Tag: business economics and quantitative methods

Questions Related to business economics and quantitative methods

If x, y are independent variable, then

  1. $Cov\left ( x, y \right )=1$

  2. $r _{xy}=0$

  3. $r _{xy}=1$

  4. $Cov\left ( x, y \right )=0$


Correct Option: B,D
Explanation:

Fact. If the variables are uncorrelated or independent then covariance
and coefficient of correlation between the variable both are equal to 0
i.e. $r _{xy}=Cov\left ( x, y \right )=0$ 

If $n=10, \sum x=4,\sum y=3, \sum x^2=8,\sum y^2=9$ and $\sum xy=3,$ then the coefficient of $r _{x,y}$ is

  1. $\frac{3}{4}$

  2. $\frac{1}{5}$

  3. $\frac{1}{6}$

  4. $\frac{1}{4}$


Correct Option: D
Explanation:

Correlation coefficient 
${ r } _{ x,y }=\dfrac { n\sum { xy } -\sum { x } \sum { y }  }{ \sqrt { \left[ n\sum { { x }^{ 2 }-{ \left( \sum { x }  \right)  }^{ 2 } }  \right] \left[ n\sum { { y }^{ 2 }-{ \left( \sum { y }  \right)  }^{ 2 } }  \right]  }  } $

$=\displaystyle\frac { 30-12 }{ \sqrt { 64\times 81 }  } $
$\Rightarrow r _{x,y}=\dfrac{1}{4}$

FInd the rank correlation from the following data:

S. No. 1 2 3 4 5 6 7 8 9 10
Rank Differences -2 -4 -1 3 2 0 -2 3 3 -2
  1. 0.64

  2. 0.50

  3. 0.45

  4. 0.34


Correct Option: A
Explanation:

Rank Difference $(d)$ | $d^2$ | | --- | --- | --- | | 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. | -2 -4 -1 3 2 0 -2 3 3 -2 | 4 16 1 9 4 0 4 9 9 4 |

 $\sum d^2=60,\quad n=10$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}$

$r=1-\cfrac{6(60)}{10(10^2-1)}$

$r=1-\cfrac{360}{990}$

$r=0.6363....\approx 0.64$

The marks obtained by nine students in physics and Mathematics are given below:

Physics 48 60 72 62 56 40 39 52 30
Mathematics 62 78 65 70 38 54 60 32 31

calculate spearman's coefficient.

  1. $r=0.66$

  2. $r=0.32$

  3. $r=0.53$

  4. $r =0.28$


Correct Option: A
Explanation:

Descending order arranged data will be as follows:

Physics: $72,62,60,56,52,48,40,39,30$
MAthematics: $78,70,65,62,60,54,38,32,31$
Thus data will be

Mathematics $(M)$ | Rank $(P)$ | Rank $(P)$ | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 48 60 72 62 56 40 39 52 30 | 62 78 65 70 38 54 60 32 31 | 6 3 1 2 4 7 8 5 9 | 4 1 3 2 7 6 5 8 9 | 2 2 2 0 3 1 3 3 0 | 4 4 4 0 9 1 9 9 0 |

$n=9,\quad \sum d^2=40$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{40\times 6}{9(9^2-1)}=1-\cfrac{240}{720}=0.66$

Find the spearman's rank coefficient of correlation from the following data:

X 48 33 40 9 16 16 65 25 16 57
Y 13 13 24 6 15 4 20 9 6 19
  1. $0.76$

  2. $0.52$

  3. $0.61$

  4. $0.85$


Correct Option: A
Explanation:

Rank | $Y$ | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 48 33 40 9 16 16 65 25 16 57 | 3 5 4 10 7 7 1 6 7 2 | 13 13 24 6 15 4 20 9 6 19 | 5 5 1 8 4 10 2 7 8 3 | 2 0 3 2 3 3 1 1 1 1 | 4 0 9 4 9 9 1 1 1 1 |

$n=10,\quad \sum d^2=39$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 39}{10(10^2-1)}=1-\cfrac{234}{990}=0.76$

The final position of twelve clubs in a football league and the average attendance at their home matches were as follows. Calculate a coefficient of correlation by ranks.

