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Levene’s Test for Homogeneity Homework Help
Homogeinity of Variances
Q1 (1 pt). Is homogeneity of variances assumption met? Use Levene’s test and report the results.
yes. The Levene’s test for homogeneity of variance is not statistically significant (pvalue = 0.456). Hence, the assumption is met.
Q2 (1 pt). Based on the means plot, would you expect a significant interaction effect between gender and film?
Yes. We see expect a significant difference across films but not gender since we see large difference between in scores for ends of curves but not between the two lines. The difference is lower across gender as compared to movies across gender.
Q3 (1 pt). Report the significance test for each main effect and the interaction effect. (Follow APA format to write up your result.)
The main effect of the gender is statistically insignificant with Fstatistic being 2.135 (pvalue = 0.153) indicating that the difference in arousal is not statistically significant across the two genders.
However, the main affect of film is statistically significant with Fstat. = 26.785 (pvalue <0.001) indicating that there is large difference in extent of arousal between the two movies.
The interaction between film and gender is also statistically insignificant with Fstat. = 0.839 (pvalue = 0.366) indicating that film and gender do not have significant interaction.
Mean’s Plot
Q4 (1 pt). Copynpaste the means plot. Report only the significant main effect.
Q5 (1 pt). Do we need to perform post hoc comparisons?
No. we do not need to perform post hoc analysis as there were only two classes and only one pair of comparison is possible.
Q6 (2 pts). Report the significance test for each main effect and the interaction effect. (Follow APA format to write up your result.)
Main effect 
Pvalue 
Alcohol 
0.000 
Gender 
0.161 
Q7 (2 pts). Report significant simple effects (refer to univariate tests).
The 4 pints contrast has significant effect with significance <0.001.
Q8 (1 pt). Based on pairwise comparisons for simple effect analysis, which pairs have significantly different mean values?
The pair which has significant difference is 4pint alcohol groups with respect to gender as well as other alchohol group.
Bonus Question (1 pt): for the pairwise comparisons in Q8, use 95% confidence interval for difference to tell the significance of mean differences between 2 pints and 4 pints for both male and female.
The difference between 2pints and 4pints are significant for males but not for females as the intervals do not overlap for males but they overlap for females.
Independent–samples Design and Measures Design
Q1 (2 pts). a. How many subjects/participants in this study?
Total of 8 subjects are in this study.
Is this an independentsamples design or repeatedmeasures design?
This is repeated measures design not the independent samples design.
Independent Variable (IV) = Animal Type
Dependent Variable (DV) = Time taken to retch
What is the total sample size? Show your calculation.
Descriptive Statistics
Q2 (1 pt). Copynpaste the descriptive statistics table. Which animal induced retching the quickest?
Descriptive Statistics 

Mean 
Std. Deviation 
N 
Stick Insect 
8.13 
2.232 
8 
Kangaroo Testicle 
4.25 
1.832 
8 
Fish Eyeball 
4.13 
2.748 
8 
Witchetty Grub 
5.75 
2.915 
8 
Fish Eyeball induced retching quickest.
Q3 (1 pt). Is the assumption of sphericity violated? How do you know? Copynpaste the corresponding SPSS output table.
Yes. The assumption of sphericity is violated. We know this from Mauchly’s test of Sphericity which yields significance of 0.047 (<0.05).
Q4 (1 pt). Copynpaste the estimated marginal means plot of Time against Animals. Based on the plot, does the time to retch seem to differ across different animals?
Q5 (2 pts). Refer to the output table – Tests of WithinSubjects Effects, choose one Fstatistic to use based on Mauchly’s test and Epsilon values. Which Fstatistic did you choose, Sphericity assumed, GreenhouseGeisser, HuynhFeldt, or LB? Why? Report the Fstatistic, pvalue, and effect size in APA format for the RepeatedMeasures ANOVA omnibus test. What is your conclusion?
We chose the GreenhouseGeisser Fstatistic because the sphericity assumption was violated. The Fstat is 3.794 (pvalue =0.063) which means we do not observe significant differences between the mean time within subjects.
Q6 (1 pt). Show how the degrees of freedom for the GreenhouseGeisser Fstatistic were calculated.
We multiply the degrees of freedom by the GreenhouseGeisser epsilon estimated from Mauchly’s test. Hence, DOF for GreenhouseGeisser is
Q7 (2 pts). Copynpaste Pairwise Comparison table. Based on the results, which pairs of animals do the time to retch significantly differ? Interpret the confidence interval provided for these comparisons.
Pairwise Comparisons 
Measure: Time 
(I) Animal 
(J) Animal 
Mean Difference (IJ) 
Std. Error 
Sig.^{b} 
95% Confidence Interval for Difference^{b} 
Lower Bound 
Upper Bound 
1 
2 
3.875^{*} 
.811 
.002 
1.956 
5.794 
3 
4.000^{*} 
.732 
.001 
2.269 
5.731 
4 
2.375 
1.792 
.227 
1.863 
6.613 
2 
1 
3.875^{*} 
.811 
.002 
5.794 
1.956 
3 
.125 
1.202 
.920 
2.717 
2.967 
4 
1.500 
1.336 
.299 
4.660 
1.660 
3 
1 
4.000^{*} 
.732 
.001 
5.731 
2.269 
2 
.125 
1.202 
.920 
2.967 
2.717 
4 
1.625 
1.822 
.402 
5.933 
2.683 
4 
1 
2.375 
1.792 
.227 
6.613 
1.863 
2 
1.500 
1.336 
.299 
1.660 
4.660 
3 
1.625 
1.822 
.402 
2.683 
5.933 
Based on estimated marginal means 
*. The mean difference is significant at the .05 level. 
b. Adjustment for multiple comparisons: Least Significant Difference (equivalent to no adjustments). 
The confidence interval for these pairwise comparison shows the 95% CI of difference in time to retch due to different animals. If the interval contains 0 that means, there was no significant difference between the effect of these two animals at 5% level of significance. If 0 is not contained in the interval, we conclude that the difference is significant at 5% level of significance.
We have pairs (1,2), (1,3), pairs turned out to be significant.
Q 8 The author of our textbook, Andy, would like to empirically research whether he wrote better songs than his old band mate Malcolm, and whether this depended on the type of songs (a symphony or song about flies). The outcome variable is the number of screams elicited by audience during listening to the songs.
a. (12 pts) Download data Escape From Inside.sav, and open in SPSS. Fill in the shaded cells in Table 1.
Table 1. Summary of Variables
Variables 
Labels 
Measures 
Numeric Values 
Alphabet Values 
Song_Type 
Type of Song 
Nominal 
0 
Symphony 
1 
Fly Song 
Songwriter 
Songwriter 
Nominal 
0 
Malcolm 
1 
Andy 
Screams 
Number of Screams Elicited by the Song 
Scale 
Integer 
 
