Question
Do the data suggest a linear . quadratic or an exponential function ? Use regression to find a model for each data set.
Hint:
1. When the difference between 2 consecutive output values (y values) for a given constant change in the input values (x values) is constant. i.e. y(n)- y(n-1) is constant for any value of n, the function is known as a linear function.
2. Regression is a statistical tool used to find a model that can represent the relation between a given change in dependent variable (output values/ y values) for a given change in independent variable (input values/ x values).
Linear Equation using regression can be represented as-
Y = a + bX, where-
a = 
b = 
The correct answer is: The given function is a quadratic function and using Regression, the given function can be modelled into the equation- Y = - X2 + 20 X - 103.
Step-by-step solution:-
From the given table, we observe the following readings-
x1 = 6, y1 = -19;
x2 = 7, y2 = -12;
x3 = 8, y3 = -7;
x4 = 9, y4 = -4;
x5 = 10, y5 = -3
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 7 - 6 = 1
dx2 = x3 - x2 = 8 - 7 = 1
dx3 = x4 - x3 = 9 - 8 = 1
dx4 = x5 - x4 = 10 - 9 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = -12 - (-19) = -12 + 19 = 7
dy2 = y3 - y2 = -7 - (-12) = -7 + 12 = 5
dy3 = y4 - y3 = -4 - (-7) = -4 + 7 = 3
dy4 = y5 - y4 = -3 - (-4) = -3 + 4 = 1
We observe that the difference for every consecutive x values is constant i.e. 1 but for y values the difference is not constant.
Hence, the given function is not a linear function.
b). Now, difference between 2 consecutive differences for y values-
dy2 - dy1 = 7 - 5 = 2
dy3 - dy2 = 5 - 3 = 2
dy4 - dy3 = 3 - 1 = 2
We observe that the difference of differences of 2 consecutive y values are constant i.e. 2.
Hence, the given function is a quadratic function.
Using Quadratic Regression formula and values from the adjacent table-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
∴ -45 = 5c + b(40) + a(330)
∴ -45 = 5c + 40b + 330a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ -320 = c(40) + b(330) + a(2,800)
∴ -320 = 40c + 330b + 2,800a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ -2,344 = c(330) + b(2,800) + a(24,354)
∴ -2,344 = 330c + 2,800b + 24,354a ....................... (Equation iii)
Dividing Equation 2 by 8, we get-
350a + 41.25b + 5c = -40 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
350a + 41.25b + 5c = -40 …............................................... (Equation iv)
- 330a + 40b + 5c = -45 …............................................... (Equation i)
20a + 1.25b = 5 .................................................. (Equation v)
Multiplying Equation ii with 8.25, we get-
23,100a + 2,722.5b + 330c = -2,640 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
24,354a + 2,800b + 330c = -2,344 ......................... (Equation iii)
- 23,100a + 2,722.5b + 330c = -2,640 ......................... (Equation vi)
1,254a + 77.5b = 296 ......................... (Equation vii)
Multiplying Equation v with 62, we get-
1,240a + 77.5b = 310 ............................................... (Equation viii)
Subtracting Equation vii from Equation viii, we get-
1,240a + 77.5b = 310 ............................................... (Equation viii)
- 1,254a + 77.5b = 296 ............................................... (Equation vii)
-14a = 14
i.e. -14a = 14
∴ a = 14/ -14 ................................... (Dividing both sides by -14)
∴ a = -1
Substituting a = -1 in Equation v, we get-
20a + 1.25b = 5 .................................................. (Equation v)
∴ 20(-1) + 1.25b = 5
∴ -20 + 1.25b = 5
∴ 1.25b = 5 + 20 ........................................ (Taking all constants together)
∴ 1.25b = 25
∴ b = 25/1.25 ............................................ (Dividing both sides by 1.25)
∴ b = 20
Substituting a = -1 and b = 20 in Equation i, we get-
330a + 40b + 5c = -45 .............................. (Equation i)
∴ 330 (-1) + 40 (20) + 5c = -45
∴ -330 + 800 + 5c = -45
∴ 470 + 5c = -45
∴ 5c = -45 - 470 ..................... (Taking all constants together)
∴ 5c = -515
∴ c = -515/5 ........................... (Dividing both sides by 5)
∴ c = -103
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = -1 X2 + 20 X + (-103)
∴ Y = - X2 + 20 X - 103
Final Answer:-
∴ The given function is a quadratic function and using Regression, the given function can be modelled into the equation- Y = - X2 + 20 X - 103.
