how do you do math sample to predict?
481083
It depends what you want to predict.  For example, you could want to know how many people at a school go out for lunch instead of eat at the cafeteria.  Instead of asking everyone, you could choose to ask 30 people at random in the school and use that sample to predict the total school population who go out for lunch.\n\nFor example, if 12 out of the 30 people go out to eat lunch and there are 800 people in the school, then you could predict the total number of persons going out to lunch by taking 12/30 and multiplying that to the total population.\n\n(12/30)*800=320\n\nYou use the sample to predict the behavior of the larger group.  The bigger the sample, usually the more accurate the prediction is.
674991
For one-sample t-test, \nt = (sample mean - population mean) / estimated standard error. \n\nIn math notation, t = (X bar - mu not) / (s/sqrt(n))\n\nAssumptions: (normal population or sample size, n, is at least 30) and population standard deviation is unknown.\n\nThere are other variations of t-tests like two-sample(pooled and unpooled) and paired ...
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.
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Use your sample as if it is your population.  From your sample of size n, take, say, 1000 with-replacement samples of size n.  That means n ordered pairs for each sample.  Compute the correlation coefficient of each sample.  The bootstrap correlation coefficient would be mean correlation coefficient of the 1000 samples.  You can make a 95% confidence interval by taking the 25th and 975th smallest sample correlation coefficient as the end points.\n\nI don't know if any programs use this, so I would program it in R.  It is free to download using the link below.  Click on the CRAN link there.
977052
You need a variable in any experiment to compare results. Perhaps you could use cold (ice) water in one container, warm in another and hot or boiling in another. You can then make your prediction on which will dissolve first and have an experiment that will verify whether you have made an accurate prediction.
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A representative sample should have the same characteristics of the population from which it is drawn.  For example, if you wanted to find out the average height of all Americans, a sample that included mostly basketball players would NOT be representative, nor would a sample that included only women. Instead, a representative sample would include people of various races, both sexes, and all ages--so it would have the same variety of heights that can be found in the whole population.\n\nStatistical theory says that the type of sample most likely to be representative is a random sample, a sample where random numbers or games of chance assure that every member in the population has an equal chance of being selected.  Since this is often impractical to do, other types of "good" (usually representative) samples include systematic samples (where you number off and, for instance, call every 100th name in the phone book) and stratified samples (where you carefully include various sub-groups, like people of all different races in your sample).\n\nAmong the least representative samples are internet polls, where anyone who wants to can answer.  Usually only those with strong opinions on a subject bother responding to such polls.
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It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
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There r many types of models. AR, MA, ARMA, ARIMA, ARCH, GARCH, and lots of other which names I cannot recall. For the prediction to be of any use, the correct model must be used.\n\nAssume u have decided which model to use, u now need to estimate the parameters. The idea is simply to choose the parameter to produce the smallest possible deviation of the model and the data.\n\nYou said u used Excel. so, i guess the method u used is simply trial and error. u try a range of values for the parameters and then pick the best one.\n\nYou want to do it by hand. This is extremely tedious. One way to do it is simply to do the thousands of calculations by hand and compare. Another way to do it is by taking derivatives.\n\nThe idea is simple. u have a model with certain parameter. Say, X(t+1)=aX(t)+et is your model where e is the random component with E(e)=0. So, your forecast of X(t+1) given information up to time t is X(t+1, forecast)=aX(t). The error of your forecast is know when at time t+1 when X(t+1) happens. So, the forecast error is X(t+1, real)-X(t+1, forecast)=(aX(t)+et) - aX(t) = et\n\nYou do this for a few t's and u collect a bunch of et's for these t's. You then square the et's and sum them and call it S.\n\nSt=(e1)^2 +(e2)^2 +e3)^2 +....+(et)^2\n=(X(1)-aX(0))^2 +(X(2)-aX(1))^2+.....\n\n (squaring them is simply to remove the sign of the et's. There r other ways, for example, |et|). You want to pick an 'a' that minimise S. To get this a, take the first derivative of S wrt a and equate it to 0.\n\ndS/da = 0\n\nSolve the equation to find possible values of a.\n\nTake the second derivative of S,\n\nd(dS/da)/da. Substitute possible values of 'a' into it. Those 'a' that make d(dS/da)/da<0 is the local point of maximum. To get the largest maximum, simply substitute these a into St to to find the one that gives the lowest St. This will be your best estimate of parameter.\n\nHowever, when u have more than 1 parameter, thing can get complicated. u would have more than 1 equation to solve and sometimes, u need to use numerical methods to solve them.... which will require the use od Excel again!\n\n\nAs for forecasting... it depends on what model u r using. Generally, for those values u have not observe, u simply use the forecast.