Comparing linear regression coefficients to a slope of 1
On Nov 24, 2012, at 6:05 PM, Catriona Hendry wrote:
Hi, @ Albyn, David.. No, its not homework. Its basic groundwork for testing allometric relationships for a graduate project I am working on. I read the guide before posting, I spent half the day trying to understand how I am going wrong based on the advice given to others. @Bert, David... I apologise for the lack of code, I wasn't sure how to explain my problem and I guess I went about it the wrong way. I do think this is what I need to be doing, I am testing allometric relationships of body size against a predicted isometric (1:1) relationship. So I would like to know if the relationship between my variables deviates from that. Hopefully the information below will be what is needed. Here is the part of the code relevant to the regression and plot:
plot(Contrast_log_MTL_ALL, Contrast_log_FTL_ALL)
It's kind of a pain that you spell these different than they are named below.
Regression_PhyloContrasts_ALL <- lm(Contrast_log_FTL_ALL ~
Contrast_log_MTL_ALL, offset=1*Contrast_log_MTL_ALL)
That looks fine. What's the problem? (Now you need to be examining the lm-object with the usual R tools. )
abline(Regression_PhyloContrasts_ALL)
abline should be given an argument of 0 for "a" and 1 for "b" if the goal were to plot a line for a predicted regression result of unity slope. If that's not the goal then you need to be more forthcoming.
the plot that resulted is attached as an image file.
No, it's not. Somewhere (not very prominently displayed) on either the Posting Guide or on the information page for Rhelp is it stated that most attachements are thrown away. They need to be .txt, .png, .pdf, or .ps. They cannot be .csv, .xls, .dat, .sas7dat or anything else.
Below are the vectors of my variables. The are converted from other values imported and indexed from a csv file, so unfortunately I don't have matrix set up for them. Contrast_log_FTL_ALL Contrast_Log_MTL_ALL 83 0.226593 0.284521 84 0.165517 0.084462 85 -0.1902 -0.0055 86 0.585176 0.639916 87 -0.01078 0.118011 88 0.161142 0.073762 89 -0.08566 -0.04788 90 -0.13818 -0.0524 91 -0.02504 -0.21099 92 -0.05027 -0.07594 93 -0.11399 -0.07251 94 -0.07299 -0.08247 95 -0.09507 -0.04817 96 0.207591 0.151695 97 -0.14224 -0.05097 98 0.06375 -0.0229 99 0.04607 0.06246 100 0.257389 0.190531 101 -0.0612 -0.10902 102 -0.1981 -0.24698 103 -0.12328 -0.36942 104 0.269877 0.341989 105 0.125377 0.227183 106 0.087038 -0.05962 107 0.114929 0.096112 108 0.252807 0.305583 109 -0.0895 -0.08586 110 -0.38483 -0.20671 111 -0.72506 -0.63785 112 -0.37212 -0.21458 113 0.010348 0.117577 114 -0.09625 -0.0059 115 -0.26291 -0.25986 116 0.056922 0.064041 117 0.051472 -0.09747 118 -0.05691 0.075005 119 0.117095 -0.15497 120 -0.01329 -0.12473 121 0.098725 0.020522 122 -0.0019 -0.01998 123 -0.12446 -0.02312 124 0.019234 0.031391 125 0.385366 0.391766 126 0.495518 0.468946 127 -0.09251 -0.08045 128 0.147965 0.139117 129 -0.03143 -0.02319 130 -0.19801 -0.14924 131 0.014104 -0.01917 132 0.031872 -0.01381 133 -0.01412 -0.04381 134 -0.12864 -0.08527 135 -0.07179 -0.03525 136 0.31003 0.29553 137 -0.09347 -0.11903 138 -0.10706 -0.16654 139 0.078655 0.065509 140 0.08279 -0.00766 141 0.181885 0.001414 142 0.345818 0.496323 143 0.235044 0.095073 144 -0.03022 0.039918 145 0.042577 0.136586 146 0.064208 0.001379 147 -0.02237 -0.03009 148 -3.55E-05 0.040197 149 0.011168 0.087116 150 0.019964 0.071822 151 -0.04602 -0.06616 152 0.083087 0.038592 153 0.032078 0.107237 154 -0.21108 -0.22347 155 0.122959 0.297917 156 -0.05898 0.012547 157 -0.07584 -0.21588 158 -0.00929 -0.06864 159 -0.01211 -0.04559 160 0.090948 0.136582 161 0.016974 0.018259 162 -0.04083 0.016245 163 -0.20328 -0.31678
Lean to use dput()
I needed to use these commands to read this irregularly dispaly list
of numbers:
:
dat <- scan()
mat <- matrix(dat, ncol=3, byrow=TRUE)
datf <- as.data.frame(mat)
names(datf) <- c("index", "Contrast_log_FTL","Contrast_Log_MTL")
David. > > > > > On Sat, Nov 24, 2012 at 8:22 PM, Bert Gunter > <gunter.berton at gene.com> wrote: > >> 1. The model is correct : lm( y~ x + offset(x)) >> ( AFAICS) >> >> 2. Read the posting guide, please: Code? I do not know what you >> mean by: >> >> " this resulted in a regression line that was plotted perpendicular >> to >> the data when added with the abline function." >> >> Of course, maybe someone else will groc this. >> >> 3. I wonder if you really want to do what you are doing, anyway. For >> example, in comparing two assays to see whether they give "similar" >> results, you would **not** do what you are doing. If you care to >> follow up >> on this, I suggest you post complete context to a statistical >> mailing list, >> not here, like stats.stackexchange .com. Also, feel free to ignore >> me, of >> course. I'm just guessing. >> >> Cheers, >> Bert >> >> Cheers, >> Bert >> >> >> On Sat, Nov 24, 2012 at 4:27 PM, Catriona Hendry <hendry at gwmail.gwu.edu >> >wrote: >> >>> Hi! >>> >>> I have a question that is probably very basic, but I cannot figure >>> out how >>> to do it. I simply need to compare the significance of a >>> regression slope >>> against a slope of 1, instead of the default of zero. >>> >>> I know this topic has been posted before, and I have tried to use >>> the >>> advice given to others to fix my problem. I tried the offset >>> command based >>> on one of these advice threads as follows: >>> >>> Regression <- lm(y~x+offset(1*x)) >>> >>> but this resulted in a regression line that was plotted >>> perpendicular to >>> the data when added with the abline function. >>> >>> I would be extremely grateful for your help!! >>> >>> Thanks!! David Winsemius, MD Alameda, CA, USA