups 62.550 215 .291
Total 62.703 217

Manual Calculation

ANOVA


Manual Calculation

4 Notexpected to be calculated manually
by medical students.


Example:
Time To Complete
Analysis

45 samples were
analysed using 3 different
blood analyser (Mach1,
Mach2 & Mach3).

15 samples were placed
into each analyser.

Time in seconds was
measured for each
sample analysis.

Example:
Time To Complete
Analysis

The overall mean of the
entire sample was 22.71
seconds.

This is called the “grand”
mean, and is often
denoted by X .

If H0 were true then we’d
expect the group means
to be close to the grand
mean.

Example:
Time To Complete
Analysis
The ANOVA test is
based on the combined
distances from X .

If the combined
distances are large, that
indicates we should
reject H0.

The Anova Statistic

To combine the differences from the grand mean we
• Square the differences
• Multiply by the numbers of observations in the groups
• Sum over the groups

( )2
( )
2
(
SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X )
2

where the X * are the group means.

“SSB” = Sum of Squares Between groups

The Anova Statistic

To combine the differences from the grand mean we
• Square the differences
• Multiply by the numbers of observations in the groups
• Sum over the groups

( )2
( )
2
(
SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X )
2

where the X * are the group means.

“SSB” = Sum of Squares Between groups

Note: This looks a bit like a variance.

Sum of Squares Between

( )
2
( )2
(
SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X )2

4 Grand Mean = 22.71
4 Mean Mach1 = 24.93; (24.93-22.71)2=4.9284
4 Mean Mach2 = 22.61; (22.61-22.71)2=0.01
4 Mean Mach3 = 20.59; (20.59-22.71)2=4.4944
4 SSB = (15*4.9284)+(15*0.01)+(15*4.4944)
4 SSB = 141.492


How big is big?

4 For the Time to Complete, SSB = 141.492

4 Is that big enough to reject H0?

4 As with the t test, we compare the statistic to
the variability of the individual observations.

4 InANOVA the variability is estimated by the
Mean Square Error, or MSE

MSE
Mean Square Error

The Mean Square Error
is a measure of the
variability after the
group effects have
been taken into
account.

∑∑ (x − X j)
1 2
MSE = ij
N −K j i

where xij is the ith
observation in the jth
group.

MSE
Mean Square Error

The Mean Square Error
is a measure of the
variability after the
group effects have
been taken into
account.

∑∑ (x − X j)
1 2
MSE = ij
N −K j i

∑∑ (xij − X j )
1 2
MSE =
N −K j i
Mach1 (x-mean)^2 Mach2 (x-mean)^2 Mach3 (x-mean)^2
23.73 1.4400 21.5 1.2321 19.74 0.7225
23.74 1.4161 21.6 1.0201 19.75 0.7056
23.75 1.3924 21.7 0.8281 19.76 0.6889
24.00 0.8649 21.7 0.8281 19.9 0.4761
24.10 0.6889 21.8 0.6561 20 0.3481
24.20 0.5329 21.9 0.5041 20.1 0.2401
25.00 0.0049 22.75 0.0196 20.3 0.0841
25.10 0.0289 22.75 0.0196 20.4 0.0361
25.20 0.0729 22.75 0.0196 20.5 0.0081
25.30 0.1369 23.3 0.4761 20.5 0.0081
25.40 0.2209 23.4 0.6241 20.6 0.0001
25.50 0.3249 23.4 0.6241 20.7 0.0121
26.30 1.8769 23.5 0.7921 22.1 2.2801
26.31 1.9044 23.5 0.7921 22.2 2.5921
26.32 1.9321 23.6 0.9801 22.3 2.9241
SUM 12.8380 9.4160 11.1262

∑∑ (xij − X j )
1 2
MSE =
N −K j i

4 Note that the variation of the means
(141.492) seems quite large (more likely
to be significant???) compared to the
variance of observations within groups
(12.8380+9.4160+11.1262=33.3802).
4 MSE = 33.3802/(45-3) = 0.7948


Notes on MSE

4 Ifthere are only two groups, the MSE is equal
to the pooled estimate of variance used in the
equal-variance t test.
4 ANOVA assumes that all the group variances
are equal.
4 Other options should be considered if group
variances differ by a factor of 2 or more.
4 (12.8380 ~ 9.4160 ~ 11.1262)

ANOVA F Test

4 The ANOVA F test is based on the F statistic
SSB (K − 1)
F=
MSE
where K is the number of groups.

4 Under H0 the F statistic has an “F” distribution,
with K-1 and N-K degrees of freedom (N is the
total number of observations)

Time to Analyse:
F test p-value
To get a p-value we
compare our F statistic
to an F(2, 42)
distribution.

Time to Analyse:
F test p-value
To get a p-value we
to an F(2, 42)
distribution.

In our example

141.492 2
F= = 89.015
33.3802 42

We cannot draw the line
since the F value is so
large, therefore the p
value is so small!!!!!!

Refer to F Dist. Table (α=0.01).
We don’t have df=2;42,
so we use df=2;40 instead.
F = 89.015, larger than 5.18
(p=0.01)
Therefore if F=89.015, p<0.01.

Why use df=2;42?
We have 3 groups
so K-1 = 2
We have 45
samples therefore
N-K = 42. ©drtamil@gmail.com 2012

Time to Analyse:
F test p-value
To get a p-value we
to an F(2, 42)
distribution.

