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From |
Steve Samuels <sjsamuels@gmail.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: RE: generate random sample |

Date |
Mon, 16 Jan 2012 00:20:26 -0500 |

Well, that's possible. But Ning is a survey methodologist and he asked about generating "a sample from a population with replacement", so I assumed that he was asking about a survey sample problem. The toy example doesn't appear like a real problem to me. If I'm wrong, then Phil's method applies. One variant is to expand to single-digit frequencies (multiply the proportions by 10, in this case). This necessitates a change to Phil's -list- statement. ************************* clear * set seed <seednum> input x prop 1 .2 2 .1 3 .3 4 .3 5 .1 end gen int fw = 10*prop expand fw bsample 3, weight(fw) list x if fw>0 ************************** Steve sjsamuels@gmail.com On Jan 15, 2012, at 11:47 PM, David Hoaglin wrote: Steve, Isn't it simply a discrete distribution that has the values 1 through 5 with the indicated probabilities? And the aim is to generate three random observations from that distribution? David Hoaglin On Jan 15, 2012, Steve Samuels wrote: Ning Li: Your example is too unrealistic and your question too vague for us to answer. Is N (population size) really 5? If so, then the "probability distribution" cannot be that of the population. If population N>5 and there are five "types" of population elements, with the distribution shown, then there is not enough information for Phil's (or anybody's) algorithm to work. Suppose Phil's algorithm draws a selection with x = 3. With the information you have provided, there is no way to link this selection to a specific population element.: But perhaps you want to draw the sample of n = 3 from N =5 with probability proportional to size, in which case the size measures can be arbitrary. But such size measures will rarely look like the proportions that you show. So please describe your problem better. . Steve sjsamuels@gmail.com On Jan 15, 2012, at 6:54 PM, Philip Ryan wrote: The following will randomly sample 3 observations on variable x with replacement from 10,000 observations with the distribution you specified. Un-comment the -set seed- line (and provide a numeric seed) if you want reproducibility between runs. see -help bsample- clear * set seed <seednum> input x fw 1 2000 2 1000 3 3000 4 3000 5 1000 end expand fw bsample 3, weight(fw) list x if fw==1 Phil Philip Ryan Professor and Director Data Management & Analysis Centre [DMAC] University of Adelaide -----Original Message----- From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of ningli_yang@fastmail.fm Sent: Monday, 16 January 2012 9:22 AM To: statalist@hsphsun2.harvard.edu Subject: st: generate random sample Dear listers I am using Stata/MP 11 Windows. What is the good way to generate a sample from a population with replacement? Say, how to generate 3 independent random numbers from population (1,2,3,4,5) with probability distribution (0.2, 0.1, 0.3, 0.3,0.1)? Many thanks for the help. Ning -- Ning Li Survey Methodologist Melbourne Institute of Applied Economic and Social Research The University of Melbourne Tel: +61 3 9035 4949 Fax: +61 3 8344 2111 * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: generate random sample***From:*ningli_yang@fastmail.fm

**Re: st: RE: generate random sample***From:*Steve Samuels <sjsamuels@gmail.com>

**Re: st: RE: generate random sample***From:*David Hoaglin <dchoaglin@gmail.com>

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