\n\nSay, you have a model that uses t-1, t-2 and t-3 to forecast t.... the history for past 3 periods. So, at time t, if u want to forecast X(t+1), u simply pluck in the observed X(t), X(t-1) and X(t-2): {Forecast of X(t+1) at time t }=F{X(t), X(t-1), X(t-2)}...where F denotes a function.\n\nSimilarly, if u want to forecast X(t+3) at time t, u have {Forecast of X(t+3) at time t} = F{X(t+2,forecast), X(t+1, forecast), X(t)}. Or u want to forecast, say, t+100, {Forecast of X(t+100) at time t} = F{X(t+99,forecast), X(t+98, forecast), X(t+97, forecast)}. Hope u know what I mean.\n\nThe difference between a forecast X(t, forecast) and a realisation in the real world X(t) is the random conponent. Without much information, all I can do is to use my best knowledge(the parameter and correct model) to forecast ahead. But as time goes by and i gather more data, I'll reset and improve my prediction everytimes I get more info. So, to predict how long u will need for the process to reach a certain level, u simply use the model(the formula), substitute the X(t) with data u observed(if it's observed already) and forecasted(if has not happened). Let the model evolve until it reach your required level, N. Then look at what the t is.. then this is the best estimate of how long it takes.\n\nIf u make assumption about the distribution(s) of et's, the random conponent, u can even predict probabilities of the accuracy of your predictions. And also, the estimated parameters has uncertainty in them too, and taking these into account, u can calculate the probability of your forecast being right, allowing for the random fluctuation et and sampling error of estimating parameters.
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The probability will completely depend on the system.  If the system was perfect, and conditions were always favourable, then you would have P(H) = 1 all of the time.  Equations of probability are based on past performance to predict the future, and the more you know about the past, the better you can predict the future.  Predicting the probability of something occurring is really a matter of statistics, and I don't know how you could obtain an answer any other way given the parameters of the system.  If the system was perfectly designed, as I said you would always have a hit, and if it is not, then you cannot know what makes some bursts miss or you cannot control what you know the problem is, or you would fix it to get a perfect system.\n\nI would think that you are going to need historical data to statistically determine the probability.  Run a number of experiments, recording how many bursts were fired and how many has at least one of the 3 rounds hit the target.  Your "estimated" probability of a hit P(H), is then obviously number of hits divided by total number of busts fired.  Remember that conditions such as weather (wind) affect the system, so by controlling these variables you can construct several probabilities for varying conditions, or you can lump them all together by using a random sampling method.\n\nMore useful probably, a confidence interval can be stated to a certain degree of......well confidence.  If you wanted to get a range of probabilities of a hit that is 95% certain to occur for any given firing, then using a Gaussian Distibution table find 0.95, or whatever confidence level you want (0.95 is 1.96 btw).  then, since Gaussian is symmetric, you have:\n\nlet P = probability desired to get the interval for\nlet P_est = estimated probability from sample (the dividing thing from above)\nlet Sigma = variability\n\n-1.96 <  [(P_est) - (P)]/{(Sigma)*sqrt[{(P_est)*(1 - P_est)}/N]} < 1.96\n\nSorry thats not fun to write in a text box.  This is based on a mathematical model for probability.  There is a slight approximation in the above method, but it is often quite satisfactory, and the exact method requires an inequality quadratic solution.  Then re-arrage, remembering the rules of arithmetic with inequalities, and since you know everything except P bsaed on your data, you will have \n\n(some number) < P < (some other number)\n\nWhich is the range of probabilities for hits that, based on your data, will be valid for any given sample of firings you conduct, whatever percentage of the time that you selected as your level of confidence.\n\nThe more historical data (sample size) you have, the better you can predict the future probability, and the smaller the range of your confidence intervals will be.\n\nHope this is at least somewhat helpful to you.  This is just some of the basic statistics that I know.\n\nCheers,
140577
Buy a TI-89 graphing calculator, that'll do the trick.
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n = N/(1+Ne(e))\n\nWhere:\nn = number of samples\nN = total population\ne = margin of error\n\nI could not find the squared key on my board so i just manipulated it, however it is the same.
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.
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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'
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MINITAB is free for 30 days, provides a reasonable graphical interface and has some good statistical methods.\n\nAlternatively use the excel add-in Analysis ToolPak from the 'Tools->Add-ins' menu. Then click  Tools->Data Analysis and use one of the methods there. (Moving average / exponential smoothing & regression are good simple methods for forecasting)\n\nThe best software is probably ForecastX Wizzard but is very expensive
552270
would calculate your chance of me calculating a statistics question for you at 1234.367:1
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
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.