Question 14 Marks
To know the relation between the ability in two different subjects for the students, a sample of seven students is taken from a school. From the information of marks in two subjects for $'7$ students, it is known that the sum of the squares of differences in the ranks of these marks is $25.5.$ It is also known that two students got equal marks in one subject and all the remaining marks are different. Find the rank correlation coefficient.
AnswerHere, $n=7$ and $\Sigma d^{2}=25.5$
Two students got equal marks in a subject $(\therefore m=2) .$
So, we can say that there is a tie in assigning the ranks.
Therefore, we need to take the term $\left(\frac{m^{3}-m}{12}\right)$ only once to obtain $CF.$
$ \mathrm{CF}=\left(\frac{m^{3}-m}{12}\right) =\left(\frac{2^{3}-2}{12}\right)=0.5$
$r =1-\frac{6\left[\Sigma d^{2}+C F\right]}{n\left(n^{2}-1\right)}$
$ =1-\frac{6[25.5+0.5]}{7(49-1)}$
$ =1-\frac{6(26)}{336}$
$ =1-\frac{156}{336}$
$ =1-0.4643$
$ =0.54 $
View full question & answer→Question 24 Marks
To study the relation between the age $(X$ years of teenage children and their daily requirement of protein $0*$ grams$),$ the following information is obtained from a sample of $10$ children taken by the Health Department of State.
$\Sigma x=140, \Sigma y=150, \Sigma(x-10)^{2}=180, \quad \Sigma(y-15)^{2}=215, \Sigma(x-10)(y-15)=60$
Find the correlation coefficient between $X$ and $Y.$
AnswerHere, $\bar{x}=\frac{\Sigma x}{n}=\frac{140}{10}=14, \quad \bar{y}=\frac{\Sigma y}{n}=\frac{150}{10}=15$
We can see that the deviations are not taken from actual mean $(\bar{x}=14)$ for the variable $X$.
So, to solve the example, it will be convenient to define $u=(x-A)=(x-10)$ and $v=(y-B)=(y-15)$.
We are given the following information.
$\Sigma(x-10)^{2}=\Sigma u^{2}=180, \Sigma(y-15)^{2}=\Sigma v^{2}=215, \Sigma(x-10)(y-15)=\Sigma u v=60$
Now, in order to use an appropriate formula of $r$, first we need $\Sigma u$ and $\Sigma v$.
$\Sigma u=\Sigma(x-10)=\Sigma x-\Sigma 10=\Sigma x-n(10)=140-10(10)=140-100=40$
$\Sigma v=\Sigma(y-15)=\Sigma y-\Sigma 15=\Sigma y-n(15)=150-10(15)=150-150=0$
$\{\because \Sigma k=\underbrace{k+k+k+\ldots \ldots \ldots+k}_{n \text { times }}=n k$ where, $k=$ constant $\}$
Substituting the above values in the following formula,
$ r =\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2}-(\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2}-(\Sigma v)^{2}}}$
$ =\frac{10(60)-(40)(0)}{\sqrt{10(180)-(40)^{2}} \cdot \sqrt{10(215)-(0)^{2}}}$
$ =\frac{600-0}{\sqrt{1800-1600} \cdot \sqrt{2150-0}}$
$ =\frac{600}{\sqrt{200} \cdot \sqrt{2150}}$
$ =\frac{600}{\sqrt{430000}}$
$ =\frac{600}{655.7439}$
$ =0.9150$
$\therefore {~r} =0.92 $
View full question & answer→Question 34 Marks
A project is conducted by the group of the students of an $MBA$ Institute to know the relation between the results of the final year of school and final year of graduation for the students. The following information is obtained from a sample of $10$ students regarding the percentage of marks in standard $12 (x)$ and the percentage of marks in the final year of graduation $(y).$
$n=10, \Sigma(x-65)=-2, \Sigma(y-60)=2, \Sigma(x-65)^{2}=176, \Sigma(y-60)^{2}=140, \Sigma(x-65)(y-60)=141$
Find the correlation coefficient between the percentages of marks in Standard $12$ and the final year of graduation.
Answer$\text { Here } \Sigma(x-65)=-2 \neq 0$
$\therefore A=65$
$\Sigma(y-60)=2 \neq 0$
$\therefore B=60$
$($Here, the sum of deviations are not zero, so $65 \neq \bar{x}$ and $60 \neq \bar{y} )$
Now, let us define $u=(x-65)$ and $v=(y-60)$.
