How can i calculate variable adjusted percentiles (e.g. age, race , sex adjusted X) in SAS ?
347691
Just use a by or class statement before the model statement.  I need more information about exactly what you are looking for and the research question to help with the program.  If you are doing Chisq just stratify.
455303
To obtain 95% confidence intervals for the population variance we use the result:\n\n[(n-1) S²] ÷ (sigma²) ~ chi-square(n-1)\n\nWhere sigma² is the population variance\nand S² is the sample variance\n\nFor the males we have n = 40, so using the result we have a chi-square(39).\n\nFor a 95% confidence interval we want the values that contain 95% of the distribution between them (the lower limit will have 2½% below it and the upper limit will have 2½% above it).\n\nUnfortunately my tables don't have a chi-square 39 listed - so you will have to look these up in your tables - I will use the chi-square 40 results just to show you the method.  The chi-square 40 results are 24.43 and 59.34.\n\nNow we have these values (called "critical values") - we stick them into our thrilling result:\n\n[39 × 11.3²]/(sigma²) = 24.43\n\nand:\n\n[39 × 11.3²]/(sigma²) = 59.34\n\nRearranging each of these gives us the sigma²:\n\n(sigma²) = [39 × 11.3²]/24.43 = 203.844\n\n(sigma²) = [39 × 11.3²]/59.34 = 83.922\n\nNow you want the confidence interval for the standard deviation (not the variance), so we need to root these:\n\nsigma = 14.3\n\nsigma = 9.16\n\nSo our 95% confidence interval is (9.16, 14.3).\n\nYou can do a very similar thing for the females and then can compare these two limits to see if there is any overlap between them.\n\nHope this helps.
252390
I'm going to e-mail you an Excel spreadsheet that will do it.   Don't change anything except the dates and the yellow cells and everything else will automatically calculate.
1133778
http://online.unitconverterpro.com/metric-conversion-tables/convert-alpha/density.html
680316
I doubt that you can.  Use MINTAB or SPSS.
274067
based on your height, gender, or can be uder a certain body-fat percentage
494209
are u kidding, just use the dam formulas for confidence interval and z scores it is not hard just plug them in
1151182
While not having done any research on SAS programming prior to this, I did manage to pull up a couple of web addresses and references that will hopefully put you in the right direction. Feel free to IM/email me if my answer (or potentially someone elses that may be more qualified in the field) should it not quite be enough and I will see what I can do. And as a general reference (I didn't see if they have one for SAS, but...) the 'for Dummies' books are exceptionally well written, allowing people to absorb as much or as little, at once, as they want. They also manage to further educate most all highly-experienced people as well. I have yet to personally find a more well-written off-the-presses set of instructional guides for the Computer Science field. Good luck!\n\nhttp://www.ats.ucla.edu/stat/sas/\nhttp://www.courseadvisor.com/site/72053001\nhttp://www.vdestiny.com/courses.html
163979
Try the Bureau of Labor Statistics :)
1127129
you should be able to get the total through a "x" but otherwise, check the site
561597
Yes SAS rocks especially the logistic regressions!
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...
865854
I assume that you mean that 1.06*se was added to xbar.  I think that's what you mean.  Otherwise your interval was not well defined.  The minimum value is larger than the maximum.\n\nSince you have a normally distributed population with a known sd sigma, then xbar would have a normal distribution.  You subtracted 0.96 standard errors and added 1.06 standard errors, so you would need to look up the area between -0.96 and 1.06 on the standard normal table.\n\nP(xbar - 0.96(sigma/sqrt(n)) < mu < xbar + 1.06(sigma/sqrt(n)))\n= P(- 0.96(sigma/sqrt(n)) < mu - xbar < 1.06(sigma/sqrt(n)))\n= P(-0.96 < (mu-xbar)/(sigma/sqrt(n)) < 1.06)\n= P(-0.96 < z < 1.06)\n\nThat would turn out to be 0.8554-0.1685 = 0.6869.  That would be the probablity that mu would be in one of these confidence intervals.\n\nTo find the probability that mu will be in at least 690 of the 1000 intervals, you would need to recognize that this is a binomial.  There are n = 1000 trials, each trial has probability of success p = 0.6869, and the trials are independent.  np and n(1-p) are both greater than or equal to 5, so the sample size is large enough.  Using the normal approximation to the binomial, the mean would be 686.9 and the standard deviation would be sqrt(1000*.6869*(1-0.6869)) = 14.67.  Using the continuity correction, I would need to start at a value of 689.5.\n\nz = (689.5-686.9)/14.67 = 0.18\n\nLooking that up on your table, the area to the right is 1-0.5714 = 0.4286.  There is a 42.86% chance of getting at least 690 intervals that contain mu.
424848
http://www.utexas.edu/its/rc/tutorials/stat/spss/spss1/
623677
Analysis of variance or ANOVA checks for the variation contributed by teh population in question
979610
Well, you've got the averages. You can now find the total points my multiplying the mean with 10. You can also determine the mean deviation, standard deviation, variance, co-variance etc of the data of each group and determine the relative levels of consistancy of each group and compare the spread of the data.\nFor further stuff...look it up in wikipedia.
19993
You can download Megastat, which is a statistical add-in for Excel  at the following website :\n\nhttp://blue.butler.edu/~orris/megastat/index.html