Samacheer Kalvi Class 11 Business Maths Solutions Chapter 9 Correlation and Regression Analysis Ex 9.1

 

Question 1. Calculate the correlation co-efficient for the following data:

Answer: First, we need to calculate the mean for X and Y values. After that, we find the deviations from the mean, square them, and then calculate their products. Finally, we use the formula for the coefficient of correlation to get the result.

Answer:
\(N = 10\)
\( \sum X = 60 \), \( \sum Y = 40 \)
\( \sum x^2 = 134 \), \( \sum y^2 = 52 \)
\( \sum xy = 48 \)
Now, we find the mean for X and Y:
\( \overline{X} = \frac{\sum X}{N} = \frac{60}{10} = 6 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{40}{10} = 4 \)
Next, we apply the formula for the Coefficient of correlation:
\( r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} \)
So, we substitute the values into the formula:
\( r = \frac{48}{\sqrt{134 \times 52}} \)
\( r = \frac{48}{\sqrt{6968}} \)
\( r = \frac{48}{83.474} \)
\( r = 0.575 \)
This positive correlation value shows that as X increases, Y tends to increase as well, indicating a moderate relationship between the variables.
In simple words: First, we find the average for X and Y. Then we use these averages to fill in the table by finding the difference from each value, squaring these differences, and multiplying them. Finally, we put the total sums into a special formula to find the correlation number, which is 0.575.

🎯 Exam Tip: Always make sure to calculate the means correctly, as errors there will affect all subsequent calculations for deviations and the final correlation coefficient.

 

Question 2. Find the coefficient of correlation for the following:

Answer: To find the coefficient of correlation, we first calculate the means for Cost (X) and Sales (Y). Then, we determine the deviations from these means and use them to fill out a table for \(x^2\), \(y^2\), and \(xy\). Finally, we apply the correlation formula using the sums from the table.

Answer:
\(N = 9\)
\( \sum X = 180 \), \( \sum Y = 360 \)
\( \sum x^2 = 120 \), \( \sum y^2 = 346 \)
\( \sum xy = 193 \)
Now, we find the mean for X and Y:
\( \overline{X} = \frac{\sum X}{N} = \frac{180}{9} = 20 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{360}{9} = 40 \)
Next, we apply the formula for the Coefficient of correlation:
\( r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} \)
So, we substitute the values into the formula:
\( r = \frac{193}{\sqrt{120 \times 346}} \)
\( r = \frac{193}{\sqrt{41520}} \)
\( r = \frac{193}{203.76} \)
\( r = 0.947 \)
The very high positive correlation of 0.947 indicates a strong direct relationship, meaning as cost increases, sales also tend to increase significantly.
In simple words: We find the average cost and average sales. Then, we create a table to calculate how much each cost and sales number is different from its average, square those differences, and multiply them. We add up all these numbers and use a special formula to get the correlation, which is 0.947.

🎯 Exam Tip: When dealing with cost and sales data, a strong positive correlation like this suggests a clear market trend, which is a useful insight for businesses.

 

Question 3. Calculate the coefficient of correlation for the ages of husbands and their respective wives:

Answer: We need to calculate the sum of husbands' ages (X), wives' ages (Y), their squares (\(X^2\), \(Y^2\)), and their product (\(XY\)). Then, we will use the combined formula for the coefficient of correlation that doesn't rely on deviations.

