[quote=“skoster”][quote=“superking”][quote=“skoster”][quote=“tomthorne”]I am not a statistician, but I have a feeling that statistical significance should come into play here. Three, possible, examples of people outside the 7 nations working legally would probably be put down to chance. Would they not?
To even mildly suggest that people should base their careers based on these odds seems to me to be irresponsible.[/quote]
Depends entirely on the population that sample comes from. If there are 4 posters on Forumosa who are not from the 7/10 countries and have attempted to gain the work permit, then 3 is definitely statistically significant to the population of non-7/10 Forumosa posters who’ve attempted to get a work permit.
I doubt there are enough representatives of the population posting on Forumosa to get a usable sample of all people non-7/10 who have attempted to get a work permit based on TEFL. If you figure a reasonably large population (unknown actual number) and want a proportional statistic with 95% confidence level and margin of error of 5% you’ll need about 386 representatives in your sample based on: n = ((z^2pq)+ME^2)/(ME^2)[/quote]
I’d be interested to know how you made that power calculation, skoster. How did you estimate population variability and how did you get the Z score? Stats are so confusing for me! :bluemad:[/quote]
I treated it as a static, unknown sized population. First I got my alpha using 1 minus the confidence interval. For the z I took 1-alpha/2 and plugged it into an online Normal Calculator (link below). I chose a relatively high confidence interval and used p and q both at 0.5 because with an unknown population size I wanted to be conservative and land on the higher sample size. I’m no statistician, I can only do the basics.
Which part of the equation takes into account smurf frequency?