Curve Fitting
- Curve fitting is the method of finding an approximate curve that satisfies the given data.
- There are two types of curves to be fitted:
- Exact fit - Passes through all the points, suitable for a small number of data points.
- Best fit - Does not pass through all the points, suitable for a large number of data points.
- Why are we learning curve fitting?
- To perform trend analysis, such as exponential or linear trends.
- To carry out interpolation and optimization.
- If you want to know the value of properties at any intermediate point, curve fitting helps to estimate those properties at intermediate points.
- There are three different curves that can be fitted using the least squares method:
- ☆ Straight line
- Parabola
- ☆ Exponential curve
Fitting of a Straight Line Using the Least Squares Method
Working Rule (Steps)
- Let y = ax + b (Equation 1) be the curve of best fit, where 'a' and 'b' are the constants to be determined.
- Write the normal equations:
There are two constants in y = ax + b, so there are two normal equations:
∑y = a∑x + b∑1 (Note: ∑1 = n, so it is ∑y = a∑x + bn)
where n is the number of given data points.
∑xy = a∑x² + b∑x - Using tabular calculation with the given data, find the unknowns 'a' and 'b'.
- Substitute these values of 'a' and 'b' into Equation 1 to get the required curve of best fit.
Question: By the method of least squares, find the best fitting straight line to the data given below:
x | 5 | 10 | 15 | 20 | 25
y | 15 | 19 | 23 | 26 | 30
Solution:
- Let y = ax + b (Equation 1) be the straight line of best fit where 'a' and 'b' are the constants to be determined.
- The normal equations are:
∑y = a∑x + bn (Equation 2)
where n is the number of given data points
∑xy = a∑x² + b∑x (Equation 3)
The number of given data points, n = 5. - Tabular calculations as we need ∑x, ∑y, ∑x², and ∑xy for finding the values of 'a' and 'b'
through the normal equations:
x | y | x² | xy -------------------------- 5 | 15 | 25 | 75 10 | 19 | 100 | 190 15 | 23 | 225 | 345 20 | 26 | 400 | 520 25 | 30 | 625 | 750∑x = 75, ∑y = 113, ∑x² = 1375, and ∑xy = 1880
Now substitute these summation values into the normal equations to get:
Equation 2: 113 = 75a + 5b (Equation 4)
Equation 3: 1880 = 1375a + 75b (Equation 5)
On solving Equation 4 and Equation 5 we get:
a = 0.74
b = 11.5 - Now substitute the values of 'a' and 'b' into Equation 1 to get:
y = 0.74x + 11.5
Fitting of Exponential Curves by the Method of Least Squares
Example: y = aebx (exponential), y = axb (power)
Question: Fit y = aebx to the following data using the method of least squares:
x | 0 | 5 | 8 | 12 | 20
y | 3 | 1.5 | 1 | 0.55 | 0.18
- Let y = aebx (Equation 1) be the curve to be fitted where 'a' and 'b' are constants to be determined.
- Taking the logarithm on both sides, we get:
log(y) = log(aebx)
log(y) = log(a) + log(ebx) [since log(AB) = log(A) + log(B)]
log(y) = log(a) + bx log(e) [since log(Am) = m log(A)]
Therefore, Y = A + Bx where Y = log(y), A = log(a), and B = b log(e)
Now, the normal equations are:
∑Y = An + B∑x (Equation 2)
∑xY = A∑x + B∑x² (Equation 3)
The number of data points, n = 5 - Tabular calculation:
x | y | Y = log(y) | xY | x² --------------------------------------------- 0 | 3 | 0.4771 | 0 | 0 5 | 1.5 | 0.1761 | 0.8805 | 25 8 | 1 | 0.0000 | 0 | 64 12 | 0.55 | -0.2596 | -3.1152 | 144 20 | 0.18 | -0.7447 | -14.894 | 400∑x = 45, ∑Y = -0.3512, ∑xY = -17.1292, and ∑x² = 633 - Now we will substitute these values into Equations 2 and 3 to get A and B:
Equation 2: -0.3512 = 5A + 45B (Equation 4)
Equation 3: -17.1292 = 45A + 633B (Equation 5)
Upon solving, we get:
B = -0.0612
A = 0.48056
But A = log10(a)
log10(a) = 0.48056
a = 100.48056
a ≈ 3.0238
And B = b log(e)
b log10(e) = -0.0612
b = -0.0612 / log(e)
b ≈ -0.1409 - Now Equation 1 becomes:
y = 3.0238 e-0.1409x
Fit a curve y = axb to the following data:
x | 1 | 2 | 3 | 4 | 5 | 6
y | 2.98 | 4.26 | 5.21 | 6.10 | 6.80 | 7.50
Solution:
- Let y = axb (Equation 1) be the curve to be fitted where 'a' and 'b' are constants to
