Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF
ID: 3352281 • Letter: C
Question
Consider the partial ANOVA table shown below. Let a = .01
Source of Variation
DF
SS
MS
F
Between Treatments
3
180
Within Treatments (Error)
Total
19
380
If all the samples have five observations each:
there are 10 possible pairs of sample means.
the only appropriate comparison test is the Tukey-Kramer.
all of the absolute differences will likely exceed their corresponding critical values.
there is no need for a comparison test – the null hypothesis is not rejected.
2 points
QUESTION 2
Consider the partial ANOVA table shown below. Let a = .01
Source of Variation
DF
SS
MS
F
Between Treatments
3
180
Within Treatments (Error)
Total
19
380
The mean square within treatments (MSE) is
12.5
200
16
60
2 points
QUESTION 3
Consider the partial ANOVA table shown below:
Source of Variation
DF
SS
MS
F
Between Treatments
64
8
Within Treatments (Error)
2
Total
100
The number of degrees of freedom corresponding to within treatments is
20
18
4
5
2 points
QUESTION 4
Consider the following ANOVA table.
Source of Variation
SS
df
MS
F
Treatment
49411.11
2
24705.56
10.4304
Error
35529.17
15
2368.611
Total
84940.28
17
The appropriate null hypothesis is:
mu1 = mu2
mu1 = mu2 = mu3
At least one mean is different.
The populations are independent.
2 points
QUESTION 5
The following information is from a 1-way ANOVA involving three treatments:
Treatment 1
Treatment 2
Treatment 3
Sample Size
5
10
5
Sample Mean
4
8
9
The overall mean for all the treatments is
7.25
4.89
6.67
7.00
Source of Variation
DF
SS
MS
F
Between Treatments
3
180
Within Treatments (Error)
Total
19
380
Explanation / Answer
1)
there is no need for a comparison test – the null hypothesis is not rejected.
2)
mean square within treatments (MSE) is =12.5
3)
number of degrees of freedom corresponding to within treatments is =18
4)
mu1 = mu2 = mu3
5)
overall mean for all the treatments is =(4*5+8*10+9*5)/(5+10+5)=7.25
source df SS MS F Fcrit between 3 180 60 4.8 5.29 within 16 200 12.5 total 19 380Related Questions
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