ETS 85 Report

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PREPARED BY: REX MABANTA

RALPH STEPHEN BARTOLO

REYNANTE LUMAWAN

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Analysis of Variance ANOVA) is a hypothesis-testing technique used to test the equality oftwo or more population (or treatment) meansby examining the variances of samples that

are taken.

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Most of the time ANOVAis used to compare theequality of three or more means, however whenthe means from two samples are compared using

ANOVA it is equivalent to using a t-test tocompare the means of independent samples.

ANOVA is based on comparing the variance (orvariation) between the data samples to variationwithin each particular sample. If the betweenvariation is much larger than the within variation,the means of different samples will not be equal.If the between and within variations areapproximately the same size, then there will beno significant difference between sample means.

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Steps in hypothesis testing for ONE WAYANOVA1. Formulate the null and the appropriatealternative hypothesis.2. Specify the level of significance to be used.3. Critical value

4. Establish the critical regions.5. Compute the actual value.6. Make a statistical decision, which is to rejectthe null hypothesis when the computed valueof f distribution is

( k-1,N-k); otherwise

null hypothesis is not rejected.7. Draw the appropriate conclusion.

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Formulas for ONE WAY ANOVA:

Ti=

=1ij

T. = =1 i

MS=and Fcomp =

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Balanced ONE WAY ANOVA

N=kn

SSA= . . .=1=1 2 SSbet= 1 =1 i2.

SSE=

x .

=1

=12 SSerror=

2

=1

=1ij -

1 =1 i2

SST=

x . .=1

=1

2 SST=

2

=1=1

ij -.

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Unbalanced ONE WAY ANOVA

=1

SSbet=

=1

-.

SSerror= 2=1=1 ij-

=1

SST= 2=1=1 ij- .

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Observation

No.

Treatment 1 Treatment 2 Treatment 3

1 54 31 46

2

46 38 42

3 68 42 38

4 58 44 55

5

60 26 50

6 52 36 48

T

i

Mean

X

2

T.

N

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ANOVA Table

Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments SSA k-1 MSA=1

f=MSA/MSE

Error SSE N-k MSE=

Total SST N-1

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Sample problem for a balanced ANOVA

Observation

No.

Treatment 1 Treatment 2 Treatment 3

1 54 31 46

2 46 38 42

3 68 42 38

4

58 44 55

5 60 26 50

6 52 36 48

Given the hypothetical data below, test the hypothesis Ho:1= 2= 3 vs. the

alternative H1: at least one pair of means is not equal, using an = 0.05 level of

significance.

l

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Solution:

1. Hypothesis: Ho:1= 2= 3vs. the alternativeH1: at least one pair of means is not equal

2. Significance Level: = 0.05

3. Critical value: (v1,v2)Observation No. Treatment 1 Treatment 2 Treatment 3

1 54 31 46

2

46 38 423 68 42 38

4 58 44 55

5 60 26 50

6 52 36 48

T.=834 338 217 279

Mean=46.33 56.33 36.17 46.50

X

2

=40554 19326 8077 13153

T

i

2

=239174 114244 47089 77841

N=18

n=6 n=6 n=6

bl

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ANOVA Table

ource of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments SSA 2 MSA=1

f=MSA/MSE

Error SSE 15 MSE=

Total SST 17

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4. Critical Region: ff0.05(2,15)= 3.685. Computation:

SSbet= 1 =1 i2 . =2391746 -834218 =1220.33

SSerror=

2

=1

=1ij -

1

=1i2

=40554-39862.33=691.67

SST= SSbet+ SSerror=1220.33+691.67=1912

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Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments 1220.33 2 610.165 f=13.233

Error 619.67 15 46.111

Total 1912 17

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6. Decision:

Since f=13.2333.68, the hypothesis of

equal means is rejected.7. Conclusion: There is enough evidence to

conclude that the means are not equal.

