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Correlation Test - 13

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Correlation Test - 13
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  • Question 1
    1 / -0
    Following are the rank obtained by 10 students in two subjects , Statistics and Mathematics . To what extent the knowledge of the students in the two subjects is related?
    Statistics13345678910
    Mathematics24153971068
    Solution

    Statistic $$(X)$$

    Mathematics $$(Y)$$

    $$XY$$

    $$X^2$$

    $$Y^2$$

    1

    3

    3

    4

    5

    6

    7

    8

    9

    10

    2

    4

    1

    5

    3

    9

    7

    10

    6

    8

    2

    12

    3

    20

    15

    54

    49

    80

    54

    80

    1

    9

    9

    16

    25

    6

    49

    64

    81

    100

    4

    16

    1

    25

    9

    81

    49

    100

    36

    64

     $$\sum X=56,\quad \sum Y=55,\quad \sum XY=369,\quad \sum X^2=390,\quad \sum Y^2=385$$

    $$N=10$$

    Cov$$(x,y)=\cfrac{\sum XY}{N}-\cfrac{\sum X}{N}.\cfrac{\sum Y}{N}=\cfrac{369}{10}-\cfrac{56}{10}.\cfrac{55}{10}=6.1$$

    $$\sigma_x=\sqrt{\cfrac{\sum X^2}{N}-\left(\cfrac{\sum X^2}{N}\right)^2}=\sqrt{\cfrac{390}{10}-\left(\cfrac{56}{10}\right)^2}=2.76$$

    $$\sigma_y=\sqrt{\cfrac{\sum Y^2}{N}-\left(\cfrac{\sum Y^2}{N}\right)^2}=\sqrt{\cfrac{385}{10}-\left(5.5\right)^2}=2.87$$

    $$r=\cfrac{Cov(x,y)}{\sigma_x.\sigma_y}=\cfrac{6.1}{2.87\times 2.76}=0.77$$


  • Question 2
    1 / -0

    Directions For Questions

    Study the following table and answer the questions based on it.
    Expenditure of a Company (in Lakh Rupees) per Annum Over the given years.
    YearSalaryFuel and TransportBonusInterest on LoansTaxes
    $$1998$$$$288$$$$98$$$$3.00$$$$23.4$$$$83$$
    $$1999$$$$342$$$$112$$$$2.52$$$$32.5$$$$108$$
    $$2000$$$$324$$$$101$$$$3.84$$$$41.6$$$$74$$
    $$2001$$$$336$$$$133$$$$3.68$$$$36.4$$$$88$$
    $$2002$$$$420$$$$142$$$$3.96$$$$49.4$$$$98$$

    ...view full instructions

    Total expenditure on all these items in $$1998$$ was approximately what percent of the total expenditure in $$2002$$?
    Solution
    Required percentage $$= \left [\dfrac {(288 + 98 + 3.00 + 23.4 + 83)}{(420 + 142 + 3.96 + 49.4 + 98)} \times 100\right ]$$%
    $$= \left [\dfrac {485.4}{713.36}\times 100\right ]$$%
    $$\approx 69.45$$%. 
  • Question 3
    1 / -0
    If increase in value of one variable is accompnied by the proportional decrease in second variable, then corrrelation between two variables is
    Solution
    If increase in value of one variable is accompnied by the proportional decrease in second variable, then corrrelation between two variables is perfect negative.
  • Question 4
    1 / -0
    The correlation between number of hours and efficiency of work is
    Solution
    relationship between their output and their working hours
    $$\therefore$$ The correlation between number of hours and efficiency of work is imperfect negative
    Option D is correct

  • Question 5
    1 / -0
    Find Karl pearson's correlation coefficient.
    Marks in english10251325221112252120
    Marks in hindi12221615181817232417
    Solution
    $$\bar { x } =\cfrac { \sum { x }  }{ n  }=\cfrac{184}{10}=18.4 $$
    $$\bar { y } =\cfrac { \sum {y }  }{ n  }=\cfrac{182}{10}=18.2 $$
    $$r=\cfrac { \sum { (x-\bar { x } )(y-\bar { y } ) }  }{ \sqrt { \sum { { (x-\bar { x } ) }^{ 2 } } \sum { { (y-\bar { y } ) }^{ 2 } }  }  } =\cfrac { 97.44 }{ \sqrt { 348.4\times 127.50 }  } =0.462$$

