Y and X axis Least Squares regression

[Einstein]

Hi,

Say the graphs y axis begins at 60, how do i determine when it crosses the y intercept? even for 3 median line etc..
Do i select two points and put on my calc and calc lin reg line? also what happens if i my x axis doesnt begin at 0?

bit confused with those type of questions.

Thanks

[Zealous]

Hi,

Say the graphs y axis begins at 60, how do i determine when it crosses the y intercept? even for 3 median line etc..
Do i select two points and put on my calc and calc lin reg line? also what happens if i my x axis doesnt begin at 0?

bit confused with those type of questions.

Thanks

So your calculator is just moving the graphing window so you can't see the origin (0,0) - or at least the x and y axis.

Do a linear regression of your data. You'll get an equation of a line that fits the data in the form of y=mx+c or sometimes y=ax+b. The y intercept will be the c or b value - or the constant (not a coefficient of x).

[Einstein]

what i meant was, Say the equation is  y= -27.88-1.17x and my x axis starts at 155, how do i plot the line ? the answer doesn't start at -27.88 it cuts the y axis (which is visible on my graph) at about 15.50

[~V]

what i meant was, Say the equation is  y= -27.88-1.17x and my x axis starts at 155, how do i plot the line ? the answer doesn't start at -27.88 it cuts the y axis (which is visible on my graph) at about 15.50

Are you provided with those set of axis and you need to start at 155? If not, just go ahead and plot the least square regression above with -27.88 cutting the y axis.
Edit: Or maybe you need to zoom out and you'll see the yint at -27.88? LOL

[Einstein]

no its on a exam paper, y axis starts at 0 ends at 30, x axis starts at 155 ends at 185, y int is -27 and it cuts the y int on the test paper at about 15

[The Brightest Witch]

Is this a specific question you could show us or refer to?

[~V]

This is really confusing now, yeah, please post a picture, that'd be very helpful. Are you given a table of values instead of just the scatterplot? Because you could've calculated least square regression equation wrong, especially if it's reading off a graph.

[Einstein]

Alright, here it is have a look at the table of values, graph, question 1 g and 1f, this is the answer book.

[~V]

Photon, i've sketched the scatterplot onto the CAS and it did come out the same as the graph in that image. When putting a line on a scatterplot, you need two points right? And the first thing you would do, well, what i would do is to mark the y-intercept right? But in this question, it seems as though they didn't use the y-int, but they used the lowest interval instead. So they subbed in 155 as x and got 17.64, so their line will cross (155,17.64) and the other point they chose was the highest interval value,185. So sub 185 as x and you'll get 26.34 as y. So its (185,26.34). Mark those two points and draw the line.That's why your CAS is handy because you go to data and statistics and see  if where the smallest interval is and just sub in equation to see where to draw it. As for question 1g, it just tells you to describe the meaning of the slope (gradient), just describe for every y (depending on the context), x is increasing/decreasing(depending if m is -ve or +ve) by gradient.My explanations are pretty dodgy, but hope it helped.

[Einstein]

Yeap i know what you mean, Thanks !