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Problem:Susan Sound predicts that students will learn mosteffectively with a constant background sound, as opposed to anunpredictable sound or no sound at all. She randomly dividestwenty-four students into three groups of eight. All students

study a passage of text for 30 minutes. Those in group 1 studywith background sound at a constant volume in thebackground. Those in group 2 study with noise that changesvolume periodically. Those in group 3 study with no sound atall. After studying, all students take a 10 point multiple choicetest over the material, using an = 0.05 level of significance..Their scores follow:

group test scores

1) constant sound

74 6 8 6 6 2 9

2) random sound 55 3 4 4 7 2 2

3) no sound 24 7 1 2 1 5 5

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x

1

x2

x3

7 5 2

4

5 4

6

3 7

8 4 1

6 4 2

6

7 1

2 2 5

9 2 5

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Solution:

Hypothesis: Ho:1= 2= 3

H1: at least one pair of means is not equal


3. Critical value: (v1,v2)x

1

x2

x3

7 5 2

4 5 4

6 3 7

8 4 1

6 4 2

6 7 1

2 2 5

9 2 5

T.=107

48 3227

T

i

2

=4057

2304

1024 729

=13.375 6 4 3.375x

2

=595322

148 125

N=24

n=8 n=8 n=8

ANOVA T bl

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ANOVA Table

ource of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f


f=MSA/MSE

Error SSE 21 MSE=

Total SST 23

4 C iti l R i f f (2 21) 3 47

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SSbet=

1

=1i2

.

=4057

8-107

24=30.084

SSerror= 2=1=1 ij - 1 =1 i2=595-507.125=87.875

SST= SSbet+SSerror=30.084+87.875=117.959

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Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments 30.084 2 15.042 f=3.594

Error 87.875 21 4.185

Total 117.959 23

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6. Decision:

Since f=3.5943.47, the hypothesis ofequal means is rejected.

7. Conclusion: There is enough evidence toconclude that the means are as she predicted,in that the constant music group has thehighest score.

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A researcher is concerned about the level of knowledgepossessed by university students regarding United Stateshistory. Students completed a high school senior levelstandardized U.S. history exam. Major for students was also

recorded. Data in terms of percent correct is recorded below for32 students. Compute the appropriate test for the dataprovided below, using =.01

Education Business/Management Behavioral/Social Science Fine Arts

62 72 42 80

81 49 52 57

75 63 31 87

58 68 80 64

67 39 22 28

48 79 71 29

26 40 68 62

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Education

Business/Manag

ement

Behavioral/Social

ScienceFine Arts

62 72 42 80

81 49 52 57

75 63 31 87

58 68 80 64

67 39 22 28

48 79 71 29

26 40 68 62

T.=1600 417 410 366 407

Ti2=641594 173889 168100 133956 165649

X2=101164 26863 25540 21978 26783

Mean=228.56 59.57 58.57 52.28 58.14

N=28 n=7n=7 n=7 n=7

Solution:

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Solution:

1. Hypothesis: Ho:1= 2= 3= 4H1: at least one pair of means

is not equal


3. Critical value: (v1,v2)Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f


f=MSA/MSE

Error

SSE 24 MSE=

Total SST 27

4 Critical Region: f f (3 24)=4 72

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4. Critical Region: ff0.01(3,24)=4.725. Computation:

SSbet=

1

=1i2

.

=641594

7-

1600

28=91656.29-91428.57

=227.72

SSerror= 2=1=1 ij - 1 =1 i2=101164-91656.29=9507.71

SST= SSbet+SSerror=227.72+9507.71=9735.43

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Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments 227.72 3 75.90 f=0.19

Error 9507.71 24 396.15

Total 9735.43 27

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6. Decision:

Since f=0.194.72, the hypothesis of

equal means is not rejected.7. Conclusion: There is insufficient evidenceat the 0.01 level of significance to reject theclaim that the means are equal.

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Problem for unbalanced ONE WAY ANOVABelow are the test scores from one of my algebra classes. The scores for each examhave been ranked numerically, just so no one tries to figure out who got what scoreby finding a list of students and comparing alphabetically.