Club A B C D E F G H I J K L
Position 1 2 3 4 5 6 7 8 9 10 11 12
Attendance (thousands) 27 30 18 25 32 12 19 11 32 12 12 15
  1. 0.34

  2. 0.56

  3. 0.32

  4. 0.48


Correct Option: D
Explanation:

Attendance | Rank | Position | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 27 30 18 25 32 12 19 11 32 12 12 15 | 4 3 7 5 1 9 6 12 1 9 9 8 | 1 2 3 4 5 6 7 8 9 10 11 12 | 3 1 4 1 4 3 1 4 8 1 2 4 | 9 1 16 1 16 9 1 16 64 1 4 16 |

$n=12,\quad \sum d^2=154$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 154}{12(12^2-1)}=1-\cfrac{924}{1716}=0.48$

Find the rank correlation coefficient between the heights of fathers and sons from the following data:

Heights of fathers in inches  65 66 67 67 68 69 70 72
Height of sons in inches 67 68 65 68 72 72 69 71
  1. $0.67$

  2. $0.58$

  3. $0.42$

  4. $0.92$


Correct Option: B
Explanation:

Rank | Height(Son) | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 65 66 67 67 68 69 70 72 | 8 7 5 5 4 3 2 1 | 67 68 65 68 72 72 69 71 | 7 5 8 5 1 1 4 3 | 1 2 3 0 3 2 2 2   | 1 4 9 0 9 4 4 4 |

$n=08,\quad \sum d^2=35$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 35}{8(8^2-1)}=1-\cfrac{210}{504}=0.58$

The marks in History and Mathematics of twelve students in a public examination are given below. Calculate a coefficient of correlation by ranks.

Student $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$ $K$ $L$
History $69$ $36$ $39$ $71$ $67$ $76$ $40$ $20$ $85$ $65$ $55$ $34$
Mathematics $33$ $52$ $71$ $25$ $79$ $22$ $83$ $81$ $24$ $35$ $46$ $64$

Interpret the result.

  1. A very good student of history is a very bad student in mathematics.

  2. A bad student of history is even bad student in mathematics.

  3. This is positive correlation

  4. None of the above


Correct Option: A
Explanation:

History $(H)$ | Mathematics$(M)$ | Rank $(H)$ | Rank$(M)$ | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 69 36 39 71 67 76 40 20 85 65 55 34 | 33 52 71 25 79 22 83 81 24 35 46 64 | 4 10 9 3 5 2 8 12 1 6 7 11 | 9 6 4 10 3 12 1 2 11 8 7 5 | 5 4 5 7 2 10 7 10 10 2 0 6 | 25 6 25 49 4 100 49 100 100 4 0 36   |

$n=12,\quad \sum d^2=508$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 508}{12(12^2-1)}=1-\cfrac{3048}{1716}=-0.77$

Since $r<0,$ we can say that a very good student of history is a very bad student in mathematics.

The marks in history and mathematics of twelve students in a public examination are given below. Calculate a coefficient of correlation by ranks.

Student A B C D E F G H I J K L
History 69 36 39 71 67 76 40 20 85 65 55 34
Mathematics 33 52 71 25 79 22 83 81 24 35 46 64


  1. -0.77

  2. -0.92

  3. 0.77

  4. 0.92


Correct Option: A
Explanation:

History $(H)$ | Mathematics$(M)$ | Rank $(H)$ | Rank$(M)$ | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 69 36 39 71 67 76 40 20 85 65 55 34   | 33 52 71 25 79 22 83 81 24 35 46 64 | 4 10 9 3 5 2 8 12 1 6 7 11 | 9 6 4 10 3 12 1 2 11 8 7 5 | 5 4 5 7 2 10 7 10 10 2 0 6   | 25 6 25 49 4 100 49 100 100 4 0 36   |

$n=12,\quad \sum d^2=508$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 508}{12(12^2-1)}=1-\cfrac{3048}{1716}=-0.77$

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?

Statistics 1 3 3 4 5 6 7 8 9 10
Mathematics 2 4 1 5 3 9 7 10 6 8
  1. 0.76

  2. 0.66

  3. 0.56

  4. 0.48


Correct Option: A
Explanation:

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

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

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

$r=\cfrac{Cov(x,y)}{\sigma _x.\sigma _y}=\cfrac{6.1}{2.87\times 2.76}=0.77$