b. (8 pts) Report sample size for each group and total sample size. Fill in the shaded cells in Table 2.
Table 2. Summary of Cases in Each Condition


Songwriter 

Sample size (N) 

Malcolm 
Andy 
Sum 
Song_Type 
Symphony 
17 
17 
34 
Fly Song 
17 
17 
34 

Sum 
34 
34 
68 
c. (4 pts) How many factors in this experiment? This is a _2___by__2__ design (fill in levels of each factor). Is this design balanced?
Yes. This design is balanced.
d. (16 pts) Obtain descriptive statistics for dependent variable within each experimental group.
Table 3. Summary of Descriptive Statistics for Screams by Groups


Mean 
Std. Deviation 
Skewness 
Kurtosis 
Song_Type 
Symphony 
8.29 
2.125 
0.193 
0.160 

Fly Song 
6.21 
1.997 
0.572 
0.730 
Songwriter 
Malcolm 
6.53 
1.911 
0.257 
0.599 

Andy 
7.97 
2.455 
0.062 
0.572 
Note: Std. Deviation – standard deviation of a group.
e. (20 pts) I plotted the mean and its 95% confidence interval (CI) of screams by songwriter and song type as shown in Figure 1. Describe this figure by visually comparing the means. (List 23 key observations).
Figure 1. Error Bar Graph for Screams by Groups
Note: circle indicates the mean number of screams in each condition, and the bar shows the 95% confidence interval of the mean.
The plot shows that the mean difference of screams between the two songwriters are significantly different for the song Symphony but not for the Fly Song at 5% level of significance. This is because the 95% confidence interval for ANDY in song Symphony does not overlap with any other confidence interval. The difference in screams for two songs is also statistically significant at 5% level of significance.
There does seems to be significant interaction between song type and signers as we do not see similar trend in the two types of songs from two different songwriters
f. (20 pts) Use Univariate function under General Linear Model in SPSS to conduct factorial ANOVA.
i. (5 pts) What is the null hypothesis for assumption of homogeneity of variance? What does the result of Levene’s test indicate?
The null hypothesis for homogeneity of variance is that the variance in number of screamers is same for each main effect group.
ii. (5 pts) What are the null hypotheses being tested by ANOVA model?
The null hypothesis tested by ANOVA for main effects are:
iii. (5 pts) Fill in the necessary cells in the ANOVA summary table (Table 4).
Table 4. ANOVA Summary Table
Source 
SS 
df 
MS 
F 
p 
Partial 
Song Type 
74.132 
1 
74.132 
20.874 
0.000 
0.246 
Songwriter 
35.309 
1 
35.309 
9.942 
0.002 
0.134 
Song Type*Songwriter 
18.015 
1 
18.015 
5.072 
0.028 
0.073 
Residual/Error 
227.294 
64 
3.551 



Total 
3929 
68 




iv. (5 pts) Write up the results based on APA format.
From the ANOVA table, we find out that both the main effects as well as interaction effects are statistically significant at 5% level of significance. The main effect of Songwriter has Fstat = 7.742 (sig <0.001) and the same for the Song Type is Fstat = 20.874 (sig =0.002). The interaction effect was also found out to be statistically significant with Fstat = 5.072 (sig = 0.028).
g. (10 pts) Conduct simple effect analysis if necessary. Report and interpret the results of simple effect analysis. Please also include your SPSS syntax here.
The significant simple effect for the song type is 2.088 and same for the song writer is 1.442
Estimates 
Dependent Variable: Number of Screams Elicited by the Song 
Type of Song 
Mean 
Std. Error 
95% Confidence Interval 
Lower Bound 
Upper Bound 
Symphony 
8.294 
.333 
7.629 
8.959 
Fly Song 
6.206 
.333 
5.541 
6.871 
Estimates 
Dependent Variable: Number of Screams Elicited by the Song 
Songwriter 
Mean 
Std. Error 
95% Confidence Interval 
Lower Bound 
Upper Bound 
Malcolm 
6.529 
.333 
5.864 
7.195 
Andy 
7.971 
.333 
7.305 
8.636 
h. (10 pts) Do we need to perform post hoc comparisons for main effects? Why or why not?
There is no need to perform the post hoc analysis as there are only 2 levels for each main effect and hence, only 1 pair of effects to compare which is trivial.