Question

Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF SS...

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

Homework Answers

Answer #2
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

answered by: anonymous
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