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 | df | SS | MS | F | Fcrit |
between | 3 | 180 | 60 | 4.8 | 5.29 |
within | 16 | 200 | 12.5 | ||
total | 19 | 380 |
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)
source | df | SS | MS | F | Fcrit |
between | 4 | 64 | 16 | 8 | 4.58 |
within | 18 | 36 | 2 | ||
total | 22 | 100 |
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
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