When pulling a stastical sample with sample size unknown, how do you determine margin of error. n=((z*s)/e)^2?
446411
Usually when  you pull a sample, you know the size of the sample you pulled.  Otherwise, how would you calculate any of the test statistics much less the margin of error?
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
454512
population sample size (n) = 12\nsquare root of sample size = 3.464\nmean of the data = 11.692\nstandard deviation = .777\n\nIt's too hard to type in math notation so, in words, use the formula: \nmean plus or minus critical k-value multiplied by (standard deviation divided by the square-root of n)\n\nSo for your example, it would be:\n11.692 +/- k(.777/3.464) --> take .777 divided by 3.464\n= 11.692 +/- k(.224) so solving for k -->\n11.692 + k(.224) = 13 \nk = 5.84 \n11.692 - k(.224) = 10\nk = 7.6\n\nThis formula is for the confidence interval for the mean of a normal population. It assumes that this was a simple random sample and is just one of the formulas that you could possibly use. You didn't specify if the data was Normally distributed (I'm not going to check and do everything for you). Just to put this answer in perspective, a confidence level of 99.9 % corresponds to a "k-value" of 3.291
327228
Sum of\n(freq.observed - freq.expected)^2 / freq.expected\n\nWhere freq.expected=\nRowTotalxColumnTotal / Total Sample Size.\n\nThe chisq value you get is checked against the table of critical values where df=(r-1)(c-1). \n\nr=number of rows\nc=number of columns
885649
The bigger the sample size, the more accurate the estimate.\n\nSay I was estimating the percentage of red m&m's in a pack.\n\nIf I sampled 10 m&m's, I might I might find that the sample contains 1 red one. This could lead me to think that a pack contained 10% red m&m's.\n\nHowever, if I sampled 100 m&m's, I might find that the sample contains 15 red m&m's, leading me to think that a pack contains 15% red m&m's. This answer is impossible using the sample size of 10, as you can't get 1.5 m&m's of a particular colour.\n\nIf I sampled 1000 m&m's I might find that I have 143 red ones, leading me to think that the pack contains 14.3% red m&m's.\n\nSo, as you can see, the bigger the sample size, the closer the estimation comes to the actual answer.
182111
Tracking error is the standard deviation of the portfolio of differences:\nLet X be your portfolio; let B be the chosen benchmark.\nThen d=X-B is the portfolio of differences (which adds to zero)\nthen TE=std(d) is the tracking error. If you have a covariance matrix S for your assets, then \nTE=((X-B)'S(X-B))^(1/2).\n\nTracking error "interpretation" is the uncertainty of the excess returns (that is, of the return of the portfolio vs the benchmark's)
760097
Try this: http://en.wikipedia.org/wiki/Confidence_interval
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
308472
Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
611847
Depends on the sample size and population and things.  Generally, a z-table is good enough.  If the area outside is .0002, then that's around 3.7 sigmas.
882187
You've already got the answer: the margin of error is 4%, at the 95% confidence level. That means that, whatever the results of the poll shows, we can be 95% confident that the poll result will be within 4% of the population mean -- regardless of what the population mean happens to be.\n\nIn other words, suppose we were to sample the same population 100 times with 100 different polls, each of which sampled 500 people. And then suppose we had an actual vote, where we canvassed every single person in the population and recorded the results, and determined the true population mean. Then we would expect that in our 100 polls, 95% of them would have a result that falls within 4% of the true population mean.
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
303069
Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
195594
heres 1 from my university. dont know how much itll help.\n\nhttp://www.upscale.utoronto.ca/PVB/Harrison/ErrorAnalysis/index.html
165255
Take a look at the error function, and modify it so that you are dealing with distributions in 2 dimensions, using polar coordinates.\n\nFor example, the counterpart of the error function in 2 dimensions, in polar coordinates can actually be integrated, to produce the comulative distribution function\n\nF(R,s)=1-e^[-(R^2)/(2*s^2)]\n\nwhere s is the standard deviation, and R is the radius of a circle centered on the origin. \n\nAs R--> infinitity, then F(R,s)----> 1\nF(0,s)=0\n\nSuppose that the target has an edge R0 such that\n\nF(R0,s)=.5 for a given value of s. \n\nThen, you have to find the value of S for which F(R0,S)=0.9.
828638
Assuming the population is normal, to calculate the percentage of subjects obtaining 65 or below, we find the Z score that corresponds to 65:\n\nZ score \n= ( score - mean ) / standard deviation\n= (65 - 80) / 6\n= -2.5\n\nLooking up a Z-score of -2.5 in a 'table of standard normal probabilities' (or using statistical software), gives 0.0062 which is 0.62%. \n\nAs for the confidence interval.. it is 'estimate' +/- 'margin of error'\n80 +/- 1.95 * 6 / sqrt(100)\n80 +/- 1.176\nThe 95% confidence interval (78.824,81.176)\n\n(I.e we can be 95% confident that the true population mean lies between 78.824 and 81.176, and this is based our the sample of 100 subjects which had a mean of 80)
168454
WilliamLCB is correct about the when.  The why is described in some detail at the Mathworld link below.  The summary is that dividing by (n-1) for the sample variance is required to give you a *unbiased* estimator for the population variance.