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Confidence Intervals
J.D. Bramble, Ph.D.
Creighton University Medical Center
Med 483 -- Fall 2005
Confidence Intervals

Data can be described by point estimates



Point estimates from a sample are not always
equal to population parameters
Data can be described by interval estimates


mean, standard deviation, etc.
shows the variability of the estimate.
Using the standard error we can see the amount
that the estimate will vary from the true value.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Confidence Intervals

Interval estimates are called confidence intervals (CI).

CI define the an upper limit and lower limit associated
with a known probability.

These limits are known as confidence limits.

The associated probability of the CI is most commonly
95%, but may be 99% or 90%
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Confidence Intervals

Confidence limits set the boundaries that
are likely to include the population mean.

Thus, we can conclude that in general, we
are 95% confident that the true mean of the
population is found within these limits.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Standard Error



s
n
The standard error is defined as
We expect that the mean is within one
standard error of m quite often.
SE is a measure of the precision of x as an
estimate of m.


The smaller SE the more precise the estimate
SE includes two factors that affect the
precision of the measurement n and sd
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Standard Deviation vs Standard Error

Standard deviation describes the dispersion
of the data.


The variability from one data point to the next
Standard error (SE) describes the
uncertainty in the mean of the data that is a
result of sampling error.

The variability associated with the sample mean
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Calculating Confidence Intervals

Recall that 95% of the area under a standard
curve is between z = ±1.96.
95.45%
-1.96
1.96
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Calculating Confidence Intervals

 


The general formula is:  x  z
n

 

x  z
=
n 


P
0.95

Lower limit = x - 1.96( n )

Upper limit = x + 1.96 ( n )
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Calculating CI of two samples



We use the t-distribution.
The t distribution describes the distribution
of the sample mean when the variance is
also estimated from sample data.
Thus, the formula for the CI in these cases
is:
s 

*
 x t

n

J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example: Problem

To assess the effectiveness of hormone replacement
therapy on bone mineral density, 94 women
between the age of 45 and 64 were given estrogen
medication. After taking the medication for 36
months the bone mineral density was measured for
each of the women in the study. The average density
was 0.878 g/cm2 with a standard deviation of 0.126
g/cm2. Calculate a 95% CI for the mineral bone
density of this population.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example: SE and t

Recall that SE is:
s
0.126

 0.013
n
94

t(a, df) = t(0.025, 93) = 1.990
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example: Calculations
0.126 

 0.878  1.99

94 

0.878  0.2586
0.852 to 0.904
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example: Conclusion

The 95% confidence limits are:


lower: 0.852 g/cm2; upper: 0.904 g/cm2
We are 95% confident that the average bone
density of all women age 45 to 64 who take
this hormone replacement medication is
between 0.852 g/cm2 and 0.904 g/cm2.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example: Conclusions (cont’d)

For a 95% confidence intervals we believe
that 95% of the samples drawn form the
population would have a mean that fall
within the confidence limits
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Other Confidence Limits



For a 99% or 90% CI the calculations and
interpretations are similar.
What CI is going to give the widest or
narrowest interval?
CI can be established for any parameter

mean, proportion, relative risk, odds ratio, etc.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Using CI to Test Hypotheses





Diastolic blood pressure of 12 people before and
after administration of a new drug.
Paired t-test
Hypotheses: H0: md > 0; Ha: md < 0
xd = -3.1
sd = 4.1
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Using CI to Test Hypotheses
xd  t a
( , n 1)
2
sd
n
4.1
 3.1  (1.795)
 3.1  2.12
12
 5.22  m d  0.98
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Using CI to Test Hypotheses


Conclusion – since zero does not fall within
the interval we can conclude with 95%
certainty that there is a significant decrease
in blood pressure after taking the new drug.
If we did a paired t-test the conclusions
would be the same.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Visual Representation of CI
True Population Mean (m)
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CI for different samples
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67
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11
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: Analysis of Variance
Single Factor
J.D. Bramble, Ph.D.
Creighton University Medical Center
Med 483 -- Fall 2005
Objectives






