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To determine the strength of the correlation between the total cost of items and the shipping costs, we need to calculate the Pearson correlation coefficient. The Pearson correlation coefficient measures the linear relationship between two datasets.
Here is a step-by-step guide:
1. List Down Data Points:
- Total cost of items: 25, 45, 50, 70
- Shipping costs: 5.99, 8.99, 8.99, 10.99
2. Calculate Means:
- Mean of total cost of items [tex]\[(\bar{x}) = \frac{25 + 45 + 50 + 70}{4} = \frac{190}{4} = 47.5\][/tex]
- Mean of shipping costs [tex]\[(\bar{y}) = \frac{5.99 + 8.99 + 8.99 + 10.99}{4} = \frac{34.96}{4} = 8.74\][/tex]
3. Calculate Deviations:
- Deviations from the mean for total cost of items:
- [tex]\(25 - 47.5 = -22.5\)[/tex]
- [tex]\(45 - 47.5 = -2.5\)[/tex]
- [tex]\(50 - 47.5 = 2.5\)[/tex]
- [tex]\(70 - 47.5 = 22.5\)[/tex]
- Deviations from the mean for shipping costs:
- [tex]\(5.99 - 8.74 = -2.75\)[/tex]
- [tex]\(8.99 - 8.74 = 0.25\)[/tex]
- [tex]\(8.99 - 8.74 = 0.25\)[/tex]
- [tex]\(10.99 - 8.74 = 2.25\)[/tex]
4. Calculate Covariance:
[tex]\[ \text{Cov}(X, Y) = \frac{1}{n-1} \sum{(x_i - \bar{x})(y_i - \bar{y})} \][/tex]
[tex]\[ \text{Cov}(X, Y) = \frac{1}{3}((-22.5 \cdot -2.75) + (-2.5 \cdot 0.25) + (2.5 \cdot 0.25) + (22.5 \cdot 2.25)) \][/tex]
[tex]\[ = \frac{1}{3}(61.875 + -0.625 + 0.625 + 50.625) = \frac{1}{3}(112.5) = 37.5 \][/tex]
5. Calculate Standard Deviations:
- Standard deviation of total cost of items ([tex]\( \sigma_X \)[/tex]):
[tex]\[ \sigma_X = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} = \sqrt{\frac{1}{3}((-22.5)^2 + (-2.5)^2 + (2.5)^2 + (22.5)^2)} \][/tex]
[tex]\[ = \sqrt{\frac{1}{3}(506.25 + 6.25 + 6.25 + 506.25)} = \sqrt{341.67} = 18.48 \][/tex]
- Standard deviation of shipping costs ([tex]\( \sigma_Y \)[/tex]):
[tex]\[ \sigma_Y = \sqrt{\frac{1}{n-1} \sum (y_i - \bar{y})^2} = \sqrt{\frac{1}{3}((-2.75)^2 + (0.25)^2 + (0.25)^2 + (2.25)^2)} \][/tex]
[tex]\[ = \sqrt{\frac{1}{3}(7.5625 + 0.0625 + 0.0625 + 5.0625)} = \sqrt{4.25} = 2.06 \][/tex]
6. Calculate Correlation Coefficient:
[tex]\[ r = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y} = \frac{37.5}{18.48 \cdot 2.06} = \frac{37.5}{38.07} \approx 0.984 \][/tex]
7. Interpret the Correlation Coefficient:
- If [tex]\( r \)[/tex] is 0.984, this indicates a strong positive correlation (since 0.984 is much closer to 1).
Hence, the correct description for the strength of the model is "a strong positive correlation".
Here is a step-by-step guide:
1. List Down Data Points:
- Total cost of items: 25, 45, 50, 70
- Shipping costs: 5.99, 8.99, 8.99, 10.99
2. Calculate Means:
- Mean of total cost of items [tex]\[(\bar{x}) = \frac{25 + 45 + 50 + 70}{4} = \frac{190}{4} = 47.5\][/tex]
- Mean of shipping costs [tex]\[(\bar{y}) = \frac{5.99 + 8.99 + 8.99 + 10.99}{4} = \frac{34.96}{4} = 8.74\][/tex]
3. Calculate Deviations:
- Deviations from the mean for total cost of items:
- [tex]\(25 - 47.5 = -22.5\)[/tex]
- [tex]\(45 - 47.5 = -2.5\)[/tex]
- [tex]\(50 - 47.5 = 2.5\)[/tex]
- [tex]\(70 - 47.5 = 22.5\)[/tex]
- Deviations from the mean for shipping costs:
- [tex]\(5.99 - 8.74 = -2.75\)[/tex]
- [tex]\(8.99 - 8.74 = 0.25\)[/tex]
- [tex]\(8.99 - 8.74 = 0.25\)[/tex]
- [tex]\(10.99 - 8.74 = 2.25\)[/tex]
4. Calculate Covariance:
[tex]\[ \text{Cov}(X, Y) = \frac{1}{n-1} \sum{(x_i - \bar{x})(y_i - \bar{y})} \][/tex]
[tex]\[ \text{Cov}(X, Y) = \frac{1}{3}((-22.5 \cdot -2.75) + (-2.5 \cdot 0.25) + (2.5 \cdot 0.25) + (22.5 \cdot 2.25)) \][/tex]
[tex]\[ = \frac{1}{3}(61.875 + -0.625 + 0.625 + 50.625) = \frac{1}{3}(112.5) = 37.5 \][/tex]
5. Calculate Standard Deviations:
- Standard deviation of total cost of items ([tex]\( \sigma_X \)[/tex]):
[tex]\[ \sigma_X = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} = \sqrt{\frac{1}{3}((-22.5)^2 + (-2.5)^2 + (2.5)^2 + (22.5)^2)} \][/tex]
[tex]\[ = \sqrt{\frac{1}{3}(506.25 + 6.25 + 6.25 + 506.25)} = \sqrt{341.67} = 18.48 \][/tex]
- Standard deviation of shipping costs ([tex]\( \sigma_Y \)[/tex]):
[tex]\[ \sigma_Y = \sqrt{\frac{1}{n-1} \sum (y_i - \bar{y})^2} = \sqrt{\frac{1}{3}((-2.75)^2 + (0.25)^2 + (0.25)^2 + (2.25)^2)} \][/tex]
[tex]\[ = \sqrt{\frac{1}{3}(7.5625 + 0.0625 + 0.0625 + 5.0625)} = \sqrt{4.25} = 2.06 \][/tex]
6. Calculate Correlation Coefficient:
[tex]\[ r = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y} = \frac{37.5}{18.48 \cdot 2.06} = \frac{37.5}{38.07} \approx 0.984 \][/tex]
7. Interpret the Correlation Coefficient:
- If [tex]\( r \)[/tex] is 0.984, this indicates a strong positive correlation (since 0.984 is much closer to 1).
Hence, the correct description for the strength of the model is "a strong positive correlation".
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