t test and anova
DESCRIPTION
TTest and ANOVATRANSCRIPT
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Inferential Statistics, T Test, ANOVA & Proportionate Test
Assoc. Prof . Dr Azmi Mohd Tamil
Dept of Community Health
Universiti Kebangsaan Malaysia
FF2613
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Inferential Statistic
4When 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"
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Drug A Better Than Drug B?
4Drug 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)
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Null Hypothesis
4Null Hyphotesis;
“no difference of effectiveness between
drug A and drug B in treating disease D"
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Null Hypothesis
4H0 is 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.
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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!
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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.
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Confidence interval
4Confidence 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.
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What is level of significance? Chance?
What is level of significance? Chance?
t0 2.0639-2.0639
.025
Reject H0
Reject H0
.025
t0 2.0639-2.0639
.025
Reject H0
Reject H0
.025
-1.96 1.96
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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.
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Error
4Although we have determined the level
of significance and confidence interval,
there is still a chance of error.
4There are 2 types;
• Type I Error
• Type II Error
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Treatments are
not different
Treatments are
different
Conclude
treatments are
not different
Conclude
treatments are
different
DECISION
REALITY
Correct DecisionType I error
α error
Type II error
β error
Correct Decision
(Cell b)(Cell a)
(Cell c) (Cell d)
Error
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Error
Test of Correct Null Hypothesis
Incorrect Null
Hypothesis
Significance (Ho not rejected) (Ho rejected)
Null Hypothesis
Not Rejected Correct Conclusion Type II Error
Null Hypothesis
Rejected Type I Error Correct Conclusion
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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).
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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.
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Type of treatment * Pain (2 hrs post-op) Crosstabulation
8 7 15
53.3% 46.7% 100.0%
4 11 15
26.7% 73.3% 100.0%
12 18 30
40.0% 60.0% 100.0%
Count
% within Type
of treatment
Count
% within Type
of treatment
Count
% within Type
of treatment
Pethidine
Cocktail
Type of treatment
Total
No pain In pain
Pain (2 hrs post-op)
Total
Example of Type II Error
Data of a clinical trial on 30 patients on comparison of pain control between
two modes of treatment.
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?
Chi-square =2.222, p=0.136
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Not significant since power of the study is less than 80%.
Power is only
32%!
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Check for the errors
4You can check for type II errors of your
own data analysis by checking for the
power of the respective analysis
4This can easily be done by utilising
software such as Power & Sample Size
(PS2) from the website of the Vanderbilt
University
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Determining the appropriate statistical test
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Data Analysis
4Descriptive – summarising data
4Test of Association
4Multivariate – controlling for confounders
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Test of Association
4To study the relationship between one
or more risk variable(s) (independent)
with outcome variable (dependent)
4For example; does ethnicity affects the
suicidal/para-suicidal tendencies of
psychiatric patients.
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Problem Flow Chart
Marital Status
Suicidal Tendencies
Ethnicity
Independent Variables
Dependent Variable
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Multivariat
4Studies the association between
multiple causative factors/variables
(independent variables) with the
outcome (dependent).
4For example; risk factors such as
parental care, practise of religion,
education level of parents & disciplinary
problems of their child (outcome).
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Hypothesis TestingHypothesis Testing
4Distinguish parametric & non-parametric
procedures
4Test two or more populations using
parametric & non-parametric procedures
• Means
• Medians
• Variances
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Parametric Test Procedures
Parametric Test Procedures
4 Involve population parameters
• Example: Population mean
4Require interval scale or ratio scale
• Whole numbers or fractions
• Example: Height in inches: 72, 60.5, 54.7
4Have stringent assumptions
• Example: Normal distribution
4Examples: Z test, t test
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Nonparametric Test Procedures
Nonparametric Test Procedures
4Statistic does not depend on population
distribution
4Data may be nominally or ordinally
scaled
• Example: Male-female
4May involve population parameters such
as median
4Example: Wilcoxon rank sum test
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Parametric Analysis –Quantitative
Qualitative
Dichotomus
Quantitative Normally distributed data Student's t Test
Qualitative
Polinomial
Quantitative Normally distributed data ANOVA
Quantitative Quantitative Repeated measurement of the
same individual & item (e.g.
