What is the quadratic regression equation that fits these data?
| x | y |
| --- | -- |
| -4 | 40 |
| -3 | 28 |
| -2 | 10 |
| -1 | 8 |
| 0 | 7 |
| 1 | 10 |
| 2 | 16 |
| 3 | 26 |
| 4 | 40 |
A. [tex]y=1.64 x^2+1.24 x+8.08[/tex]
B. [tex]y=16.76 \cdot 1.02^x[/tex]
C. [tex]y=2.13 x^2+0.13 x+6.39[/tex]
Answer (2)
Calculate the necessary sums from the given data points.
Use these sums to compute the intermediate values required for the least squares method.
Calculate the coefficients a, b, and c using the least squares formulas.
The quadratic regression equation is y = 2.13 x 2 + 0.13 x + 6.39 .
Explanation
Understanding the Problem We are given a set of data points (x, y) and asked to find the quadratic regression equation that best fits the data. The general form of a quadratic equation is y = a x 2 + b x + c , where a, b, and c are coefficients we need to determine.
Method of Least Squares To find the best-fit quadratic equation, we can use the method of least squares. This involves minimizing the sum of the squares of the differences between the observed y-values and the y-values predicted by the quadratic equation. We can use the following formulas to calculate the coefficients a, b, and c:
a = ( S x 2 x 2 − S x 2 x 2 S xx ) ( S x y S x 2 x − S x 2 y S xx ) b = S x 2 x 2 ( S x 2 y − a S x 2 x ) c = n ( ∑ y − a ∑ x 2 )
Where: S xx = ∑ x 2 − (( ∑ x ) 2 / n ) S x y = ∑ x y − (( ∑ x ∑ y ) / n ) S x 2 x 2 = ∑ x 4 − (( ∑ x 2 ) 2 / n ) S x 2 y = ∑ x 2 y − (( ∑ x 2 ∑ y ) / n ) S x 2 x = ∑ x 3 − (( ∑ x 2 ∑ x ) / n )
Calculating Sums Given the data points: x = [-4, -3, -2, -1, 0, 1, 2, 3, 4] y = [40, 28, 10, 8, 7, 10, 16, 26, 40] n = 9
We calculate the following sums: ∑ x = − 4 − 3 − 2 − 1 + 0 + 1 + 2 + 3 + 4 = 0 ∑ y = 40 + 28 + 10 + 8 + 7 + 10 + 16 + 26 + 40 = 185 ∑ x 2 = 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 = 60 ∑ x 3 = − 64 − 27 − 8 − 1 + 0 + 1 + 8 + 27 + 64 = 0 ∑ x 4 = 256 + 81 + 16 + 1 + 0 + 1 + 16 + 81 + 256 = 708 ∑ x y = − 160 − 84 − 20 − 8 + 0 + 10 + 32 + 78 + 160 = 0 $\sum x^2y = 640 + 252 + 40 + 8 + 0 + 10 + 64 + 234 + 640 = 1888
Calculating Intermediate Values Now we can calculate the intermediate values: S xx = 60 − ( 0 2 /9 ) = 60 S x y = 0 − ( 0 ∗ 185 ) /9 = 0 S x 2 x 2 = 708 − ( 6 0 2 /9 ) = 708 − 400 = 308 S x 2 y = 1888 − ( 60 ∗ 185 ) /9 = 1888 − 1233.33 = 654.67 $S_{x^2x} = 0 - (60 * 0) / 9 = 0
Calculating Coefficients Now we can calculate the coefficients: a = ( 0 ∗ 0 − 654.67 ∗ 60 ) / ( 0 2 − 308 ∗ 60 ) = ( − 39280 ) / ( − 18480 ) = 2.12554 b = ( 654.67 − 2.12554 ∗ 0 ) /308 = 654.67/308 = 0.13333 $c = (185 - 2.12554 * 60) / 9 = (185 - 127.5324) / 9 = 6.38528
Final Equation Therefore, the quadratic regression equation is approximately: y = 2.13 x 2 + 0.13 x + 6.39
Comparing this with the given options, we find that option C is the closest.
Conclusion The quadratic regression equation that fits the data is y = 2.13 x 2 + 0.13 x + 6.39 .
Examples
Quadratic regression is used in various fields, such as physics to model projectile motion, in economics to model cost curves, and in data science to find non-linear relationships between variables. For example, if you want to model the relationship between the amount of fertilizer used and crop yield, you might find that the relationship is quadratic. Initially, increasing fertilizer increases yield, but beyond a certain point, adding more fertilizer decreases yield due to over-saturation. This can be modeled using a quadratic regression equation, allowing farmers to optimize fertilizer usage for maximum yield.
The quadratic regression equation that best fits the given data points is approximately y = 2.13 x 2 + 0.13 x + 6.39 , which corresponds to option C. This equation represents the relationship between the x and y values provided. Therefore, the answer is option C.
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