Know the assumptions for an ANOVA
When is ANOVA used rather than a t-test
Set up ANOVA tables and understand the relationships
between the values within the table
Compute the F-ratio and appropriate degrees of
freedom
Know how and when to use a two factor ANOVA
Apply Tukey’s multiple comparison procedure
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA vs. t-test




A statistical method of comparing means of
different groups.
A single factor ANOVA for two groups produces
the same p-value as an independent t-test
The t-test is inappropriate for more than two
groups – increases probability of a Type I error
Using a t-test to test the means of each pair leads
to problems regarding the proper level of
significance.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA vs. t-test




ANOVA is not limited to two groups
Can appropriately handle comparisons of
several means from several groups
Thus, ANOVA overcomes the difficulty of
doing multiple t-tests
The sampling distribution used is the F
distribution.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: assumptions

The observations are independent

one observation is not correlated with another observation.

Variance of the various groups are homogeneous

ANOVA is a robust test that is not as sensitive to
departures from normality and homogeneity,
especially when sample sizes are large and nearly
equal for each group.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: Characteristics


ANOVA analyzes the variance of the groups to
evaluate differences in the mean.
Within group



Measures the variance of observations within each
group
variance due to “chance”
Between groups


measure the variance between the groups
variance due to treatment or chance
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: Characteristics

It can be shown that when means of each
group are equal, the within and the
between group variance is equal.
treatment  chance
F
chance

The F-statistic is the ratio of the estimated
variance
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA : the F distribution


The ratio follows an F distribution
The F statistic has two sets of degrees of
freedom.


For between groups -- (I - 1); where I is the
number of groups
For within groups -- I(J - 1); where J is the
number of observations in each group
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: single factor




Let I = the number of population samples
Let J = the number of observations in each sample
Thus the data consist of IJ observations
The overall or grand mean is:
I
X 
J
  x jk
I 1 J 1
IJ
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: single factor




Now it is necessary to compute the sums of
squares for the treatment -- SSTr (between group);
error--SSE (within group), and the total-- SST.
Sum of the squared deviations between groups
The total sums of squares measures the amount of
variation about the grand mean
With algebraic manipulation we find that:
SST = SSTr + SSE
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: sums of squared
I
J
1
1 2
2
2
SSTr  ( x IJ  x)    xIJ 
x
J I 1 J 1
IJ
1 2
SST  ( x IJ  x)    x  x
I 1 J 1
IJ
2
I
J
2
IJ
When completing the ANOVA table usually only SSTr and SST are
calculated. SSE is found by SSE = SST - SSTr
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: mean sums of square


After calculating the sums of squares, F is
simple the ratio of the mean squares of both
the treatment and error.
The mean squares is the sums of squares
divided by the appropriate degrees of
freedom.
SSTr
MSTr 
I 1
SSE
MSE 
I ( J  1)
MSTr
F
MSE
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: single factor table
Sources of
variation
Degrees of
freedom
SS
MS
F
MSTr
MSE
Treatment
I-1
SSTr
MSTr
Error
I(J - 1)
SSE
MSE
Total
IJ - 1
SST
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: Example

An experiment was conducted to examine various
modes of medication delivery. A total of 15 subjects
diagnosed with the flu were enrolled and the length
of time until alleviation of major symptoms was
measured for three groups:



Group A received an inhaled version,
Group B received an injection, and
Group C received an oral dose.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Single factor example
Groups
Time (min)
Average
A
B
C
56
62
72
102
58
100
90
78
117
87
68
109
94
87
103
85.8
70.6
100.2
x=
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Single factor example: set up

H0: all three means are equal or m1 = m2 = m3

Ha: at least one mean is different

a = 0.05

Critical value: F(a, df) given I-1 = 2 and
I(J-1) = 12, F(0.05, 2,12) = 3.89
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Single factor example:
calculating sums of squares
SST  114,893  (1283) / 15  114,893  109,739.3  5,153.7
2
1
1
SSTr  [( 429) 2  (353) 2  (501) 2 ]  12832  2,962.8
5
15
SSE  5,153.3  2,962.8  2,190.9
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Single factor example:
completing the table
Sources of
variation
Degrees of
freedom
SS
MS
F
4.44
Treatments
2
2,190.3
1,095.5
Error
12
2,962.8
246.9
Total
14
5,153.7
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Single factor example:
decision and conclusions

Compare Fstat to Fcrit: 4.45 > 3.89, therefore
fail to reject H0.

