If for 25 observations of pairs (x,y), `Sigmax=200,Sigmay=150,Sigmax^(2)=3000 and Sigmaxy=1500` then the equation of line of regression of y on x is

Options

A. `A) 3x+14y=60`
B. B) `3x-14y+60=0`
C. `C) 14x-3y=60`
D. `D) 3x-14y=60`

Answer

To find the equation of the line of regression of \( y \) on \( x \) given the data, we will follow these steps:

### Step 1: Identify the given values
We have the following values:
- Number of observations, \( n = 25 \)
- \( \Sigma x = 200 \)
- \( \Sigma y = 150 \)
- \( \Sigma x^2 = 3000 \)
- \( \Sigma xy = 1500 \)

### Step 2: Calculate the means \( \bar{x} \) and \( \bar{y} \)
The means are calculated as follows:
\[ \bar{x} = \frac{\Sigma x}{n} = \frac{200}{25} = 8 \]
\[ \bar{y} = \frac{\Sigma y}{n} = \frac{150}{25} = 6 \]

### Step 3: Calculate the slope \( b_{yx} \)
The formula for the slope \( b_{yx} \) of the regression line of \( y \) on \( x \) is:
\[ b_{yx} = \frac{\Sigma xy - n \cdot \bar{x} \cdot \bar{y}}{\Sigma x^2 - n \cdot \bar{x}^2} \]

Substituting the known values:
\[ b_{yx} = \frac{1500 - 25 \cdot 8 \cdot 6}{3000 - 25 \cdot 8^2} \]
Calculating the terms:
\[ = \frac{1500 - 1200}{3000 - 1600} \]
\[ = \frac{300}{1400} = \frac{3}{14} \]

### Step 4: Write the equation of the regression line
The equation of the regression line can be expressed as:
\[ y - \bar{y} = b_{yx}(x - \bar{x}) \]
Substituting the values of \( \bar{y} \), \( b_{yx} \), and \( \bar{x} \):
\[ y - 6 = \frac{3}{14}(x - 8) \]

### Step 5: Rearranging the equation
Multiplying through by 14 to eliminate the fraction:
\[ 14(y - 6) = 3(x - 8) \]
Expanding both sides:
\[ 14y - 84 = 3x - 24 \]
Rearranging gives:
\[ 3x - 14y + 60 = 0 \]

### Final Answer
Thus, the equation of the line of regression of \( y \) on \( x \) is:
\[ 3x - 14y + 60 = 0 \]

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