Hb level before & after
treatment). Normally
distributed data
Paired t Test
Quantitative -
continous
Quantitative -
continous
Normally distributed data Pearson Correlation
& Linear
Regresssion
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non-parametric tests
Variable 1 Variable 2 Criteria Type of Test
Qualitative
Dichotomus
Qualitative
Dichotomus
Sample size < 20 or (< 40 but
with at least one expected
value < 5)
Fisher Test
Qualitative
Dichotomus
Quantitative Data not normally distributed Wilcoxon Rank Sum
Test or U Mann-
Whitney Test
Qualitative
Polinomial
Quantitative Data not normally distributed Kruskal-Wallis One
Way ANOVA Test
Quantitative Quantitative Repeated measurement of the
same individual & item
Wilcoxon Rank Sign
TestQuantitative -
continous
Quantitative -
continous
Data not normally distributed Spearman/Kendall
Rank Correlation
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Statistical Tests - Qualitative
Variable 1 Variable 2 Criteria Type of Test
Qualitative Qualitative Sample size > 20 dan no
expected value < 5Chi Square Test (X
2)
Qualitative
Dichotomus
Qualitative
Dichotomus
Sample size > 30 Proportionate Test
Qualitative
Dichotomus
Qualitative
Dichotomus
Sample size > 40 but with at
least one expected value < 5X2 Test with Yates
Correction
Qualitative Quantitative Normally distributed data Student's t TestQualitative
Dichotomus
Qualitative
Dichotomus
Sample size < 20 or (< 40 but
with at least one expected
value < 5)
Fisher Test
Qualitative Quantitative Data not normally distributed Wilcoxon Rank Sum
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Data Analysis
4Using SPSS;
http://161.142.92.104/spss/
4Using Excel;
http://161.142.92.104/excel/
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T Test, ANOVA & Proportionate Test
Assoc. Prof . Dr Azmi Mohd Tamil
Dept of Community Health
Universiti Kebangsaan Malaysia
FF2613
© d rtam il@ g m ail.co m 2012
T - Test
Independent T-Test
Student’s T-Test
Paired T-Test
ANOVA
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Student’s T-test
William Sealy Gosset @
“Student”, 1908. The Probable
Error of Mean. Biometrika.
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Student’s T-Test
4To 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 =
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Examples: Student’s t-test
4Comparing the level of blood cholestrol
(mg/dL) between the hypertensive and
normotensive.
4Comparing the HAMD score of two
groups of psychiatric patients treated
with two different types of drugs (i.e.
Fluoxetine & Sertraline
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Example
Group Statistics
35 4.2571 3.12808
32 3.8125 4.39529
DRUG
F
S
DHAMAWK6
N Mean Std. Deviation
Independent Samples Test
.48 65 .633 .4446Equal variances
assumed
DHAMAWK6
t df
Sig.
(2-tailed)
Mean
Difference
t-test for Equality of Means
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Assumptions of T test
4Observations are normally distributed in
each population. (Explore)
4The 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)
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Manual Calculation
4 Sample size > 30 4 Small sample size,
equal variance
1 2
2 2
1 2
1 2
X Xt
s s
n n
−=
+
1 2
0
1 2
2 22 1 1 2 20
1 2
1 1
( 1) ( 1)
( 1) ( 1)
X Xt
sn n
n s n ss
n n
−=
+
− + −=
− + −
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Example – compare cholesterol level
4Hypertensive :Mean : 214.92s.d. : 39.22n : 64
4Normal :Mean : 182.19s.d. : 37.26n : 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.
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Calculation
4 t = (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
1 2
2 2
1 2
1 2
X Xt
s s
n n
−=
+
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=0.00006Therefore 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.
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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.
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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.
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Exercise (answer)
4Null hypothesis rejected
4There is a difference of marks between
UKM and ACMS students. UKM marks
higher than AUCMS
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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…
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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.
