SOLUTION: Please i need help i am so confused. My teacher assigned me this today and it's due Monday. If i have data points: (2,1), (4,3), (5,5), (7,6), (3,18) 1. How can i find a

Algebra ->  Test -> SOLUTION: Please i need help i am so confused. My teacher assigned me this today and it's due Monday. If i have data points: (2,1), (4,3), (5,5), (7,6), (3,18) 1. How can i find a      Log On


   

Question 672408: Please i need help i am so confused. My teacher assigned me this today and it's due Monday.
If i have data points: (2,1), (4,3), (5,5), (7,6), (3,18)
1. How can i find and graph a line of best fit that includes the data points?
2. How can i find and graph a line of best fit that does not include the data points ?
3. Which equation would better model the data and why ?
Please please please i honestly don't understand what he means in these questions. Please I'd appreciate the help.
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
x|y
2|1
4|3
5|5
7|6
3|18
...
Line that include data points, chose (2,1) and (7,6) from visual quick plot of the points.
slope = %281-6%29%2F%282-7%29+=+-5%2F-5+=+1
y - 1 = 1(x-2)
y = x - 2 + 1
highlight_green%28y+=+x-1%29
...
Line that does not include the data points
Using the "least square method" for best fit line
Find xy and x^2
x|y|xy|x^2
2|1|2|4
4|3|12|16
5|5|25|25
7|6|42|49
3|18|54|9
..........
21|33|135|103 ==> Sums of each column
...
slope = %28135-%28%2821%2A33%29%2F5%29%29%2F%28103-%2821%5E2%2F5%29%29
= %28135-%28693%2F5%29%29%2F%28103-%28441%2F5%29%29%29
= -3.6+%2F+14.8+=+-.243
...
y-intercept
6.6 - 4.2(-.243) = 7.62
...
Best fit line
highlight%28y+=+-.243x+%2B+7.62%29
...
For an explanation of the forumula
http://hotmath.com/hotmath_help/topics/line-of-best-fit.html
...
The first equation is a better model, as there are only two outliers from the line, one more extreme than the other. This line thereby captures on it three of the 5 data points. The second line, while a true "best fit line" captures none of the data points and represents too wide of a variance resulting from one extreme outlier. Not to mention the negative slope, whereas, the data represents a positive slope trend, not a negative one.
...
graph%28300%2C200%2C-10%2C10%2C-10%2C10%2C-.243x%2B7.62%2Cx-1%29
.....................
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-Reading Boosters
Email: HomeworkHelpers@ReadingBoosters.com
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