What is a large enough Sample Size for a Kolmogorov-Smirnov goodness of fit test?
915189
To my knowledge, there is no min sample size.  In fact, that is one reason K-S might be used, in that Chi Square g-o-f tests can require a relatively larger sample size.
551595
Gaussian distribution
732858
in testing the hypothesis > is the right side, < is the left side, not equal to is a two tailed test....for whatever test your using find the critical value...if it is greater than the alpha reject otherwise fail to reject
477523
The no of degrees of freedom for a chi-square distribution ia n-1
380438
Seems to be the right sample size to get their point across.
1070837
9".\n\nThis data has been compiled with a sample size of 1.
683899
The most common F test statistic is the one-way ANOVA.  That test statistic is F = sB^2/sW^2, where sB^2 is the between groups variance\n\nsB^2 = sum(i=1 to k)n_i(xbar_i - xbar)^2/(k-1), where n_i is the sample size for group i, xbar_i is the sample mean of group i, xbar is the mean of all observations regardless of sample, and k is the number of groups,\n\nand sW^2 is the within groups variance\n\nsW^2 = sum(i=1 to k)(n_i-1)s^2_i/(n-k) where s^2_i is the sample variance of group i and n = sum(i=1 to k)n_i is the total number of observations regardless of what sample it comes from.\n\nIf the null is true (meaning the mu_1 = mu_2 = ... = mu_k), then sB^2 and sW^2 will be about the same, meaning that the F statistic will be around 1.  If the null is not true (meaing that at least one of mu_1, mu_2, ..., mu_k is different from the others), then sB^2 will be larger than sW^2, making F larger than 1.  So therefore you should reject the null if F is too large.\n\nUnder the null, the F statistic will have an F distribution with k-1 numerator degrees of freedom and n-k denominator degrees of freedom.  So take your F statistic and find the area in the tail of that F distribution to get the p-value.
840031
Trial division is fine for numbers up to maybe 12 or 14 digits, but after that it starts to lose out badly to much faster and more mysterious methods which you need several years' deep study of number theory to understand.\n\nVariations of the Goldwasser-Killian algorithm seem to be the favourite for numbers up to a couple of thousand digits or so.  The Miiller-Rabin test is reckoned to be perfectly satisfactory for numbers of around 100 to 150 digits, the size used for Internet encryption.  A completely new test "AKS" was discovered in 2002 by three Indian researchers.
658731
Slovin's formula is used to determine the ideal sample size out of a population (in other words, how large of a percentage of the whole must I sample so that it adequately represents the whole).  The formula is:\n\nn = N/(1 + N * e^2)\n\nwhere n is the sample size, N is the population, and e is the margin of error involved.
648428
www.startsampling.com
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Good character\nIntelligence\nStrength
319924
It depends on the distribution.  Some probability distributions have multiple parameters (such as the mean and standard deviation for the Normal distribution).  Phi2 would be one parameter of the particular distribution used in the problem.
969083
That statement is false.\n\nIf those statements above are true and assuming n is large, then the standard error would be 5, because by the empirical rule, it would take two standard errors above and below expected to get 95% in the middle.\n\nThe standard error for the sum is equal to sqrt(n)*SD, so in order to double the standard error, you would have to actually multiply the sample size by 4.  That is because\n\nsqrt(4n)*SD = sqrt(4)sqrt(n)*SD\n= 2*sqrt(n)*SD\n\nSo doubling the sample size would not be enough to double the standard error for the sum.  Since the standard error for the second part of the problem would be 10 (using the same reasoning as before, and using the fact that doubling the sample size would double the expected sum), then doubling the sample size would not be enough.\n\nIn other words, in order for that statement to be true, you would have to quadruple the sample size, not double.\n\nedit: The population should be large as compared to the sample size also.  That probably should have been stated in the problem, but, given the wording of the problem, it is probably safe to assume that.\n\nUm, rev, 400 is the expected sum, not the number of draws.  There's a difference.  Way to pretend you know what is going on here.
323296
First you need to ask yourself if you have any reason to believe that measurements of nitrogen oxide [chemical formula -- Just say "NO"] emissions across different cars should be normally distributed.  Since this is a statistics problem, we assume that the answer is yes.\n\nThen you compute the sample mean (the sum of those seven measurements, divided by seven) and the sample variance (the sum of (x - sample mean)^2 for those seven measurements x, all divided by six).  The sample standard deviation is then the square root of the sample variance.\n\nFinally, you look up on your table or with your calculator or wherever what "Z score" corresponds to 98% of the area under the normal pdf curve being between\n\n(mean - Z standard deviations, mean + Z standard deviations)\n\nThey want to know if that upper bound is less than 0.165 grams / mile [a peculiar mix of metric and non-metric units].\n\nThat's probably how they want you to solve it.  Compare that to the Bayesian approach (Google for it).
458039
In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
570786
First of all, if the population is normal, the distribution of sample means is normal regardless of sample size.  So I. is true.\n\nSecond, if the population is not normal, the distribution of sample means is approximately normal if the sample size is large.  So III. is also true.\n\nII. cannot be determined to be true unless we know what the population looks like.\n\nSo I would go with C.\n\nUPDATE: Apparently ghakh hasn't heard of the Central Limit Theorem.  You know, the "Fundamental Theorem of Statistics"?  In layman's terms, it states that if the sample size is large enough, then the sample means will be approximately normal if the sample size is large enough REGARDLESS OF WHAT THE POPULATION LOOKS LIKE!!!!!!  All you need is a finite variance, which for elementary statistics students is not something they need to be concerned with.\n\nMaybe you should take my statistics class, ghakh.  You might learn something!
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.