Description
Write a 2- to 3-paragraph analysis of your one-way ANOVA results for your research question.
Include any post-hoc tests with an analysis of the strength of any relationship found (effect size).
Also, in your analysis, display the data for the output.
Based on your results, provide an explanation of what the implications of social change might be.
Use proper APA format, citations, and referencing for your analysis, research question, and display of output.
I am attaching an example of a perfect paper to help guide you. It is called WK7Assgn1-example-1.docx
I am also attaching the notes from my one-way anova test which will be what you need to write the paragraph. It is called week7.docx
I am also going to attach a small video from the professor with some good notes on how to word. It is called WAL_RSCH8210_07_A_EN-DL.zip
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Explanation & Answer
Hello, please see the following file below.However, I cannot elaborate until you give me the source of data or study of the summary tables that you have.If you will give it to me, I can elaborate more the paper here with 1 - 2 paragraphs.
With the results summarized in the ANOVA table, the p-value was found to be at p = 0.000.
This is shown in Table 1 below where the sum of squares, degrees of freedom and mean squares
are shown.
Table 1. Analysis of Variance Results Summary
ANOVA
Q1. Age
Sum of Squares
df
Mean Square
F
Sig.
14489.903
4
3622.476
17.151
.000
Within Groups
10626193.940
50311
211.210
Total
10640683.840
50315
Between Groups
The results here signifies that the null hypothesis is rejected at the said p-value. The low F
value here also supports the claim that equality of means is rejected. Table 2 shows a test of
homogeneity using Levene test (Levene 1960) where it can be seen that the level of significance
is essentially zero. The null hypothesis here is that variances are equal and the alternative
hypothesis assumes unequal variances.
Table 2. Test of Homogeneity of Variances
Test of
Homogeneity of
Variances
Q1. Age
Levene Statistic
df1
df2
Sig.
7.959
4
50311
0.000
From Table 1, since the p-value is a lot lower than the common threshold of...