Regression / correlation coefficient

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rexmorgan

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I am given the following example in a book. I cannot seem to make the leap from step 1 to step 2. Any help is greatly appreciated.

Example 4.3
Fit the following rainfall data to determine the 10-year intensity-duration-frequency curve.

t = duration (min) 5 10 15 30 60 120
i = intensity (in/hr) 7.1 5.9 5.1 3.8 2.3 1.4
1/i 0.14 0.17 0.20 0.26 0.43 0.71

Solution
1. A model of the form i = A/(t + B) can be expressed in linear form as 1/i = t/A + B/A.
2. The regression of 1/i versus t yields 1/i = 0.005t + 0.12, from which A = 200 and B = 24.
3. Thus the rainfall formula is i = 200/(t +24). The correlation coefficient is -0.007.

Thanks in advance. I am really having a problem understanding where the regression of 1/i came from.
 
A transformation is often useful for turning non-linear problems into linear problems. If there is a simplifying assumption that loses NOTHING, why would we not do it?

i = A/(t+B)

That 't' in the denominator could present a bit of a problem. It may still be possible, but a reciprocal transformation creates wonderful simplicity.

1/i = (t+B)/A = t/A + B/A = mt+b

Now, you translate all your data.

Take all your "i" and turn them into 1/i. (You can rename this to 'y' if that will be of any benefit.)
Take all your A and trun them into 1/A (then call them 'm' if that will make it less confusing.)
Take all your A and B and turn them into B/A (than call them "b" if that will make it less confusing.)

If you did all the renaming, your new model is y = mx + b, which I believe will be familiar. Of course, when you are done with this regression, you will have y, m, and b. You will have to use these to get i, A, and B before you think you are done.

Let's see what you get.
 
So are you saying that if I take the first 1/i, which is 0.14. How do I get the m or the b?

0.14 = 1/A + B/A
or
0.14 = m + b

I am given the i, but I am uncertain on how to go about solving for m or b. I know that I could say

0.14 - m = b
or
0.14 - b = m

But that does not really get me anywhere. I am sure that I am missing something very simple, but for some reason I am missing what this book is trying to teach me. BTW this is a self-taught book. No instructor. It seems like I remember doing this sort of thing in college but it has been a while since I have done any of it. Thanks again for any help.:confused:
 
Do this:

Example 4.3
Fit the following rainfall data to determine the 10-year intensity-duration-frequency curve.

t = duration (min) 5 10 15 30 60 120
y = 0.14 0.17 0.20 0.26 0.43 0.71

How would you solve this problem, using the model y = mt + b?

Ignore all the previous discussion. One piece at a time.
 
y = mt + b
0.14 = m*5 + b
This is where I am stuck. Not sure where to go from here. This is what I have done. Not sure if this is the right method or the best method.

0.14 = m*5 +b ; 0.14 - b = m*5 ; (0.14-b)/5 = m
0.17 = m*10 + b ; 0.17-b = m*10 ; (0.17-b)/10 = m
(0.17 -b)/10 = (0.14-b)/5 because they both equal m, they also equal each other
0.17/10 - b/10 = 0.14/5 - b/5 rewriting to make solving easier
0.017 - b/10 = 0.028 - b/5 remove denominator for numbers
-b/10 + b/5 = 0.011 subtracted both sides by 0.017 & added (b+5) to both sides
-b/10 + 2b/10 = 0.011 multiply b/5 times 2/2
b/10 = 0.011 subtract like terms
b = 0.11 multiply both sides by 10 to solve

Now taking the value for b substitute into original formula
0.14 = m*5 +0.11
0.03 = m*5
0.006 = m

Now taking that value plug it into original formula to see if it is true.
0.14 = 0.006*5 + 0.11
0.14 = 0.03 + 0.11 (this is true)
Now for the other formula
0.17 = 0.006*10 + 0.11
0.17 = .06 .11 (this is true)

Now what next.
 
Thanks for the reply Dennis. I wanted to show all steps just in case I was going down the wrong path. Just want to make sure I not missing something or headed off in the wrong direction. I still am not following the example from my original post. Any ideas on pointing me in the right direction?
 
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