x2 = 7, y2 = -12;
x3 = 8, y3 = -7;
x4 = 9, y4 = -4;
x5 = 10, y5 = -3
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 7 - 6 = 1
dx2 = x3 - x2 = 8 - 7 = 1
dx3 = x4 - x3 = 9 - 8 = 1
dx4 = x5 - x4 = 10 - 9 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = -12 - (-19) = -12 + 19 = 7
dy2 = y3 - y2 = -7 - (-12) = -7 + 12 = 5
dy3 = y4 - y3 = -4 - (-7) = -4 + 7 = 3
dy4 = y5 - y4 = -3 - (-4) = -3 + 4 = 1
We observe that the difference for every consecutive x values is constant i.e. 1 but for y values the difference is not constant.
Hence, the given function is not a linear function.
b). Now, difference between 2 consecutive differences for y values-
dy2 - dy1 = 7 - 5 = 2
dy3 - dy2 = 5 - 3 = 2
dy4 - dy3 = 3 - 1 = 2
We observe that the difference of differences of 2 consecutive y values are constant i.e. 2.
Hence, the given function is a quadratic function.
Using Quadratic Regression formula and values from the adjacent table-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
∴ -45 = 5c + b(40) + a(330)
∴ -45 = 5c + 40b + 330a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ -320 = c(40) + b(330) + a(2,800)
∴ -320 = 40c + 330b + 2,800a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ -2,344 = c(330) + b(2,800) + a(24,354)
∴ -2,344 = 330c + 2,800b + 24,354a ....................... (Equation iii)
Dividing Equation 2 by 8, we get-
350a + 41.25b + 5c = -40 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
350a + 41.25b + 5c = -40 …............................................... (Equation iv)
- 330a + 40b + 5c = -45 …............................................... (Equation i)
20a + 1.25b = 5 .................................................. (Equation v)
Multiplying Equation ii with 8.25, we get-
23,100a + 2,722.5b + 330c = -2,640 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
24,354a + 2,800b + 330c = -2,344 ......................... (Equation iii)
- 23,100a + 2,722.5b + 330c = -2,640 ......................... (Equation vi)
1,254a + 77.5b = 296 ......................... (Equation vii)
Multiplying Equation v with 62, we get-
1,240a + 77.5b = 310 ............................................... (Equation viii)
Subtracting Equation vii from Equation viii, we get-
1,240a + 77.5b = 310 ............................................... (Equation viii)
- 1,254a + 77.5b = 296 ............................................... (Equation vii)
-14a = 14
i.e. -14a = 14
∴ a = 14/ -14 ................................... (Dividing both sides by -14)
∴ a = -1
Substituting a = -1 in Equation v, we get-
20a + 1.25b = 5 .................................................. (Equation v)
∴ 20(-1) + 1.25b = 5
∴ -20 + 1.25b = 5
∴ 1.25b = 5 + 20 ........................................ (Taking all constants together)
∴ 1.25b = 25
∴ b = 25/1.25 ............................................ (Dividing both sides by 1.25)
∴ b = 20
Substituting a = -1 and b = 20 in Equation i, we get-
330a + 40b + 5c = -45 .............................. (Equation i)
∴ 330 (-1) + 40 (20) + 5c = -45
∴ -330 + 800 + 5c = -45
∴ 470 + 5c = -45
∴ 5c = -45 - 470 ..................... (Taking all constants together)
∴ 5c = -515
∴ c = -515/5 ........................... (Dividing both sides by 5)
∴ c = -103
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = -1 X2 + 20 X + (-103)
∴ Y = - X2 + 20 X - 103
Final Answer:-
∴ The given function is a quadratic function and using Regression, the given function can be modelled into the equation- Y = - X2 + 20 X - 103.
,where H is the height in centimeters and M is the mass in kilograms. A doctor calculates a particular dose of medicine for a patient whose BSA is less than 1.9. If the patient is 160 cm tall, what must the mass of the person be for the dose to be appropriate?
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