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To find the correlation coefficient for the given data points, we need to use the correlation formula, specifically Pearson’s correlation coefficient. Here’s a step-by-step method to determine this:
1. List the given data points:
- [tex]\( x \)[/tex] values: 0, 5, 10, 15
- [tex]\( y \)[/tex] values: 15, 10, 5, 0
2. Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- Mean of [tex]\( x \)[/tex], [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = \frac{0 + 5 + 10 + 15}{4} = \frac{30}{4} = 7.5 \][/tex]
- Mean of [tex]\( y \)[/tex], [tex]\(\bar{y}\)[/tex]:
[tex]\[ \bar{y} = \frac{15 + 10 + 5 + 0}{4} = \frac{30}{4} = 7.5 \][/tex]
3. Compute the deviations from the means:
- Deviations of [tex]\( x \)[/tex] from the mean:
[tex]\[ x_1 - \bar{x} = 0 - 7.5 = -7.5 \][/tex]
[tex]\[ x_2 - \bar{x} = 5 - 7.5 = -2.5 \][/tex]
[tex]\[ x_3 - \bar{x} = 10 - 7.5 = 2.5 \][/tex]
[tex]\[ x_4 - \bar{x} = 15 - 7.5 = 7.5 \][/tex]
- Deviations of [tex]\( y \)[/tex] from the mean:
[tex]\[ y_1 - \bar{y} = 15 - 7.5 = 7.5 \][/tex]
[tex]\[ y_2 - \bar{y} = 10 - 7.5 = 2.5 \][/tex]
[tex]\[ y_3 - \bar{y} = 5 - 7.5 = -2.5 \][/tex]
[tex]\[ y_4 - \bar{y} = 0 - 7.5 = -7.5 \][/tex]
4. Calculate the products of the deviations for each pair:
[tex]\[ (-7.5 \times 7.5) = -56.25 \][/tex]
[tex]\[ (-2.5 \times 2.5) = -6.25 \][/tex]
[tex]\[ (2.5 \times -2.5) = -6.25 \][/tex]
[tex]\[ (7.5 \times -7.5) = -56.25 \][/tex]
5. Summing up these products:
[tex]\[ -56.25 + (-6.25) + (-6.25) + (-56.25) = -125 \][/tex]
6. Calculate the sum of squared deviations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- For [tex]\( x \)[/tex]:
[tex]\[ (-7.5^2) + (-2.5^2) + (2.5^2) + (7.5^2) = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
- For [tex]\( y \)[/tex]:
[tex]\[ (7.5^2) + (2.5^2) + (-2.5^2) + (-7.5^2) = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
7. Putting it all into the Pearson correlation coefficient formula:
[tex]\[ r = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sqrt{\sum{(x_i - \bar{x})^2}} \times \sqrt{\sum{(y_i - \bar{y})^2}}} \][/tex]
Substituting in the sums from above:
[tex]\[ r = \frac{-125}{\sqrt{125} \times \sqrt{125}} = \frac{-125}{125} = -1 \][/tex]
Therefore, the correlation coefficient for the data shown in the table is [tex]\( -1 \)[/tex], indicating a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. List the given data points:
- [tex]\( x \)[/tex] values: 0, 5, 10, 15
- [tex]\( y \)[/tex] values: 15, 10, 5, 0
2. Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- Mean of [tex]\( x \)[/tex], [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = \frac{0 + 5 + 10 + 15}{4} = \frac{30}{4} = 7.5 \][/tex]
- Mean of [tex]\( y \)[/tex], [tex]\(\bar{y}\)[/tex]:
[tex]\[ \bar{y} = \frac{15 + 10 + 5 + 0}{4} = \frac{30}{4} = 7.5 \][/tex]
3. Compute the deviations from the means:
- Deviations of [tex]\( x \)[/tex] from the mean:
[tex]\[ x_1 - \bar{x} = 0 - 7.5 = -7.5 \][/tex]
[tex]\[ x_2 - \bar{x} = 5 - 7.5 = -2.5 \][/tex]
[tex]\[ x_3 - \bar{x} = 10 - 7.5 = 2.5 \][/tex]
[tex]\[ x_4 - \bar{x} = 15 - 7.5 = 7.5 \][/tex]
- Deviations of [tex]\( y \)[/tex] from the mean:
[tex]\[ y_1 - \bar{y} = 15 - 7.5 = 7.5 \][/tex]
[tex]\[ y_2 - \bar{y} = 10 - 7.5 = 2.5 \][/tex]
[tex]\[ y_3 - \bar{y} = 5 - 7.5 = -2.5 \][/tex]
[tex]\[ y_4 - \bar{y} = 0 - 7.5 = -7.5 \][/tex]
4. Calculate the products of the deviations for each pair:
[tex]\[ (-7.5 \times 7.5) = -56.25 \][/tex]
[tex]\[ (-2.5 \times 2.5) = -6.25 \][/tex]
[tex]\[ (2.5 \times -2.5) = -6.25 \][/tex]
[tex]\[ (7.5 \times -7.5) = -56.25 \][/tex]
5. Summing up these products:
[tex]\[ -56.25 + (-6.25) + (-6.25) + (-56.25) = -125 \][/tex]
6. Calculate the sum of squared deviations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- For [tex]\( x \)[/tex]:
[tex]\[ (-7.5^2) + (-2.5^2) + (2.5^2) + (7.5^2) = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
- For [tex]\( y \)[/tex]:
[tex]\[ (7.5^2) + (2.5^2) + (-2.5^2) + (-7.5^2) = 56.25 + 6.25 + 6.25 + 56.25 = 125 \][/tex]
7. Putting it all into the Pearson correlation coefficient formula:
[tex]\[ r = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sqrt{\sum{(x_i - \bar{x})^2}} \times \sqrt{\sum{(y_i - \bar{y})^2}}} \][/tex]
Substituting in the sums from above:
[tex]\[ r = \frac{-125}{\sqrt{125} \times \sqrt{125}} = \frac{-125}{125} = -1 \][/tex]
Therefore, the correlation coefficient for the data shown in the table is [tex]\( -1 \)[/tex], indicating a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
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