Line of Best Fit (Least Square Method)
A
line of best fit
is a straight note that is the best approximation of the given sic of data .
It is used to study the nature of the relation back between two variables. ( We ‘re merely considering the two-dimensional case, hera. )
A line of best fit can be approximately determined using an eyeball method by drawing a straight agate line on a scatter plot so that the number of points above the course and below the line is about equal ( and the line passes through arsenic many points as possible ) .
A more accurate way of finding the line of best fit is the
least square method
.
Use the stick to steps to find the equality of line of best fit for a laid of ordered pairs ( adam 1, yttrium 1 ), ( ten 2, yttrium 2 ), … ( x north, y normality ) .
tone 1 : Calculate the entail of the x -values and the entail of the yttrium -values .
adam ¯ = ∑ one = 1 north ten i n Y ¯ = ∑ one = 1 n y i n
step 2 : The watch formula gives the slope of the course of best fit :
megabyte = ∑ i = 1 north ( x i − X ¯ ) ( y one − Y ¯ ) ∑ iodine = 1 n ( ten iodine − X ¯ ) 2
step 3 : Compute the yttrium -intercept of the note by using the formula :
barn = Y ¯ − thousand X ¯
pace 4 : Use the gradient m and the y -intercept barn to form the equality of the line .
Example:
Use the least square method acting to determine the equation of line of best fit for the data. then plot the cable .
|
x |
8 |
2 |
11 |
6 |
5 |
4 |
12 |
9 |
6 |
1 |
|
y |
3 |
10 |
3 |
6 |
8 |
12 |
1 |
4 |
9 |
14 |
Solution:
Plot the points on a align plane.
Calculate the means of the x -values and the yttrium -values .
x ¯ = 8 + 2 + 11 + 6 + 5 + 4 + 12 + 9 + 6 + 1 10 = 6.4 Y ¯ = 3 + 10 + 3 + 6 + 8 + 12 + 1 + 4 + 9 + 14 10 = 7
now calculate x i − X ¯, yttrium one − Y ¯, ( x one − X ¯ ) ( y iodine − Y ¯ ), and ( adam i − X ¯ ) 2 for each one .
|
i |
x i |
y i |
x i − X ¯ |
y i − Y ¯ |
( x i − X ¯ ) ( y i − Y ¯ ) |
( x i − X ¯ ) 2 |
|
1 |
8 |
3 |
1.6 |
− 4 |
− 6.4 |
2.56 |
|
2 |
2 |
10 |
− 4.4 |
3 |
− 13.2 |
19.36 |
|
3 |
11 |
3 |
4.6 |
− 4 |
− 18.4 |
21.16 |
|
4 |
6 |
6 |
− 0.4 |
− 1 |
0.4 |
0.16 |
|
5 |
5 |
8 |
− 1.4 |
1 |
− 1.4 |
1.96 |
|
6 |
4 |
12 |
− 2.4 |
5 |
− 12 |
5.76 |
|
7 |
12 |
1 |
5.6 |
− 6 |
− 33.6 |
31.36 |
|
8 |
9 |
4 |
2.6 |
− 3 |
− 7.8 |
6.76 |
|
9 |
6 |
9 |
− 0.4 |
2 |
− 0.8 |
0.16 |
|
10 |
1 |
14 |
− 5.4 |
7 |
− 37.8 |
29.16 |
|
∑ i = 1 n ( x i − X ¯ ) ( y i − Y ¯ ) = − 131 |
∑ i = 1 n ( x i − X ¯ ) 2 = 118.4 |
Calculate the slope .
m = ∑ one = 1 newton ( ten one − X ¯ ) ( y one − Y ¯ ) ∑ i = 1 n ( ten i − X ¯ ) 2 = − 131 118.4 ≈ − 1.1
Calculate the yttrium -intercept .
Use the formula to compute the yttrium -intercept. bel = Y ¯ − m X ¯ = 7 − ( − 1.1 × 6.4 ) = 7 + 7.04 ≈ 14.0
Use the slope and yttrium -intercept to form the equation of the line of best match. The slope of the line is − 1.1 and the yttrium -intercept is 14.0.
consequently, the equation is yttrium = − 1.1 x + 14.0 .
Draw the line on the break up plot .
