Given 12 samples of a large population, I want to determine what a critical k-value would be as well as determine how to calculate the k-value so as to see if I can be confident the answers will lead me to the assumption that all samples would fall within a certain range.\n\ngiven 11.2, 10.8, 11.9, 12.4, 12.8, 11.0, 10.5, 12.8, 12.2, 12.0, 11.1, 11.6\n\nI want to see if given this population I can be certain that all samples would fall within 10 - 13.\n\nI don't know much about stats.\n\nThanks!!
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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
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As it happens, I just got back from a test on statistics. We covered that very question. \n\nHere goes:\nthe sample size is dependent on how confident you want to be about your results. In the following formula, then Z is the confidence level. Use 1.65 for 90% confidence level, 1.96 for 95%, 2,33 for 98% and 2.58 for 99%. \nYou also need to know the standard deviation, which should be given. and the error allowable, for instance 3%, or 5% or such. That number goes into the formula as E. \n\nso number n equals = (Z * standard deviation/E)^2
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They give you mu the mean of the population as 7\nThey give you sigma as the std dev = 1\n\nYou are working with a sampling distribution of x bar the sample mean with the sample size of n = 9\n\nThe mean of this sampling distribution is also 7 but the std dev of the sample mean is sigma/sqrt(9) = 1/3\n\nP(6.4<sample mean<7.2)\nP(6.4-7/(1/3) < Z< 7.2-7/(1/3)), look this in a table of std normal values.\n\nYou need to understand how to use this table.\n\nRemember that when you draw from a normal population a sample of size n, the sample mean is normally distributed with a std dev of sigma/sqrt(n), also called the standard error of the mean..\n\nTo solve problems like this you convert the sample varible to a standard normal distribution by z = (sample mean - mu)/std error\n\nii) Just find the z value which is the 85 percentile and work your way back  z(85th percentile) = (x_bar - mu)/std error using algebra to find the value of x_bar, the sample mean.
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The problem with your problem is that Chi Square analysis is typically inaccurate with just a few data points (four in your case).  I suggest you look up the Chi Square limitations in your textbook.  (I've forgotten what the limits are, but I do know a good Chi Square test needs more than just a few data points.)\n\nIf your results were valid, look up the table critical value (CV) under df = 3 and between .100<=p<=.050.  If 6.85 exceeds the table CV, then you can posit that the differences between the expected values and the actual values are significant.  If you have a value p = .077, in between .100 and .050, you can find the CV for your problem between the table's CV at p = .100 and at p = .05 by ordinary arithmetic.\n\nThe degree of freedom (df) simply means you have N - 1 options on selecting your data points, where N = the number of data points (e.g., four in you case).  There are N - 1 options (degrees of freedom) because, for example, if you select 6, 9, 1...3 has to be the next one.   This results from the fact that the total (sum) for your data points is 19, and if you preselect 6, 9, and 1, for the sum to stay at 19, 3 has to be the fourth data point.
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Put the data of each set into bins according to their leading digit. Then according to Benford's law the distribution should be as follows:\nLeading digit Probability\n1 30.1%\n2 17.6%\n3 12.5%\n4 9.7%\n5 7.9%\n6 6.7%\n7 5.8%\n8 5.1%\n9 4.6%\n\nUsing a chi-squared or Kolmogorov test it is possible to determine which of the four data sets differs significantly from Benford's distribution.
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the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
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really good or really bad, depending on how you want to look at it... and don't try telling me that wasn't a statistical answer!\n\np <0.0000001
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a.  The standardized version of the mean is computed\n\n(sample mean - population mean)/(population sd/sqrt(sample size))\n\nso\n\n(108 - 100)/(16/sqrt(4))\n= 8/8\n= 1\n\nb.  The studentized version of the mean is computed\n\n(sample mean - population mean)/(sample sd/sqrt(sample size))\n\nThe only difference is that you use sample sd as opposed to population sd.\n\n(108-100)/(12/sqrt(4)) = 8/6\n= 1.33
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If the population has a mean of 58 and a sd of 8, then the sample means of size 100 will have a mean of 58 and a sd of 8/sqrt(100) = 0.8\n\na) Within 3 of 58 means that the sample mean will be between 58-3 = 55 and 58+3 = 61\n\nz = (55-58)/0.8 = -3.75\nz = (61-58)/0.8 = 3.75\n\nFrom the table, the area between the two is almost 1.\n\nb) The mean weight is less than 2.5 of 58 or more means less than 58-2.5 = 55.5\n\nz = (55.5-58)/0.8 = -3.13\n\nFrom the table, the are greater than that is almost 0.\n\nc) z = (57.5-58)/0.8 = -0.63\nz = (58.7-58)/0.8 = 0.88\n\nFrom the table, the area between is 0.811 - 0.264 = 0.547.\n\nd) 56 +- (1.96)(0.8)\n56 +- 1.568\n(54.432,57.568)
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b is the most convincing.\n\n90 is the value farthest from the mean, and a small sample variance indicates that it's unlikely that you just somehow ended up with an unusually high (or low) population.
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Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
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In the experiment of testing for the number of students out of 37 who consider statistics exciting\n\nNo. of trials made \n==> No. of students tested = 37\n==>n = 37\n\nLet the event of a student tested being intersted in the subject of statistics be a success\n\nProbability of success in each trial\n= probability that a student tested is interested in the subject of statistics\n==> p = 0.75\n[Given that 75% of the population of all business statistics 1 students consider statistics exciting]\n\nTherefore, \nq = 1 - p ==> q = 1 - 0.75 ==> q = 0.25\n\nLet "x" indicate the number of successes in "n" trials\nsince the number of students who are interested in statistics can be either 0, 1, 2, .... 37\nThe values "x" can carry are 0, 1, 2, .... 37\n==> "X" is a discrete random variate with range = {0, 1, 2, .... 37}\n\nThe probability distribution of "x" is a binomial distribution..\n\n[Since there a number of trials all of which are identical and independent with only two possible outcomes in each trial either a success i.e. the student being interested or a failure i.e. the student not being interested, with the probability of success in each trial remaining the same all throught]\n\nThe probability function of a binomial distribution\n==> P(X=x) = nCx * p^x * q^(n-x)\n\nTherefore, for the above distribution\nP(X=x) = 37Cx * (0.75)^x * (0.25)^(37-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * (0.25)^(-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * 1/(0.25)^(x)\n==>P(X=x) = 37Cx * (0.75/0.25)^x * (0.25)^(37)\n==>P(X=x) = 37Cx * (3)^x * (0.25)^(37)\n\nTherefore, \nProbability of observiing 36 or more of the students who consider statistics to be an exciting subject\n==> P(X >= 36) = P(X=36) + P(X=37)\n\n\nP(X = 36) = C(37,36) * (3)^36 * (0.25)^(37)\nP(X = 37) = C(37,37) * (3)^37 * (0.25)^(37)\n\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
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Wouldn't the probability than 5 of the 38 thought it was exciting be\n\n38C5(.79^5)(.21^33) = 6.63x10^-18 : basically impossible?  Even if you added in the probabilities of 4, 3, 2, 1, and 0, it's still tiny.\n\nSo I would infer that either the 79% is wrong or the sample was not independently chosen.
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I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
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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.
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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.
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http://cis.jhu.edu/~bruno/s06-311/\n\nhttp://www.eng.wayne.edu/page.php?id=1434\n\nhttp://web.njit.edu/~dhar/math333/fin333_sp02A.pdf#search='sample%20Probability%20and%20Statistics%20exam'