# How to draw regression line? S1Watch

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#1
Hi,

Been doing regression lines in S1, and have a set of coordinates which I've plotted, and now have to draw the regression line.

My teacher said to do it using the coordinates of mean of x and mean of y, but not sure exactly what to do for that?

Thanks
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#2
anyone?
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11 years ago
#3
Use a calculator to find the gradient and the y intercept. sorted.
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11 years ago
#4
http://www.ocr.org.uk/Data/publicati...lae_tables.pdf

Page 9, (method of least squares)
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8 years ago
#5
I used a calculator to calculate y=a+bx
I'm given: y=509+3.25x
If I rearrange that to the form of y=mx+c; y=3.25x+509
I know the c-intercept but I'm not sure how to draw the gradient of the line, should I just aim to go through the intercept and go for a line of best fit?
[June 10 paper, Q6 (B) (i)]
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8 years ago
#6
basically what razor 94 said, i dont know how to draw the gradient
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8 years ago
#7
You have the mean of y and x. y = b + ax, so b = (mean y) - a(mean x), using your data.

On during LOBF, make sure that it goes through mean of y and mean of x : (mean x, mean y) and the y-intercept. This will form a straight line between two points so no need to worry about the gradient as it will be represented there.
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8 years ago
#8
If you're trying to draw a line of regression, I'm assuming you have worked out the equation first.
y = a + bx
Using your data, put the lowest x and the highest x into the equation and you'll get values for y.
Using said values of x and y, plot them on the graph and simply join up the points.
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2 years ago
#9
(Original post by razor94)
I used a calculator to calculate y=a+bx
I'm given: y=509+3.25x
If I rearrange that to the form of y=mx+c; y=3.25x+509
I know the c-intercept but I'm not sure how to draw the gradient of the line, should I just aim to go through the intercept and go for a line of best fit?
[June 10 paper, Q6 (B) (i)]
maebe if youman had a brizzy you could work it oot
0
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