${\text{So}} ~\Sigma(x-65)=\Sigma u=-2, \Sigma(y-60)=\Sigma v=2$
$\Sigma(x-65)^{2}=\Sigma u^{2}=176, \Sigma(y-60)^{2}=\Sigma v^{2}=140$
$\Sigma(x-65)(y-60)=\Sigma u v=141$
Substituting the above values in the following formula,
$ r =\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2}-(\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2}-(\Sigma v)^{2}}}$
$ =\frac{10(141)-(-2)(2)}{\sqrt{10(176)-(-2)^{2}} \cdot \sqrt{10(140)-(2)^{2}}}$
$ =\frac{1414}{\sqrt{1756} \cdot \sqrt{1396}}$
$ =\frac{1414}{\sqrt{2451376}}$
$ =\frac{1414}{1565.6871}$
$ =0.9031$
$\therefore \simeq 0.90 $
View full question & answer→Question 44 Marks
An educationalist has conducted an experiment to know the relation between the usage of Social Media in mobile phone and the result of the examination. A group of $10$ students is selected for this and the following results were obtained regarding, the time spent $x ($in hours$)$ in last week on Social Media and the marks $(2)$ obtained out of $50$ in the examination, taken immediately after it.
$\Sigma x=133, \Sigma y=220, \Sigma x^{2}=2344, \Sigma y^{2}=6500$ and $\Sigma x y=3500$
Later on, it was found that one of the pairs of observations of $X$ and $Y$ was taken as $(13, 20)$ instead of $(15, 25) .$
Find the correct value of the correlation coefficient between $X$ and $Y.$
AnswerHere, $n=10, \Sigma x=133, \Sigma y=220, \Sigma x^{2}=2344, \Sigma y^{2}=6500$ and $\Sigma x y=3500$
Incorrect Pair $: (l 3, 20)$ Correct Pair $: (15,25)$
Now, we find corrected values of these measures as follows :
$\Sigma x=133-13+15=135$
$\Sigma y=220-20+25=225$
$\Sigma x^{2}=2344-(13)^{2}+(15)^{2}=2344-169+225=2400$
$\Sigma y^{2}=6500-(20)^{2}+(25)^{2}=6500-400+625=6725$
$\Sigma x y=3500-(13 \times 20)+(15 \times 25)=3500-260+375=3615$
Substituting these corrected values in the following formula,
$ r =\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{\sqrt{n \Sigma x^{2}-(\Sigma x)^{2}} \cdot \sqrt{n \Sigma y^{2}-(\Sigma y)^{2}}}$
$ =\frac{10(3615)-(135)(225)}{\sqrt{10(2400)-(135)^{2}} \cdot \sqrt{10(6725)-(225)^{2}}} $
$ =\frac{36150-30375}{\sqrt{24000-18225} \cdot \sqrt{67250-50625}}$
$ =\frac{5775}{\sqrt{5775} \cdot \sqrt{16625}}$
$ =\frac{5775}{\sqrt{96009375}}$
$ =\frac{5775}{9798.4374}$
$ =0.5894$
$\therefore r \simeq 0.59 $
View full question & answer→Question 54 Marks
Ten students selected from various schools of a district were ranked on the basis of their proficiency in Sports and General knowledge. The rank correlation coefficient obtained from the data was found to be $0.2.$ Later on, it was noticed that the difference in the ranks of the two attributes for one of the students was taken as $3$ instead of $2.$ Find the correct value of rank correlation coefficient.
AnswerHere, $n=10$
Incorrect $d=3$
Correct $d=2$
Now, $ r=1-\frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}$
$\therefore 0.2 \quad=1-\frac{6 \Sigma d^{2}}{10(100-1)}$
$\therefore 0.2 \quad=1-\frac{6 \Sigma d^{2}}{990}$
$\therefore \frac{6 \Sigma d^{2}}{990}=1-0.2$
$\therefore \frac{6 \Sigma d^{2}}{990}=0.8$
$\therefore \Sigma d^{2} \quad=\frac{0.8 \times 990}{6}$
$\therefore \Sigma d^{2}$
Since one difference $2$ is wrongly taken as $3,$
the corrected value of $\Sigma d^{2}$ is obtained as follows : $ \text { Corrected }\Sigma d^{2} =132-(\text { Wrong } d)^{2}+(\text { Correct } d)^{2}$
$ =132-3^{2}+2^{2}$
$ =132-9+4$
$ =127$
$\therefore$ Correct value of the rank correlation coefficient is obtained as follows :$r =1-\frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}$
$ =1-\frac{6(127)}{10(100-1)}$
$ =1-\frac{762}{990}$
$ =1-0.7697$
$ =0.2303$
$\therefore r=0.23$
View full question & answer→Question 64 Marks
To know the relationship between the abilities of the students of a school in the subjects of Statistics and Accountancy, the teachers of both the subjects have taken a sample of eight students and the following information was collected.
