Dear Cat
My apologies for presuming...
Here's a "primitive" solution: compute a t-statistic or CI.
t = (beta-hat - 1)/SE(beta-hat), compare to qt(.975, res.df)
Or Better, compute the 95% confidence interval
beta-hat + c(-1,1)*qt(.975, res.df)*SE(beta-hat)
albyn
On 2012-11-24 18:05, 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)
Regression_PhyloContrasts_ALL <- lm(Contrast_log_FTL_ALL ~
Contrast_log_MTL_ALL, offset=1*Contrast_log_MTL_ALL)
abline(Regression_PhyloContrasts_ALL)
the plot that resulted is attached as an image file.
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
On Sat, Nov 24, 2012 at 8:22 PM, Bert Gunter <gunter.ber...@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
<hen...@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!!
Cat
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--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
