GSEB Class 12 Statistics Solutions Chapter 2 Linear Correlation Exercise 2.3

GSEB Solutions Class 12 Statistics Part 1 Chapter 2 Linear Correlation Ex 2.3

 

Question 1. Six companies are ranked by two market analysts based on their recent growth. Find the rank correlation coefficient between the evaluations given by these two analysts.

Company	A	B	C	D	E	F
Rank by Analyst 1	5	2	1	4	3	6
Rank by Analyst 2	6	4	3	2	1	5

Answer:Here, the number of companies \( n = 6 \). \( R_x \) represents the ranks given by Analyst 1, and \( R_y \) represents the ranks given by Analyst 2.
The following table helps calculate the rank correlation coefficient:

Company	\(R_x\)	\(R_y\)	\(d = R_x - R_y\)	\(d^2\)
A	5	6	-1	1
B	2	4	-2	4
C	1	3	-2	4
D	4	2	2	4
E	3	1	2	4
F	6	5	1	1
Total	\(n = 6\)	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 18 \)

The rank correlation coefficient \( r \) is calculated using the formula:
\( r = 1 - \frac{6 \Sigma d^2}{n(n^2-1)} \)
Substitute \( n = 6 \) and \( \Sigma d^2 = 18 \) into the formula:
\( r = 1 - \frac{6(18)}{6(6^2-1)} \)
\( = 1 - \frac{108}{6(36-1)} \)
\( = 1 - \frac{108}{6 \times 35} \)
\( = 1 - \frac{108}{210} \)
\( = 1 - 0.51 \)
\( = 0.49 \)
Therefore, the rank correlation coefficient between the evaluations from the two analysts is \( 0.49 \).
In simple words: We calculated how much the two analysts' rankings agree. We found the differences in ranks, squared them, summed them up, and then used a formula. The result shows a moderate positive relationship between their rankings.

🎯 Exam Tip: Remember to clearly show the calculation of \(d\) and \(d^2\) in a table. Ensure the formula for rank correlation is correctly applied and calculations are accurate to avoid common errors.

 

Question 2. An official ranked nine sample villages based on their work in the 'Swachhata Abhiyan' and 'Beti Bachavo Abhiyan' programs. The ranks are provided below. Calculate the rank correlation coefficient between the villages' performance in these two initiatives.

Village	1	2	3	4	5	6	7	8	9
Rank for Swachhata Abhiyan	4	8	7	1	9	5	6	2	3
Rank for Beti Bachavo Abhiyan	6	8	5	1	9	7	3	4	2

Answer:Here, the number of villages \( n = 9 \). \( R_x \) represents the ranks for Swachhata Abhiyan, and \( R_y \) represents the ranks for Beti Bachavo Abhiyan.
The table below is set up for calculating the rank correlation coefficient:

Village	\(R_x\)	\(R_y\)	\(d = R_x - R_y\)	\(d^2\)
1	4	6	-2	4
2	8	8	0	0
3	7	5	2	4
4	1	1	0	0
5	9	9	0	0
6	5	7	-2	4
7	6	3	3	9
8	2	4	-2	4
9	3	2	1	1
Total	\(n = 9\)	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 26 \)

The rank correlation coefficient \( r \) is calculated using the formula:
\( r = 1 - \frac{6 \Sigma d^2}{n(n^2-1)} \)
Substitute \( n = 9 \) and \( \Sigma d^2 = 26 \) into the formula:
\( r = 1 - \frac{6(26)}{9(9^2-1)} \)
\( = 1 - \frac{156}{9(81-1)} \)
\( = 1 - \frac{156}{9 \times 80} \)
\( = 1 - \frac{156}{720} \)
\( = 1 - 0.22 \)
\( = 0.78 \)
Therefore, the rank correlation coefficient between the villages' performance in the two Abhiyans is \( 0.78 \).
In simple words: We assessed how well the villages performed similarly in two different programs by comparing their ranks. After calculating the differences and using a specific formula, we found a strong positive connection, meaning villages that ranked high in one program also tended to rank high in the other.