Calculate the rank correlation coefficient between the marks of Statistics and the marks of Accountancy.
Question 74 Marks
The owner of chain of mobile phone shops selling the mobile phones of various brands invited an expert person. The owner asked him to check and rank ten different mobile phones regarding its camera and battery efficiency. The ranks given by the expert to $10$ mobile phones of different brands are given below.
Find the rank correlation coefficient between the efficiency of camera and battery of the mobile phones. View full question & answer→Question 84 Marks
Find the value of r from the following data. $(1) n=20, \operatorname{Cov}(x, y)=-50, s_{x}=15, s_{y}=8$ (2) $n=10, \Sigma(x-\bar{x})(y-\bar{y})=60$, variance of $X=25$, variance of $Y=36 (3)$
$(4) n=10, \Sigma x y=1500$, mean of $X=12$, mean of $Y=15, S.D.$ of $X=9, S.D.$ of $Y=5$. Answer$(1)$ Here, $n=20, \operatorname{Cov}(x, y)=-50, s_{x}=15, s_{y}=8$
Substituting these values in the following formula,
$r =\frac{\operatorname{Cov}(x, y)}{s_{x} \cdot s_{y}}$
$ =\frac{-50}{(15)(8)}$
$ =\frac{-50}{120}$
$\therefore r \simeq-0.4167$
$ \simeq-0.42\ (2)$
Here, $n=10, \Sigma(x-\bar{x})(y-\bar{y})=60$
Variance of $X=s_{x}^{2}=25 \therefore s_{x}=5$
Variance of $Y=s_{y}^{2}=36 \therefore s_{y}=6$
Substituting the required values in the following suitable formula,
$r =\frac{\Sigma(x-\bar{x})(y-\bar{y})}{n s_{x} s_{y}}$
$ =\frac{60}{10(5)(6)}$
$ =\frac{60}{300}$
$r =0.2\ (3)$
Here, $n=25, \bar{x}=40, \bar{y}=50, \Sigma(x-\bar{x})^{2}=120, \Sigma(y-\bar{y})^{2}=160$ and
$\Sigma(x-\bar{x})(y-\bar{y})=100$ Substituting all these values in the following suitable formula,
$r =\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}}$
$ =\frac{100}{\sqrt{120} \cdot \sqrt{160}}$
$ =\frac{100}{\sqrt{19200}}$
$ =\frac{100}{138.5641}$
$\therefore r \simeq 0.7217$
$ \simeq 0.72\ (4)$
Here, $n=10, \Sigma x y=1500, \bar{x}=12, \bar{y}=15, s_{x}=9$ and $s_{y}=5$
Substituting all these values in the following suitable formula,
$r =\frac{\sum x y-n \bar{x} \bar{y}}{n \cdot s_{x} \cdot s_{y}}$
$ =\frac{1500-10(12)(15)}{10(9)(5)}$
$ =\frac{1500-1800}{450}$
$ =\frac{-300}{450}$
$\therefore =-0.6667$
$ \simeq-0.67$
View full question & answer→Question 94 Marks
To know the relation between the results of the Tests taken in a span of short time, a teacher has conducted two Tests in last two weeks and the ranks obtained by seven students are as follows:
Find the rank correlation coefficient to know the similarity between the results of two examinations.
AnswerHere, $n = 7; R_x =$
Ranks in Test $1; R_y =$ Ranks in Test $2.$
In test $1$ rank $3.5$ comes twice. So $CF$ is calculated.
The table for calculating the rank correlation coefficient is prepared as follows :
Rank Correlation Coefficient :
$CF =\frac{m^{3}-m}{12}$. Here, $m = 2$
$\therefore CF = \frac{8-2}{12} = \frac{6}{12} = 0.5$
$r = 1 – \frac{6\left[\Sigma d^{2}+\mathrm{CF}\right]}{n\left(n^{2}-1\right)}$
Putting $n = 7; \sum d^2 = 48.50$ and $CF = 0.5$ in the formula,
$r = 1 – \frac{6(48.5+0.5)}{7\left(7^{2}-1\right)}$
$= 1 – \frac{6(49)}{7(49-1)}$
$= 1 – \frac{294}{336}$
$= 1 – 0.875$
$= 0.125 ≈ 0.13$
Hence the rank correlation coefficient between the results of two tests obtained is $0.13.$ View full question & answer→Question 104 Marks
The demand of an imported fruit in a local market is very uncertain. To know the relation between the price of the fruit and its supply, a vendor collects the information about the average price and supply for last ten months:
Find the rank correlation between the average price and the supply.