be determined. Taking the logarithm on both sides of Equation 1, we get:
log(y) = log(axb)
log(y) = log(a) + log(xb) [since log(AB) = log(A) + log(B)]
log(y) = log(a) + b log(x)
Therefore, Y = A + bX where Y = log(y), X = log(x), and A = log(a)
Now, the normal equations are:
∑Y = An + b∑X (Equation 2)
∑XY = A∑X + b∑X² (Equation 3)
The number of given data points, n = 6 -
Tabular calculation:
x | y | X = log(x) | Y = log(y) | X² | XY ------------------------------------------------------------------- 1 | 2.98 | 0 | 0.4742 | 0 | 0 2 | 4.26 | 0.3010 | 0.6294 | 0.0906 | 0.1894 3 | 5.21 | 0.4771 | 0.7168 | 0.2276 | 0.3419 4 | 6.10 | 0.6021 | 0.7853 | 0.3625 | 0.4727 5 | 6.80 | 0.6990 | 0.8325 | 0.4886 | 0.5819 6 | 7.50 | 0.7782 | 0.8751 | 0.6056 | 0.6809 ------------------------------------------------------------------- ∑ | | 2.8574 | 4.3133 | 1.7743 | 2.2668 - Now we will substitute these values into Equations 2 and 3 to get A and B:
Equation 2: 4.3133 = 6A + 2.8574b (Equation 4)
Equation 3: 2.2668 = 2.8574A + 1.7743b (Equation 5)
On solving these equations, we get:
b = 0.5140
A = 0.4741
Now, A = log(a)
So, log(a) = 0.4741
a = 100.4741
a ≈ 2.9792 - Equation 1 becomes:
y = 2.9792x0.5140
Correlation
- Correlation refers to the relationship between two or more variables. The degree of strength of the
relationship between the two variables is given by the coefficient of correlation, denoted by 'r'.
The formula for 'r' is:
Where X = x - mean(x) and Y = y - mean(y). Here, mean(x) and mean(y) are the means of x and y, respectively.
mean(x) = (sum of x) / n and mean(y) = (sum of y) / n, where n is the total number of given data points. - 'r' is also known as Karl Pearson's coefficient of correlation.
- If the values of x and y are numerically large, we apply the following formula:
- The coefficient of correlation (r) has values between -1 and 1:
- If r = 0, there is no correlation.
- If r = +1, the relationship is perfectly positive.
- If r = -1, the relationship is perfectly negative.
Karl Pearson's coefficient of correlation
Example Question: Find the coefficient of correlation for the given data:
x | 6 | 2 | 10 | 4 | 8
y | 9 | 11 | 5 | 8 | 7
Solution:
- There are numerically small values, so we are going to use the following formula:
Where X = x - mean(x) and Y = y - mean(y) - The number of given data points, n = 5
Let's find mean(x) = sum of x / n = (6 + 2 + 10 + 4 + 8) / 5 = 6
mean(y) = sum of y / n = (9 + 11 + 5 + 8 + 7) / 5 = 8
Therefore, X = x - mean(x) = x - 6
Therefore, Y = y - mean(y) = y - 8 - Tabular calculation:
x | y | X | Y | X² | Y² | XY ------------------------------------------ 6 | 9 | 0 | 1 | 0 | 1 | 0 2 | 11 | -4 | 3 | 16 | 9 | -12 10 | 5 | 4 | -3 | 16 | 9 | -12 4 | 8 | -2 | 0 | 4 | 0 | 0 8 | 7 | 2 | -1 | 4 | 1 | -2 ------------------------------------------ ∑ | | | | 40 | 20 | -26 - r = -26 / sqrt(40 * 20)
Therefore, r = -0.9192
Question: Find the correlation for the following data:
x | 122 | 135 | 143 | 103 | 156 | 178 | 190
y | 46 | 78 | 89 | 46 | 82 | 52 | 96
Solution:
- The number of given data points, n = 7
mean(x) = 146.71
mean(y) = 69.85
As mean(x) and mean(y) are not integers, we use the method of assumed means where we round them off. So now, mean(x) = 147 and mean(y) = 70
Now, X = x - mean(x) = x - 147
and Y = y - mean(y) = y - 70
As the data points are larger, we are going to use the following formula:
- Tabular calculation:
x | y | X | Y | X² | Y² | XY --------------------------------------------------- 122 | 46 | -25 | -24 | 625 | 576 | 600 135 | 78 | -12 | 8 | 144 | 64 | -96 143 | 89 | -4 | 19 | 16 | 361 | -76 103 | 46 | -44 | -24 | 1936 | 576 | 1056 156 | 82 | 9 | 12 | 81 | 144 | 108 178 | 52 | 31 | -18 | 961 | 324 | -558 190 | 96 | 43 | 26 | 1849 | 676 | 1118 ---------------------------------------------------- ∑ | | -2 | -17 | 5612 | 2712 | 2152 - After putting all the values in the formula, we get r = 0.5506
Spearman's Rank Correlation
- Determine the correlation coefficient between two variables that cannot be measured quantitatively.