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T-Test Results
4Compare the mean+sd of both groups.
• Normal 58.7+11.2 kg
• SGA 51.0+ 9.4 kg
4Apparently there is a difference of
weight between the two groups.
Group Statistics
108 58.666 11.2302 1.0806
109 51.037 9.3574 .8963
SGA
Normal
SGA
Weight at first ANC
N Mean Std. Deviation
Std. Error
Mean
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Results & Homogeneity of Variances
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.
Independent Samples Test
1.862 .174 5.439 215 .000 7.629 1.4028 4.8641 10.3940
5.434 207.543 .000 7.629 1.4039 4.8612 10.3969
Equal variances
assumed
Equal variances
not assumed
Weight at first ANC
F Sig.
Levene's Test for
Equality of Variances
t df Sig. (2-tailed)
Mean
Difference
Std. Error
Difference Lower Upper
95% Confidence
Interval of the
Difference
t-test for Equality of Means
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How to present the result?
Group N Mean test p
Normal 108 58.7+11.2 kg
T test
t = 5.439<0.0005
SGA 109 51.0+ 9.4
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Examples of paired t-test
4Comparing the HAMD score between
week 0 and week 6 of treatment with
Sertraline for a group of psychiatric
patients.
4Comparing the haemoglobin level
amongst anaemic pregnant women after
6 weeks of treatment with haematinics.
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Example
Paired Samples Statistics
13.9688 32 6.48315
3.8125 32 4.39529
DHAMAWK0
DHAMAWK6
Pair
1
Mean N Std. Deviation
Paired Samples Test
10.1563 6.75903 8.500 31 .000DHAMAWK0 -
DHAMAWK6
Pair
1
Mean
Std.
Deviation
Paired Differences
t df
Sig.
(2-tailed)
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M a n u a l C a l c u l a t i o n
D The measurement of the systolic and diastolic
blood pressures was done two consecutive
times with an interval of 10 minutes. You want
t o d e t e r m in e w h e t h e r t h e r e w a s a n y
difference between those two measurements.
4H0:There is no difference of the systolic blood
pressure during the first (time 0) and second
measurement (time 10 minutes).
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Calculation
4Calculate the difference between first &
second measurement and square it.
Total up the difference and the square.
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Calculation
4∑ d = 112 ∑ d2 = 1842 n = 36
4Mean d = 112/36 = 3.11
4sd = ((1842-1122/36)/35)0.5
sd = 6.53
4 t = 3.11/(6.53/6)
t = 2.858
4df = np – 1 = 36 – 1 = 35.
4Refer to t table;
( )22
0
1
1
d
i
d
p
dt
s
n
dd
nsn
df n
−=
−=
−
= −
∑∑
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.
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Conclusion
with t = 2.858, 0.005<p<0.01Therefore 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.
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Paired T-Test In SPSS
4 For this exercise, we will be using the data from the CD, under Chapter 7, sgapair.sav
4 This data came from a controlled trial on haematinic effect on Hb.
4 Open the data & select ->Analyse>Compare Means
>Paired-Samples T
Test…
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Paired T-Test In SPSS
4 We want to see whether there is any association 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.
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Paired T-Test Results
4This shows the mean & standard
deviation of the two groups.
Paired Samples Statistics
10.247 70 .3566 .0426
10.594 70 .9706 .1160
HB2
HB3
Pair
1
Mean N Std. Deviation
Std. Error
Mean
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Paired T-Test Results
4This shows the mean difference of Hb
before & after treatment is only 0.347
g%.
4Yet the t=3.018 & p=0.004 show the
difference is statistically significant.
Paired Samples Test
-.347 .9623 .1150 -.577 -.118 -3.018 69 .004HB2 - HB3Pair 1
Mean Std. Deviation
Std. Error
Mean Lower Upper
95% Confidence
Interval of the
Difference
Paired Differences
t df Sig. (2-tailed)
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How to present the result?
