If the predicted value of y = 3x, then for x=2.6, y = 3*2.6 = 7.8
If in fact y=3x is the regression equation, then the slope = 3, and y-intercept=0
(in general, the regression equation is y = b + ax, with y-intercept b and slope a)
I'm having a little trouble with the following question:
Four pairs of data yield r = 0.942 and the regression equation (predicated value of y =3x. Also, the mean of y=12.75. What is the best predicted value of y for x=2.6?
Can you help? I'm not sure what formula to use. I'm trying to figure out if I'm suppose to use the regression equation. However, I don't know how to find the slope and y-intercept.
Thanks,
If the predicted value of y = 3x, then for x=2.6, y = 3*2.6 = 7.8
If in fact y=3x is the regression equation, then the slope = 3, and y-intercept=0
(in general, the regression equation is y = b + ax, with y-intercept b and slope a)
John,
Thanks, I got 7.8. However, I'm still not sure what formula to use. In the question it gives r = 0.942 and the mean of y to be 12.75. How do I incorporate thoses values in the equation since the predicted value of y is now 7.8?
Now I'm a bit confused about what the question is asking for....
Could you post exactly how the entire question is worded?
Thanks,
John
Here is the question again:
Four pairs of data yield r = 0.942 and the regression equation (predicted value of y =3x. Also, the mean of y=12.75. What is the best predicted value of y for x=2.6?
Last edited by 1NAMILL; 11-19-2005 at 01:57 PM.
The best predicted value of y for x=2.6 is 3*2.6 = 7.8
Sometimes they will give you irrelevant info in the question to see if you can cut through the nonsense and answer the question.....
Thanks John. I got confused with all the extra stuff.
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