Statistics and Correlation

Calculation of Regression Coefficients and Correlation Coefficient (2 marks)

Given regression equations:

[ Y=2X \quad \text{...(i)} ] [ 4X=3Y \quad \text{...(ii)} ]

Write both in the form (Y=mX):

[ \text{(i)}\quad Y=2X \quad \Rightarrow m_1=2 ] [ \text{(ii)}\quad 4X=3Y \quad \Rightarrow Y=\frac{4}{3}X \quad \Rightarrow m_2=\frac{4}{3} ]

In the form (Y=mX), the regression line of (X) on (Y) is steeper than the regression line of (Y) on (X). Since (2>\frac{4}{3}), equation (i) is the regression line of (X) on (Y), and equation (ii) is the regression line of (Y) on (X).

Therefore,

[ b_{yx}=\frac{4}{3} ]

From (Y=2X), we get (X=\frac{1}{2}Y), so

[ b_{xy}=\frac{1}{2} ]

Now,

[ r=\pm\sqrt{b_{yx}\times b_{xy}} ]

[ r=\sqrt{\frac{4}{3}\times \frac{1}{2}} =\sqrt{\frac{2}{3}} =0.8165 ]

Since both regression coefficients are positive, (r) is positive.

[ \boxed{r=0.8165} ]

Check:
[ b_{yx}\times b_{xy}=\frac{4}{3}\times \frac{1}{2}=\frac{2}{3}<1 ] So the result is valid. If we had taken (b_{yx}=2) and (b_{xy}=\frac{3}{4}), the product would be (\frac{3}{2}>1), which is impossible. Hence the correct identification is as above.

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Teaching Correlation Coefficient Using These Equations (8 marks)

1. Review and Motivation

• Recall scatter diagrams, positive and negative correlation, and regression lines.
• Ask: “If two regression lines are given, can we find the correlation coefficient?” This creates interest.

2. Introduce Regression Coefficients

Define:

• (b_{yx}) = regression coefficient of (Y) on (X)
• (b_{xy}) = regression coefficient of (X) on (Y)

Explain the relation:

[ b_{yx}=r\frac{\sigma_y}{\sigma_x},\quad b_{xy}=r\frac{\sigma_x}{\sigma_y} ]

Multiplying,

[ b_{yx}\times b_{xy}=r^2 ]

Therefore,

[ r=\pm\sqrt{b_{yx}\times b_{xy}} ]

The sign of (r) is the same as the sign of the regression coefficients.

3. Presentation Using the Given Equations

Write on the board:

[ Y=2X \quad \text{and} \quad 4X=3Y ]

Ask students to write both as (Y=mX):

[ Y=2X \quad \text{and} \quad Y=\frac{4}{3}X ]

Explain that in (Y=mX) form, the line of (X) on (Y) is steeper. Since (2>\frac{4}{3}), the steeper line (Y=2X) is the regression line of (X) on (Y). So,

[ X=\frac{1}{2}Y \Rightarrow b_{xy}=\frac{1}{2} ]

The less steep line (Y=\frac{4}{3}X) is the regression line of (Y) on (X). So,

[ b_{yx}=\frac{4}{3} ]

4. Calculate and Interpret

[ r=\sqrt{\frac{4}{3}\times \frac{1}{2}} =\sqrt{\frac{2}{3}} =0.8165 ]

Since (r>0), there is a high positive correlation between (X) and (Y). The value (0.8165) is close to (1), so the variables are strongly positively correlated.

5. Verify and Conclude

Check:

[ b_{yx}\times b_{xy}=\frac{4}{3}\times \frac{1}{2}=\frac{2}{3}<1 ]

This is valid. If the lines were wrongly identified, the product would exceed (1), which is impossible. Hence the correct regression coefficients are:

[ b_{yx}=\frac{4}{3},\quad b_{xy}=\frac{1}{2},\quad r=0.8165 ]

6. Teaching Aids/Activities

• Draw the two lines on the same graph to show the steeper line is (X) on (Y).
• Ask students to compute (r) for another pair of equations and check (b_{yx}b_{xy}\le 1).
• Use questioning: “Which line is (Y) on (X)? Why?” “What does the sign of (r) tell us?”
• Evaluation: Give a short problem with two regression equations and ask for (b_{yx}, b_{xy}, r).

Total number of students: $N=490$

Median

Median position: $\frac{N}{2}=\frac{490}{2}=245$

From the cumulative frequency table, 245 lies in the class 40–60.

Median formula: $M=L+\frac{\frac{N}{2}-cf}{f}\times h$

Where:

• $L=40$
• $cf=210$
• $f=130$
• $h=20$

$M=40+\frac{245-210}{130}\times 20$

$M=40+\frac{35}{130}\times 20$

$M=40+5.3846$

$M=45.38 \text{ approximately}$

Median marks = 45.38

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First quartile $(Q_1)$

$Q_1=L+\frac{\frac{N}{4}-cf}{f}\times h$

$\frac{N}{4}=\frac{490}{4}=122.5$

This lies in the class 20–40.

$Q_1=20+\frac{122.5-50}{160}\times 20$

$Q_1=20+\frac{72.5}{160}\times 20$

$Q_1=20+9.0625$

$Q_1=29.06 \text{ approximately}$

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Third quartile $(Q_3)$

$Q_3=L+\frac{\frac{3N}{4}-cf}{f}\times h$

$\frac{3N}{4}=\frac{3\times490}{4}=367.5$

This lies in the class 60–80.

$Q_3=60+\frac{367.5-340}{90}\times 20$

$Q_3=60+\frac{27.5}{90}\times 20$

$Q_3=60+6.1111$

$Q_3=66.11 \text{ approximately}$

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Number of students between $Q_1$ and $Q_3$

Since quartiles divide the total frequency into four equal parts, the number of students between the first and third quartile is:

$\frac{N}{2}=\frac{490}{2}=245$

Hence, 245 students get marks between the first and third quartile.