In our example

141.492 2
F= = 89.015
33.3802 42

The p-value is really
P(F (2,42) > 89.015) = 0.00000000000008

ANOVA Table
Results are often displayed using an ANOVA Table

Sum of Mean
Squares df Square F Sig.
Between
141.492 2 40.746 89.015 .0000000
Groups

Total 174.872 44

ANOVA Table

Sum of Mean
Between
141.492 2 40.746 89.015 .0000000
Groups

Total 174.872 44

Pop Quiz!: Where are the following quantities presented in this table?

Sum of Squares Mean Square F Statistic p value
Between (SSB) Error (MSE)

ANOVA Table

Sum of Mean
Between
141.492 2 40.746 89.015 .0000000
Groups

Total 174.872 44

Sum of Squares Mean Square F Statistic p value
Between (SSB) Error (MSE)

ANOVA In SPSS

7, sga-bab7.sav
case-control study on
factors affecting SGA in
Kelantan.
>Analyse
>Compare Means
>One-Way
ANOVA…

ANOVA in SPSS

between the babies’ weight
and mothers’ type of work.
So select the risk factor
(typework) into ‘Factor’ & the
outcome (birthwgt) into
‘Dependent’.
4 Now click on the ‘Post Hoc’
button. Select Bonferonni.
4 Click the ‘Continue’ button &
then click the ‘OK’ button.
4 Then click on the ‘Options’
button.


ANOVA in SPSS

4 Select ‘Descriptive’,
‘Homegeneity of
variance test’ and
‘Means plot’.
4 Click ‘Continue’ and
then ‘OK’.


ANOVA Results
Descriptives

Birth weight
95% Confidence Interval for
Mean
N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum
Housewife 151 2.7801 .52623 .04282 2.6955 2.8647 1.90 4.72
Office work 23 2.7643 .60319 .12577 2.5035 3.0252 1.60 3.96
Field work 44 2.8430 .55001 .08292 2.6757 3.0102 1.90 3.79
Total 218 2.7911 .53754 .03641 2.7193 2.8629 1.60 4.72

4 Compare the mean+sd of all groups.
4 Apparently there are not much
difference of babies’ weight between the
groups.


Results & Homogeneity of
Variances
Test of Homogeneity of Variances

Birth weight
Levene
Statistic df1 df2 Sig.
.757 2 215 .470

4 Look at the p value of Levene’s Test. If p
is not significant then equal variances is
assumed.


ANOVA Results
ANOVA

Birth weight
Sum of
Squares df Mean Square F Sig.
Between Groups .153 2 .077 .263 .769
Total 62.703 217

4 Sothe F value here is 0.263 and p =0.769.
The difference is not significant. Therefore
there is no association between the
babies’ weight and mothers’ type of work.


How to present the
result?

Type of Work Mean+sd Test p

Office 2.76 + 0.60

ANOVA
Housewife 2.78 + 0.53 0.769
F = 0.263

Farmer 2.84 + 0.55


Proportionate Test


Proportionate Test

4 Qualitativedata utilises rates, i.e. rate of
anaemia among males & females
4 To compare such rates, statistical tests
such as Z-Test and Chi-square can be
used.


Formula
p1 − p2 • where p1 is the rate for
z= event 1 = a1/n1
1 1 
p0 q0  +  • p2 is the rate for event 2
= a2/n2
 n1 n2 
• a1 and a2 are frequencies
of event 1 and 2

p1n1 + p2 n2 4 We refer to the normal
p0 = distribution table to
n1 + n2 decide whether to reject
or not the null
hypothesis.
q0 = 1 − p0


http://stattrek.com/hypothesis-
test/proportion.aspx

4 ■The sampling method is simple random
sampling.
4 ■Each sample point can result in just two
possible outcomes. We call one of these
outcomes a success and the other, a failure.
4 ■The sample includes at least 10 successes
and 10 failures.
4 ■The population size is at least 10 times as
big as the sample size.


Example

4 Comparison of worm infestation rate
between male and female medical
students in Year 2.
4 Rate for males ; p1= 29/96 = 0.302
4 Rate for females;p2 =24/104 = 0.231
4 H0: There is no difference of worm
infestation rate between male and
female medical students in Year 2


Cont.
p1 p2

p0 q0


Cont.

4 p0 = (29/96*96)+(24/104*104) = 0.265
96+104

4 q0 = 1 – 0.265 = 0.735


Cont.

4z = 0.302 - 0.231 = 1.1367
((0.735*0.265) (1/96 + 1/104))0.5

4 From the normal distribution table (A1), z value
is significant at p=0.05 if it is above 1.96. Since
the value is less than 1.96, then there is no
difference of rate for worm infestatation
between the male and female students.

Refer to Table A1.
We don’t have 1.1367 so we
use 1.14 instead. If z = 1.14,
then p=0.1271x2=0.2542
Therefore if z=1.14,
p=0.2542. H0 not rejected

Exercise (try it)

4 Comparison of failure rate between
ACMS and UKM medical students in
Year 2 for minitest 1 (MS2 2012).
4 Rate for UKM ; p1= 42/196 = 0.214
4 Rate for ACMS;p2 = 35/70 = 0.5

Answer

4 P1 = 0.214, p2 = 0.5, p0 = 0.289, q0 = 0.711
4 N1 = 196, n2 = 70, Z = 20.470.5 = 4.52
4 p < 0.00006

T test and ANOVA

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