If you have the equation for the line of best fit, then just substitute the x value to be 0.
If you're going to use the form y=mx+c then you'll need to know the gradient and a point on the line of best fit.
For example if you know your gradient to be 0.5, and you know that the point (100,5) falls on the line of best fit, then all you'd need to do is rearrange the equation 5=(0.5*100)+c to make c the subject of the formula!
yup - the equation is on your graph: (assuming that`s not the computer`s own- which it may well be)
y=(106757)x+14.26
If not, then i don`t really know since i don`t much practice stats.
Thank you very much!
My maths skills are very weak. Please can you explain where the 10-6 came from in the brackets? Also, how might I multiply by x when I don't yet know it (I'm trying to find x when I know y)?
If you have the equation for the line of best fit, then just substitute the x value to be 0.
If you're going to use the form y=mx+c then you'll need to know the gradient and a point on the line of best fit.
For example if you know your gradient to be 0.5, and you know that the point (100,5) falls on the line of best fit, then all you'd need to do is rearrange the equation 5=(0.5*100)+c to make c the subject of the formula!
I hope that's right, and that it helps!
Thank you very much, tedsimonds19!
How would I find x if (for example) y is 20.0, if the intercept is 14.26 and the gradient is 7.57x10-4?
My maths skills are very weak. Please can you explain where the 10-6 came from in the brackets? Also, how might I multiply by x when I don't yet know it (I'm trying to find x when I know y)?
I`d probably go with the other contributors on this one - i know very little of statistics (my own, deliberate choice!), and they will know what they`re talking about...
Well if this is the case, y=mx+c would become 20=(7.57*10^-4)x+14.26 20-14.26=(7.57*10^-4)x 7.74/(7.57*10^-4)=x x=10224.5706737
I think this is right, but I'm not sure of the context, or indeed of my own mathematical skills!
Hope it seems alright?
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OK! Just looked back at the original post, my sums DO NOT look right (oops!) I'd better stay away from maths
Your last post looks good to me; if I use 5.74 instead of 7.74 then it gives me 7582.56 which looks about right when I use the line of best fit from 20.0.
Also, do you happen to know how to find x when y = 0 using just the line of best fit? i.e. how do I draw on the graph from 0 on the y axis to meet the line of best fit?
Your last post looks good to me; if I use 5.74 instead of 7.74 then it gives me 7582.56 which looks about right when I use the line of best fit from 20.0.
Also, do you happen to know how to find x when y = 0 using just the line of best fit? i.e. how do I draw on the graph from 0 on the y axis to meet the line of best fit?
I really appreciate your help!
Hahahaha yes of course 5.74 is correct, I blindly typed the wrong thing into my calculator.
So if we want to find out the value of x when y=0, we would use this again, but substituting 0 for y.
That looks right to me, because the trendline shows a positive correlation, when y=0, the x value will be waaaaay over to the left, in the negative side.
That looks right to me, because the trendline shows a positive correlation, when y=0, the x value will be waaaaay over to the left, in the negative side.
Hope this helped!!
This has helped like I can't even express! Thank you! Is it possible to use no equation at all and to just draw from y=0 using the line of best fit?If so, how can I do this?
This has helped like I can't even express! Thank you! Is it possible to use no equation at all and to just draw from y=0 using the line of best fit?If so, how can I do this?
Again, I appreciate your help so much.
It's really no problem at all! It's been nice to help
If you really wanted to draw a line from y=0, you'd have to extend the y axis and the line of best fit to where the line crosses the x axis. Which would be the point at which the line will have the value (-18837.5,0).
You could do this, but the accuracy may not be as great as using a formula.
It's really no problem at all! It's been nice to help
If you really wanted to draw a line from y=0, you'd have to extend the y axis and the line of best fit to where the line crosses the x axis. Which would be the point at which the line will have the value (-18837.5,0).
You could do this, but the accuracy may not be as great as using a formula.
Ah, I see, it all makes sense now! I've practiced using the line of best fit for positive values like 20.0 (as for the other example) before, where the line of best fit is positive too. But never for negative, so I really didn't know what to do.
Your help has been invaluable. I'm doing a science course and really struggle with the maths side of it. I feel I've learned a lot from this and am very grateful!
Ah, I see, it all makes sense now! I've practiced using the line of best fit for positive values like 20.0 (as for the other example) before, where the line of best fit is positive too. But never for negative, so I really didn't know what to do.
Your help has been invaluable. I'm doing a science course and really struggle with the maths side of it. I feel I've learned a lot from this and am very grateful!
Thank you again .
Aah I see! Yes, if it's a negative value you'd have to use the negative side of the y-axis.
It's really been a pleasure! I study maths on the IB (however only maths studies), if you ever need any help again feel free to inbox me on here! I'm not good at science-mathsy things usually but have been doing this kind of thing in class lately.
Aah I see! Yes, if it's a negative value you'd have to use the negative side of the y-axis.
It's really been a pleasure! I study maths on the IB (however only maths studies), if you ever need any help again feel free to inbox me on here! I'm not good at science-mathsy things usually but have been doing this kind of thing in class lately.
Take care!
That's really kind of you. Thank you very much! Best of luck with your maths studies and take care .