Answer :
Sure, I'll walk you through the step-by-step process to calculate the Pearson correlation coefficient between the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] variables, given the incomplete dataset and assuming that the missing value in [tex]\( y \)[/tex] can be imputed based on the remaining values.
### Step 1: Identify the missing value in [tex]\( y \)[/tex]
First, we need to impute the missing [tex]\( y \)[/tex] value using the available data in the [tex]\( y \)[/tex] column.
Given [tex]\( y = [9, 11, ?, 8, 7] \)[/tex], let's denote the missing value as [tex]\( y_3 \)[/tex].
We assume that the missing value in [tex]\( y \)[/tex] is equal to the mean of the rest of the [tex]\( y \)[/tex] values:
[tex]\[ \text{Mean of remaining } y = \frac{9 + 11 + 8 + 7}{4} = \frac{35}{4} = 8.75 \][/tex]
So, [tex]\( y_3 = 8.75 \)[/tex].
### Step 2: Complete the data
Now, we have the complete dataset:
[tex]\[ x = [6, 1, 10, 4, 8] \][/tex]
[tex]\[ y = [9, 11, 8.75, 8, 7] \][/tex]
### Step 3: Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Mean of } x = \frac{6 + 1 + 10 + 4 + 8}{5} = \frac{29}{5} = 5.8 \][/tex]
[tex]\[ \text{Mean of } y = \frac{9 + 11 + 8.75 + 8 + 7}{5} = \frac{43.75}{5} = 8.75 \][/tex]
### Step 4: Calculate the covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Cov}(x, y) = \frac{1}{N} \sum_{i=1}^{N} \left( (x_i - \bar{x})(y_i - \bar{y}) \right) \][/tex]
[tex]\[ \text{Cov}(x, y) = \frac{1}{5} \left( (6 - 5.8)(9 - 8.75) + (1 - 5.8)(11 - 8.75) + (10 - 5.8)(8.75 - 8.75) + (4 - 5.8)(8 - 8.75) + (8 - 5.8)(7 - 8.75) \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( 0.2 \times 0.25 + (-4.8) \times 2.25 + 4.2 \times 0 + (-1.8) \times -0.75 + 2.2 \times -1.75 \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( 0.05 + (-10.8) + 0 + 1.35 + (-3.85) \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( -13.25 \right) = -2.65 \][/tex]
### Step 5: Calculate the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Std}(x) = \sqrt{ \frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2 } \][/tex]
[tex]\[ \text{Std}(x) = \sqrt{ \frac{1}{5} \left( (6-5.8)^2 + (1-5.8)^2 + (10-5.8)^2 + (4-5.8)^2 + (8-5.8)^2 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( 0.04 + 23.04 + 17.64 + 3.24 + 4.84 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{48.8}{5} } = \sqrt{9.76} \approx 3.12 \][/tex]
[tex]\[ \text{Std}(y) = \sqrt{ \frac{1}{N} \sum_{i=1}^{N} (y_i - \bar{y})^2 } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( (9-8.75)^2 + (11-8.75)^2 + (8.75-8.75)^2 + (8-8.75)^2 + (7-8.75)^2 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( 0.0625 + 5.0625 + 0 + 0.5625 + 3.0625 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{8.75}{5} } = \sqrt{1.75} \approx 1.32 \][/tex]
### Step 6: Calculate the Pearson correlation coefficient
[tex]\[ r = \frac{\text{Cov}(x, y)}{\text{Std}(x) \times \text{Std}(y)} \][/tex]
[tex]\[ r = \frac{-2.65}{3.12 \times 1.32} \approx \frac{-2.65}{4.12} \approx -0.641 \][/tex]
### Summary
The Pearson correlation coefficient between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is approximately [tex]\( -0.641 \)[/tex]. Other intermediate results are:
- The imputed missing value in [tex]\( y \)[/tex] is [tex]\( 8.75 \)[/tex].
- The mean of [tex]\( x \)[/tex] is [tex]\( 5.8 \)[/tex].
- The mean of [tex]\( y \)[/tex] is [tex]\( 8.75 \)[/tex].
- The covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is [tex]\( -2.65 \)[/tex].
- The standard deviation of [tex]\( x \)[/tex] is [tex]\( 3.12 \)[/tex].
- The standard deviation of [tex]\( y \)[/tex] is [tex]\( 1.32 \)[/tex].
