Group 5: Apple, Billy, Erin, Lulu, Sally
2012 / 3 / 28
Homogeneity of Variance
Q1: FlegeYKLiu.sav file
Answer 1: The boxes have different lengths. Therefore, we can know
that the assumption of the equality of variances is violated.
Group 5: Apple, Billy, Erin, Lulu, Sally
2012 / 3 / 28
Q2: DeKeyser2000.sav file
Answer 2: The boxes have different lengths. Therefore, we can know
that the assumption of the equality of variances is violated.
Group 5: Apple, Billy, Erin, Lulu, Sally
2012 / 3 / 28
Q3: Lyster.Written.sav file
(1) Numerical Output:
a. Data→ split file→ “con”
b. Analyze → descriptive statistics → CompGain2
Descriptive Statistics
Cond
FFIrecast
FFIprompt
FFIonly
Comparison
N
Mean
Std. Deviation
Statistic
Statistic
Statistic
CompGain2
38
Valid N (listwise)
38
CompGain2
49
Valid N (listwise)
49
CompGain2
42
Valid N (listwise)
42
CompGain2
51
Valid N (listwise)
51
Skewness
Statistic
Std. Error
4.2105
5.18917
.531
.383
8.7755
6.78314
-.424
.340
2.1905
4.76900
.855
.365
.4118
4.64834
-.077
.333
Answer:
Looking at SD of each group. Their SD are ranged from 4.64~ 6.78. It shows
that there is no significant difference.
Group 5: Apple, Billy, Erin, Lulu, Sally
2012 / 3 / 28
(2) Box-plot
Graphs → Legacy Dialogs→ Bosplot → Variable (CompGains2) &
Category Axis (cond)
Answer:
(1) Although the means of the second group (FFlprompt) is slightly higher than
that of the others, the lengths of the boxes are not substantially different
among the four groups. Therefore, variances are considered to be
homogeneous; that is, variances are quite equal.
(2) It is noteworthy that there are several outliers: No.37 (18), No.41 (-5),
No. 67 (-8), and No.105 (15). These cases might be excluded in the further
statistics process.
Group 5: Apple, Billy, Erin, Lulu, Sally
2012 / 3 / 28
(3) Levene’s test:
Analyze → compare means → One-way ANOVA → dependent list
(CompGains2) & factor (cond) → “Options” tick “Homogeneity of
Variances”
Test of Homogeneity of Variances
CompGain2
Levene Statistic
2.346
df1
df2
3
Sig.
176
.075
Ans:
As can be seen, p > .05 (0.075). As a result, the variances of the groups are equal.