One-Way Analysis of Variance (ANOVA): Between …howardsei.org/uploads/One-way_Between_ANOVA.pdf · Systematic Variance Systematic variance (or between-groups variance) is that part
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Systematic Variance
Systematic variance (or between-groups
variance) is that part of the total variance in
participants’ responses that differs between
the experimental groups.
If the independent variable (IV) has an
effect, then dependent variable (DV) should
show different scores in different
experimental conditions.
Measure: Experimental
Random
Assignment
group A
group B
EQUAL
No GRE
Prep class
GRE Prep
class
Manipulation (IV) Measure (DV)
1000 M+V GRE
1200 M+V GRE
UNEQUAL
Error variance (or within-groups variance)
is the portion of the total variance in
participants’ responses that remains
unaccounted for after systematic variance
due to the independent variable is removed.
total variance = systematic variance + error variance
Error Variance
group A
mean
GRE = 1000
group B
mean
GRE = 1200
total variance = systematic variance + error variance
overall mean
GRE = 1100
. error variance systematic variance
. .
950
Ex: a researcher is interested in comparing behavioral health interventions in the Affordable Care Act in terms of total work days for populations which are high risk:
Level 1: No intervention
Level 2: Mandated Physical & Health Screening
Level 3: Physical, Screening & Health Coach
Why not do a series of 3 individual comparisons (t-test)?
1) μ1 = μ2
2) μ1 = μ3
3) μ2 = μ3
Family-wise error: it increases the likelihood that we will think we have a reliable difference when we do not.
ANOVA makes one simultaneous comparison, keeping the family-wise error rate equal to a
7
Between-Groups Variance: scores differ
because they are from different conditions,
(i.e., systematic variance)
Within-Groups Variance: scores may differ from
each other even when participants are in the
same condition (i.e., error variance).
anceerror vari
variancesystematic
wn
bt
MS
MSF
As variance between groups > variance within groups, F
increases.
MSbt : Difference between means
– The mean square between groups estimates the
variability in scores between levels of the factor
MSwn
– The mean square within groups estimates the
variability in scores within the levels of the factor
– Actually use pooled variance
anceerror vari
variancesystematic
wn
bt
MS
MSF
Hypotheses for ANOVA
H0: μ1 = μ2 = μ3 … μk
Ha: not all means equal in population
Degrees of freedom k levels of factor, N sample size
dfbt = k – 1
dfwn = N – k
dftotal = N – 1
9
Distribution of F When H0 Is True
F-distribution is one-tailed.
Differences in group means in any direction lead to greater F, so the test
is 2-tailed.
Assumptions
ANOVA determines whether significant
differences occur across a categorical
factor with two or more conditions
Assumptions
• Independent random samples
• Interval / ratio level data (i.e. parametric)
• Normally distributed
• Homogeneity of Variance
11
Formulas
13
2)(2 need weso
2)(2......
22
2
2
1
1
total
totaltotaltotal
bttotalwn
wn
wnwn
total
total
k
kbt
bt
btbt
wn
bt
N
XXSS
SSSSSS
df
SSMS
N
X
n
X
n
X
n
XSS
df
SSMS
MS
MSF
Note N = total number in sample, n1 etc is
the number in each group and k
is the number of groups (levels of IV)
Step by step computation of F
Step a: ∑X by group and overall ∑X and ∑X2
Step b: calculate SSbt and then MSbt
Step c: calculate SStotal
Step d: calculate SSwn and then MSwn
Step e: calculate F
Source SS df MS F p
Between SSbt dfbt MSbt F p
Within SSwn dfwn MSwn
Total SStotal dftotal
An example
What impact do preventive care provisions
in the Affordable Care Act have on work
days for patients with behavioral risk
factors?