There evidence suggest that the time it takes
to alleviate major flu symptoms differed
significantly due to the mode of medication
delivery.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Where is the Difference?

Recall the hypotheses of the ANOVA




If we fail to reject Ho the analysis is complete.
What does it mean when Ho is rejected


Ho is that all the means are equal
Ha is that at least on is not.
at least one mean is different
Which m's are different from one another.


if only two treatment levels.
three or more treatment levels
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Finding the difference

We must do a post hoc analysis.




a test that is done after the ANOVA
The purpose is to determine the location of
the difference.
Different of post hoc test are available and
are discussed in the text.
These test include Bonferroni, Sceffe,
Student Newman-Keuls, and Tukey' HSD.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Tukey’s HSD
MSE
w  Qa , I , I ( J 1)) *
J

Where,
a = significance level
I = number of groups
J = number of observation per treatment
MSE = mean square error (or within group MS)
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Using the Tukey’s HSD





All the information, except Q, needed to find w is
located in the ANOVA table
Q is determined by using the studentized range
distribution, a, I and dfwithin.
Once w is determined order all treatment level
means in ascending order
Underline those values that differ by less than w.
Treatment means not underlined correspond to
treatments that are significantly different.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example


Using the previous example, we no want to find
which form(s) of medication really is different
form the others.
To start we will order the means
Groups
Average
B
70.6
A
85.8
C
100.2
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
The ANOVA

Does this data indicate that the amount of time it takes a
student to nod off is dependent on the statistical topic
being studied?
Source
Treatment (i.e., between)
Error (i.e., within)
Total
df
SS
MS
F
3
5882.4
1960.8
21.09
16
19
1487.4
7369.8
93
Since the computed F-statistic of 21.09 is greater than the
critical value of F(0.05, 3, 16) = 3.24 we reject Ho.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Computing the Tukey’s


There are I = 3 treatments and the degrees of
freedom for the error is 12; thus, from the
table Q(0.05, 3, 12) = 3.77.
Computing the Tukey value we get:
MSE
246.9
Q(a , I , I ( J 1) *
 3.77 *
 26.5
J
5
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
And the Difference is…

Ordering the treatment level means and
underscoring those that differ by less that 26.5
Groups
Average

B
70.6
A
85.8
C
100.2
We conclude that only significant difference is
between group B (injection) and group C (oral).
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANOVA: Analysis of Variance
Two Factor
J.D. Bramble, Ph.D.
Creighton University Medical Center
Med 483 -- Fall 2005
Two-Factor ANOVA

Single factor ANOVAs


Two factor ANOVAs


subjects or treatments are categorized in only
one way (i.e., type of treatment)
subjects or treatments are categorized in two
ways (i.e., type of treatment and gender)
Two factor ANOVAs test the influence of
both factors.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Examples of Two Factor Designs

An experiment is designed to test if there is a
difference in how fast 3 different antacid brands (Acid
Eater, Relieve the Burn, and Blah Stomach) dissolve in
male and female stomachs.

What type of study techniques (1 hour every day, 3
hours once a week, an "all-nighter" prior to the exam, a
late night party before the exam) results in better test
scores while controlling for the person's age (<17, 1820, 21-23, 24-26, 27 >).
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Advantages of Two-way ANOVAs



Economy
In a two-factor analysis we can test
interactions.
Testing for an interaction allows us to
determine whether the variation of the
treatment varies by the conditions in which
the treatment is applied
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Example
Instruction
Ability
Computer Classroom
Whiz
90
82
Novice
Means
80
88
85
85
Means
86
84
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Three Research Hypotheses

Is there a significant difference between those
taught by computer and those taught in the
classroom?

