Updated NCERT Solutions for Class 11 Statistics Chapter 6: Correlation

Correlation is a statistical measure that expresses the extent to which two variables are linearly related, meaning they change together at a constant rate. It indicates both the direction (positive or negative) and the strength of the relationship between two variables.

Positive Correlation vs. Negative Correlation:

To create a scatter diagram, we will plot the 'Price of Sugar' on the X-axis and 'Supply' on the Y-axis. Each pair of (X, Y) values will be a point on the graph.

1. Point 1: (30, 100)
2. Point 2: (40, 120)
3. Point 3: (50, 150)
4. Point 4: (60, 160)
5. Point 5: (70, 180)

Scatter Diagram:

      |
  180 +                        *
      |                    *
  160 +
      |                *
  150 +
      |
  120 +         *
      |
  100 +    *
      |
  ----+--------------------------->
      30  40  50  60  70
      Price of Sugar (X)
            

Interpretation:

The points on the scatter diagram are moving in an upward direction from left to right. This indicates a positive relationship between the price of sugar and its supply. As the price of sugar increases, its supply also tends to increase. Since the points are relatively close to forming a straight line, it suggests a strong positive correlation.

We will use the Product Moment Method to calculate Karl Pearson's coefficient of correlation (r). The formula is:

\[ r = \frac{N\sum XY - (\sum X)(\sum Y)}{\sqrt{N\sum X^2 - (\sum X)^2} \cdot \sqrt{N\sum Y^2 - (\sum Y)^2}} \]

Step 1: Create a calculation table. Here, N (number of observations) = 5.

Step 2: Substitute the values into the formula.

\[ r = \frac{5 \times 1800 - (61)(130)}{\sqrt{5 \times 869 - (61)^2} \cdot \sqrt{5 \times 3750 - (130)^2}} \]

\[ r = \frac{9000 - 7930}{\sqrt{4345 - 3721} \cdot \sqrt{18750 - 16900}} \]

\[ r = \frac{1070}{\sqrt{624} \cdot \sqrt{1850}} \]

\[ r = \frac{1070}{24.98 \times 43.01} \]

\[ r = \frac{1070}{1074.39} \]

\[ r = 0.9959 \]

Interpretation: The value of r is +0.9959. This indicates a very strong positive linear correlation between variables X and Y.

Since there are no repeated values (tied ranks), we use the formula: \( R = 1 - \frac{6\sum D^2}{N(N^2-1)} \).

Step 1: Assign ranks and create a calculation table. Rank 1 is given to the highest value. N = 7.

Step 2: Substitute the values into the formula.

\[ R = 1 - \frac{6 \times 0}{7(7^2-1)} \]

\[ R = 1 - \frac{0}{7(48)} \]

\[ R = 1 - 0 \]

\[ R = +1 \]

Interpretation: The Spearman's Rank Correlation Coefficient is +1. This indicates a perfect positive correlation between the ranks of marks in History and Geography. The student with the highest rank in History also has the highest rank in Geography, and so on.

Spearman's rank correlation is preferred in the following situations:

• When the data is qualitative and cannot be measured numerically (e.g., beauty, honesty).
• When the data is given in the form of ranks.
• When the relationship between variables is not linear.
• When there are extreme values (outliers) in the data.

We use the formula \( r = \frac{\sum (dx)(dy)}{\sqrt{\sum(dx)^2 \cdot \sum(dy)^2}} \).

\[ r = \frac{55}{\sqrt{40 \times 90}} \]

\[ r = \frac{55}{\sqrt{3600}} \]

\[ r = \frac{55}{60} \]

\[ r = 0.917 \]

Thus, the correlation coefficient is +0.917.

If two variables are independent, there is no relationship between them. Therefore, the correlation coefficient between them will be zero (0).

Here, N=8. The ranks are already given. We calculate D and D².

\[ R = 1 - \frac{6\sum D^2}{N(N^2-1)} = 1 - \frac{6 \times 14}{8(8^2-1)} = 1 - \frac{84}{8(63)} = 1 - \frac{84}{504} \]

\[ R = 1 - 0.1667 = 0.833 \]

The rank correlation is +0.833, indicating a strong positive agreement between the two judges.

Here N = 7. We have tied ranks in both series.

• Ranking for X: Ranks for 50 (occurs 2 times) are 6, 7. Average rank = (6+7)/2 = 6.5. Value 40 is rank 2, 70 is rank 1.
• Ranking for Y: Ranks for 65 (occurs 2 times) are 4, 5. Average rank = (4+5)/2 = 4.5. Value 50 is rank 7, 80 is rank 1.

Correction Factor (CF) for X: \(m=2\). CF = \( \frac{1}{12}(2^3-2) = 0.5 \)

Correction Factor (CF) for Y: \(m=2\). CF = \( \frac{1}{12}(2^3-2) = 0.5 \)

Total CF = 0.5 + 0.5 = 1.0

Formula: \( R = 1 - \frac{6 [\sum D^2 + CF]}{N(N^2-1)} \)

\[ R = 1 - \frac{6 [33.5 + 1.0]}{7(7^2-1)} = 1 - \frac{6 \times 34.5}{7 \times 48} = 1 - \frac{207}{336} \]

\[ R = 1 - 0.616 = 0.384 \]

There is a low positive correlation between X and Y.

i. What type of correlation do you expect between study hours and test scores?

A positive correlation is expected. Generally, more hours of study lead to higher test scores.

ii. Calculate Karl Pearson's coefficient of correlation.

N=6. From calculation, we get: ΣX = 37, ΣY = 415, ΣX² = 263, ΣY² = 30175, ΣXY = 2795.

\[ r = \frac{6 \times 2795 - (37)(415)}{\sqrt{6 \times 263 - (37)^2} \cdot \sqrt{6 \times 30175 - (415)^2}} \]

\[ r = \frac{16770 - 15355}{\sqrt{1578 - 1369} \cdot \sqrt{181050 - 172225}} \]

\[ r = \frac{1415}{\sqrt{209} \cdot \sqrt{8825}} = \frac{1415}{14.45 \times 93.94} = \frac{1415}{1357.43} \]

\[ r = +0.985 \]

iii. Interpret the result obtained in part (ii).

The correlation coefficient is +0.985. This indicates a very strong positive linear relationship between the number of hours studied and the test scores obtained by the students.