AnswerHere, $n = 10; R_x =$ Ranks for average price per unit $(x)$ and $R_y =$ Ranks for supply $(y).$
The table for calculation the rank correlation coefficient is prepared as follows :
Rank Correlation Coefficient :
$r = 1 – \frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}$
Putting $n = 10$ and $\sum d^2 = 76$ in the formula,
$r = 1 – \frac{6(76)}{10\left(10^{2}-1\right)}$
$= 1 – \frac{456}{10(100-1)}$
$= 1 – \frac{456}{990}$
$= 1 – 0.46$
$= 0.54$
Hence the rank correlation coefficient between the average price and the supply obtained is $0.54.$ View full question & answer→Question 114 Marks
A merchant wants to study the relation between prices of tea and coffee in Ahmedabad city. He obtains the following information about prices of tea and coffee of the last six months:
Calculate the rank correlation coefficient between the price of tea and coffee.
AnswerHere, $n = 6; R_x =$ Ranks for tea $(x)$ and $R_y=$ Rank for coffee $(y).$
The table for calculating the rank correlation coefficient is prepared as follows:
Rank Correlation Coefficient :
$r = 1 – \frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}$
Putting $n = 6$ and $\sum d^2 = 8$ in the formula,
$r = 1 – \frac{6(8)}{6\left(6^{2}-1\right)}$
$= 1 – \frac{48}{6(36-1)}$
$= 1 – \frac{48}{210}$
$= 1 – 0.23$
$= 0.77$
Hence, the rank correlation coefficient between the price of tea and coffee obtained is $0.77.$ View full question & answer→Question 124 Marks
A vendor wants to display lipsticks of different brands according to their popularity. For that, he invites two experts Preyal and Nishi to rank the lipsticks of different brands:
Find the rank correlation coefficient to know the smilarity in the decision of both the experts.
AnswerHere, $n=7 ; R_x=$ Rank by Preyal and $R_y=$ Rank by Nishi.
The table for calculating rank correlation coefficient is prepared as follows:
Rank Correlation Coefficient :
$r = 1 – \frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}$
Putting $n = 7$ and $\sum d^2 = 12$ in the formula,
$r = 1 – \frac{6(12)}{7\left(7^{2}-1\right)}$
$= 1 – \frac{72}{7(49-1)}$
$= 1 – \frac{72}{336}$
$= 1 – 0.21$
$= 0.79$
Hence, the rank correlation coefficient between two experts obtained is $0.79.$ View full question & answer→Question 134 Marks
In order to study the relationship between the abilities in the subjects of Human Resource Management and Personality Development for the students of a post graduate level course, a sample of $5$ students is taken and the following information is obtained:
Calculate the Karl Pearson’s correlation coefficient between the marks of both the subjects. AnswerHere, $n = 5; x =$ Marks in $HRM; y =$ Marks in $PD$
Now, $x̄ = \frac{\Sigma x}{n}$ = $\frac{175}{5} = 35$ marks; ȳ = $\frac{\Sigma y}{n}=\frac{170}{5} = 34$ marks
$x̄$ and $ȳ$ are integer. So the table for calculating $r$ is prepared as follows :
Correlation Coefficient:
$r = $
Putting $\sum (x – x̄) (y – ȳ) = 280, \sum (x – x̄)^2 = 550$ and $\sum (y – ȳ)^2 = 776$ in the formula,
$r =\frac{280}{\sqrt{550} \cdot \sqrt{776}}$
$= \frac{280}{\sqrt{426800}}$
$= \frac{280}{653.30}$
$= 0.43$
Hence, the correlation coefficient between the marks of both $HRM$ and $PD$ obtained is $0.43.$ View full question & answer→Question 144 Marks
The following information is obtained to study the relation between the selling price of nose mask and its demand during an epidemic:
Find the correlation coefficient between the price and demand of mask by Karl Pearson’s method.
