I am trying to calculate z scores from p-values to get SEs using a logistic regression summary statistics data, a part of which is presented below.
SNP | P | P(R) | OR | OR(R) | Q | I |
rs3094315 | 0.4913 | 0.4913 | 0.9420 | 0.9420 | 0.3873 | 0.00 |
rs12562034 | 0.9737 | 0.9737 | 1.0022 | 1.0022 | 0.5991 | 0.00 |
P is p value and (R) indicates random effect models. Q is Q statistic and I I2 index to assess the heterogeneity of the data.
Since they did not provide SEs or 95% CIs in this summary statistics, I am using the following formulas to get SEs:
SE=beta/z=ln(OR)/invnormal(p)
beta=ln(OR)
z=invnormal(p)
Then, I got these results.
SNP | β | z | z(R) | SE | SE(R) |
rs3094315 | -0.05975 | -0.021809 | -0.021809 | 2.739642 | 2.739642 |
rs12562034 | 0.0021976 | 1.938190 | 1.938190 | 0.001134 | 0.001134 |
Code:
logistic foreign price mpg rep78 Logistic regression Number of obs = 69 LR chi2(3) = 34.08 Prob > chi2 = 0.0000 Log likelihood = -25.362394 Pseudo R2 = 0.4018 ------------------------------------------------------------------------------ foreign | Odds Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- price | 1.000141 .0001379 1.03 0.305 .9998712 1.000412 mpg | 1.18063 .0966693 2.03 0.043 1.005583 1.386148 rep78 | 5.321595 2.656576 3.35 0.001 2.000398 14.15687 _cons | .0000112 .000032 -4.00 0.000 4.23e-08 .0029816 ------------------------------------------------------------------------------
I will really appreciate any help and comments.
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