Is there a significant difference between computer
whizzes and computer novices

Is there a significant interaction between type of
instruction and computer ability of the subject
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Two Factor ANOVA Example

Researchers are interested on the effect of caffeine and
performance. Controlling for the students academic
program, subjects were given 3 different levels of
caffeine for two weeks prior to taking a standard
aptitude test. They record the test scores below.
Under grad
Med
Pharm
Nur
Law
None
76
67
81
56
51
Low
82
69
96
59
70
Med
68
59
67
54
42
High
63
56
64
58
37
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Writing the hypothesis


Hypotheses are written the same as a single factor
ANOVA with the exception of adding a second set of
hypotheses for the second factor.
For Factor A (Caffeine level) (I = # treatment levels)


Ho: mnone = mlow = mmed = mhigh
Ha: at least on m is different
For Factor B (Program) (J = # treatment levels)

Ho: munder grad = mmed = mpharm = mnur = mlaw
Ha: at least one m is different
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Critical Values


Critical values for a two factor ANOVA are found
by looking on an F-table at the appropriate degrees
of freedom.
Degrees of freedom for a two factor ANOVA are
found for all sources of variation and the total




For factor A: I-1
For factor B: J-1
For error: (I-1)(J-1)
For total: I(J-1)
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Calculating Degrees of Freedom

For our example I = 4 and J = 5; thus, the df are:





dftask = 2-1 = 1
dfdose = 3 –1 = 2
dferror = 3 * 4 = 12
dftotal = (4 *5)-1 = 19
Notice the relationship between the degrees of
freedom is the same as a single factor ANOVA
dfFactor 1 + dfFactor 2 + dferror = dftotal;
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Calculating Critical Values

With the df known we can now find the critical values


The critical values are found by looking on an F-table at
the appropriate alpha and degrees of freedom for each
factor.



one for Factor 1and one for Factor 2.
For factor A:F(a, dfFactor A, dferror)
For Factor B: F(a, dfFactor B, dferror).
For our example


Factor 1 is F(0.05, 3, 12) = 3.49
Factor 2 is F(0.05, 4, 12) = 3.26
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Computing the Test Statistic

First compute the sums of squares for the different
sources of variation.





SST -- sums of square for the total
SSA -- sums of square for factor A
SSB -- sums of square for factor B
SSE -- sums of square for the error
The relationship still holds that if you add all the sums
of squares you get the sums of squares of the total.
Thus, SST = SSA + SSB + SSE
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Computing the Test Statistic

The mean sums of squares for Factor 1,
Factor 2, and the Error can be computed by
dividing the sums of squares by the
appropriate degrees of freedom.

The F-statistic is calculated by dividing the
mean sums of square for each factor by the
mean sums of square of the error.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
The ANOVA Table
Sources of
Variation
df
Factor 1
I-1
Factor 2
J-1
Error
(I-1)(J-1)
Total
IJ-1
Sources of
Variation
Month
Lot
Error
Total
SS
SSA
SSB
SSE
SST
MS
MSA
MSB
MSE
F
MSA/MSE
MSB/MSE
df
SS
MS
F
3
4
12
19
1182.95
1947.5
441.3
3571.75
394.32
486.88
36.78
10.72
13.24
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Decision and Conclusion

The relationships for making the decision is the
same.

For factor A, F(0.05, 3, 12) = 3.49 and Fstat = 10.72.
Since Fstat > Fcrit we reject Ho

For factor B F(0.05, 4, 12) = 3.26 and Fstat = 13.24.
Since Fstat > Fcrit we again reject Ho
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Where is the difference?

