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Certainly! Let's calculate the correlation coefficient between the given prices and demands using Karl Pearson's method. The correlation coefficient (denoted as [tex]\( r \)[/tex]) measures the strength and direction of a linear relationship between two variables.
Below are the given data points:
- Prices, [tex]\( x \)[/tex]: 100, 120, 190, 200, 240, 260
- Demands, [tex]\( y \)[/tex]: 20, 15, 16, 12, 14, 13
### Step 1: Calculate Means
First, we need to calculate the mean of the prices ([tex]\( \bar{x} \)[/tex]) and the mean of the demands ([tex]\( \bar{y} \)[/tex]).
[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{100 + 120 + 190 + 200 + 240 + 260}{6} = \frac{1110}{6} = 185.0 \][/tex]
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{20 + 15 + 16 + 12 + 14 + 13}{6} = \frac{90}{6} = 15.0 \][/tex]
### Step 2: Calculate Covariance
Next, we calculate the covariance between the prices and the demands. The covariance is given by:
[tex]\[ \text{Cov}(X, Y) = \frac{\sum (x - \bar{x})(y - \bar{y})}{n} \][/tex]
Substituting [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex]:
[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]
[tex]\[ = \frac{(-85)(5) + (-65)(0) + (5)(1) + (15)(-3) + (55)(-1) + (75)(-2)}{6} \][/tex]
[tex]\[ = \frac{-425 + 0 + 5 - 45 - 55 - 150}{6} = \frac{-670}{6} = -111.66666666666667 \][/tex]
### Step 3: Calculate Standard Deviations
Now, calculate the standard deviations for both the prices ([tex]\( \sigma_x \)[/tex]) and the demands ([tex]\( \sigma_y \)[/tex]).
The standard deviation is given by:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n}} \quad \text{and} \quad \sigma_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n}} \][/tex]
### Standard Deviation of Prices:
[tex]\[ \sigma_x = \sqrt{\frac{(100-185)^2 + (120-185)^2 + (190-185)^2 + (200-185)^2 + (240-185)^2 + (260-185)^2}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{7225 + 4225 + 25 + 225 + 3025 + 5625}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{20650}{6}} = \sqrt{3441.6666666666665} \approx 58.23801736552049 \][/tex]
### Standard Deviation of Demands:
[tex]\[ \sigma_y = \sqrt{\frac{(20-15)^2 + (15-15)^2 + (16-15)^2 + (12-15)^2 + (14-15)^2 + (13-15)^2}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{25 + 0 + 1 + 9 + 1 + 4}{6}} = \sqrt{\frac{40}{6}} = \sqrt{6.666666666666667} \approx 2.581988897471611 \][/tex]
### Step 4: Calculate the Correlation Coefficient
Finally, the correlation coefficient [tex]\( r \)[/tex] is given by:
[tex]\[ r = \frac{\text{Cov}(X, Y)}{\sigma_x \sigma_y} \][/tex]
Substituting the values obtained:
[tex]\[ r = \frac{-111.66666666666667}{58.23801736552049 \times 2.581988897471611} \approx -0.7426130900234668 \][/tex]
### Conclusion
The correlation coefficient [tex]\( r \)[/tex] is approximately [tex]\(-0.743\)[/tex]. This indicates a strong negative linear relationship between the prices and the demands.
Below are the given data points:
- Prices, [tex]\( x \)[/tex]: 100, 120, 190, 200, 240, 260
- Demands, [tex]\( y \)[/tex]: 20, 15, 16, 12, 14, 13
### Step 1: Calculate Means
First, we need to calculate the mean of the prices ([tex]\( \bar{x} \)[/tex]) and the mean of the demands ([tex]\( \bar{y} \)[/tex]).
[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{100 + 120 + 190 + 200 + 240 + 260}{6} = \frac{1110}{6} = 185.0 \][/tex]
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{20 + 15 + 16 + 12 + 14 + 13}{6} = \frac{90}{6} = 15.0 \][/tex]
### Step 2: Calculate Covariance
Next, we calculate the covariance between the prices and the demands. The covariance is given by:
[tex]\[ \text{Cov}(X, Y) = \frac{\sum (x - \bar{x})(y - \bar{y})}{n} \][/tex]
Substituting [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex]:
[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]
[tex]\[ = \frac{(-85)(5) + (-65)(0) + (5)(1) + (15)(-3) + (55)(-1) + (75)(-2)}{6} \][/tex]
[tex]\[ = \frac{-425 + 0 + 5 - 45 - 55 - 150}{6} = \frac{-670}{6} = -111.66666666666667 \][/tex]
### Step 3: Calculate Standard Deviations
Now, calculate the standard deviations for both the prices ([tex]\( \sigma_x \)[/tex]) and the demands ([tex]\( \sigma_y \)[/tex]).
The standard deviation is given by:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n}} \quad \text{and} \quad \sigma_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n}} \][/tex]
### Standard Deviation of Prices:
[tex]\[ \sigma_x = \sqrt{\frac{(100-185)^2 + (120-185)^2 + (190-185)^2 + (200-185)^2 + (240-185)^2 + (260-185)^2}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{7225 + 4225 + 25 + 225 + 3025 + 5625}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{20650}{6}} = \sqrt{3441.6666666666665} \approx 58.23801736552049 \][/tex]
### Standard Deviation of Demands:
[tex]\[ \sigma_y = \sqrt{\frac{(20-15)^2 + (15-15)^2 + (16-15)^2 + (12-15)^2 + (14-15)^2 + (13-15)^2}{6}} \][/tex]
[tex]\[ = \sqrt{\frac{25 + 0 + 1 + 9 + 1 + 4}{6}} = \sqrt{\frac{40}{6}} = \sqrt{6.666666666666667} \approx 2.581988897471611 \][/tex]
### Step 4: Calculate the Correlation Coefficient
Finally, the correlation coefficient [tex]\( r \)[/tex] is given by:
[tex]\[ r = \frac{\text{Cov}(X, Y)}{\sigma_x \sigma_y} \][/tex]
Substituting the values obtained:
[tex]\[ r = \frac{-111.66666666666667}{58.23801736552049 \times 2.581988897471611} \approx -0.7426130900234668 \][/tex]
### Conclusion
The correlation coefficient [tex]\( r \)[/tex] is approximately [tex]\(-0.743\)[/tex]. This indicates a strong negative linear relationship between the prices and the demands.
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