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  • 7/28/2019 Correlation & Regression - 3 Presentation - Unitedworld School of Business

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    Correlation-Regression

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    Methods of Measuring correlation

    1. Graphic Method2. Diagramatic Method- Scatter Diagram

    3. Algebraic method

    a. Karl Pearsons Coefficient of correlationb. Spearmans Rank Co-efficient Correlationc. Coefficient of Concurrent deviations

    d. Least Squares Method

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    Karl Pearsons Coefficient of Correlation

    dx dy ( Gamma) = -------------------------

    dx2 dy2

    dx dy= -------------------------

    N xy

    dx = x-xbardy = y- ybar

    dx dy = sum of products of deviations from respective

    arithmetic means of both series

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    Karl Pearsons Coefficient of Correlation

    After calculating assumed or working mean Ax & Ay dx dy ( dx) *( dy) ( Gamma) = --------------------------------

    [ N dx2 - ( dx)2 * [ Ndy2 - ( dy)2 ]

    dx dy = total of products of deviation from assumedmeans of x and y series

    dx = total of deviations of x series dy = total of deviations of y series

    dx2 = total of squared deviations of x series dy2 = total of squared deviations of y series

    N= No. of items ( no. of paired items

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    Karl Pearsons Coefficient of CorrelationAfter calculating assumed or working mean Ax &

    Ay

    dx x dy

    dx dy - ----------------N

    ( Gamma) = -------------------------

    ( dx)2

    ( dy)2

    [ dx2 - --------- ] x [ dy2 - ------------]N N

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    Assumptions of Karl Pearsons Coefficient of Correlation

    1. Linear relationship exists between the variablesProperties of Karl Pearsons Coefficient of Correlation1.value lies between +1 & - 1

    2.Zero means no correlation

    3. ( Gamma) = bxy X byxWhere bxy X byx are regression coefficicent

    Merit

    Convenient for accurate interpretation as it gives degree &

    direction of relationship between two variables

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    Limitations

    1. Assumes linear relationship , even though itmay not be

    2. Method & process of calculation is difficult &

    time consuming3. Affected by extreme values in distribution

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    Probable Error of Karl Pearsons Coefficient of

    Correlation

    1- 2

    Probable Error of ( Gamma) = 0.6745 -------- N

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    Q7.Calculate coefficient of correlation for following data

    X65 63 67 64 68 62 70 66 68 67 69 71

    Y 68 66 68 65 69 66 68 65 71 67 68 70

    Ans dx dy ( Gamma) = ------------------------- dx2 dy2

    dx dy= -------------------

    N xy

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    1 2 3 4 5 6 7 8 9 10 11 12

    Su

    mX Xbar

    X 65 63 67 64 68 62 70 66 68 67 69 71 800 66.67

    Y 68 66 68 65 69 66 68 65 71 67 68 70 811 67.58

    dx=x-xbar -1.67 -3.67 0.33 -2.67 1.33 -4.67 3.33 -0.67 1.33 0.33 2.33 4.33

    dx2 2.78 13.44 0.11 7.11 1.78 21.78 11.11 0.44 1.78 0.11 5.44 18.78

    84.

    67

    dx.dy -0.69 5.81 0.14 6.89 1.89 7.39 1.39 1.72 4.56 -0.19 0.97 10.47

    40.

    33

    dy=y-ybar 0.42 -1.58 0.42 -2.58 1.42 -1.58 0.42 -2.58 3.42 -0.58 0.42 2.42

    dy2 0.17 2.51 0.17 6.67 2.01 2.51 0.17 6.67 11.67 0.34 0.17 5.84

    38.

    92

    dx dysum dx2*

    sumdy2

    3294.

    9

    dx2 dy2 57.40

    coeff of

    correlation = 0.70

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    Q8. following information about age of husbands

    & wives. Find correlation coefficient

    Husband

    23 27 28 29 30 31 33 35 36 39

    Wife 18 22 23 24 25 26 28 29 30 32

    ( Gamma) =0.99

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    1 2 3 4 5 6 7 8 9 10

    Sum

    X Xbar

    X 23 27 28 29 30 31 33 35 36 39 311 31.10

    Y 18 22 23 24 25 26 28 29 30 32 257 25.70

    dx=x-

    xbar -8.10 -4.10 -3.10 -2.10 -1.10 -0.10 1.90 3.90 4.90 7.90

    dx2 65.61 16.81 9.61 4.41 1.21 0.01 3.61 15.21 24.01 62.41

    202.

    9

    dx.dy 62.37 15.17 8.37 3.57 0.77 -0.03 4.37 12.87 21.07 49.77

    178.

    3

    dy=y-

    ybar -7.70 -3.70 -2.70 -1.70 -0.70 0.30 2.30 3.30 4.30 6.30

    dy2 59.29 13.69 7.29 2.89 0.49 0.09 5.29 10.89 18.49 39.69

    158.

    1

    dx dy sum dx2* sumdy232078.

    49

    dx2 dy2 179.10

    coeff of correlation

    = 1.00

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    Q9. ten competitors in a cooking competition are

    ranked by three judges in the following way .byusing rank coorelation method find out which

    pair of judges have nearest approach

    Ans P&Q= -0.21 , Q &R=--0.3 P &R = +0.64

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    Q9. ten competitors in a cooking competition are ranked

    by three judges in the following way .by using rank

    coorelation method find out which pair of judges havenearest approach

    P Q R

    1 1 3 6

    2 6 5 43 5 8 9

    4 10 4 8

    5 3 7 1

    6 2 10 27 4 2 3

    8 9 1 10

    9 7 6 5

    10 8 9 7

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    Rank coefficient of correlation

    6 d2 (rho) = 1 - -------------------

    N3-N

    d2 = total of squared differenceN = number of items

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    P Q R

    Rp-

    Rq dpq2

    Rq-

    Rr dqr2

    Rp-

    Rr dpr2

    1 1 3 6 -2 4 -3 9 -5 25

    2 6 5 4 1 1 1 1 2 4

    3 5 8 9 -3 9 -1 1 -4 16

    4 10 4 8 6 36 -4 16 2 4

    5 3 7 1 -4 16 6 36 2 4

    6 2 10 2 -8 64 8 64 0 0

    7 4 2 3 2 4 -1 1 1 1

    8 9 1 10 8 64 -9 81 -1 1

    9 7 6 5 1 1 1 1 2 4

    10 8 9 7 -1 1 2 4 1 1

    1000 200 214 0 60

    6Sigma d2 1200 1284 360

    N3-N 990 6Sigma d2/N3-N 1.21 1.297 0.3636

    P= -0.21 -0.297 0.636364

    C O i

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    Campus Overview

    907/A Uvarshad,

    Gandhinagar

    Highway, Ahmedabad382422.

    Ahmedabad Kolkata

    Infinity Benchmark,

    10th Floor, Plot G1,

    Block EP & GP,

    Sector V, Salt-Lake,

    Kolkata700091.

    Mumbai

    Goldline Business Centre

    Linkway Estate,

    Next to Chincholi Fire

    Brigade, Malad (West),

    Mumbai400 064.

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    Thank You