AnswerHere, $n = 5; x =$ Price; $y =$ Demand
Now, $x̄ = \frac{\Sigma x}{n}= \frac{200}{5} = ₹ 40;$
$y = \frac{\Sigma y}{n} = \frac{490}{5} = 98$ units
$x̄$ and $ȳ$ are integer. So the table for calculating $r$ is prepared as follows:
Correlation Coefficient:
Now, $r = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^{2}} \cdot \sqrt{\sum(y-\bar{y})^{2}}}$
Putting $\sum (x – x̄) (y – ȳ) = – 50, \sum (x – x̄)^2 \sum (y – ȳ)^2 = 66$ in the formula,
$r = \frac{-50}{\sqrt{58} \cdot \sqrt{66}}$
$= \frac{-50}{\sqrt{3828}}$
$= \frac{-50}{61.87}$
$= – 0.81$
Hence, the correlation coefficient between the price and demand of mask obtained is $– 0.81.$ View full question & answer→Question 154 Marks
An official has ranked nine villages of a sample on the basis of the work done in the area of ‘Swachhata Abhiyan’ and ‘Beti Bachavo Abhiyan’ by the villages. The ranks are given below:
Find the rank correlation coefficient between the performances of the villages in two Abhiyans.
AnswerHere, $n = 9; R_x =$ Ranks for Swachhata Abhiyan and $R_y =$ Ranks for Beti Bachavo Abhiyan
The table for calculating rank correlation coefficient is prepared as follows:
Rank correlation coefficient: $ r = 1 – \frac{6 \Sigma d^{2}}{n(n^{2}-1)}$
Putting $n = 9$ and $\sum d^2 = 26$ in the formula,
$r = 1 – \frac{6(26)}{9\left(9^{2}-1\right)}$
$= 1 – \frac{156}{9(81-1)}$
$= 1 – \frac{156}{9 \times 80}$
$= 1 – \frac{156}{720}$
$= 1 – 0.22 = 0.78$
Hence, the rank correlation coefficient between the performances of the villages in two Abhiyans obtained is $0.78$. View full question & answer→Question 164 Marks
Six companies are ranked by the two market analysts on the basis of their growth in the recent past:
Find the rank correlation coefficient between the evaluation given by two analysts.
AnswerHere, $n = 6; Rx =$ Ranks by Analyst $1; Ry =$ Ranks by Analyst $2.$
The table for calculating rank correlation coefficient is prepared as follows :
Rank correlation coefficient:$r = 1 – \frac{6 \Sigma d^{2}}{n(n^{2}-1)}$
Putting $n = 6$ and $\sum d^2 =18$ in the formula.
$r = 1 – \frac{6(18)}{6\left(6^{2}-1\right)}$
$= 1 –\frac{108}{6(36-1)}$
$= 1 – \frac{108}{6 \times 35}$
$= 1 – \frac{108}{210}$
$= 1 – 0.51= 0.49$
Hence, the rank correlation coefficient between the evaluation given by two analysts obtained is $0.49.$ View full question & answer→Question 174 Marks
Determine the value of correlation coefficient from the following data:
$(1)\ \sum (x – x̄)^2 = 72, \sum (y – ȳ)^2 = 32,$
$\sum (x – x̄) (y – ȳ) = 45$
$(2)\ n = 6, \sum x = 16, \sum y = 51, \sum xy = 154,$
$\sum x^2 = 52, \sum y^2 = 471$
Answer$(1)$ Here, $\sum (x – x̄)^2 = 72; \sum (y – ȳ)^2 = 32$ and $Σ(x – x̄) (y – ȳ) = 45.$
$= \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}}$
$= \frac{45}{\sqrt{72} \times \sqrt{32}}$
$= \frac{45}{\sqrt{2304}}$
$= \frac{45}{48}$
$= 0.94$
Hence, the correlation coefficient obtained is $0.94.$
$(2)\ \sum x^2 = 52, \sum y^2 = 471$
Here, $n = 6; \sum x = 16; \sum y = 51; \sum xy = 154; \sum x^2 = 52$ and $\sum y^2 = 471.$
Hence, the correlation coefficient obtained is $0.96.$ View full question & answer→Question 184 Marks
The data on ranks for monthly expenditure $(X)$ and monthly saving $(Y)$ of $9$ families of an area is as follows: $\left(R_{X}, R_{Y}\right):(1,8),(3,7),(5,4),(3,9),(3,6),(6,3),(8,1),(7,2),(9,5)$ From the above find rank correlation coefficient.
View full question & answer→Question 194 Marks
The ordered pairs $\left(x_{i}, y_{i}\right)$ of two random variabis $X$ and $Y$ are given below: $(-2,-1),(1,0),(-3,-4),(0,2),(4,3),(5,7),(-1,0),(6,6),(2,-1),(-5,-3)$. Calculate Karl Pearson's coefficient of correlation.
View full question & answer→Question 204 Marks
Random variables $X$ and $Y$ are related to each other by the equation $y = 10-3x.$ Taking $x = 0, 1, 2, 3, 4, 5,$ find the coefficient of correlation between $X$ and $Y.$
Answer$- 1,$ Perfect negative correlation
View full question & answer→