  • Question 6
    1 / -0
    HeightLong JumpHigh Jump
    A158324175
    B165365185
    C162380180
    D170400184
    E175350199
    F163350172
    G178425189
    Find rank correlation between height & long jump and height & high jump.
    Solution
    $$Height$$ $$Rank(Height)$$ $$Long\,Jump$$ $$Rank(Long\,jump)$$ $$d$$ $$d^2$$ 
    $$158$$ $$7$$ $$324$$ $$6$$ $$1$$ $$1$$ 
    $$165$$ $$4$$ $$365$$ $$4$$ $$0$$ $$0$$ 
    $$162$$ $$6$$ $$380$$ $$3$$ $$3$$ $$9$$ 
    $$170$$ $$3$$ $$400$$ $$2$$ $$1$$ $$1$$ 
    $$175$$ $$2$$ $$350$$ $$6$$ $$4$$ $$16$$ 
    $$163$$ $$5$$ $$350$$ $$5$$ $$0$$ $$0$$ 
    $$178$$$$1$$$$425$$ $$1$$ $$0$$ $$0$$ 
         $$\sum d^2=26$$ 
    $$\rho=1-\dfrac{6\sum d^2}{n(n^2-1)}$$

        $$=1-\dfrac{6\times 26}{7((7)^2-1)}$$

        $$=0.52$$

    $$Height$$ $$Rank(Height)$$ $$High\,jump$$ $$Rank(High\,jump)$$ $$d$$ $$d^2$$ 
    $$158$$ $$7$$ $$175$$ $$6$$ $$1$$ $$1$$ 
    $$165$$ $$4$$ $$185$$ $$3$$ $$1$$ $$1$$ 
    $$162$$ $$6$$ $$180$$ $$5$$ $$1$$ $$1$$ 
    $$170$$ $$3$$ $$184$$ $$4$$ $$1$$ $$1$$ 
    $$175$$ $$2$$ $$199$$ $$1$$ $$1$$ $$1$$ 
    $$163$$ $$5$$ $$172$$ $$7$$ $$2$$ $$4$$ 
    $$178$$ $$1$$ $$189$$ $$2$$ $$1$$ $$1$$ 
         $$\sum d^2=10$$ 
    $$\rho=1-\dfrac{6\sum d^2}{n(n^2-1)}$$

        $$=1-\dfrac{6\times 10}{7((7)^2-1)}$$

        $$=0.86$$
  • Question 7
    1 / -0
    Chemistry2869431017472335
    Biology8491243130334523
    The marks obtained in Chemistry and Biology are as follows;
    Solution
    $$\rho=1-\cfrac{6\sum{{d}^{2}}}{n({n}^{2}-1)}=\cfrac{6\times{162}^{2}}{9\times 80}=-0.35$$

  • Question 8
    1 / -0
    When two regression coefficients bear same algebraic signs, then correlation coefficient is: 
    Solution
    The '+' or '-' sign is given to the correlation coefficient based on the signs of the two regression coefficients.

    $$r=\pm  \sqrt{b_{yx}*b_{xy}}$$

    where $$r$$ is the correlation coefficient
    $$b_{yx},b_{xy}$$ are the regression coefficients.
  • Question 9
    1 / -0
    Calculate Spearman's rank coefficient for the data:
    Mathematics12345678910
    Statistics24153971068
    Solution
    $$\rho=1-\cfrac{6\sum{{d}^{2}}}{n({n}^{2}-1)}=1-\cfrac{6\times 40}{10(100-1)}=1-\cfrac{240}{990}=0.758\approx 0.76$$ 

  • Question 10
    1 / -0
    It is high degree of relation if 
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