SPSS Module 6 Assignment-ANOVA General Instructions: In this assignment, we will
ID: 3328655 • Letter: S
Question
SPSS Module 6 Assignment-ANOVA
General Instructions: In this assignment, we will be examining 2 One-Way ANOVAs and interpreting the results.
ANOVA
The researchers want to compare the mean rating of Emotional Support of the groups of where students currently live (1=Dorm, 2=Apartment or House with Peers or Alone, 3=Parents). Also, researchers think that the family composition of the house people grow up in will influence the amount of loyalty females will reported with their closest female friends. As we learned from the text and the PPTs, we know these research situations require the use of the ANOVA. Here are the researchers’ hypotheses: 1. The researchers think that students who live at home will report different levels of emotional support from their closest female friends than those students who live either in the dorms or in an apartment or house with peers or alone. 2. The researchers also believe that females will report differing levels of loyalty with their closest female friends based on the family composition of the house in which they grew up.
Use the SPSS data to and make conclusions.
Part 1—The Analyses
1.For this assignment, we will need to run 2 One Way ANOVA Tests: one comparing levels of where students currently live and one comparing the family composition of the house in which they grew up. Unlike the last assignment, our groups are already formed.
2.We also need to run the Tukey multiple comparison procedures.
3.Additionally, we need to create a graph for each hypothesis.
Hypotheses:
1.The researchers think that students who live at home will report different levels of emotional support from their closest female friends than those students who live either in the dorms or in an apartment or house with peers or alone.
2.The researchers also believe that females will report differing levels of loyalty with their closest female friends based on the family composition of the house in which they grew up.
Hypothesis One Output
Descriptives
EmoSupport
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Min
Max
Lower Bound
Upper Bound
Dorm
34
57.44
7.166
1.229
54.94
59.94
37
64
Apartment or House with Peers or Self
34
56.09
7.012
1.203
53.64
58.53
40
64
Parents
32
55.75
10.479
1.852
51.97
59.53
9
64
Total
100
56.44
8.272
.827
54.80
58.08
9
64
Test of Homogeneity of Variances
EmoSupport
Levene Statistic
df1
df2
Sig.
.191
2
97
.827
ANOVA
EmoSupport
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
53.522
2
26.761
.386
.681
Within Groups
6721.118
97
69.290
Total
6774.640
99
Multiple Comparisons
EmoSupport
Tukey HSD
(I) Where do you live
(J) Where do you live
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Dorm
Apartment or House with Peers or Self
1.353
2.019
.781
-3.45
6.16
Parents
1.691
2.050
.689
-3.19
6.57
Apartment or House with Peers or Self
Dorm
-1.353
2.019
.781
-6.16
3.45
Parents
.338
2.050
.985
-4.54
5.22
Parents
Dorm
-1.691
2.050
.689
-6.57
3.19
Apartment or House with Peers or Self
-.338
2.050
.985
-5.22
4.54
Hypothesis Two Output:
Descriptives
Loyalty
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Min
Max
Lower Bound
Upper Bound
single parent
32
60.34
5.637
.997
58.31
62.38
48
64
two parent no step-parents
42
61.81
3.583
.553
60.69
62.93
50
64
two parent with step-parent
26
58.77
8.392
1.646
55.38
62.16
28
64
Total
100
60.55
5.880
.588
59.38
61.72
28
64
Test of Homogeneity of Variances
Loyalty
Levene Statistic
df1
df2
Sig.
5.933
2
97
.004
ANOVA
Loyalty
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
150.440
2
75.220
2.230
.113
Within Groups
3272.310
97
33.735
Total
3422.750
99
Multiple Comparisons
Loyalty
Tukey HSD
(I) Household you grew up in
(J) Household you grew up in
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
single parent
two parent no step-parents
-1.466
1.363
.532
-4.71
1.78
two parent with step-parent
1.575
1.534
.562
-2.08
5.22
two parent no step-parents
single parent
1.466
1.363
.532
-1.78
4.71
two parent with step-parent
3.040
1.449
.096
-.41
6.49
two parent with step-parent
single parent
-1.575
1.534
.562
-5.22
2.08
two parent no step-parents
-3.040
1.449
.096
-6.49
.41
Part 2—The APA Write-Up Instructions
As with previous assignments, the American Psychological Association (APA) has standards for how statistical results should be presented. While the actual word choice varies, there are several essential components that are common among all presentations of statistical results and interpretations. Use the following instructions for ALL APA write-ups required for this course:
For each analysis, use the following steps to write a complete description of results in proper APA format.
