Regression results confusion. Help!

[DomiDarko]

Hello and please help,

I am testing biodiversity population reduction to see if its related to spatial scale (further north of the equator = less species).
My linear regression line is quite significant showing a steep drop but my data analysis figures are not. What am i doing wrong? Should i accept the null hypothesis that predicts their is no correllation as the P value is 4.02 (isnt that 402% probablility that the null is correct) or is their a correlation after all? I'm so confused. Please explain!

Here are my results.
R Square = 0.12
Standard Error = 181577.40
F number = 18.08
Significance F = 4.04
Observations = 131

Intercept = 355574.75 standard error = 22707.34 t-stat = 15.65 p-value 1.26 Lower 95% = 310647.71 Upper 95% = 400501.79
No of species = -46096.56 standard error = 10840.04 t-stat = -4.25 p-value = 4.02 Lower 95% = -67543.85 Upper 95% = -24649.27

 

[Ben Nevis]

The R Square indicates a very low level of correlation.
Did you do an X Y Scatter plot of the data? With R Square = 0.12 the data points should look fairly randomly distributed. If the data are well correlated you should see a distinct pattern.

 

[tybaltlives]

what Ben said. Looks like you can plot your data. Also, i'm not certain that a LINEAR regression is appropriate for a biodiversity model. Seems to me that there would be a drop off the further you got from the equator but it might follow a different function, even a NON-LINEAR one. Maybe something like this:

Excel 2010
	K	L	M	N	O
27	b+(b + (a-b) * x^c / (d + x^ c))
28				Distance	Species
29				1	10
30	a	20		2	9.85796
31	b	10		3	9.150787
32	c	4.5		4	7.460239
33	d	1500		5	5.180252
34				6	3.211012
35				7	1.911506
36				8	1.147094
37				9	0.708587
38				10	0.453162
Sheet1

Cell Formulas
Range	Formula
O29	=$L$31+($L$31-($L$30-$L$31)*N29^$L$32)/($L$33+N29^$L$32)
O30	=$L$31+($L$31-($L$30-$L$31)*N30^$L$32)/($L$33+N30^$L$32)
O31	=$L$31+($L$31-($L$30-$L$31)*N31^$L$32)/($L$33+N31^$L$32)
O32	=$L$31+($L$31-($L$30-$L$31)*N32^$L$32)/($L$33+N32^$L$32)
O33	=$L$31+($L$31-($L$30-$L$31)*N33^$L$32)/($L$33+N33^$L$32)
O34	=$L$31+($L$31-($L$30-$L$31)*N34^$L$32)/($L$33+N34^$L$32)
O35	=$L$31+($L$31-($L$30-$L$31)*N35^$L$32)/($L$33+N35^$L$32)
O36	=$L$31+($L$31-($L$30-$L$31)*N36^$L$32)/($L$33+N36^$L$32)
O37	=$L$31+($L$31-($L$30-$L$31)*N37^$L$32)/($L$33+N37^$L$32)
O38	=$L$31+($L$31-($L$30-$L$31)*N38^$L$32)/($L$33+N38^$L$32)

 

[tybaltlives]

There was something else I wanted to mention. In my experience, p-values are typically expressed as ".05" (95% confidence) or ".01" (99% confidence). So if you get a p-value of .2 then you reject the null hypothesis. but if it is .04 then you cannot reject the null hypothesis. I've never seen a P value of "4.02", for example.