# Levene’s Test for Homogeneity Homework Help

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## 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 (p-value = 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 F-statistic being 2.135 (p-value = 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 F-stat. = 26.785 (p-value <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 F-stat. = 0.839 (p-value = 0.366) indicating that film and gender do not have significant interaction.

## Mean’s Plot

Q4 (1 pt). Copy-n-paste 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 P-value 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 4-pint 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 2-pints and 4-pints 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 independent-samples design or repeated-measures 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). Copy-n-paste 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? Copy-n-paste 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). Copy-n-paste 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 Within-Subjects Effects, choose one F-statistic to use based on Mauchly’s test and Epsilon values. Which F-statistic did you choose, Sphericity assumed, Greenhouse-Geisser, Huynh-Feldt, or L-B? Why? Report the F-statistic, p-value, and effect size in APA format for the Repeated-Measures ANOVA omnibus test. What is your conclusion?

We chose the Greenhouse-Geisser F-statistic because the sphericity assumption was violated. The F-stat is 3.794 (p-value =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 Greenhouse-Geisser F-statistic were calculated.

We multiply the degrees of freedom by the Greenhouse-Geisser epsilon estimated from Mauchly’s test. Hence, DOF for Greenhouse-Geisser is

Q7 (2 pts). Copy-n-paste 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 (I-J) Std. Error Sig.b 95% Confidence Interval for Differenceb
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 2-3 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 F-stat = 7.742 (sig <0.001) and the same for the Song Type is F-stat = 20.874 (sig =0.002). The interaction effect was also found out to be statistically significant with F-stat = 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.