1.State what hypothesis was tested.
2.State what test was used.
3.What decision did you make? Reject the null or retain (fail to reject) the null.
4.Were the groups significantly different from each other?
5.Report the means and standard deviations for each group.
6.Put numbers in APA format:
a.General Format: symbol for the test (df)= obtained value, p> or < significance level
b.Specific for F Tests: F(2, 99)=3.98, p<.05
i.Now you need to discuss which of the mean comparisons done by the Tukey were significant. You do not need to report numbers just which groups were different from each other, if any.
7.Report Effect Size (if known)
Example:
Freshmen and Sophmores were expected to score significantly higher on a measure of depression than were Juniors and Seniors. A One-Way ANOVA was used to test the hypothesis leading to the rejection of the null hypothesis, indicating a significant difference between the means, F(3, 96)=4.61, p<.05. Tukey post-hoc tests revealed that Freshmen (M=9.65, SD=1.59) were significantly more depressed than were Juniors (M=4.59, SD=.98) and Seniors (M=5.68, SD=1.22) but not significantly different from Sophomores (M=8.95, SD=1.43). Also, Sophomores were significantly more depressed than Juniors and Seniors but not Freshman.
Since we ran 2 analyses, you will do 2 write-ups for this assignment.
Hypothesis One: The researchers think that students who live at home will report different levels of emotional support from their closest female friends than those students who live either in the dorms or in an apartment or house with peers or alone.
1.Answer the following questions
a.What is the null hypothesis tested?
b.Can we assume equal variances? (Was the Levene’s Test for Equality of Variance significant or not significant?)
c.Write up the results for the ANOVA and the Tukey procedure for Hypothesis One below:
Hypothesis Two: The researchers also believe that females will report differing levels of loyalty with their closest female friends based on the family composition of the house in which they grew up.
2.Answer the following questions
a.What is the null hypothesis tested?
b.Can we assume equal variances? (Was the Levene’s Test for Equality of Variance significant or not significant?)
c.Write up the results for the ANOVA and the Tukey procedure for Hypothesis One below:
Descriptives
EmoSupport
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Min
Max
Lower Bound
Upper Bound
Dorm
34
57.44
7.166
1.229
54.94
59.94
37
64
Apartment or House with Peers or Self
34
56.09
7.012
1.203
53.64
58.53
40
64
Parents
32
55.75
10.479
1.852
51.97
59.53
9
64
Total
100
56.44
8.272
.827
54.80
58.08
9
64
57.5 O 57 E 56.5 E 56 55.5 Parents Apartment or House with Peers or Self Dorm Where do you liveExplanation / Answer
1.
a. Null Hypothesis H0: The level of emotional support from their closest female friends for students who live either in the home, dorms or in an apartment or house with peers or alone are same.
b.
Yes, we can assume equal variances, as the p-value of Levene's test for equality of variances is greater than 0.05 and we fail to reject the null hypothesis and is significant.
c.
The students who live at home are expected to have different level of emotional support from their closest female friends than those students who live either in the dorms or in an apartment or house with peers or alone.
A one-way ANOVA was used to test the hypothesis leading to the acceptance of null hypothesis, indicating no difference between the means, F(2,97)=0.386, p > 0.05. Tukey post-hoc tests revealed that for all house types, the confidence interval contains 0, so emotional support from their closest female friends has the same levels for students living in different home types.
2.
a. Null Hypothesis H0: The levels of loyalty of females with their closest female friends based on the family composition of the house in which they grew up are same.
b.
No, we cannot assume equal variances, as the p-value of Levene's test for equality of variances is less than 0.05 and we reject the null hypothesis and is not significant.
c.
The females are expected to have different levels of loyalty with their closest female friends based on the family composition of the house in which they grew up. A one-way ANOVA was used to test the hypothesis leading to the acceptance of null hypothesis, indicating no difference between the means, F(2,97)=2.230, p > 0.05. Tukey post-hoc tests revealed that for all house types, the confidence interval contains 0, so levels of loyalty of females with their closest female friends based on the family composition of the house in which they grew up are same.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.