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### (a) Compute the correlation coefficient and Interpret the result
The correlation coefficient [tex]\( r \)[/tex] measures the strength and direction of a linear relationship between the number of employees required [tex]\( y \)[/tex] and the number of departments [tex]\( x \)[/tex].
To find the correlation coefficient [tex]\( r \)[/tex], we use the formula:
[tex]\[ r = \frac{n \sum xy - (\sum x)(\sum y)}{\sqrt{ \left[ n \sum x^2 - (\sum x)^2 \right] \left[ n \sum y^2 - (\sum y)^2 \right] }} \][/tex]
Given:
- [tex]\( n = 10 \)[/tex]
- [tex]\( \sum x = 20 \)[/tex]
- [tex]\( \sum y = 40 \)[/tex]
- [tex]\( \sum x^2 = 100 \)[/tex]
- [tex]\( \sum y^2 = 120 \)[/tex]
- [tex]\( \sum xy = 180 \)[/tex]
Plugging in these values, we calculate the correlation coefficient as follows:
[tex]\[ r = \frac{10 \cdot 180 - 20 \cdot 40}{\sqrt{ \left[ 10 \cdot 100 - (20)^2 \right] \left[ 10 \cdot 120 - (40)^2 \right] }} \][/tex]
Simplifying the numerator:
[tex]\[ 10 \cdot 180 = 1800 \][/tex]
[tex]\[ 20 \cdot 40 = 800 \][/tex]
[tex]\[ 1800 - 800 = 1000 \][/tex]
Simplifying the denominator:
[tex]\[ n \sum x^2 - (\sum x)^2 = 10 \cdot 100 - 20^2 = 1000 - 400 = 600 \][/tex]
[tex]\[ n \sum y^2 - (\sum y)^2 = 10 \cdot 120 - 40^2 = 1200 - 1600 = -400 \][/tex]
Now:
[tex]\[ r = \frac{1000}{\sqrt{600 \cdot (-400)}} \][/tex]
Since we have a negative [tex]\( \sqrt{} \)[/tex] term (which results in an imaginary number), ultimately:
[tex]\[ r \approx (1.2498999054348719e-16 - 2.041241452319315j) \][/tex]
### Interpretation of the Result
The result is a complex number, which indicates that the relationship and the data provided may not suit typical linear correlation analysis, suggesting potential calculation/data issues or the need for different analytical methods.
### (b) Develop a linear regression equation
The linear regression line is given by:
[tex]\[ y = b_0 + b_1 x \][/tex]
Where:
[tex]\[ b_1 = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2} \][/tex]
[tex]\[ b_0 = \frac{\sum y}{n} - b_1 \left( \frac{\sum x}{n} \right) \][/tex]
Calculating [tex]\( b_1 \)[/tex]:
[tex]\[ b_1 = \frac{1000}{600} \approx 1.6666666666666667 \][/tex]
Calculating [tex]\( b_0 \)[/tex]:
[tex]\[ b_0 = \frac{40}{10} - 1.6666666666666667 \left( \frac{20}{10} \right) \][/tex]
[tex]\[ b_0 = 4 - 1.6666666666666667 \cdot 2 \approx 0.6666666666666665 \][/tex]
So the regression equation is:
[tex]\[ y = 0.6666666666666665 + 1.6666666666666667 x \][/tex]
### (c) Forecast the number of employees for 16 departments
To forecast the number of employees [tex]\( y \)[/tex] when [tex]\( x = 16 \)[/tex], we use the regression equation:
[tex]\[ y = 0.6666666666666665 + 1.6666666666666667 \cdot 16 \][/tex]
Calculating:
[tex]\[ y = 0.6666666666666665 + 26.666666666666668 \approx 27.333333333333336 \][/tex]
Therefore, the forecasted number of employees for 16 departments is 27.33 (approximately 27 employees).
In summary:
1. The correlation coefficient is a complex number, highlighting potential issues with data or method.
2. The regression equation is [tex]\( y = 0.6666666666666665 + 1.6666666666666667 x \)[/tex].
3. Forecast for 16 departments is approximately 27 employees.
### (a) Compute the correlation coefficient and Interpret the result
The correlation coefficient [tex]\( r \)[/tex] measures the strength and direction of a linear relationship between the number of employees required [tex]\( y \)[/tex] and the number of departments [tex]\( x \)[/tex].
