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    ANOVA Test

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    lacey

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    ANOVA Test

    Uploaded by

    lacey
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    ANOVA-test

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    5 views4 pages

    ANOVA Test

    Uploaded by

    lacey

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    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
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    of 4

    Republic of the Philippines

    Department of Education
    REGION X – NORTHERN MINDANAO
    DIVISION OF GINGOOG CITY
    GINGOOG CITY COMPREHENSIVE NATIONAL HIGH SCHOOL

    Statistics: ANOVA (Analysis of Variance)

    Submitted By:

    Gabby B. Estroga

    Kamhylle Caszandra N. Abanggan

    Jemairi C. Felisilda

    Maeve Bethany T. Bailo

    Jems Audray R. Toquero

    Matt Dhentville R. Ovar

    Submitted To:

    Mrs. Fatima A. Bucio


    REPLICATES X1 X2 X3

    R1 13 26 5

    R2 10 27 7

    R3 10 19 9

    R4 11 16 8

    R5 13 22 4

    X 1 = no. of the first letter of the first name K= no. of groups N= no. of data

    X 2 = birth date

    X 3 = no. of letters in last names

    I. Null Hypothesis

    There is no significant difference between the treatments.

    II. Alternative Hypothesis

    There is significant difference between the treatments.

    III. Test of Statistics

    F<FC: NS: Accept H 0

    F>FC: S: Accept H 1

    IV. Rejection Region

    Significance level: 3. 89

    df 1 = 3-1=2

    df 2 = 15-3=12

    V. Calculation of Test Statistics

    X1 X2 X3 X1 2 X2 2 X3 2

    13 26 5 169 676 25

    10 27 7 100 729 49
    10 19 9 100 361 81

    11 16 8 121 256 64

    13 22 4 169 484 16

    ⅀ x 1=57 ⅀ x 2=110 ⅀ x 3=33 ⅀ x 1 =659


    2 ⅀ x 2 =2,506
    2 ⅀ x 3 =235
    2

    VI. Solving for Variance

    5(659)−572 5(2,506)−110 2 5(235)−332


    S1 = 2 S2 = 2 S3 = 2
    5 (5 −1) 5(5− 1) 5(5 −1)

    3,295 −3,249 12,530 −12,100 1,175 −1,089


    S1 =
    2 S2 = 2 S3 =
    2
    5(4) 5( 4) 5(4)

    46 430 86
    S1 =
    2 S2 =
    2 S3 =
    2
    20 20 20

    S1 =2.3
    2 S2 =21.5
    2 S3 =4.3
    2

    VII. Solving for Sum of Squares between VIII. Solving for Sum of Squares within
    Groups Groups

    ⅀X= 57+110+33=200 N=15

    2 2 2 2
    (57) (110) (33) (200) SSW =(5 −1)2.3+(5 − 1)21.5+(5 −)4.3
    SS B= + + −
    5 5 5 15
    SSW =(4)2.3+(4)21.5+(4) 4.3
    3,249 12,100 1,089 40,000
    SS B= + + −
    5 5 5 15 SSW =9.2+86+17.2

    SS B=649.8+2,420+ 217.8− 2,666.67 SSW =112.4

    SS B=3,287.6 −2,666.67

    SS B=620.93

    IX. Solving Mean Squares X. Solving Mean Squares XI. Solve for F Value
    between Groups within Groups
    310.47
    F=
    620.93 112.4 9.37
    MS B = MS W =
    3−1 15 −3
    F=33.13
    620.93 112.4 F> Fc
    MS B = MS W =
    2 12
    33.13>3.89 : Significant
    MS B =310.47 MS W =9.37

    XII.Set-Up ANOVA Table

    Source of Variance SS Df MS F

    Between Groups 620.93 2 310.47

    Within Groups 112.4 12 9.37 33.13

    TOTAL: 733.33 14 319.84

    XIII. CONCLUSION

    Therefore, the F value is greater than FC. Which means, reject null hypothesis( H 0) and accept alternative
    hypothesis( H 1). These results shows that there is significant difference among the treatments.

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