Group NMean D
(Diff.)Test p
Before
treatment
(HB2) vs
After
treatment
(HB3)
70 0.35 + 0.96
Paired T-
test
t = 3.018
0.004
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ANOVA
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ANOVA –Analysis of Variance
4Extension of independent-samples t test
4Compares the means of groups of
independent observations
• Don’t be fooled by the name. ANOVA does
not compare variances.
4Can compare more than two groups
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One-Way ANOVA F-Test
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
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Examples
4Comparing the blood cholesterol levels
between the bus drivers, bus conductors
and taxi drivers.
4Comparing the mean systolic pressure
between Malays, Chinese, Indian &
Others.
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One-Way ANOVA F-Test Assumptions
One-Way ANOVA F-Test Assumptions
4Randomness & independence of errors
• Independent random samples are drawn
4Normality
• Populations are normally distributed
4Homogeneity of variance
• Populations have equal variances
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Example
Descriptives
Birth weight
151 2.7801 .52623 1.90 4.72
23 2.7643 .60319 1.60 3.96
44 2.8430 .55001 1.90 3.79
218 2.7911 .53754 1.60 4.72
Housewife
Office work
Field work
Total
N Mean Std. Deviation Minimum Maximum
ANOVA
Birth weight
.153 2 .077 .263 .769
62.550 215 .291
62.703 217
Between Groups
Within Groups
Total
Sum of
Squares df Mean Square F Sig.
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Manual Calculation
4Not expected 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 .
If H0 were true then we’d
expect the group means
to be close to the grand
mean.
X
Example: Time To Complete Analysis
The ANOVA test is
based on the combined
distances from .
If the combined
distances are large, that
indicates we should
reject H0.
X
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
where the are the group means.
“SSB” = Sum of Squares Between groups
( ) ( ) ( )23
2
2
2
1 151515 SSB XXXXXX MachMachMach −+−+−=
*X
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
where the are the group means.
“SSB” = Sum of Squares Between groups
Note: This looks a bit like a variance.
*X
( ) ( ) ( )23
2
2
2
1 151515 SSB XXXXXX MachMachMach −+−+−=
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4Grand Mean = 22.71
4Mean Mach1 = 24.93; (24.93-22.71)2=4.9284
4Mean Mach2 = 22.61; (22.61-22.71)2=0.01
4Mean Mach3 = 20.59; (20.59-22.71)2=4.4944
4SSB = (15*4.9284)+(15*0.01)+(15*4.4944)
4SSB = 141.492
( ) ( ) ( )23
2
2
2
1 151515 SSB XXXXXX MachMachMach −+−+−=
Sum of Squares Between
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 In ANOVA the variability is estimated by the
Mean Square Error, or MSE
MSEMean Square Error
The Mean Square Error
is a measure of the
variability after the
group effects have
been taken into
account.
where xij is the ith
observation in the jth
group.
( )∑∑ −−
=j i
jij XxKN
MSE21
MSEMean Square Error
The Mean Square Error
is a measure of the
variability after the
group effects have
been taken into
account.
where xij is the ith
observation in the jth
group.
( )∑∑ −−
=j i
jij XxKN
MSE21
MSEMean Square Error
The Mean Square Error
is a measure of the
variability after the
group effects have
been taken into
account.
( )∑∑ −−
=j i
jij XxKN
MSE21
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( )∑∑ −−
=j i
jijXx
KNMSE
21
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
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4Note 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).
4MSE = 33.3802/(45-3) = 0.7948
( )∑∑ −−
=j i
jijXx
KNMSE
21
Notes on MSE
4 If there 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
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)
MSE
KSSBF
)1( −=
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
compare our F statistic
to an F(2, 42)
distribution.
In our example
We cannot draw the line
since the F value is so
large, therefore the p
value is so small!!!!!!
015.89423802.33
2492.141==F
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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 = 2We have 45 samples therefore N-K = 42.
Time to Analyse:F test p-value
To get a p-value we
compare our F statistic
to an F(2, 42)
distribution.
