can some one explain bootstrapping and maximum likelyhood! i have an exam on Wednesday. thanx?
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Hi, here are some good websites that explain these methods better than I could:\nhttp://www.dbbm.fiocruz.br/james/GlossaryB.html \n(short definition of bootstrapping)\nhttp://virgil.ruc.dk/kurser/Sekvens/bootstr.htm\nhttp://gogarten.uconn.edu/mcb221/class30.html\n\n+++++++++++++++++++++++++++++++++++++++++++++++\nhttp://www.dbbm.fiocruz.br/james/GlossaryM.html (short definition of maximum likelihood)\nhttp://statgen.iop.kcl.ac.uk/bgim/mle/sslike_1.html (explanations)\n\nBoth are used for phylogenetic analysis. Maximum likelihood is one of the criteria for inferring phylogenetic relationships, based on a given model of evolution. It's used for protein or DNA sequence characters. \n\nJacknifing is used to evaluate equally parsimonious phylogenetic trees; basically it involves repeated random resamplings of data to see if the resulting trees are reliable.
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A hypothesis test starts from the assumption that the null hypothesis is correct (c) and then you test whether the data is likely under this assumption.  If the data refutes the null hypothesis to the chosen significance level than you are said to have proven the alternative hypothesis to be likely.\n\n\nIs this for highschool homework?  If you need help I have a degree in statistics and can help you understand this stuff but if you just want someone to do your homework you should re-think this.  Trying to get to the end result without learning the process won't get you anywhere in life.  Even if you don't use this in your future career it is still worth learning.
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This is a Two Tailed test ... in other words, you are trying to determine if you have statistical evidence to suggest that the average change was something other than zero disruptions different between weeks 1 and 2 ... just because you saw a change of 5, doesn't mean anything ... it could just be a random fluke !\n\nHo (Null Hypothesis):  Sample Aveage change in distruptions was zero -- no change, or X-Bar = 0\nHa (alternative):  Sample Average change was not 0, or X-Bar <> 0\n\nAt the 0.05 (5%) level of significance (in total), you would have 0.025 probability in the upper end of the distribution and another 0.025 probability in the other end ... that's a total of 0.05 or 5%\n\nUsing your Z-Score table, you find the Z-Score (the number of Standard Deviations in a Standardized Normal Distribution with mean of 0 and Standard Deviation of 1) associated with 0.025 or 2.5% ... this is:\n\n-1.959961082 Standard Deviations from the mean for the low tail and 1.959961082 standard deviations from the mean for the high tail.\n\nSo, the number of standard deviations from a presumed mean of zero (under the null hypothesis) for the sample at hand is:\n\n( 5 - 0 ) / 2  = 2.5\n\nThat is to say, that if the Null Hypothesis of NO CHANGE in behavior were true, a legitimate value of 5 drawn purely randomly could still happen, but that would be 2.5 Standard Deviations away.\n\nBut, your upper and lower limits were only +/- 1.959961082 !!!\n\nSo 95% of the time, you would expect to see something between +/- 1.959961082 standard deviations from zero (within this range) ... but this time, you saw a value associated with a whopping 2.5 standard deviations from the presumed mean of zero ... (we'll call 1.959961082 = 1.96 here)\n\n<< I ACTUALLY MADE A LITTLE GRAPHIC HERE, BUT IT WON'T DISPLAY PROPERLY >>\n\nSo you would reject Ho.\n\nConclusion:  Based upon the sample taken, beleieved to be truely random and under the assumptions of normality, we have statistical evidience which would tend to suggest that the observed sample difference of 5 is statistically significant at the 95% confidence level, thereby, leading us to conclude, based upon the evidence observed, that the incentive system may have played a roll in the positive inprovement of the respondents.\n\nDoes this help ?   My contact information is in my profile.\n\nRegards,\n\nAlan
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i've based my answer on a probability tree i've drawn.\n(i may not be right!)\n\nProbability both stocks go up\n= 17/25 x 10/17\n=2/5\n\nProbability neither stock goes up\n=8/25 X 6/8\n=6/25\n\ni'm not too sure about the following answers (haven't learnt conditional probability):\n\nIf Gateway goes up, what is the conditional probability Micron goes up?\n=(10/17) / (17/25)\n=250/289\n\nIf Micron goes up, what is the conditional probability Gateway goes up?\n=114/272 x 17/25\n=57/200\n\nIf Micron does not go up, what is the conditional probability Gateway will go up?\n=155/272 x 17/25\n=31/80\n\n\nAgain i may not be right!
