Stats - Confidence Intervals?
970404
The problem with that statement is that you are making a statement about the sample, not the population.  You know for a fact that of the 1026 people in you sample, 57% said no.  There is no doubt about that.  There is no need to make a confidence statement about a sample.  Confidence intervals are guesses as to what the value of the population parameter is.  So you need to be making a statement about the population percent, not the sample percent.\n\nNow there are many ways to interpret confidence.  Think about taking many samples of size 1026 from the population.  If you make a confidence interval with all of these, 95% of these confidence intervals will contain the true population proportion; 5% won't.\n\nAlso, the confidence interval should be 54% to 60%, and this is for the percentage who said no.
262206
range = mean +- (t/sqrt(N))*st.dev.\n\nWhere t is found in a chart according to the confidence level in question, N is the number of data points, and st.dev. is the standard deviation of the data set.
458114
The confidence interval is the gap you set to determine a level of certainty in your answer. For example, in public health, we usually set alpha=0.05, meaning we have a confidence interval arbitrarily set to 95%.\n\nIf the p-value we get for whatever statistic we just crunched is less than 0.05, then we say that we are at least 95% certain of the validity of our answer. The lower out p-value is, the more certain we are. If we get p=0.00002, then that's 99.99998% confidence, which is higher than the 95% threshold we set.\n\nIn short, it's a measure of validity for our results.
830991
I'm wondering why you created the data 1-3500?  I would think you would use the set of data available and figure the standard deviation and average.  With that information, you would be able to predict tomorrows value at a certain confidence level.  I forget is that the p test or the t test.  its been a while since I took statistics.
1166976
As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
1180637
I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
480069
It takes only one example to prove the statement to be false, and I provided one:\n\n(Now, while you may never naturally encounter the set of numbers I used, the definitions of mean and median fit, so the statement is false.)
189610
Look at the table of values for the normal distribution.  95% of the values are included within +/- 1.9 SD of the mean.  So:  1.9 x 6.2 = 11.78 and the interval is\n\n     26 - 11.78 <= Mean <= 26 + 11.78, or\n           14.22<= Mean >= 37.78
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
91852
Confidence limits are also called confidence intervals.  What do you need to calculate a confidence interval for?  There are different formulae for proportions, means, difference of two means, etc.  Sample size also affects the formulae for confidence levels, we use the t distribution for small samples (<30) and the z for large samples.  To further complicate things confidence intervals can be one or two sided.\nA confidence interval of 95% for a mean means that if we created 100 intervals with different data sets, 95 would contain the true mean.  \n95% is called the confidence level and is used in the formula to find the appropriate z or t value.\nHere is the formula for a two sided 95% confidence interval for a mean with a large sample:\nxbar +or- z(alpha/2)*stdev/sqrt(n)\nwhere xbar is the sample mean, stdev is the standard deviation, and n is the sample size.  You can look up z on a normal distribution table, the z value for a 95% interval is 1.96.  \nThere are many different formulae though, so depending on your problem this may not be the correct one.  \nFor a small sample size the formula is:\nxbar +or- t(alpha/2, n-1)*s/sqrt(n)\nwhere s is your sample standard deviation.  Use a t-table to find the value of t you need for the equation.\nI cannot include all the formulae here due to time and space constraints but perhaps this site will offer some insights:\nhttp://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm\nhttp://techniques.geog.ox.ac.uk/mod_2/glossary/confint.html#conflim
181649
what statistical informations can be extracted from difference between actual and estimate data?and how?
598935
Used to set insurance premiums, the probability of getting into accidents based on your car type/age\n\nFor evaluating worker performance, ie you're on time 95% of the time or only 80%\n\nHow can you actually tell when two sets of data are different? Need statistics.\n\n A lot of jobs in stats!!
837986
http://www.statsoft.com/textbook/stathome.html
437525
When you compute a mean from a sample, you are estimating the mean of the population from which you sampled.  The population mean is sometimes refered to as the true mean and the mean you compute from your sample is called the sample mean.\n\nWith repeated sampling, you'll come up with a different sample mean each time.  This distribution of sample means is refered to as the sample distribution and it is approximately Normal in distribution.\n\nWhen you compute the standard error using the standard deviation and the sample size, you are estimating the variability in the sample mean.\n\nWhen you estimate confidence intervals, you are estimating the range of possible values where the population may lie relative to your sample mean.\n\nFor instance, your sample mean plus or minus 1.96 standard errors represents an estimate of the 95% confidence interval or range that may contain the population mean relative to your sample mean.  There is a 5% chance that the population mean is actually higher or lower than this range.\n\nLet's say you toss a coin twenty times and measure 9 heads and 11 tails.  The average number of coin tosses at 0.45 heads (which implies 0.55 tails).  The standard error estimate is 0.055343. Multipling the standard error estimate by 1.96 and adding and subtracting the result from the mean yields our 95% confidence interval. There is a 95% chance that the population mean for this process is between 0.341528 and 0.558472.
793664
What's your problem - your in college - do your own homework (this son of a statistician is unimpressed).  If you don't know how to construct a confidence interval look at http://facweb.bhc.edu/EastCampus/leeb/ma108/06_01w/tsld032.htm\n\nGood Luck!!!
986574
if you are using a TI-83, use the Z-Interval test funcition.\n\nManual working:\n\nSampling error = 2.06(SD/sqrt n)\n\n\nlower boundary =  20 - 2.06(3/sqrt 25) = 18.764\nupper boundary =  20 + 2.06(3/sqrt 25) = 21.236\n\n96% confidence limits is\n18.764 < mean < 21.236
674991
For one-sample t-test, \nt = (sample mean - population mean) / estimated standard error. \n\nIn math notation, t = (X bar - mu not) / (s/sqrt(n))\n\nAssumptions: (normal population or sample size, n, is at least 30) and population standard deviation is unknown.\n\nThere are other variations of t-tests like two-sample(pooled and unpooled) and paired ...