One other idea, the LSMESTIMATE went out the window, as it looks like the inverse link is applied to the difference, so you would get something like an odds ratio. So, with an alpha=.05, it looks like tinv=1.96 within rounding error, which suggests df for each mean of infinity. That won't work, so I am going to pick a large n for each group, say n=100. Thus upper minus lower should be about 3.92 "standard errors". Infected: 33721 - 19944 = 13777 13777 / 3.92 = 3514.54 (as a standard error) Stddev1 = sqrt(100) * 3514.54 = 35145.4 Non-infected: 17117 - 10139 = 6978 6978 / 3.92 = 1780.10 (as a standard error) Stddev2 = sqrt(100) * 1780.10 = 17801.0 Pooled standard error of the difference = sqrt ( (35145.4)*(35145.4)/100 + (17801)*(17801) / 100) = 3939.64 (formula from page 81 of Steel and Torrie, Principles and Procedures of Statistics, 1960) Thus an approximate 95% confidence interval on the difference of 25933 - 13174 = 12759 would be Lower 12759 - 3939.64 * 1.972 = 12759-7769.04 = 4989.96 Upper: 12759 + 3939.64 * 1.972 = 12759 + 7769.04 = 20528.04, where 1.972 is the t value for a 2 tailed distribution with 198 degrees of freedom. I note that the standard errors I came up with from the asymmetric confidence bounds are not much different from what is reported, so you could probably use an ODS statement to capture all of the needed values in a single dataset, never bother with the upper minus lower business, and get what you need fairly easily. I would recommend moving from GENMOD to GLIMMIX, in case you want to regard any of your factors as random in a hierarchical sense, and to get the degrees of freedom for the estimates in the dataset (as it looks as though GENMOD doesn't print them). Good luck, Steve Denham
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