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p is actual parameter of distribution, and is unknown to you.\np-hat is it's estimate, based on observed data.
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Oy that is a great question.  I take it you taking a risk managment or a derivative course! ;)  Basically you would have to put in functions (functions of interest and volatility variations) for both and then evaluate it accordingly.  It is nearly impossible to do by hand but computer can provide you with some good insight :).  That's when 3D probability that engineers love so much kicks in.  \n\nRival
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1.) Ms. Jordan 275 / 2500 = 0.11 \n     0.11  x  100 = 11%\n\n2.) 900 students x 1.07 = 963 students .\n\n3.) Virus Scan  300 sec / 0.40 = 750 sec for 100%\n\n4.) Gasoline... 8882 / 0.85 = 10,449 BBLs per day.\n\n5.) Jerry's new salary is\n     18,500 x 1.105 = $ 20,442.50\n\n6.) Chen's Investment wound up being \n     6400 x 0.918 = $ 5875.20\n\nHave a nice day,\nZahbudar
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Is the term you are looking for "probability"?\n\nI hope this helps.
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I know this! I did a problem like this a while ago! It's an exponential equation. It can be used to predict exponential increases or decreases, like bacteria or double/triple/quadrupled information.
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I predict the value will increase without bound. One factor in  any potential success is the way you integrate around the pole. There could be a strong singularity!
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1. probabilty of getting a value that extreme in truth if the null hypothesis is true\n2. 1. null hypothesis is false \n    2. null hypothesis is true and there's another factor at work\n3. conclusion: it is likely that you should reject the null hypothesis\n4. wtf
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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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try\nhttp://www.peterwebb.co.uk/probability.htm
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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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A really good book is Bayesian Data Analysis, Second Edition, by Andrew Gelman, et al.  I found it to be one of the easier to comprehend books on Bayesian statistics.  It was as "rigorous" as you might like, but I have found it absolutely useful in day to day analysis tasks.\n\nHowever, I have to admit that until I had a formal course in Bayesian Statistics, reading all the books I tried to read on the subject just didn't get to the fundamental gist.  David Draper out of the University of California, Santa Cruz is about one of the best mathematics / statistics lecturers I've ever encountered.  He put together the foundation that helped me understand the Gelman work.  But reading Gelman first helped me appreciate what I was learning from Draper.  If Draper ever finishes his book (the basis for the course notes), I'd recommend that as a good foundation for more rigorous understanding.
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Conditional probability just means you restrict yourself to those cases where the first event happens.\n\nYou need to first draw a Venn diagram.  Put two overlapping circles labelled G (Gateway goes up) and M (Micron goes up).\n\nThe intersection is 10 (because they said that on the 10 of the 17 days both stocks went up).  That leaves 7 left to put in the G circle.  The next part says that on 2 days when Gateway *didn't* go up, Micron did.  So put 2 in the M circle.\n\nThat's 7 + 10 + 2 = 19.  That leaves 6 days when neither went up.\n\nI've added a picture below so you can see the diagram.  Now you should be able to answer the questions.\n\n1) If Gateway goes up, what is the conditional probability Micron goes up?  (10/17)\n2) If Micron goes up, what is the conditional probability Gateway goes up? (10/12 or 5/6)\n3) If Micron does not go up, what is the conditional probability Gateway will go up?  Here you are considering the area *outside* the M circle.  That is 13 days.  On 7 of those days, Gateway goes up... so the probability is:  (7/13)
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One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.