How do I determine what a critical k-value is given a 12 sample population.  How do I determine k-value?
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
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.
266823
Enter the number into MS Excel, then use the standard deviation formula from the "function" option. Your answer will depend on whether those numbers represent the population or just a sample from a population.
266644
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.
374562
I think that you would need at least 2 values in each cell so that the within groups SD can be estimated, so if each of the two ways has two factors then you would need a sample of at least 8, so that there is two for each cell.  Could be wrong though.
716901
You can use polynomial long division to find the value of k. Be very careful! For the first expression, after dividing the numerator by the denominator a remainder of  8 - 4k  is obtained. Set this equal to zero to find the desired value of k.\n\nTo review polynomial long division, see:\n\nhttp://en.wikipedia.org/wiki/Polynomial_long_division
792033
The critical region of the sampling distribution of a statistics is also known as the alpha region.\n\nIt is the area, or areas, of the sampling distribution of a statistic that will lead to the rejection of the hypothesis tested when the hypothesis is true.\n\nI hope this is simple to understand.\n\nAll the best.
1153151
not sure what critical number is, but by this logic, you can make any  number critical as long as the function has at least one critical number
500573
If you are doing a paired-sample t-test, the rule of thumb would be a minimum of 30 participants, assuming the responses are normally distributed on both the pre- and post-tests.\n\nMy understanding is that in computer simulations drawing lots of different samples from a virtual population, n=30 is the smallest sample that has an acceptable probability of approximating a normal distribution.\n\nIf you want to do any regression or factor analyses, you should have a least n=100.
744818
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n'
242094
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/
767594
The critical value of r is wherever there is a change in the root structure of the rhs.  Set your rhs to zero, and look at the values of r where you go from two real to two complex roots (you have a quadratic rhs, in this case).  That is the critical value of r.\n\naçafrão341@yahoo.com
900568
K(e) = 14%\nK(d) = 9% x (100%-34%) = 5.94%\nWACC = W(e)xK(e)+W(d)xK(d)\n(75%/175%)x5.94% + (100%/175%)x14% = WACC\n2.546% + 8.000% = 11.546%
714755
A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.
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
300365
Percentiles are fairly simple to find. If you have N numbers, you first sort them in ascending order.  To find the 50th percentile, take the middle number (average them if there are 2). In general to get the k-th percentile, simply take the value of the number in the sorted array after going through the first k% of the numbers.  Percentiles are a great way to visualize distribution as its immune to outliers in the data.
842271
My rule of thumb is that the skewness should be less than 2 in absolute value.  Other people might have different rules of thumb.\n\nYou may find the Jarque-Bera test of interest.  It is a test for normally that uses both the sample skewness and sample kurtosis.  I have posted a link about it below.