Re: How calculate 95%=1.96 stdv
Hi Stefan, s.petersson [EMAIL PROTECTED] wrote in message news:XBE07.7641$[EMAIL PROTECTED]... Let's say I want to calculate this constant with a security level of 93.4563, how do I do that? Basically I want to unfold a function like this: f(95)=1.96 Where I can replace 95 with any number ranging from 0-100. To Eric's reply I'd just add that use of a table is unnecessary. Especially in a computer program, it is easier to use a numerical function to calculate the confidence interval. The tables you've seen are for the cumulative probabilities of the standard normal curve--otherwise known as the standard normal cumulative density function (cdf). The standard normal cdf is the function: +infinity p = PHI(z) = INTEGRAL phi(z) -infinity where: z = standard normal deviate PHI(z) = is the probability (p) of observing a score at or below z phi(z) = is the formula for the standard normal curve: 1/sqrt(2*pi) * exp(-z^2/2) Note that PHI() and phi() -- (these mean the greek letters, upper-case and lower-case, respectively) are different. PHI() is the cumulant of phi(). With the function above, one supplies a value for z, and is given a cumulative probability. You seek the inverse function for PHI(), sometimes called the probit function. With the probit function, one supplies a value for p and is returned the value of z such that the area under the standard normal curve from -inf to z equals p. (As Eric noted, you may need to adjust p to handle issues of 1- vs 2-tailed intervals.) Both the PHI() and probit() functions are well approximated in simple applications (such as calculating confidence intervals) by simple polynomial formulas of a few terms. Some of these take as few as 2 or 3 lines of code. A good reference for such approximations is: Abramowitz, M., and I. A. Stegan, 1972: Handbook of Mathematical Functions. Dover. Hope this helps. John Uebersax = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
distribution algorithms (fwd)
At the following URL, you will find some public domain fortran programs for calculating varius cumulative and inverse distributions. http://blas.mcmaster.ca/~fred/statf.html Mark Eakin [EMAIL PROTECTED] = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Propensity Analysis with Stata 6.0
Does Stata 6 support propensity analysis (I'm trying to test some data for selection/sampling bias)? If not, any ideas how I can do it? Any help would be appreciated! = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Double mediation
Hello, I'm hoping someone can help me with this. I have looked at a multitude of resources including the David Kenny page, this and other newsgroups, Pedhazur (1982), Cohen Cohen (1983), and Darlington (1990?), to no avail. I am hoping someone can direct me to the right resource. I am trying to conduct a test of double mediation. In other words, I am trying to test the hypothesis that x--z1--z2--y. Is there a way to do this (and if so, what is it?), or must I result to a path analysis or a structural equation model? Thanks in advance for any help. = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: How to calculate on-line the SD of a population?
Luigi Bianchi wrote in message 9i2doj$r61$[EMAIL PROTECTED]... Hi to all, it's the first time that I post to this NG, so I hope it is the right place. I have the following problem: I read data from an A/D board and I have to provide an estimation of the SD of the population on-line, that is each time I read a sample I have to update the mean and SD. While it is really easy to update the mean, I don't remember how to do the same thing with the SD. I remember that there was a formula, but I don't remeber it. Anyone could help me? Thanks in advance, Luigi -- Luigi Bianchi http://www.luigbianchi.com [EMAIL PROTECTED] Programming, C++, OWL, VCL, SDK, Dfm2API You update the mean and the sum of squares of deviations from the mean: For each new case (new value x) n = n + 1 dev = x - mean mean = mean + dev/n ssq = ssq + dev*(x - mean) Then the usual st.devn. estimate is: sd = sqrt(ssq/(n-1)) If you want an approximately unbiased estimate of the std. devn., use sd = sqrt(ssq/(n-1.5)) -- Alan Miller (Honorary Research Fellow, CSIRO Mathematical Information Sciences) http://www.ozemail.com.au/~milleraj http://users.bigpond.net.au/amiller/ = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
