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To find the correlation coefficient for the data shown in the table, we will follow these steps:
1. List the data points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 15 \\ \hline 5 & 10 \\ \hline 10 & 5 \\ \hline 15 & 0 \\ \hline \end{array} \][/tex]
2. Calculate the sums and means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- Sum of [tex]\( x \)[/tex]: [tex]\( 0 + 5 + 10 + 15 = 30 \)[/tex]
- Sum of [tex]\( y \)[/tex]: [tex]\( 15 + 10 + 5 + 0 = 30 \)[/tex]
- Mean of [tex]\( x \)[/tex] ([tex]\( \bar{x} \)[/tex]): [tex]\( \frac{30}{4} = 7.5 \)[/tex]
- Mean of [tex]\( y \)[/tex] ([tex]\( \bar{y} \)[/tex]): [tex]\( \frac{30}{4} = 7.5 \)[/tex]
3. Calculate the deviations from the means:
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & y & (x - \bar{x}) & (y - \bar{y}) \\ \hline 0 & 15 & 0 - 7.5 = -7.5 & 15 - 7.5 = 7.5 \\ \hline 5 & 10 & 5 - 7.5 = -2.5 & 10 - 7.5 = 2.5 \\ \hline 10 & 5 & 10 - 7.5 = 2.5 & 5 - 7.5 = -2.5 \\ \hline 15 & 0 & 15 - 7.5 = 7.5 & 0 - 7.5 = -7.5 \\ \hline \end{array} \][/tex]
4. Calculate the products of the deviations:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & y & (x - \bar{x}) & (y - \bar{y}) & (x - \bar{x})(y - \bar{y}) \\ \hline 0 & 15 & -7.5 & 7.5 & (-7.5) \times 7.5 = -56.25 \\ \hline 5 & 10 & -2.5 & 2.5 & (-2.5) \times 2.5 = -6.25 \\ \hline 10 & 5 & 2.5 & -2.5 & 2.5 \times (-2.5) = -6.25 \\ \hline 15 & 0 & 7.5 & -7.5 & 7.5 \times (-7.5) = -56.25 \\ \hline \end{array} \][/tex]
5. Calculate the sum of the products of the deviations:
[tex]\[ \sum (x - \bar{x})(y - \bar{y}) = -56.25 + (-6.25) + (-6.25) + (-56.25) = -125 \][/tex]
6. Calculate the squared deviations and their sums:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & y & (x - \bar{x})^2 & (y - \bar{y})^2 \\ \hline 0 & 15 & (-7.5)^2 = 56.25 & 7.5^2 = 56.25 \\ \hline 5 & 10 & (-2.5)^2 = 6.25 & 2.5^2 = 6.25 \\ \hline 10 & 5 & 2.5^2 = 6.25 & (-2.5)^2 = 6.25 \\ \hline 15 & 0 & 7.5^2 = 56.25 & (-7.5)^2 = 56.25 \\ \hline \end{array} \][/tex]
[tex]\[ \sum (x - \bar{x})^2 = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
[tex]\[ \sum (y - \bar{y})^2 = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
7. Calculate the correlation coefficient [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sqrt{\sum (x - \bar{x})^2 \sum (y - \bar{y})^2}} \][/tex]
[tex]\[ r = \frac{-125}{\sqrt{125 \times 125}} = \frac{-125}{125} = -1.0 \][/tex]
The correlation coefficient for the data shown in the table is [tex]\(-1.0\)[/tex]. This value indicates a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. List the data points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 15 \\ \hline 5 & 10 \\ \hline 10 & 5 \\ \hline 15 & 0 \\ \hline \end{array} \][/tex]
2. Calculate the sums and means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- Sum of [tex]\( x \)[/tex]: [tex]\( 0 + 5 + 10 + 15 = 30 \)[/tex]
- Sum of [tex]\( y \)[/tex]: [tex]\( 15 + 10 + 5 + 0 = 30 \)[/tex]
- Mean of [tex]\( x \)[/tex] ([tex]\( \bar{x} \)[/tex]): [tex]\( \frac{30}{4} = 7.5 \)[/tex]
- Mean of [tex]\( y \)[/tex] ([tex]\( \bar{y} \)[/tex]): [tex]\( \frac{30}{4} = 7.5 \)[/tex]
3. Calculate the deviations from the means:
[tex]\[ \begin{array}{|c|c|c|c|} \hline x & y & (x - \bar{x}) & (y - \bar{y}) \\ \hline 0 & 15 & 0 - 7.5 = -7.5 & 15 - 7.5 = 7.5 \\ \hline 5 & 10 & 5 - 7.5 = -2.5 & 10 - 7.5 = 2.5 \\ \hline 10 & 5 & 10 - 7.5 = 2.5 & 5 - 7.5 = -2.5 \\ \hline 15 & 0 & 15 - 7.5 = 7.5 & 0 - 7.5 = -7.5 \\ \hline \end{array} \][/tex]
4. Calculate the products of the deviations:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & y & (x - \bar{x}) & (y - \bar{y}) & (x - \bar{x})(y - \bar{y}) \\ \hline 0 & 15 & -7.5 & 7.5 & (-7.5) \times 7.5 = -56.25 \\ \hline 5 & 10 & -2.5 & 2.5 & (-2.5) \times 2.5 = -6.25 \\ \hline 10 & 5 & 2.5 & -2.5 & 2.5 \times (-2.5) = -6.25 \\ \hline 15 & 0 & 7.5 & -7.5 & 7.5 \times (-7.5) = -56.25 \\ \hline \end{array} \][/tex]
5. Calculate the sum of the products of the deviations:
[tex]\[ \sum (x - \bar{x})(y - \bar{y}) = -56.25 + (-6.25) + (-6.25) + (-56.25) = -125 \][/tex]
6. Calculate the squared deviations and their sums:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & y & (x - \bar{x})^2 & (y - \bar{y})^2 \\ \hline 0 & 15 & (-7.5)^2 = 56.25 & 7.5^2 = 56.25 \\ \hline 5 & 10 & (-2.5)^2 = 6.25 & 2.5^2 = 6.25 \\ \hline 10 & 5 & 2.5^2 = 6.25 & (-2.5)^2 = 6.25 \\ \hline 15 & 0 & 7.5^2 = 56.25 & (-7.5)^2 = 56.25 \\ \hline \end{array} \][/tex]
[tex]\[ \sum (x - \bar{x})^2 = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
[tex]\[ \sum (y - \bar{y})^2 = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
7. Calculate the correlation coefficient [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sqrt{\sum (x - \bar{x})^2 \sum (y - \bar{y})^2}} \][/tex]
[tex]\[ r = \frac{-125}{\sqrt{125 \times 125}} = \frac{-125}{125} = -1.0 \][/tex]
The correlation coefficient for the data shown in the table is [tex]\(-1.0\)[/tex]. This value indicates a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
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