I have a randomized complete design problem where Six litters of hamsters with 2 hamsters from each litter (considered blocks) were available for an experiment examining differences between length of day light on the NI enzyme level (ENZYME). The treatments (DAYLENGTH) of short (1) and long (2) are randomly assigned within litter (LITTERS). How do I test for differences in treatments taking litter into account   I am not sure if my code is set up the code properly to answer the question.   data have; input litter daylength enzyme; datalines; 1 short 2.1 2 short 1.8 3 short 1.4 4 short 1.2 5 short 1.9 6 short 2.4 1 long 2.6 2 long 2.2 3 long 2.4 4 long 1.7 5 long 2.9 6 long 2.3 ; run; proc glm data=have plots=(meanplot(connect)); class daylength litter; model enzyme=daylength litter; means daylength / tukey; lsmeans daylength / stderr pdiff adjust=tukey; title 'RCB ANOVA'; run;  Does the above code answer the question on testing the differences in treatments while taking BLOCKS into account? ... View more

I have a data of the form below. And I want to find the best model that predicts y while considering interactions as well as quadratic terms using PROC REG. How do I go about it?   data bb; input x1 x2 y x3 x4 x5 ; datalines; 0.442 0.672 9.2 0.1962 13.0769 0.435 0.797 11.7 -0.3038 -31.9231 0.456 0.761 15.8 -0.2038 -21.9231 0.416 0.651 8.6 -0.4038 -31.9231 0.449 0.9 23.2 0.2962 -6.9231 0.431 0.78 27.4 -0.2038 13.0769 0.487 0.771 9.3 -0.3038 -26.9231 0.469 0.75 16 0.1962 23.0769 0.435 0.818 4.7 0.2962 23.0769 0.48 0.825 12.5 0.0962 -1.9231 0.516 0.632 20.1 0.2962 33.0769 0.493 0.757 9.1 0.2962 33.0769 0.374 0.709 8.1 -0.3038 -26.9231 0.424 0.782 8.6 -0.5038 -26.9231 0.441 0.775 20.3 -0.4038 -31.9231 0.503 0.88 25 0.1962 8.0769 0.503 0.833 19.2 -0.1038 -17.9231 0.425 0.571 3.3 0.9962 13.0769 0.371 0.816 11.2 -0.3038 -1.9231 0.504 0.714 10.5 0.4962 28.0769 0.4 0.765 10.1 0.1962 13.0769 0.482 0.655 7.2 0.6962 51.0769 0.428 0.728 9 0.1962 23.0769 0.559 0.721 24.6 0.5962 18.0769 0.441 0.757 12.6 -0.2038 -21.9231 0.492 0.747 5.6 -0.0038 8.0769 0.402 0.739 8.7 0.1962 -1.9231 0.415 0.713 7.7 -0.5038 -31.9231 0.492 0.742 24.1 -0.1038 23.0769 0.484 0.861 11.7 -0.2038 -26.9231 0.387 0.721 7.7 -0.6038 -36.9231 0.436 0.785 9.6 -0.6038 -19.9231 0.482 0.655 7.2 0.6962 51.0769 0.34 0.821 12.3 -0.5038 -31.9231 0.516 0.728 8.9 0.0962 28.0769 0.475 0.846 13.6 -0.2038 -1.9231 0.412 0.813 11.2 -0.8038 -51.9231 0.411 0.595 2.8 0.2962 18.0769 0.407 0.573 3.2 0.3962 33.0769 0.445 0.726 9.4 0.6962 16.0769 0.291 0.707 11.9 -0.7038 -56.9231 0.449 0.804 15.4 -0.4038 -11.9231 0.546 0.784 7.4 0.1962 23.0769 0.48 0.744 18.9 0.3962 23.0769 0.528 0.79 12.2 -0.5038 -31.9231 0.352 0.701 11 -0.9038 -26.9231 0.414 0.778 2.8 0.4962 33.0769 0.425 0.872 11.8 -0.8038 -31.9231 0.599 0.713 17.1 0.7962 28.0769 0.482 0.701 11.6 0.1962 13.0769 0.457 0.734 5.8 0.1962 3.0769 0.435 0.764 8.3 0.3962 18.0769 ; run; proc print data=bb; run;   ... View more

  Would the interaction (source*amount) in the table below be significant since the p-value (0.0545) is close to 0.05?   Source DF Type I SS Mean Square F Value Pr > F source 1 220.900000 220.900000 0.99 0.3269 amount 1 1299.600000 1299.600000 5.81 0.0211 source*amount 1 883.600000 883.600000 3.95 0.0545 ... View more

  I have a sas data (sas7bdat) of the form below:   1 0 336 1 0 346 0 1 269 0 1 346 1 0 323 1 0 309 0 1 322 0 1 316 1 0 300 1 0 309   How do I use proc format to edit the first two columns?   proc format; value fxx 1='High' 0='Low'; value fxxxx 1='Yes' 0='No'; run; libname info 'where data is stored'; data have; set info.have; input xx xxxx sbp; format xx fxx. xxxx fxxxx.; run; proc print data=have; run;   7 libname info 'where data is stored'; NOTE: Libref info was successfully assigned as follows: Engine: V9 Physical Name: where data is stored 8 9 data have; 10 *Use SET statement to set the permanent SAS data set; 11 set info.have; ERROR: File INFO.HAVE.DATA does not exist. 12 input xx xxxx sbp; 13 format xx fxx. xxxx fxxxx.; 14 run; NOTE: The SAS System stopped processing this step because of errors. WARNING: The data set WORK.HAVE may be incomplete. When this step was stopped there were 0 observations and 2 variables. NOTE: DATA statement used (Total process time): real time 0.03 seconds cpu time 0.03 seconds 15 16 *Use PROC PRINT to read the data carefully; 17 proc print data=have; NOTE: Writing HTML Body file: sashtml.htm ERROR: File WORK.HAVE.DATA does not exist. 18 run; NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE PRINT used (Total process time): real time 0.58 seconds cpu time 0.09 seconds I keep getting the same error above. I am trying to import the data from sas and then use proc format to make changes. Why do I still have column 1 and 2 not formatted? What am I missing?       ... View more

  Can someone please explain what this output means? How do I explain the values in the data especially the p-values in the second output? How do I draw conclusions based on the data above especially using the p-values?        ... View more

I have a data of this nature.  data have; input diet : $1. @; do i=1 to 7; input cholesterol @; output; end; drop i; datalines; A: 29 32 26 15 34 15 23 B: 11 30 34 13 27 24 21 C: 22 12 20 7 3 27 4 D: 16 21 13 6 0 9 25 E: 6 2 10 -3 8 1 -4 ; run; I did run the Tukey Test using the code below: proc glm data=have; class diet; model cholesterol=diet; means diet / lines tukey; run; And I got this output:     Can someone explain what this output mean? What do the blue and red lines stand for? And which diet is superior compared to the rest? ... View more

  How do I input data that is exactly as it is below in sas? I tried the command below but it didn't work. I want each group to have seven elements in a row.   data have; input group $ ddd @@; datalines; A: 29 32 26 15 34 15 23 B: 11 30 34 13 27 24 21 C: 22 12 20 7 3 27 4 D: 16 21 13 6 0 9 25 E: 6 2 10 -3 8 1 -4 ; run; ... View more