To find the correlation coefficient [tex]\( r \)[/tex], we use the formula:
[tex]\[ r = \frac{n \sum xy - (\sum x)(\sum y)}{\sqrt{ \left[ n \sum x^2 - (\sum x)^2 \right] \left[ n \sum y^2 - (\sum y)^2 \right] }} \][/tex]
Given:
- [tex]\( n = 10 \)[/tex]
- [tex]\( \sum x = 20 \)[/tex]
- [tex]\( \sum y = 40 \)[/tex]
- [tex]\( \sum x^2 = 100 \)[/tex]
- [tex]\( \sum y^2 = 120 \)[/tex]
- [tex]\( \sum xy = 180 \)[/tex]
Plugging in these values, we calculate the correlation coefficient as follows:
[tex]\[ r = \frac{10 \cdot 180 - 20 \cdot 40}{\sqrt{ \left[ 10 \cdot 100 - (20)^2 \right] \left[ 10 \cdot 120 - (40)^2 \right] }} \][/tex]
Simplifying the numerator:
[tex]\[ 10 \cdot 180 = 1800 \][/tex]
[tex]\[ 20 \cdot 40 = 800 \][/tex]
[tex]\[ 1800 - 800 = 1000 \][/tex]
Simplifying the denominator:
[tex]\[ n \sum x^2 - (\sum x)^2 = 10 \cdot 100 - 20^2 = 1000 - 400 = 600 \][/tex]
[tex]\[ n \sum y^2 - (\sum y)^2 = 10 \cdot 120 - 40^2 = 1200 - 1600 = -400 \][/tex]
Now:
[tex]\[ r = \frac{1000}{\sqrt{600 \cdot (-400)}} \][/tex]
Since we have a negative [tex]\( \sqrt{} \)[/tex] term (which results in an imaginary number), ultimately:
[tex]\[ r \approx (1.2498999054348719e-16 - 2.041241452319315j) \][/tex]
### Interpretation of the Result
The result is a complex number, which indicates that the relationship and the data provided may not suit typical linear correlation analysis, suggesting potential calculation/data issues or the need for different analytical methods.
### (b) Develop a linear regression equation
The linear regression line is given by:
[tex]\[ y = b_0 + b_1 x \][/tex]
Where:
[tex]\[ b_1 = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2} \][/tex]
[tex]\[ b_0 = \frac{\sum y}{n} - b_1 \left( \frac{\sum x}{n} \right) \][/tex]
Calculating [tex]\( b_1 \)[/tex]:
[tex]\[ b_1 = \frac{1000}{600} \approx 1.6666666666666667 \][/tex]
Calculating [tex]\( b_0 \)[/tex]:
[tex]\[ b_0 = \frac{40}{10} - 1.6666666666666667 \left( \frac{20}{10} \right) \][/tex]
[tex]\[ b_0 = 4 - 1.6666666666666667 \cdot 2 \approx 0.6666666666666665 \][/tex]
So the regression equation is:
[tex]\[ y = 0.6666666666666665 + 1.6666666666666667 x \][/tex]
### (c) Forecast the number of employees for 16 departments
To forecast the number of employees [tex]\( y \)[/tex] when [tex]\( x = 16 \)[/tex], we use the regression equation:
[tex]\[ y = 0.6666666666666665 + 1.6666666666666667 \cdot 16 \][/tex]
Calculating:
[tex]\[ y = 0.6666666666666665 + 26.666666666666668 \approx 27.333333333333336 \][/tex]
Therefore, the forecasted number of employees for 16 departments is 27.33 (approximately 27 employees).
In summary:
1. The correlation coefficient is a complex number, highlighting potential issues with data or method.
2. The regression equation is [tex]\( y = 0.6666666666666665 + 1.6666666666666667 x \)[/tex].
3. Forecast for 16 departments is approximately 27 employees.
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