No Mandated Annual Physical Physical, Screening
Intervention & Health Screening & Health Coach
143 135 224
117 160 220
128 197 214
188 172 222
168
15
Statistical Hypothesis:
Ho: µ1 = µ2 = µ3
Health interventions do not differ on number of work days
Ha: means are not equal
Health interventions differ on number of work days
16
Critical value and decision rule
α = .05
df numerator = dfbt = k-1 = 3-1 = 2
df denominator = dfwn = N-k = 13-3 = 10
If Fobt > 4.1028 we reject Ho and accept Ha and
conclude there is a statistically significant
difference in means
F critical from table:
17
Compute F
Step a: ∑X for each group, overall ∑X and ∑X2
No Mandated Annual Physical Physical, Screening
Intervention & Health Screening & Health Coach
143 135 224
117 160 220
128 197 214
188 172 222
168
∑X1= 576 ∑X2= 832 ∑X3= 880
n1= 4 n2= 5 n3= 4
∑Xtotal = 2,288 (∑Xtotal)2 = 5,234,944 ∑X2
total = 419,964
Notice Difference
Step b: calculate SSbt and then MSbt
20
4.61502
8.12300
2131
8.12300
402688)1936008.13844482944(
13
2)2288(
4
2880
5
2832
4
2576
2)(2......
22
2
2
1
1
bt
btbt
bt
total
total
k
kbt
df
SSMS
kdf
N
X
n
X
n
X
n
XSS
Step c: calculate SStotal
22
17276
13
2)2288( 419964
2)(2
total
totaltotaltotal
N
XXSS
Fill this in as we go
Source SS df MS F p
Between 12300.8 2 6150.4
Within
Total 17276
Step d: calculate SSwn and then MSwn
23
52.49710
2.4975
10313
2.4975
8.1230017276
wn
wnwn
wn
bttotalwn
wn
wnwn
df
SSMS
kNdf
SSSSSS
df
SSMS
Fill this in as we go
24
Source SS df MS F p
Between 12300.8 2 6150.4
Within 4975.2 10 497.52
Total 17276 12
36.1252.497
4.6150
wn
bt
MS
MSF
Step e: calculate F:
Finish Up
25
Source SS df MS F p
Between 12300.8 2 6150.4 12.36 p < .05
Within 4975.2 10 497.52
Total 17276 12
Fobt = 12.36 > Fcrit = 4.1028, we reject Ho and accept Ha
There is a reliable difference between means.
In text: “Health interventions have an impact on the number
of days worked, F(2,10) = 12.36, p < .05.”
Graphing the Effects
50
70
90
110
130
150
170
190
Job Loss Family Loss Survivor
Health Intervention Level and Work Days
No intervention Checkup & Screening + Health Coach
An
nu
al
Da
ys
of
Wo
rk
SPSS
Analyze → general linear model → univariate
Bring DV into dependent variable box
Bring IV into fixed factor box
Options check homogeneity of error test (Levene’s)
and descriptive statistics
Plotspull over Intervene, Click Add, Click continue
OK
SPSS Output for Health Intervention ex:
<- A non-significant
result indicates
Homogenious
error variances
in each group
consistent with
assumptions of F
This line indicates condition df, F, and p
Error df
F(2,10) = 12.36, p = .002
Write out the summary of the finding and the F statistic.
SPSS Output for Drug Treatment Example
Descriptive Statistics
Dependent Variable:Effect
Condition Mean Std. Deviation N
1.00 20.0000 14.76482 4
2.00 20.4000 5.17687 5
3.00 15.5714 3.99404 7
4.00 6.3333 2.87518 6
Total 14.9545 8.72040 22
29
Levene's Test of Equality of Error Variancesa
Dependent Variable:Effect
F df1 df2 Sig.
7.655 3 18 .002
Tests the null hypothesis that the error variance of
the dependent variable is equal across groups.
a. Design: Intercept + Condition
Tests of Between-Subjects Effects
Dependent Variable:Effect
Source
Type III Sum
of Squares df Mean Square F Sig.
Partial Eta
Squared
Noncent.
Parameter
Observed
Powerb
Corrected Model 698.707a 3 232.902 4.667 .014 .438 14.001 .813
Intercept 5110.944 1 5110.944 102.418 .000 .851 102.418 1.000
Condition 698.707 3 232.902 4.667 .014 .438 14.001 .813
Error 898.248 18 49.903
Total 6517.000 22
Corrected Total 1596.955 21
a. R Squared = .438 (Adjusted R Squared = .344)
b. Computed using alpha = .05 Write out the summary of the finding and the F statistic.
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