What is the appropriate statistical test for this data?
398063
If you have a quantitative dependent variable and a nominal independent variable, then the one-way ANOVA test is appropriate.\n\nIf you have nominal dependent and independent variables, then a chi-square test for independence is appropriate.
268119
To answer that question you would have to know if the differences between "having a variation" and "not having the variation" are normally distributed.  If so, you could make an assumption on the number to look at based on normal distribution and 3 standard deviations.\n\nIf there is no normal distribution to the data, then you would have to determine your own "statistics" for the odds that the deviation would occur, and calculate the number based on the possible number of "variations" that could occur, versus the odds that any one could occur.
571247
i wish i could help you...it looks like you need a stats book...i was tryin to find this website i used to use in my stats class...but of course when you are lookin for somethin u cant find it...sorry about that
928594
If you assume it's normal, then yeah, although I assume there is more data than you have given as you do not have the variance there, although you could treat this as a binomial distribution giving it mean 0.3, variance n0.3*0.7, although when calculating the confience interval you'll want to divide by n^2 for that variance.
418766
you're kidding right? I hope you don't use any of the data you collect on here for a school project. This is about the worst sample you can have for a survey. Your results are going to be pretty much worthless.
100009
I am not quite sure what the question is. What should the manufacturer do? Did they already do the experiments or do you want to know what sort of experiments to do? Do you want to know how to reduce the data? How to present the data? What statistical tools might be useful?
509983
How you use statistical tools to analyze your data depends entirely on what kind of data you have and the questions you are trying to answer with your data.  Some kinds of data lend themselves to descriptive analysis, in which you summarize your findings with means, medians, standard deviations, etc.  Other data are suitable for formal hypothesis testing, which calls for the application of statistical tools such as t-tests, analysis of variance, linear regression, multiple regression, proportional hazards modeling, life table analysis, etc.  Ideally, the preparatory phase of a research project would include a consultation with someone familiar with biostatistics before the data are ever collected.\n\nHowever, you can make a start by going to the following website:\n\nhttp://en.wikipedia.org/wiki/Statistics\n\nYou will learn a lot in a short period of time by going there.  But be careful!  Application of statistical tools in biomedical research is not a cookbook operation.  You would be well advised to consult with someone knowledgeable, who can inspect your data and offer advice about how to analyze them.  Such a person can also help you interpret the results of any statistical testing you do.  This is not always straightforward; misinterpretation of data and their analysis will lead to failure to get your papers published, at best, or worse, the misleading of the medical community and the general public about potentially serious issues.
272176
I'm sorry John, but your question sounds confusing. What exactly is the data that you need?
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/
560950
Variations in measurement attributable to peoples' behaviour.
1094318
It is not exactly clear what you need, but maybe you can use the Kolmogorov-Smirnov test (KS-test) for your problem. The KS-test tries to determine if two datasets differ significantly. The KS-test has the advantage of making no assumption about the distribution of data.  Note however, that this generality comes at some cost: other tests (for example Student's t-test) may be more sensitive if the data meet the requirements of the test.
862471
Your going to compare your mean with the population mean and see if your class really better than the average of the population.\nHo: u = xbar\nHa: u < xbar\n\nalpha = 0.10\none tailed test (of course)\ntherefore: z(table) = -1.28 (i'm not sure)\n\nuse z test since you know the population standard deviation.\n\nz = (x - u)/o/sqrt of n\n\nthat's it.\n\ngood luck
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.
163979
Try the Bureau of Labor Statistics :)
688420
try checking with National Statistics Office or NSO
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.
768633
order the data: 7,9,9,11,12,13,13,13,15,19\n\nfind the Median (middle point, as there are 10 data points it's the average of the 5th and 6th): 12.5\nLower Quartile: 9 (25th percentile, median of lower half, i.e. here 3rd observation)\nUpper Quartile: 13 (75th percentile, median of upper half, i.e. here 8th observation)\nmin. observation: 7\nmax observation: 19\n\nso it's:\n\n    12.5\n9          13\n7          19