Answer:
\(N = 10\)
\( \sum X = 311 \), \( \sum Y = 257 \)
\( \sum X^2 = 9875 \), \( \sum Y^2 = 6763 \)
\( \sum XY = 8171 \)
Now, we use the Coefficient of correlation formula without deviations:
\( r = \frac{N \sum XY - (\sum X)(\sum Y)}{\sqrt{N \sum X^2 - (\sum X)^2} \sqrt{N \sum Y^2 - (\sum Y)^2}} \)
So, we substitute the values into the formula:
\( r = \frac{10 \times 8171 - 311 \times 257}{\sqrt{10 \times 9875 - (311)^2} \sqrt{10 \times 6763 - (257)^2}} \)
\( r = \frac{81710 - 79927}{\sqrt{98750 - 96721} \sqrt{67630 - 66049}} \)
\( r = \frac{1783}{\sqrt{2029} \sqrt{1581}} \)
\( r = \frac{1783}{45.04 \times 39.76} \)
\( r = \frac{1783}{1791.0} \)
\( r = 0.9965 \)
This extremely high positive correlation means that husbands' and wives' ages are very closely related, suggesting that people often marry partners who are close to their own age.
In simple words: We add up all the ages for husbands (X) and wives (Y). We also add up their squared ages and the product of their ages. Then, we put all these totals into a big formula to find the correlation number, which comes out to 0.9965.

🎯 Exam Tip: For problems involving actual values (not deviations), use the formula with \(N \sum XY\) in the numerator to avoid extra steps and potential rounding errors from calculating means first.

 

Question 4. Calculate the coefficient of correlation between X and Y series from the following data:

Answer: We are given summary statistics for X and Y series, including the number of observations, means, and sums of squares of deviations. We also have the summation of product deviations. We can directly apply the formula for the coefficient of correlation using these values.

Given:
\( N = 15 \)
\( \overline{X} = 25 \), \( \overline{Y} = 18 \)
\( \sum x^2 = 136 \), \( \sum y^2 = 138 \)
\( \sum xy = 122 \)
Now, we apply the formula for the Coefficient of correlation:
\( r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} \)
So, we substitute the values into the formula:
\( r = \frac{122}{\sqrt{136 \times 138}} \)
\( r = \frac{122}{\sqrt{18768}} \)
\( r = \frac{122}{136.996} \)
\( r = 0.891 \)
This strong positive correlation means that there is a significant linear relationship where X and Y tend to move in the same direction.
In simple words: We are given the sum of the multiplied deviations (\(\sum xy\)), the sum of the squared deviations for X (\(\sum x^2\)), and the sum of the squared deviations for Y (\(\sum y^2\)). We just plug these numbers into the correlation formula and calculate the result, which is 0.891.

🎯 Exam Tip: When summary data like sum of squares of deviation (\(\sum x^2\)) and sum of product deviations (\(\sum xy\)) are provided directly, use the simpler formula \(r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}}\) to save time.

 

Question 5. Calculate the correlation coefficient for the following data:

Answer: We begin by calculating the mean for X and Y. Using these means, we then determine the deviations for each X and Y value, square them, and find their product. Finally, we sum these values and substitute them into the formula to calculate the coefficient of correlation.

Answer:
\(N = 10\)
\( \sum X = 310 \), \( \sum Y = 340 \)
\( \sum x^2 = 870 \), \( \sum y^2 = 720 \)
\( \sum xy = 178 \)
Now, we find the mean for X and Y:
\( \overline{X} = \frac{\sum X}{N} = \frac{310}{10} = 31 \)
\( \overline{Y} = \frac{\sum Y}{N} = \frac{340}{10} = 34 \)
Next, we apply the formula for the Coefficient of correlation:
\( r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} \)
So, we substitute the values into the formula:
\( r = \frac{178}{\sqrt{870 \times 720}} \)
\( r = \frac{178}{\sqrt{626400}} \)
\( r = \frac{178}{791.45} \)
\( r = 0.225 \)
This positive correlation, though not very high, suggests a weak tendency for X and Y to increase together. It indicates a slight relationship between the two data sets.
In simple words: We calculate the average of X and Y. Then we make a table to find the differences from these averages, square them, and multiply them together. We add up these numbers and use a special formula to get the correlation, which is 0.225.

🎯 Exam Tip: Always double-check your arithmetic when summing columns to prevent small errors from propagating and affecting the final correlation coefficient significantly.

 

Question 6. Find the coefficient of correlation for the following:

Answer: We will calculate the sum of X and Y values to find their respective means. Then, we will determine the deviations from these assumed means, square them (\(dx^2\), \(dy^2\)), and calculate their product (\(dxdy\)). Finally, we will use the formula for correlation coefficient based on assumed means.