🎯 Exam Tip: Always double-check your subtraction for \(d\) values and squaring for \(d^2\). A common mistake is to miss that \( \Sigma d \) should always be zero for correctly calculated ranks, which serves as a quick check.

 

Question 3. A town planning committee conducted a survey and obtained the following information. Determine the rank correlation coefficient between the population of the cities and their rate of population growth.

City	Population (in lakh) (x)	Rate of growth (per thousand) (y)
A	57	13
B	45	20
C	14	10
D	18	15
E	8	5

Answer:Here, the number of cities \( n = 5 \). \( R_x \) represents the ranks for population (x), and \( R_y \) represents the ranks for the rate of growth (y).
The following table shows the preparation for calculating the rank correlation coefficient:

City	Population (in lakh) (x)	Rate of growth (per thousand) (y)	Rank of x (\(R_x\))	Rank of y (\(R_y\))	\(d = R_x - R_y\)	\(d^2\)
A	57	13	1	3	-2	4
B	45	20	2	1	1	1
C	14	10	4	4	0	0
D	18	15	3	2	1	1
E	8	5	5	5	0	0
Total	\(n = 5\)	-	-	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 6 \)

The rank correlation coefficient \( r \) is calculated using the formula:
\( r = 1 - \frac{6 \Sigma d^2}{n(n^2-1)} \)
Substitute \( n = 5 \) and \( \Sigma d^2 = 6 \) into the formula:
\( r = 1 - \frac{6(6)}{5(5^2-1)} \)
\( = 1 - \frac{36}{5(25-1)} \)
\( = 1 - \frac{36}{5 \times 24} \)
\( = 1 - \frac{36}{120} \)
\( = 1 - 0.3 \)
\( = 0.7 \)
Therefore, the rank correlation coefficient between the population of the cities and their rate of population growth is \( 0.7 \).
In simple words: We found how population size relates to how fast a city grows by giving ranks to each and then using a formula. The result shows a fairly strong positive link, meaning larger cities tend to have higher growth rates.

🎯 Exam Tip: When ranking, assign rank 1 to the highest value, then 2 to the next highest, and so on. Be careful with ties; if there are any, use the average rank for the tied observations, which would require a correction factor in the formula.

 

Question 4. The following information was collected from a sample of ten students at a Science college. Find the rank correlation coefficient between the students' abilities in Mathematics and Statistics subjects.

Student	1	2	3	4	5	6	7	8	9	10
Marks in Mathematics	39	65	62	90	82	75	25	98	36	78
Marks in Statistics	47	53	58	86	62	68	60	91	51	84

Answer:Here, the number of students \( n = 10 \). \( R_x \) represents the ranks for marks in Mathematics (x), and \( R_y \) represents the ranks for marks in Statistics (y).
The following table illustrates the calculation for the rank correlation coefficient:

Student	Marks in Mathematics (x)	Marks in Statistics (y)	Ranks of x (\(R_x\))	Ranks of y (\(R_y\))	\(d = R_x - R_y\)	\(d^2\)
1	39	47	8	10	-2	4
2	65	53	6	8	-2	4
3	62	58	7	7	0	0
4	90	86	2	2	0	0
5	82	62	3	5	-2	4
6	75	68	5	4	1	1
7	25	60	10	6	4	16
8	98	91	1	1	0	0
9	36	51	9	9	0	0
10	78	84	4	3	1	1
Total	\(n = 10\)	-	-	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 30 \)

The rank correlation coefficient \( r \) is calculated using the formula:
\( r = 1 - \frac{6 \Sigma d^2}{n(n^2-1)} \)
Substitute \( n = 10 \) and \( \Sigma d^2 = 30 \) into the formula:
\( r = 1 - \frac{6(30)}{10(10^2-1)} \)
\( = 1 - \frac{180}{10(100-1)} \)
\( = 1 - \frac{180}{990} \)
\( = 1 - 0.18 \)
\( = 0.82 \)
Therefore, the rank correlation coefficient between students' ability in Mathematics and Statistics is \( 0.82 \).
In simple words: We checked if students who do well in Math also do well in Statistics by giving ranks for their marks in both subjects. The result shows a strong positive connection, meaning a student's performance in one subject is highly related to their performance in the other.

