On 15 Jan 2003 14:05:24 -0800, [EMAIL PROTECTED] (dennis roberts) wrote: > since i followed the model of m and m ... and, you seem to think their > wording is ok ... then, by inference ... so is mine
"followed the model"? Well, M & M used a known variance and did not make a probability statement about outcome, whereas you used an estimated variance, and stuck in a percentage -- so that what you wrote took the form of a power statement on the future variance. - I hope to get to that library today, to read M & M. > > what is puzzling about your original disagreement with my handout note is > the following: > > Where it says, > "how large of a SAMPLE would I need, > with 95% confidence, to produce an interval ...." - >>>> your words >>>>> > it ought to say, > "... with 50% confidence, ..." > > you claim that the method i illustrated essentially was working with a > fifty (50%) confidence interval (level) ... where did you come up with that? The illustrated method wasn't "working with" a 50% CI. The flat extrapolation, attempting to match prior results, implicitly *has* the 50% power (approx.) to match the narrowness of the previous 95% CI. Please notice that in your original, excerpted above - You are asking, how large a SAMPLE is needed, with 95% confidence, to produce <a Confidence Interval of a stated size (.05 gallons) (also 95%)> . I have to read that as a power statement concerning the achievement of a CI of a certain size. That is to say (given the earlier presentation), you hope to achieve the same variance as just seen with N=16. An alternate statement might have been, How large of a sample would I need to have 80% confidence of achieving a 99% CI that is no more than X gallons - the 80% is not quite the usual wording for it, but it is a power statement. Now, let's turn to the essay: It is *all* about achieving certain estimates of mu. (And the randomizations feature a fixed, known value of sigma; the consistency of estimates of sigma, which should be the focus, is never mentioned.) > > let's assume for the moment ... that we are taking samples from a > population where the mean is 100 and the sigma is 16 ... like iq scores > > i generated 1000 samples of n=16 ... found the means ... then, using a t > value based on about 15 df ... when sample mean +/- 2.1315 units of error > (16/sqrt 16 = 4) ... and found that about 96 % of these intervals captured > 100 ... ie, about 96% of the intervals USING A MARGIN OF ERROR OF 2.1315 * > 4 = 8.5 ... CAPTURED the mu value ... so, the mu value was within + OR - > 8.5 about 96% of the time ... > > NOT 50% This randomization is not strictly parallel to what you asked, but this answer still has relevance: How many of the samples achieved sigma of 16 or less? My statement (estimating power as 50%, for achieving the same standard deviation in a later sampling) was based on the common sense realization that half the outcomes would be smaller variance than the other half. Your original "power statement" was estimated from *one* random realization of sigma in a sample of 16. Yours *is* a statement about confidence in achieving a small size for the resulting CI, not about means. [snip, stuff about estimating mu from fixed variance; rather than estimating variance from an estimated variance.] Hope this helps. -- Rich Ulrich, [EMAIL PROTECTED] http://www.pitt.edu/~wpilib/index.html . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