Exam 1 2 3 4 5 6 7 8

Scores 21

35

40

42

45

57

59

6060

61

62

64

65

67

68

68

72

73

74

75

76

78

8091

17

45

49

57

57

61

62

6263

64

67

69

74

75

78

78

78

79

80

86

88

89

90

24

52

56

59

59

63

65

6768

72

73

74

75

75

76

80

82

82

83

88

90

90

37

43

52

54

58

60

61

6263

64

67

67

71

72

74

75

77

77

79

37

37

60

65

69

75

75

7676

83

84

85

87

87

89

89

90

21

43

50

51

53

69

69

7072

73

74

74

80

81

89

94

29

31

43

55

62

63

64

6669

71

75

75

77

83

86

91

23

38

43

52

53

56

57

6263

64

65

70

72

73

75

76

80

80

83

Exam 1 2 3 4 5 6 7 8

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Scores 21

35

40

42

45

57

59

60

60

61

62

64

65

67

68

68

72

73

74

75

76

7880

91

17

45

49

57

57

61

62

62

63

64

67

69

74

75

78

78

78

79

80

86

88

8990

24

52

56

59

59

63

65

67

68

72

73

74

75

75

76

80

82

82

83

88

90

90

37

43

52

54

58

60

61

62

63

64

67

67

71

72

74

75

77

77

79

37

37

60

65

69

75

75

76

76

83

84

85

87

87

89

89

90

21

43

50

51

53

69

69

70

72

73

74

74

80

81

89

94

29

31

43

55

62

63

64

66

69

71

75

75

77

83

86

91

23

38

43

52

53

56

57

62

63

64

65

70

72

73

75

76

80

80

83

N=156 n=24 n=23 n=22 n=19 n=17 n=16 n=16 n=19

T.=10379 1493 1568 1553 1213 1264 1063 1040 1185

Mean=532.97 62.21 68.17 70.59 63.84 74.35 66.44 65.00 62.37

Ti2/ni=692953.65 2229049 2458624 2411809 1471369 1597696 1129969 1081600 1404225

2 =

S l ti

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Solution:

Hypothesis: Ho: Ho:1= 2= 3= 4= 5= 6= 7= 8

H1: At least there is one pair of mean that is not equal.


3. Critical value:

(v1,v2)

Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean Square Computed f


f=MSA/MSE

Error SSE 148 MSE=

Total SST 155

4 Critical Region: f f0 05(7 148)= 2 09

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SSbet=

=1-

.

=692953.647-690536.160=2417.49

SSerror= 2

=1=1 ij-

=1

=731097-692953.647=38143.35

SST= 2=1=1 ij- .

=731097-690536.160=40560.84

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Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

Square

Computed f

Treatments 2417.49 7 345.36 f=1.340

Error 38143.35 148 257.72

Total

40560.84

155

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6. Decision:

Since f=1.3402.09, the hypothesis of equal means isnot rejected.

7. Conclusion: There is insufficient evidence at the 0.05 level ofsignificance to reject the claim that the means are equal.

Example: A firm wishes to compare four programs for training workers

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p p p g gto perform a certain manual task. Eighteen new employees areassigned to the training programs. At the end of the training period, atest is conducted to see how quickly trainees can perform the task. Thenumber of times the task is performed per minute is recorded for each

trainee, with the following results:

PROGRAM 1 PROGRAM 2 PROGRAM 3 PROGRAM 4

19 10 18 10

10 12 11 13

12 8 18 14

9 11 18

10 9

T.=222 60 51 47 64

2i=12506 3600 2601 2209 4096x2=2854 686 529 769 870

N=18 n=4 n=5 n=3 n=5

Solution:

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Solution:

Hypothesis: Ho: Ho: 1= 2= 3= 4

H1: At least there is one pair of

means that is not equal.2. Significance Level: = 0.05

3. Critical value: (v1,v2)

Solution:

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Solution:

Hypothesis: Ho: 1= 2= 3= 4

H1: At least there is one pair of means that is not equal.


3. Critical value:

(v1,v2)

Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean Square Computed f


f=MSA/MSE

Error SSE 14

MSE=

Total SST 17

4. Critical Region: ff0 05(3,14)=3.34

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4. Critical Region: ff0.05(3,14) 3.345. Computation:

SSbet=

=1-

.

=2975.63-2738=237.63

SSerror= 2

=1=1 ij-

=1

=2854-2975.63=-121.63

SST= 2=1=1 ij- .

=2854-2738=116

Source of Sum of Degrees of Mean Square

Computed f

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Variation Squares Freedom

Treatments 237.63 3 79.21 f=-9.125

Error -121.63 14 -8.68

Total 116 17

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6. Decision:

Since f=-9.1253.34, the hypothesis of

equal means is not rejected.7. Conclusion: There is insufficient evidenceat the 0.05 level of significance to reject theclaim that the means are equal.

ETS 85 Report

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