🎯 Exam Tip: Pay close attention to assigning ranks correctly. If two scores are the same (tied ranks), the problem will require a correction factor (CF). In this problem, there are no tied ranks, so the basic formula works.

 

Question 5. Given the following information on the heights of husbands and wives, calculate the rank correlation coefficient between their heights.

Height of husband (cms)	156	153	185	157	163	191	162
Height of wife (cms)	154	148	162	157	162	170	154

Answer:Here, the number of pairs \( n = 7 \). \( R_x \) represents the ranks for the height of the husband (x), and \( R_y \) represents the ranks for the height of the wife (y).
The following table shows the calculation for the rank correlation coefficient:

Height of husband (x)	Height of wife (y)	Rank of x (\(R_x\))	Rank of y (\(R_y\))	\(d = R_x - R_y\)	\(d^2\)
156	154	6	5.5	0.5	0.25
153	148	7	7	0	0
185	162	2	2.5	-0.5	0.25
157	157	5	4	1	1.00
163	162	3	2.5	0.5	0.25
191	170	1	1	0	0
162	154	4	5.5	-1.5	2.25
Total	\(n = 7\)	-	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 4.00 \)

In this problem, two observations of variable y (height of wife) are repeated twice: \(162\) cms (at ranks 2 and 3, average rank \((2+3)/2 = 2.5\)) and \(154\) cms (at ranks 5 and 6, average rank \((5+6)/2 = 5.5\)). Therefore, a correction factor (CF) must be calculated.
The CF is calculated as follows:

Repeated variable	No. of repetition (m)	\(\frac{m^3-m}{12}\)
162	2	\(\frac{2^3-2}{12} = \frac{8-2}{12} = \frac{6}{12} = 0.5\)
154	2	\(\frac{2^3-2}{12} = \frac{8-2}{12} = \frac{6}{12} = 0.5\)
Total	-	\( \text{CF} = \Sigma \left(\frac{m^3-m}{12}\right) = 0.5 + 0.5 = 1 \)

Now, the rank correlation coefficient \( r \) (with correction for tied ranks) is calculated using the formula:
\( r = 1 - \frac{6(\Sigma d^2 + \text{CF})}{n(n^2-1)} \)
Substitute \( n = 7 \), \( \Sigma d^2 = 4 \), and \( \text{CF} = 1 \) into the formula:
\( r = 1 - \frac{6(4+1)}{7(7^2-1)} \)
\( = 1 - \frac{6(5)}{7(49-1)} \)
\( = 1 - \frac{30}{7 \times 48} \)
\( = 1 - \frac{30}{336} \)
\( = 1 - 0.09 \)
\( = 0.91 \)
Therefore, the rank correlation coefficient between the heights of husbands and wives is \( 0.91 \).
In simple words: We checked how closely the heights of husbands and wives are related. Because some heights were the same, we used a special adjustment in our calculations. The result shows a very strong positive connection, meaning taller husbands tend to have taller wives, and vice-versa.

🎯 Exam Tip: When dealing with tied ranks, accurately calculate the average rank for each set of tied observations. Remember to compute the correction factor (CF) for each tied value and add them up before applying the modified rank correlation formula.

 

Question 6. Two interviewers assigned scores to candidates based on their interview performance. Calculate the rank correlation coefficient between the evaluations provided by the two interviewers.