Looking on the table to get Q



For factor 1: Q(a, I,( I-1)(J-1)) . (Notice that I is the # of
levels for Factor A and (I-1)(J-1) is the df of the error)
Thus, Q for Factor B: Q(a, J, (I-1)(J-1)).
The formula for w is

For factor A:

For factor B
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Computing the Tukey’s

For factor A:

Ordering the means and underscoring all the pairs
that differ by less than w = 11.39
High
55.6

Med
58
None
66.2
Low
75.2
There is a significant difference in test scores
between the high and medium caffeine groups and
the none.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Two Factor ANOVA:Repeated Measures

Measurements are made repeatedly on each subject
(before, during, and after the intervention)



Subjects are recruited as matched sets on variables such
as age or diagnosis
A laboratory is experiment is run several times, each
time with several parallel treatments.
When appropriate, the use of the repeated
measures ANOVA test is usually more powerful
than ordinary ANOVA.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Two Factor ANOVA Example

How does various types of music affect
agitation in Alzheimer’s patients?
Group
Piano
Mozart
Easy Listening
Early
21
24
22
18
20
9
12
10
5
9
29
26
30
24
26
Middle
22
20
25
18
20
14
18
11
9
13
15
18
20
13
19
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Writing the hypothesis

For Factor A (Music)


For Factor B (Stage)


Ho: mpiano = mmozart = measy listening
Ha: at least on m is different
Ho: mearly = mmiddle
Ha: at least one m is different
For the Interaction

Ho: no interaction between music and stage on agitation level
Ha: there is an interaction
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Two Factor Repeated Measures ANOVA


Compute SS for the total (SST), error (SSE), factor
A(SSA), factor B (SSB), and the interaction of AB
(SSAB). SST = SSA + SSB + SSAB + SSE
Each SS has associated degrees of freedom





SST = IJK
SSE = IJ(K - 1)
SSA = I-1
SSB = J - 1
SSAB = (I - 1)(J - 1)
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Two Factor Repeated Measures ANOVA


MS are computed by the appropriate SS/df
The test statistic is arrived at by the
appropriate MS divided be MSE.
Hypotheses
H01 vs. Ha1
H02 vs. Ha2
H012 vs. Ha12
Test Statistic
MSA / MSE
MSB / MSE
MSAB / MSE
Critical Value
Fa, I-1, IJ(K - 1)
Fa, J-1, IJ(K - 1)
Fa, (I-1)(J-1), IJ(K - 1)
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Repeated Measures ANOVA Table
Source
df
SS
MS
F
Factor 1
I-1
SS1
SS1 / df1
MS1/MSE
Factor 2
J-1
SS2
SS2 / df2
MS2/MSE
(I-1)(J-1)
SS1x2
IJ(K-1)
SSE
IJK-1
SST
Interaction
Within (Error)
Total
SS1x2 / df1x2 MS1x2/MSE
SSE / dfE
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
The ANOVA Table
Source
df
SS
MS
F
Music
2
740
370
48.89
Stage
1
30
30
4.05
Music x Stage
2
260
130
17.53
Error
24
178
7.42
Total
29
1208
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Repeated Measures: Tukey’s


When no significant interaction is found
For comparing levels of factor A, obtain Qa, I,
IJ(K - 1)

For comparing levels of factor B, obtain Qa,
J, IJ(K - 1)



w = Q * MSE/JK for factor 1 comparisons
w = Q * MSE/IK for factor 2 comparisons
Arrange sample means in increasing order and
underscore pairs of differing by less than w
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Multivariate Analysis of Variance




Referred to as MANOVA
Used when there is multiple dependent
variables
Dependent variables are usually releted to
one another
MANOVA helps to determine the effect of
the treatment (IV) on any one outcome (DV)
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
MANOVA Example

Does sex, race, and educational level affect how well
people deal with the pressure of a terminal disease?

IV = sex (2), race (4), education (4)

DV = Coping strategies (5)

MANOVA can estimate the effects of the IV (sex,
race, and education) for each of the five scales of
coping strategies, independent of one another.
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
Analysis of Covariance




Referred to as ANCOVA
Allows researchers to adjust or equalize
baseline differences between groups
In addition to the DV and IV a covariate is
enter into the model.
The covariate is a variable that is known to
have an effect on the DV
J.D. Bramble, Ph.D.
MED 483 – Fall 2005
ANCOVA example

Wood et al. (2002) tested an educational intervention
to promote breast self examination (BSE).

Quasi experimental design

Difference in knowledge and skill related to BSE.

Enter these covariates into the model is essential to
determine if the difference is the intervention or the
initial difference
J.D. Bramble, Ph.D.
MED 483 – Fall 2005