Hi Run the following spss program ... every time z is within 1.96 units of mu, mu is within lower and upper 95% CIs, as must be the case. And both proportions / probabilities will be very close to the expected 95%, with slight variation across runs.
input program. loop o = 1 to 100000. end case. end loop. end file. end input program. compute mu = 100. compute sigma = 15. compute m = 0. compute se = sigma/3. compute z = 0. compute lci = 0. compute uci = 0. compute pz = 0. compute pci = 0. do repeat v = v1 to v9. compute v = rv.norm(mu,sigma). end repeat. compute m = mean(v1 to v9). compute z = (m - mu)/se. compute lci = m - 1.96*se. compute uci = m + 1.96*se. if abs(z) le 1.96 pz = 1. if (mu ge lci) and (mu le uci) pci = 1. freq pz. freq pci. crosstabs table pz by pci. I'll leave it to better minds than mine to figure out why I cannot say that M and CI are both randomly selected from a hypothetical probability distribution, giving the (identical) probability assigned to each. Take care Jim James M. Clark Professor of Psychology 204-786-9757 204-774-4134 Fax [email protected] --- You are currently subscribed to tips as: [email protected]. To unsubscribe click here: http://fsulist.frostburg.edu/u?id=13090.68da6e6e5325aa33287ff385b70df5d5&n=T&l=tips&o=17431 or send a blank email to leave-17431-13090.68da6e6e5325aa33287ff385b70df...@fsulist.frostburg.edu