### Step 1: Identify the missing value in [tex]\( y \)[/tex]
First, we need to impute the missing [tex]\( y \)[/tex] value using the available data in the [tex]\( y \)[/tex] column.
Given [tex]\( y = [9, 11, ?, 8, 7] \)[/tex], let's denote the missing value as [tex]\( y_3 \)[/tex].
We assume that the missing value in [tex]\( y \)[/tex] is equal to the mean of the rest of the [tex]\( y \)[/tex] values:
[tex]\[ \text{Mean of remaining } y = \frac{9 + 11 + 8 + 7}{4} = \frac{35}{4} = 8.75 \][/tex]
So, [tex]\( y_3 = 8.75 \)[/tex].
### Step 2: Complete the data
Now, we have the complete dataset:
[tex]\[ x = [6, 1, 10, 4, 8] \][/tex]
[tex]\[ y = [9, 11, 8.75, 8, 7] \][/tex]
### Step 3: Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Mean of } x = \frac{6 + 1 + 10 + 4 + 8}{5} = \frac{29}{5} = 5.8 \][/tex]
[tex]\[ \text{Mean of } y = \frac{9 + 11 + 8.75 + 8 + 7}{5} = \frac{43.75}{5} = 8.75 \][/tex]
### Step 4: Calculate the covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Cov}(x, y) = \frac{1}{N} \sum_{i=1}^{N} \left( (x_i - \bar{x})(y_i - \bar{y}) \right) \][/tex]
[tex]\[ \text{Cov}(x, y) = \frac{1}{5} \left( (6 - 5.8)(9 - 8.75) + (1 - 5.8)(11 - 8.75) + (10 - 5.8)(8.75 - 8.75) + (4 - 5.8)(8 - 8.75) + (8 - 5.8)(7 - 8.75) \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( 0.2 \times 0.25 + (-4.8) \times 2.25 + 4.2 \times 0 + (-1.8) \times -0.75 + 2.2 \times -1.75 \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( 0.05 + (-10.8) + 0 + 1.35 + (-3.85) \right) \][/tex]
[tex]\[ = \frac{1}{5} \left( -13.25 \right) = -2.65 \][/tex]
### Step 5: Calculate the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
[tex]\[ \text{Std}(x) = \sqrt{ \frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2 } \][/tex]
[tex]\[ \text{Std}(x) = \sqrt{ \frac{1}{5} \left( (6-5.8)^2 + (1-5.8)^2 + (10-5.8)^2 + (4-5.8)^2 + (8-5.8)^2 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( 0.04 + 23.04 + 17.64 + 3.24 + 4.84 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{48.8}{5} } = \sqrt{9.76} \approx 3.12 \][/tex]
[tex]\[ \text{Std}(y) = \sqrt{ \frac{1}{N} \sum_{i=1}^{N} (y_i - \bar{y})^2 } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( (9-8.75)^2 + (11-8.75)^2 + (8.75-8.75)^2 + (8-8.75)^2 + (7-8.75)^2 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{1}{5} \left( 0.0625 + 5.0625 + 0 + 0.5625 + 3.0625 \right) } \][/tex]
[tex]\[ = \sqrt{ \frac{8.75}{5} } = \sqrt{1.75} \approx 1.32 \][/tex]
### Step 6: Calculate the Pearson correlation coefficient
[tex]\[ r = \frac{\text{Cov}(x, y)}{\text{Std}(x) \times \text{Std}(y)} \][/tex]
[tex]\[ r = \frac{-2.65}{3.12 \times 1.32} \approx \frac{-2.65}{4.12} \approx -0.641 \][/tex]
### Summary
The Pearson correlation coefficient between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is approximately [tex]\( -0.641 \)[/tex]. Other intermediate results are:
- The imputed missing value in [tex]\( y \)[/tex] is [tex]\( 8.75 \)[/tex].
- The mean of [tex]\( x \)[/tex] is [tex]\( 5.8 \)[/tex].
- The mean of [tex]\( y \)[/tex] is [tex]\( 8.75 \)[/tex].
- The covariance between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is [tex]\( -2.65 \)[/tex].
- The standard deviation of [tex]\( x \)[/tex] is [tex]\( 3.12 \)[/tex].
- The standard deviation of [tex]\( y \)[/tex] is [tex]\( 1.32 \)[/tex].
