An ANOVA test was conducted and the following results were obtained. Descriptive
ID: 3232444 • Letter: A
Question
An ANOVA test was conducted and the following results were obtained.
Descriptive Statistics
Groups
Count
Sum
Average
Variance
A
244
109831
450.13
16833.16
B
1337
689273
515.54
13858.67
C
1060
641232
604.94
12652.42
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
7164292.02
2
3582146.01
262.46
0.000
Within Groups
36004549.12
2638
13648.43
Total
43168841.14
2640
1. How many groups are in the analysis? How do you know this from degrees of freedom?
2. What was the total sample size used in the analysis? How do you know this from degrees of freedom?
3. What is your decision about the null? How would you still be able to reach this decision if neither the p-value nor the critical value were known?
4. Interpret the results of this ANOVA test (Step 5) in full sentences.
5. What would be the Type I error rate if a series of t-tests were performed instead of the ANOVA test? Show your work.
Groups
Count
Sum
Average
Variance
A
244
109831
450.13
16833.16
B
1337
689273
515.54
13858.67
C
1060
641232
604.94
12652.42
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
7164292.02
2
3582146.01
262.46
0.000
Within Groups
36004549.12
2638
13648.43
Total
43168841.14
2640
Explanation / Answer
1) There are 3 groups in the anova analysis . We know that df is given as n-1 , where n is number of groups
now n-1 = 2 , so n = 3 groups
2) we know that the df within is given as , where k is number of groups which is 3
k(N-1) = 3*(n-1) = 2638 , solve for n
n= 2638/3 +1 = 880 approx
3)
based on the table below
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
7164292.02
2
3582146.01
262.46
0.000
Within Groups
36004549.12
2638
13648.43
Total
43168841.14
2640
we see that the p value is less than 0.05 , hence we reject null hypothesis in favor of alternate hypothesis
4) The hypothesis is
Ho : There is no difference in the mean values across the groups
H1 : There is a signficant difference in the mean values across the groups , atleast for 2 groups .
As we see that the p value is 0.00 , which is less than 0.05 , hence we can reject null hypothesis in favor of alternate hypothesis and conclude that There is a signficant difference in the mean values across the groups , atleast for 2 groups . .
Remember that anova is an omnibus test , it only tells that the group means are different. it doesnt tell which group means are different
Please note that we can answer onyl 4 subparts to a question at a time , as per the answering guidelines.
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
7164292.02
2
3582146.01
262.46
0.000
Within Groups
36004549.12
2638
13648.43
Total
43168841.14
2640
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