Consider the following data for supply (X) and the price (Y) of a commodity for last six years. Year in B.S. 2075 2076 2077 2078 2079 2080 : : : : : : : : : : : : : : X 45 50 56 62 65 70 Y 65 70 75…

a) Correlation Coefficient between X and Y

The correlation coefficient $r$ between two variables $X$ and $Y$ is calculated using the formula:

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

Step 1: Compute the required sums

Given data:

Step 2: Plug into the formula

$$n=6,\sum X=348,\sum Y=480,\sum XY=28435,\sum X^2=20630,\sum Y^2=39250$$

$$r=\frac{6(28435)-(348)(480)}{\sqrt{[6(20630)-(348{)}^2][6(39250)-(480{)}^2]}}$$

$$r=\frac{170610-167040}{\sqrt{[123780-121104][235500-230400]}}$$

$$r=\frac{3570}{\sqrt{(2676)(5100)}}$$

$$r=\frac{3570}{\sqrt{13647600}}$$

$$r=\frac{3570}{3694.3}\approx 0.966$$

Correlation coefficient $r\approx 0.966$.

---

b) Calculate the supply when the price of the commodity is Rs. 150

We use the linear regression equation to predict $X$ (supply) given $Y$ (price):

$$X=a+bY$$

where: $$b=\frac{n\sum XY-\sum X\sum Y}{n\sum Y^2-(\sum Y{)}^2}$$ $$a=\bar{X}-b\bar{Y}$$

Step 1: Compute $b$ (slope)

$$b=\frac{6(28435)-(348)(480)}{6(39250)-(480{)}^2}$$

$$b=\frac{170610-167040}{235500-230400}$$

$$b=\frac{3570}{5100}=0.7$$

Step 2: Compute $a$ (intercept)

$$\bar{X}=\frac{348}{6}=58,\bar{Y}=\frac{480}{6}=80$$

$$a=58-(0.7)(80)=58-56=2$$

Step 3: Regression equation

$$X=2+0.7Y$$

Step 4: Predict $X$ when $Y=150$

$$X=2+0.7(150)=2+105=107$$

Supply when the price is Rs. 150 is 107 units.