In our example
The p-value is really
( ) 00080000000000.0015.89(2,42) => FP
015.89423802.33
2492.141==F
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
Sum of Squares
Between (SSB)
Mean Square
Error (MSE)F Statistic p value
Pop Quiz!: Where are the following quantities presented in this table?
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
Sum of Squares
Between (SSB)
Mean Square
Error (MSE)F Statistic p value
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
Sum of Squares
Between (SSB)
Mean Square
Error (MSE)F Statistic p value
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
Sum of Squares
Between (SSB)
Mean Square
Error (MSE)F Statistic p value
ANOVA Table
Sum of
Squares df
Mean
Square F Sig.
Between
Groups141.492 2 40.746 89.015 .0000000
Within Groups 33.380 42 .795
Total 174.872 44
Results are often displayed using an ANOVA Table
Sum of Squares
Between (SSB)
Mean Square
Error (MSE)F Statistic p value
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ANOVA 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>One-Way
ANOVA…
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ANOVA in SPSS
4 We want to see whether there is any association 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.
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ANOVA in SPSS
4 Select ‘Descriptive’,
‘Homegeneity of
variance test’ and
‘Means plot’.
4 Click ‘Continue’ and
then ‘OK’.
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ANOVA Results
4Compare the mean+sd of all groups.
4Apparently there are not much
difference of babies’ weight between the
groups.
Descriptives
Birth weight
151 2.7801 .52623 .04282 2.6955 2.8647 1.90 4.72
23 2.7643 .60319 .12577 2.5035 3.0252 1.60 3.96
44 2.8430 .55001 .08292 2.6757 3.0102 1.90 3.79
218 2.7911 .53754 .03641 2.7193 2.8629 1.60 4.72
Housewife
Office work
Field work
Total
N Mean Std. Deviation Std. Error Lower Bound Upper Bound
95% Confidence Interval for
Mean
Minimum Maximum
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Results & Homogeneity of Variances
4Look at the p value of Levene’s Test. If p
is not significant then equal variances is
assumed.
Test of Homogeneity of Variances
Birth weight
.757 2 215 .470
Levene
Statistic df1 df2 Sig.
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ANOVA Results
4So the 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.
ANOVA
Birth weight
.153 2 .077 .263 .769
62.550 215 .291
62.703 217
Between Groups
Within Groups
Total
Sum of
Squares df Mean Square F Sig.
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How to present the result?
Type of Work Mean+sd Test p
Office 2.76 + 0.60
ANOVA
F = 0.2630.769Housewife 2.78 + 0.53
Farmer 2.84 + 0.55
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Proportionate Test
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Proportionate Test
4Qualitative data utilises rates, i.e. rate of
anaemia among males & females
4To compare such rates, statistical tests
such as Z-Test and Chi-square can be
used.
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Formula
• where p1 is the rate for
event 1 = a1/n1
• p2 is the rate for event 2
= a2/n2
• a1 and a2 are frequencies
of event 1 and 2
4 We refer to the normal
distribution table to
decide whether to reject
or not the null
hypothesis.
1 2
0 0
1 2
1 1 2 20
1 2
0 0
1 1
1
p pz
p qn n
p n p np
n n
q p
−=
+
+=
+
= −
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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.
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Example
4Comparison of worm infestation rate between male and female medical students in Year 2.
4Rate for males ; p1= 29/96 = 0.302
4Rate for females;p2 =24/104 = 0.231
4H0: There is no difference of worm infestation rate between male and female medical students in Year 2
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Cont.
4 z = 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=0.2542Therefore if z=1.14, p=0.2542. H0 not rejected
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Exercise (try it)
4Comparison of failure rate between
ACMS and UKM medical students in
Year 2 for minitest 1 (MS2 2012).
4Rate for UKM ; p1= 42/196 = 0.214
4Rate for ACMS;p2 = 35/70 = 0.5
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Answer
4P1 = 0.214, p2 = 0.5, p0 = 0.289, q0 = 0.711
4N1 = 196, n2 = 70, Z = 20.470.5 = 4.52
4p < 0.00006