Answer:
\(N = 8\)
\( \sum X = 600 \), \( \sum Y = 724 \)
\( \sum dx^2 = 1172 \), \( \sum dy^2 = 3372 \)
\( \sum dxdy = -146 \)
Now, we apply the formula for the Coefficient of correlation using deviations from assumed mean:
\( r = \frac{N \sum dxdy - (\sum dx)(\sum dy)}{\sqrt{N \sum dx^2 - (\sum dx)^2} \sqrt{N \sum dy^2 - (\sum dy)^2}} \)
So, we substitute the values into the formula:
\( r = \frac{8(-146) - 0(4)}{\sqrt{8 \times 1172 - (0)^2} \sqrt{8 \times 3372 - (4)^2}} \)
\( r = \frac{-1168 - 0}{\sqrt{9376 - 0} \sqrt{26976 - 16}} \)
\( r = \frac{-1168}{\sqrt{9376} \sqrt{26960}} \)
\( r = \frac{-1168}{96.83 \times 164.20} \)
\( r = \frac{-1168}{15899.806} \)
\( r = -0.0735 \)
The very small negative correlation indicates that there is almost no linear relationship between X and Y, and if anything, they tend to move in opposite directions very slightly.
In simple words: We find the sum of X and Y, then pick an assumed average for each. We calculate the differences from these assumed averages, square them, and multiply them. We add up all these results and put them into a formula to get the correlation, which is -0.0735.

🎯 Exam Tip: When dealing with negative correlation, even a small negative value near zero still suggests that the variables might move in opposite directions, albeit weakly.

 

Question 7. An examination of 11 applicants for an accountant post was taken by a finance company. The marks obtained by the applicants in the reasoning and aptitude tests are given below.

Calculate Spearman's rank correlation coefficient from the data given above.

Answer: We need to assign ranks to the marks obtained in the Reasoning test (X) and Aptitude test (Y). Then, we will find the difference in ranks (\(d\)) and square these differences (\(d^2\)). Finally, we apply Spearman's rank correlation formula using the sum of \(d^2\).

Answer:
\(N = 11\)
\( \sum d^2 = 22 \)
Now, we apply the formula for Rank correlation:
\( \rho = 1 - \frac{6 \sum d^2}{N(N^2 - 1)} \)
So, we substitute the values into the formula:
\( \rho = 1 - \frac{6 \times 22}{11(11^2 - 1)} \)
\( \rho = 1 - \frac{132}{11(121 - 1)} \)
\( \rho = 1 - \frac{132}{11(120)} \)
\( \rho = 1 - \frac{132}{1320} \)
\( \rho = 1 - 0.1 \)
\( \rho = 0.9 \)
A rank correlation coefficient of 0.9 indicates a very strong positive relationship between the ranks in reasoning and aptitude tests, meaning applicants who score well in one test tend to score well in the other.
In simple words: First, we give a rank to each person's score in both tests. Then we find the difference between their ranks and square that difference. We add up all these squared differences. Finally, we put this total into a special formula to get the rank correlation, which is 0.9.

🎯 Exam Tip: Remember to assign ranks carefully; for tied scores, assign the average of the ranks that would have been given. In this problem, there are no tied ranks, simplifying the calculation.

 

Question 8. The following are the ranks obtained by 10 students in commerce and accountancy are given below:

To what extent is the knowledge of students in the two subjects related?

Answer: Since the ranks are already given, we can directly find the difference between the ranks (\(d\)) and square them (\(d^2\)). Then, we use Spearman's rank correlation formula with the sum of \(d^2\).