Candidate	A	B	C	D	E	F	G	H
Marks by first interviewer	28	44	10	28	47	35	19	40
Marks by second interviewer	32	45	25	32	41	32	24	38

Answer:Here, the number of candidates \( n = 8 \). \( R_x \) represents the ranks for marks given by the first interviewer (x), and \( R_y \) represents the ranks for marks given by the second interviewer (y).
The table for calculating the rank correlation coefficient is prepared as follows:

Candidate	Marks by first interviewer (x)	Marks by second interviewer (y)	Rank of x (\(R_x\))	Rank of y (\(R_y\))	\(d = R_x - R_y\)	\(d^2\)
A	28	32	5.5	5	0.5	0.25
B	44	45	2	1	1	1.00
C	10	25	8	7	1	1.00
D	28	32	5.5	5	0.5	0.25
E	47	41	1	2	-1	1.00
F	35	32	4	5	-1	1.00
G	19	24	7	8	-1	1.00
H	40	38	3	3	0	0.00
Total	\(n = 8\)	-	-	-	\( \Sigma d = 0 \)	\( \Sigma d^2 = 5.50 \)

Here, observation 28 of x (marks by first interviewer) is repeated twice (at ranks 5 and 6, average rank \((5+6)/2 = 5.5\)).
Observation 32 of y (marks by second interviewer) is repeated thrice (at ranks 4, 5, and 6, average rank \((4+5+6)/3 = 5\)).
The correction factor (CF) is calculated as follows:

Observation	No. of repetitions (m)	\(\frac{m^3-m}{12}\)
\(x = 28\)	2	\(\frac{2^3-2}{12} = \frac{8-2}{12} = \frac{6}{12} = 0.5\)
\(y = 32\)	3	\(\frac{3^3-3}{12} = \frac{27-3}{12} = \frac{24}{12} = 2.0\)
Total	-	\( \text{CF} = \Sigma \left(\frac{m^3-m}{12}\right) = 0.5 + 2.0 = 2.5 \)

Now, the rank correlation coefficient \( r \) (with correction for tied ranks) is calculated using the formula:
\( r = 1 - \frac{6(\Sigma d^2 + \text{CF})}{n(n^2-1)} \)
Substitute \( n = 8 \), \( \Sigma d^2 = 5.50 \), and \( \text{CF} = 2.5 \) into the formula:
\( r = 1 - \frac{6(5.50+2.50)}{8(8^2-1)} \)
\( = 1 - \frac{6(8)}{8(64-1)} \)
\( = 1 - \frac{48}{8 \times 63} \)
\( = 1 - \frac{48}{504} \)
\( = 1 - 0.10 \)
\( = 0.90 \)
Therefore, the rank correlation coefficient between the evaluations of the two interviewers is \( 0.90 \).
In simple words: We checked how much two interviewers agreed on their candidate scores. Since some candidates received the same scores, we used a special adjustment in our calculations. The result shows a very high agreement, meaning the interviewers generally ranked candidates very similarly.

🎯 Exam Tip: When multiple values are tied, ensure you calculate a separate \( \frac{m^3-m}{12} \) for each tied group and sum them up to get the total Correction Factor (CF). This step is crucial for accurate results with tied ranks.

 

Question 7. In a beauty contest, two judges ranked ten contestants. The sum of squares of differences in their ranks was found to be 214. Calculate the rank correlation coefficient.

Answer:Here, the number of contestants \( n = 10 \). The sum of squares of differences in ranks \( \Sigma d^2 = 214 \).
There is no information about tied ranks, so we will assume CF = 0.
The rank correlation coefficient \( r \) is calculated using the formula:
\( r = 1 - \frac{6(\Sigma d^2 + \text{CF})}{n(n^2-1)} \)
Substitute \( n = 10 \), \( \Sigma d^2 = 214 \), and \( \text{CF} = 0 \) into the formula:
\( r = 1 - \frac{6(214+0)}{10(10^2-1)} \)
\( = 1 - \frac{6(214)}{10(100-1)} \)
\( = 1 - \frac{1284}{10 \times 99} \)
\( = 1 - \frac{1284}{990} \)
\( = 1 - 1.30 \)
\( = -0.30 \)
Therefore, the rank correlation coefficient obtained is \( -0.30 \).
In simple words: We wanted to see how much two judges agreed on their rankings of contestants. We were given the sum of squared differences in ranks. Using the formula, we found a weak negative connection, meaning the judges had some disagreement in their rankings.

