ANALYSIS OF VARIANCE

2

SECTION A …28 marks

Instruction: Highlight the correct answer

1.

In analysis of variance, what does represent?
A. Alternative hypothesis
B. Degree of freedom
C. Null hypothesis
D. Standard deviation

Use the scenario below to answer questions 2 – 5

Given that a sample of 5 is taken from the 3 levels A, B, and C such that:
Level A 4 3 6 7 5
Level B 7 3 8 4 9
Level C 6 9 4 5 8
2. What is the mean of level B?
A.

B.

C.
D.

3. What is the grand mean of the three levels?
A.

B.

C.
D.

4.
What is the degree of freedom for the sum of squares of error, ?
A.

B.

C.
D.
5.
What is the degree of freedom for the sum of squares of treatment, ?
A.

B.
C.
D.

Use the scenario below to answer questions 6– 8

6. What is the mean square of treatment (MST)?
A.
B.
C.
D.

7. What is the mean square of errors (MSE)?
A.

B.

C.
D.

8. What is the F-statistic value?
A.
B.
C.
D.

9. Which of the following is
NOT
true for the F-distribution?
A. It is continuous
B. It is generally skewed to the left
C. It has two degrees of freedom
D. Its units are non-negative
10. The sum of squares for treatment (SST) measures:
A. the variation
above
the
k
sample means
B. the variation
below
the
k
sample means
C. the variation
between
the
k
sample means
D. the variation
within
the
k
sample means
11. The degree of freedom for the F test in a one-way ANOVA are:
A.

B.
C.
D.

12. Which of the following components in an ANOVA table are
NOT
additive?
A. It is not possible to tell
B. Degree of freedom
C. Sum of squares
D. Mean squares

Questions 13 – 16 refers to the scenario below:

In an experiment to determine the effect of nutrition on the attention spans of primary school students, a group of 15 students was randomly assigned to three varying meal plans. The table shows their attention spans (in minutes) during an afternoon reading period.

No Lunch

Light Lunch

Full Lunch

5

13

17

8

9

15

6

12

18

7

11

16

9

14

14

35

59

80

Results shows that the total sum of squares, and the sum of squares for treatment,
13.
The correction for the mean,
A. 1859.3
B. 2018.4
C. 2208.3
D. 3115
14.
The sum of squares for error,
A. 34.8
B. 71.2
C. 110.9
D. 167
15.
What is the mean square of treatment?
A. 29.3
B. 39.6
C. 101.4
D. 110.9
16.
If the mean square error, , what is the value of the F statistic?
A. 4.97
B. 35.0
C. 39.6
D. 73.9
17. In a one-way ANOVA, if the computed F statistic exceeds the critical F value, we may:
A.
accept since there is no evidence of a difference.
B.
reject since there is evidence all the means differ.
C.
accept because a mistake has been made.
D.
reject since there is evidence of a treatment effect.
18. When applying one-way analysis of variance(ANOVA), all the following are key assumptions that should be satisfied
EXCEPT
:
A. The mean of the samples taken from the population is always equal.
B. Samples are obtained independently and randomly from the population defined by the factor level
C. The population at each factor level is approximately normally distributed.
D. The population have a common variance

Questions 19 – 21 refers to the scenario below:

Consider the following ANOVA table:

Source

SS

DF

MS

F

Treatment

560

?

140

?

Error

?

19

55

Total

19. What is the degree of freedom of treatment?
A. 2
B. 3
C. 4
D. 5
20. What is the value of sum of squares of error (SSE)?
A. 52
B. 175
C. 595
D. 1045
21. What is the value of F – static?
A. 2.5
B. 3.4
C. 4
D. 8.2
22. The statistical measure that indicates how measured data vary from the given mean value of that data is known as the _________.
A. Confidence interval
B. Hypothesis
C. Sample
D. Variance

Questions 23 – 24 refers to the scenario below:

Analysis of variance

Source

Sum of squares

Degree of freedom

Mean square

F- Ratio

Between Samples

29.6

2

F

Within Sample Error

95.4

Total

125

13

23.
The values of the “within samples” degree of freedom and “within sample” mean square are:
A.

B.
C.
D.

24. The value of
F
in the table is:
A. 0.35
B. 1.71
C. 2.89
D. 5.50
25.
By using the appropriate table, the F-value for is:
A. 2.95
B. 3.16
C. 3.34
D. 8.62

26. Give that the sum of squares for treatments (SST) for an ANOVA F-test is 10000 and there are six total treatments, the mean square for treatments (MST) is:
A. 3333.3
B. 2500
C. 2000
D. 1666

Questions 27 – 30 refers to the scenario below:

Consider the ANOVA table for the following:

Source

Sum of squares

Degree of freedom

Mean square

F- Ratio

Treatment

192

5

Error

320

12

Total

27. The total sum of squares for the given data is:
A. 17
B. 204
C. 325
D. 512
28. The critical value of F for a 10% significance level is:
A. 2.39
B. 5.06
C. 8.89
D. 9.89
29. The value of the mean square of error (MSE) is:
A. 11
B. 16
C. 26.7
D. 38.4
30. The value of the F-statistic is:
A. 0.25
B. 0.6
C. 1.44
D. 4

SECTION B…30 marks

Instruction: Answer ALL questions. ALL working must be CLEARLY shown.

Chief of Police Mark Shields wants to determine whether there is a difference in the mean number of crimes committed among four islands from the Turks and Caicos Islands. He recorded the number of crimes reported on each chosen island for a sample of six weeks.

Number of Crimes

Grand Turk

Providenciales

North Caicos

South Caicos

13

21

12

16

15

13

14

17

14

18

15

18

15

19

13

15

14

18

12

20

15

19

15

18

(i). Stating the null and alternative hypotheses, Ho and Ha, respectively. [2 marks]
(ii). State the decision rule to be taken. [2 marks]
(iii). Computing SST, SSE, and SS total. [8 marks]
(iv). State the degree of freedom for the numerator and denominator. [2 marks]
(v). Calculate the between – sample (MST) and the within – sample (MSE) variances. [4 marks]
(vi). Calculate the value of the test statistic F . [2 marks]
(vii). Completing a one-way ANOVA table. [3 marks]
(viii). Find the critical value of F for α = 0.01. [2 marks]
(ix). Show the rejection and non-rejection regions on the F Distribution curve for α = 0.01. [3 marks]
(x). What is the conclusion that should be drawn about the hypotheses? [2 marks]

4.2
W
MS
(
)
nk
–
(
)
W
MS
15,7.67
W
nkMS
-==
13,6.27
W
nkMS
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11,8.67
W
nkMS
-==
11,6.85
W
nkMS
-==
(
)
0.0512
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Fvv
==
5.8
6.2
7.8
2.8
4.7
5.2
5.9
(
)
dfSSE
2
3
12
15
(
)
dfSST
1
2
3
5
()3,()12,118536
dfSSTdfSSESSTandSSE
====
5.5
8.5
39.3
44.7
44.7
27.8
8.5
5.5
1.14
1.61
0.88
0.60
(
)
(
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knk
—
(
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(
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nkk
—
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237.6
TotalSS
=
202.8
SST
=
CM
=
SSE
=
MST
2.9
MSE
=
0
H
B
MS
nk
–