Answer:
\(N = 10\)
\( \sum d^2 = 128 \)
Now, we apply the formula for Rank correlation:
\( \rho = 1 - \frac{6 \sum d^2}{N(N^2 - 1)} \)
So, we substitute the values into the formula:
\( \rho = 1 - \frac{6 \times 128}{10(10^2 - 1)} \)
\( \rho = 1 - \frac{768}{10(100 - 1)} \)
\( \rho = 1 - \frac{768}{10(99)} \)
\( \rho = 1 - \frac{768}{990} \)
\( \rho = 1 - 0.7757 \)
\( \rho = 0.2243 \)
\( \rho \approx 0.224 \)
The rank correlation coefficient of 0.224 indicates a weak positive relationship between students' knowledge in commerce and accountancy, meaning there's a slight tendency for higher ranks in one subject to be associated with higher ranks in the other, but it's not a strong link.
In simple words: We are given the ranks for students in two subjects. We find the difference between these ranks for each student and then square these differences. We add all these squared differences and use a special formula to calculate the rank correlation, which is about 0.224.

🎯 Exam Tip: When ranks are already provided, ensure you correctly calculate the differences (\(d\)) and their squares (\(d^2\)) before applying the Spearman's formula. Positive correlation means ranks move in the same direction.

 

Question 9. A random sample of recent repair jobs was selected and the estimated cost and actual cost were recorded.

Calculate the value of spearman's correlation coefficient.

Answer: We first need to assign ranks to both the estimated costs (\(R_x\)) and actual costs (\(R_y\)). After that, we calculate the difference between these ranks (\(d\)) for each pair and square the differences (\(d^2\)). Finally, we sum all the \(d^2\) values and use Spearman's rank correlation formula.

Answer:
\(N = 8\)
\( \sum d^2 = 8 \)
Now, we apply the formula for Rank correlation:
\( \rho = 1 - \frac{6 \sum d^2}{N(N^2 - 1)} \)
So, we substitute the values into the formula:
\( \rho = 1 - \frac{6 \times 8}{8(8^2 - 1)} \)
\( \rho = 1 - \frac{48}{8(64 - 1)} \)
\( \rho = 1 - \frac{48}{8(63)} \)
\( \rho = 1 - \frac{48}{504} \)
\( \rho = 1 - \frac{1}{10.5} \)
\( \rho = 1 - 0.0952 \)
\( \rho = 0.9048 \)
\( \rho \approx 0.905 \)
A rank correlation of 0.905 indicates a very strong positive relationship between estimated and actual costs, suggesting that the estimated costs are highly accurate in predicting the actual costs for repair jobs.
In simple words: We give ranks to the estimated costs and actual costs. We find the difference between these ranks and square them. Then we add up all the squared differences. Finally, we use a special formula to find the rank correlation, which is 0.905.

🎯 Exam Tip: When assigning ranks to data, ensure that the smallest value gets rank 1 (or the largest, consistently) for both series to avoid errors in rank differences.

 

Question 10. The rank of 10 students of the same batch in two subjects A and B are given rank correlation coefficient.

Answer: Given the ranks of students in two subjects, we need to find the difference between these ranks (\(d\)) and square them (\(d^2\)). After summing all \(d^2\) values, we will apply Spearman's rank correlation formula to calculate the coefficient.

Answer:
\(N = 10\)
\( \sum d^2 = 226 \)
Now, we apply the formula for Rank correlation:
\( \rho = 1 - \frac{6 \sum d^2}{N(N^2 - 1)} \)
So, we substitute the values into the formula:
\( \rho = 1 - \frac{6 \times 226}{10(10^2 - 1)} \)
\( \rho = 1 - \frac{1356}{10(100 - 1)} \)
\( \rho = 1 - \frac{1356}{10(99)} \)
\( \rho = 1 - \frac{1356}{990} \)
\( \rho = 1 - 1.3696 \)
\( \rho = -0.3696 \)
\( \rho \approx -0.37 \)
A rank correlation coefficient of -0.37 indicates a weak negative relationship between the ranks in subjects A and B. This means that students who rank higher in one subject tend to rank slightly lower in the other, though the connection is not strong.
In simple words: We are given the ranks for students in two subjects. We find the difference between these ranks and square each difference. We add up all the squared differences. Then, we use a special formula to calculate the rank correlation, which is about -0.37.

🎯 Exam Tip: A negative correlation coefficient means that as the rank in one variable increases, the rank in the other tends to decrease, indicating an inverse relationship.