🎯 Exam Tip: Even if no ties are explicitly mentioned, it's good practice to understand the formula with the CF. If \( \Sigma d^2 \) is given directly, ensure you use the correct formula variant (with or without CF) based on problem context or explicit mention of ties.

 

Question 8. The rank correlation coefficient between the marks obtained by 10 students in two subjects was found to be 0.5. Later, it was discovered that one difference in ranks was 7, but it was incorrectly taken as 3. Determine the corrected value of the correlation coefficient.

Answer:Here, the number of students \( n = 10 \). The initial rank correlation coefficient \( r = 0.5 \).
The true difference in ranks \( d_{true} = 7 \). The false difference in ranks \( d_{false} = 3 \).
First, we need to find the incorrect \( \Sigma d^2 \) using the given \( r \). Assuming no ties initially (CF = 0).
\( r = 1 - \frac{6 \Sigma d^2}{n(n^2-1)} \)
\( 0.5 = 1 - \frac{6 \Sigma d^2}{10(10^2-1)} \)
\( 0.5 = 1 - \frac{6 \Sigma d^2}{10(100-1)} \)
\( 0.5 = 1 - \frac{6 \Sigma d^2}{10 \times 99} \)
\( 0.5 = 1 - \frac{6 \Sigma d^2}{990} \)
Now, rearrange to find \( \Sigma d^2 \):
\( \frac{6 \Sigma d^2}{990} = 1 - 0.5 \)
\( \frac{6 \Sigma d^2}{990} = 0.5 \)
\( 6 \Sigma d^2 = 0.5 \times 990 \)
\( 6 \Sigma d^2 = 495 \)
\( \Sigma d^2 = \frac{495}{6} \)
\( \Sigma d^2 = 82.5 \)
This is the incorrect sum of squared differences.
Now, we correct \( \Sigma d^2 \):
\( \text{Corrected } \Sigma d^2 = \text{Incorrect } \Sigma d^2 - (\text{False } d)^2 + (\text{True } d)^2 \)
\( \text{Corrected } \Sigma d^2 = 82.5 - (3)^2 + (7)^2 \)
\( = 82.5 - 9 + 49 \)
\( = 73.5 + 49 \)
\( = 122.5 \)
Now, calculate the corrected rank correlation coefficient using the corrected \( \Sigma d^2 \). Assuming no ties (CF = 0).
\( r_{corrected} = 1 - \frac{6 (\text{Corrected } \Sigma d^2)}{n(n^2-1)} \)
Substitute \( n = 10 \) and Corrected \( \Sigma d^2 = 122.5 \) into the formula:
\( r_{corrected} = 1 - \frac{6(122.5)}{10(10^2-1)} \)
\( = 1 - \frac{735}{10(100-1)} \)
\( = 1 - \frac{735}{990} \)
\( = 1 - 0.7424... \)
\( = 1 - 0.74 \) (approximately)
\( = 0.26 \)
Therefore, the corrected value of the rank correlation coefficient is \( 0.26 \).
In simple words: We initially found a correlation of 0.5 for students' marks. But then we realized we used a wrong number for a rank difference. We fixed this error by finding the correct sum of squared differences and recalculated the correlation. The new, correct correlation is 0.26, which is lower than before.

🎯 Exam Tip: For problems involving correction of errors, first use the incorrect correlation coefficient to find the incorrect \( \Sigma d^2 \). Then, adjust \( \Sigma d^2 \) by subtracting the square of the wrong difference and adding the square of the correct difference. Finally, recalculate the correlation with the corrected \( \Sigma d^2 \).