- Qualitative features such as honesty, beauty, character, morality, etc.
- The rank correlation coefficient is given by:
- Rank correlation coefficient has three cases:
- When actual ranks are given
- When ranks are not given
- In case of equal ranks
- The value of the rank correlation coefficient (r) can range from -1 to 1, indicating the
strength and direction of the relationship:
- If r = 1, there is a perfect positive correlation.
- If 0.7 < r < 1, there is a strong positive correlation.
- If 0.3 < r ≤ 0.7, there is a moderate positive correlation.
- If 0 < r ≤ 0.3, there is a weak positive correlation.
- If r = 0, there is no correlation.
- If -0.3 ≤ r < 0, there is a weak negative correlation.
- If -0.7 ≤ r < -0.3, there is a moderate negative correlation.
- If -1 ≤ r < -0.7, there is a strong negative correlation.
- If r = -1, there is a perfect negative correlation.
Case 1: When actual ranks are given
Question: The preference of two persons A and B for different car brands are given in the following table:
Car Brand | Person A | Person B
--------------------------------
Suzuki | 5 | 1
Hyundai | 4 | 4
Tata | 1 | 2
Skoda | 3 | 5
Toyota | 2 | 3
Determine the coefficient of rank correlation between the preferences of consumers A and B and interpret
your result.
Solution:
- Tabular Calculation to find D and D²
R1 | R2 | D = R1 - R2 | D² ---------------------------- 5 | 1 | 4 | 16 4 | 4 | 0 | 0 1 | 2 | -1 | 1 3 | 5 | -2 | 4 2 | 3 | -1 | 1 ---------------------------- ∑ | | 0 | 22 - Now the formula is:
r = 1 - ((6 * 22) / (125 - 5))
r = 1 - (132 / 120)
r = 1 - 1.1
r = -0.1
The value of r shows that there is a very weak negative correlation between the preferences of persons A and B.
Case 2: When ranks are not given
Question: Calculate the rank correlation coefficient from the given data:
Marks in Computer (x): 20 30 40 50 60 70 80 Marks in English (y) : 14 5 30 32 40 45 65
Solution:
- Tabular calculation, where ranks are assigned based on the lowest values getting the lowest rank
x | Rx | y | Ry | D | D² ----------------------------- 20 | 1 | 14 | 2 | -1 | 1 30 | 2 | 5 | 1 | 1 | 1 40 | 3 | 30 | 3 | 0 | 0 50 | 4 | 32 | 4 | 0 | 0 60 | 5 | 40 | 5 | 0 | 0 70 | 6 | 45 | 6 | 0 | 0 80 | 7 | 65 | 7 | 0 | 0 ----------------------------- ∑ | | | | 0 | 2 - Using the formula:
Here n = 7
r = 1 - ((6 * 2) / (343 - 7))
r = 1 - (12 / 336)
r = 1 - 0.0357
r = 0.964
The value of r shows that there is a very strong positive correlation between the marks in Computer and English.
Case 3: Equal ranks are given
- If some item is repeated for 'p' times in a given data, then add a factor of 1/12(p3
- p) to every repeated item.
i.e.,
If 2 items are repeated, then 1/12(p3 - p) will repeat 2 times.
Question: Calculate the rank correlation coefficient for the following data:
x | 12 | 15 | 18 | 20 | 16 | 15 | 18 | 22 | 15 | 21 | 18 | 15 --------------------------------------------------------------- y | 10 | 18 | 19 | 12 | 15 | 19 | 17 | 19 | 16 | 14 | 13 | 17
Solution:
-
x | y | Rx | Ry | D | D² -------------------------------------- 12 | 10 | 1 | 1 | 0 | 0 15 | 18 | 3.5 | 9 | -5.5 | 30.25 18 | 19 | 8 | 11 | -3 | 9 20 | 12 | 10 | 2 | 8 | 64 16 | 15 | 6 | 5 | 1 | 1 15 | 19 | 3.5 | 11 | -7.5 | 56.25 18 | 17 | 8 |7.5 | 0.5 | 0.25 22 | 19 | 12 | 11 | 1 | 1 15 | 16 | 3.5 | 6 | -2.5 | 6.25 21 | 14 | 11 | 4 | 7 | 49 18 | 13 | 8 | 3 | 5 | 25 15 | 17 | 3.5 |7.5 | -4 | 16 --------------------------------------- ∑ | | | | 0 | 258- For ranking x variable:
- Rank of 15 = (2 + 3 + 4 + 5) / 4 = 3.5
- Rank of 18 = (7 + 8 + 9) / 3 = 8
- For ranking y variable:
- Rank of 17 = (7 + 8) / 2 = 7.5
- Rank of 19 = (10 + 11 + 12) / 3 = 11
- For ranking x variable:
r = 1 - ((6 * (258 + 5 + 2 + 0.5 + 2)) / 1716)
r = 1 - (1605 / 1716)
r = 0.0646
The value of r shows that there is a very weak positive correlation between the variables.