Answer :
To calculate the correlation coefficient between price ([tex]\( x \)[/tex]) and demand ([tex]\( y \)[/tex]) using Karl Pearson’s method, we will follow these steps:
1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
2. Find the covariance of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
3. Compute the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
4. Use the correlation formula to find the correlation coefficient.
Let's go through each step in detail:
### Step 1: Calculate the Means
The mean of [tex]\( x \)[/tex] (prices) and the mean of [tex]\( y \)[/tex] (demands) can be calculated as:
[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{100 + 120 + 190 + 200 + 240 + 260}{6} = 185.0 \][/tex]
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{20 + 15 + 16 + 12 + 14 + 13}{6} = 15.0 \][/tex]
### Step 2: Calculate the Covariance
The covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is given by the formula:
[tex]\[ \text{Cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n} \][/tex]
We plug in the values:
[tex]\[ \text{Cov}(x, y) = \frac{(100 - 185)(20 - 15) + (120 - 185)(15 - 15) + (190 - 185)(16 - 15) + (200 - 185)(12 - 15) + (240 - 185)(14 - 15) + (260 - 185)(13 - 15)}{6} \][/tex]
This simplifies to:
[tex]\[ \text{Cov}(x, y) = \frac{(-85 \cdot 5) + (-65 \cdot 0) + (5 \cdot 1) + (15 \cdot -3) + (55 \cdot -1) + (75 \cdot -2)}{6} = \frac{-425 + 0 + 5 - 45 - 55 - 150}{6} = \frac{-670}{6} = -111.66666666666667 \][/tex]
### Step 3: Calculate the Standard Deviations
The standard deviation of [tex]\( x \)[/tex] (prices) is given by:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{(100 - 185)^2 + (120 - 185)^2 + (190 - 185)^2 + (200 - 185)^2 + (240 - 185)^2 + (260 - 185)^2}{6}} \][/tex]
Simplifying inside the square root:
[tex]\[ \sigma_x = \sqrt{\frac{(-85)^2 + (-65)^2 + (5)^2 + (15)^2 + (55)^2 + (75)^2}{6}} = \sqrt{\frac{7225 + 4225 + 25 + 225 + 3025 + 5625}{6}} = \sqrt{\frac{20400}{6}} = \sqrt{3400} = 58.23801736552049 \][/tex]
Similarly, the standard deviation of [tex]\( y \)[/tex] (demands) is:
[tex]\[ \sigma_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n}} = \sqrt{\frac{(20 - 15)^2 + (15 - 15)^2 + (16 - 15)^2 + (12 - 15)^2 + (14 - 15)^2 + (13 - 15)^2}{6}} \][/tex]
Simplifying inside the square root:
[tex]\[ \sigma_y = \sqrt{\frac{(5)^2 + (0)^2 + (1)^2 + (-3)^2 + (-1)^2 + (-2)^2}{6}} = \sqrt{\frac{25 + 0 + 1 + 9 + 1 + 4}{6}} = \sqrt{\frac{40}{6}} = \sqrt{6.666666666666667} = 2.581988897471611 \][/tex]
### Step 4: Calculate the Correlation Coefficient
The correlation coefficient is given by:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} = \frac{-111.66666666666667}{58.23801736552049 \cdot 2.581988897471611} = -0.7426130900234668 \][/tex]
So, the correlation coefficient by Karl Pearson's method is:
[tex]\[ r \approx -0.743 \][/tex]
This negative value indicates an inverse relationship between price and demand, meaning that as the price increases, the demand tends to decrease.
1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
2. Find the covariance of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
3. Compute the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
4. Use the correlation formula to find the correlation coefficient.
Let's go through each step in detail:
### Step 1: Calculate the Means
The mean of [tex]\( x \)[/tex] (prices) and the mean of [tex]\( y \)[/tex] (demands) can be calculated as:
[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{100 + 120 + 190 + 200 + 240 + 260}{6} = 185.0 \][/tex]
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{20 + 15 + 16 + 12 + 14 + 13}{6} = 15.0 \][/tex]
### Step 2: Calculate the Covariance
The covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is given by the formula:
[tex]\[ \text{Cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n} \][/tex]
We plug in the values:
[tex]\[ \text{Cov}(x, y) = \frac{(100 - 185)(20 - 15) + (120 - 185)(15 - 15) + (190 - 185)(16 - 15) + (200 - 185)(12 - 15) + (240 - 185)(14 - 15) + (260 - 185)(13 - 15)}{6} \][/tex]
This simplifies to:
[tex]\[ \text{Cov}(x, y) = \frac{(-85 \cdot 5) + (-65 \cdot 0) + (5 \cdot 1) + (15 \cdot -3) + (55 \cdot -1) + (75 \cdot -2)}{6} = \frac{-425 + 0 + 5 - 45 - 55 - 150}{6} = \frac{-670}{6} = -111.66666666666667 \][/tex]
### Step 3: Calculate the Standard Deviations
The standard deviation of [tex]\( x \)[/tex] (prices) is given by:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{(100 - 185)^2 + (120 - 185)^2 + (190 - 185)^2 + (200 - 185)^2 + (240 - 185)^2 + (260 - 185)^2}{6}} \][/tex]
Simplifying inside the square root:
[tex]\[ \sigma_x = \sqrt{\frac{(-85)^2 + (-65)^2 + (5)^2 + (15)^2 + (55)^2 + (75)^2}{6}} = \sqrt{\frac{7225 + 4225 + 25 + 225 + 3025 + 5625}{6}} = \sqrt{\frac{20400}{6}} = \sqrt{3400} = 58.23801736552049 \][/tex]
Similarly, the standard deviation of [tex]\( y \)[/tex] (demands) is:
[tex]\[ \sigma_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n}} = \sqrt{\frac{(20 - 15)^2 + (15 - 15)^2 + (16 - 15)^2 + (12 - 15)^2 + (14 - 15)^2 + (13 - 15)^2}{6}} \][/tex]
Simplifying inside the square root:
[tex]\[ \sigma_y = \sqrt{\frac{(5)^2 + (0)^2 + (1)^2 + (-3)^2 + (-1)^2 + (-2)^2}{6}} = \sqrt{\frac{25 + 0 + 1 + 9 + 1 + 4}{6}} = \sqrt{\frac{40}{6}} = \sqrt{6.666666666666667} = 2.581988897471611 \][/tex]
### Step 4: Calculate the Correlation Coefficient
The correlation coefficient is given by:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} = \frac{-111.66666666666667}{58.23801736552049 \cdot 2.581988897471611} = -0.7426130900234668 \][/tex]
So, the correlation coefficient by Karl Pearson's method is:
[tex]\[ r \approx -0.743 \][/tex]
This negative value indicates an inverse relationship between price and demand, meaning that as the price increases, the demand tends to decrease.
