In statistics, how do I find the standard error for 100 tosses of a coin? For 2,500 tosses?
919154
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.)
1143752
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'
1022381
You can use a chi-square test for goodness of fit.  In the link below, O would be the observed number of times each card came up and E would be the expected.  (You would have to compute this 52 times and then add them up.).  So if you did this 500 times, the E for each card would be 500/52, which will be greater than 5, which is the typical way you determine if the sample size is large enough.  Under the assumption that the cards are at random, this would have a chi square distribution with 52-1 = 51 degrees of freedom.  You can do a test of significance using this test statistic and distribution.
437525
When you compute a mean from a sample, you are estimating the mean of the population from which you sampled.  The population mean is sometimes refered to as the true mean and the mean you compute from your sample is called the sample mean.\n\nWith repeated sampling, you'll come up with a different sample mean each time.  This distribution of sample means is refered to as the sample distribution and it is approximately Normal in distribution.\n\nWhen you compute the standard error using the standard deviation and the sample size, you are estimating the variability in the sample mean.\n\nWhen you estimate confidence intervals, you are estimating the range of possible values where the population may lie relative to your sample mean.\n\nFor instance, your sample mean plus or minus 1.96 standard errors represents an estimate of the 95% confidence interval or range that may contain the population mean relative to your sample mean.  There is a 5% chance that the population mean is actually higher or lower than this range.\n\nLet's say you toss a coin twenty times and measure 9 heads and 11 tails.  The average number of coin tosses at 0.45 heads (which implies 0.55 tails).  The standard error estimate is 0.055343. Multipling the standard error estimate by 1.96 and adding and subtracting the result from the mean yields our 95% confidence interval. There is a 95% chance that the population mean for this process is between 0.341528 and 0.558472.
820314
Standard error is standard deviation divided by n^1/2. In other words, it estimates the standard deviation of the sample mean _based on the population mean_.
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
883654
If the claim is correct, then the mean of samples of size 25 has a normal distribution with a mean of 10 (the population mean) ounces and a standard deviation of 0.5/sqrt(25) = 0.1.\n\nz = (10.15-10)/0.1 = 1.5\n\nAccording to the standard normal table, the probability is 0.0669.\n\nThis is due to the fact that the population is normal.\n\nedit: You guys, it is not three sd's above.  You have to think in terms of standard error.  It is 1.5 standad errors above.  Do the math.
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
452389
There are 2 rules you need to know. The addition rule and the multiplication rule\n\nIf we roll a die the chance of rolling 6 is 1/6.\n\nthe chance of rolling 5 is also 1/6\n\nthe chance of rolling either a 5 or a 6 is therefore 1/6 + 1/6 = 1/3\n\nAnd the chance of rolling 4, 5 or 6 is 3/6 or 1/2.\n\nWhen the events are alternative outcomes and you want to know the chance of the outcome being one of these outcomes, add the probability of each outcome.\n\nIf you roll a die twice (or two dice at once) the probability of rolling a par of ones is 1/36.\n\nThis involves the multiplication rule. The chance of the first die showing a 1 is 1/6, and the chance of the next die showing a 1 is also 1 in 6. \n\nOnly one-sixth of the time do you get the first die to give you the 1 you want and only one-sixth of those times will the second due give you another 1. So 1/6 x 1/6 = 1/36.\n\nHowever if the question is what is the probability of just one six being thrown in two rolls\n\nthen the probability of 1 on the first throw but not on the second throw is (1/6 x 5/6) = 5/36\n\nAnd the probability of 1 on the second throw but not on the first throw is (5/6 x 1/6) = 5/36 \n\nAnd so throwing a 1 on either roll but not the other roll has a total probability of 5/36 + 5/36 = 10/36.\n\nThe slightly different question "what is the probability of rolling one or more 1s in two rolls of the dice?" is the sum of\n\nprobability of one on the first throw but not the second\n\n+ probability of one on the second throw but not the first\n\n+ the probability of a double one being thrown\n\n5/36 + 5/36 + 1/36 = 11/36\n\nSo the rule is when it is a matter of alternative outcomes to one event, add the probability of each outcome.\n\nWhen it is a matter of the outcome of two events both happening, multiply the probability of each happening to get the probability of BOTH happening.
442828
total chances (# of times a player in the field plays a ball) when the numbers of errors is divided by total chances fielding pct is obtained
874169
There appear to be two different definitions of the standard error.\n\nThe standard error of a sample of sample size n is the sample's standard deviation divided by sqrt(n). It therefore estimates the standard deviation of the sample mean based on the population mean (Press et al. 1992, p. 465). Note that while this definition makes no reference to a normal distribution, many uses of this quantity implicitly assume such a distribution.\n\nThe standard error of an estimate may also be defined as the square root of the estimated error variance sigma^^^2 of the quantity,\ns_e=sqrt(sigma^^^2)\n\n\n\n\n    The standard error is the estimated standard deviation of a statistic. The formula depends on what statistic you are talking about. For example, the standard error of a sample mean is just the sample standard deviation divided by the square root of the sample size. For more complicated statistics, the standard error is also more complicated.\n\n    You can use the standard error in two ways. First, the statistic divided by the standard error gives you a way of testing whether the parameter that your statistic is estimating equals zero. You compare this ratio to a t-distribution. In most contexts, the ratio of the statistic divided by its standard error is called a Wald test. For many simple applications, you can also call this a t-test.\n\n    A second application of the standard error is the production of confidence intervals. You can get a crude confidence interval by taking your statistic plus or minus two standard errors. A more precise confidence interval would use percentiles from the t-distribution.\n\n    There are a lot of subtleties in the use of the standard error, especially in more complex problems. Sometimes, for example, the standard error applies not to the statistic itself, but to the logarithm of that statistic. For example, a logistic regression model will compute an odds ratio for your data, but the standard error refers not to the odds ratio, but to the log odds ratio. In this situation, you need to compute the confidence interval on the log scale and then transform the results back to the original scale of measurement.\n&#92;\n\ns
315772
You simply type: \n=rand(25,25) and then enter.
172263
(Amount of Error - Correct Answer) divided by correct answer, times 100.
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
540053
In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
232542
I did this on my calculator because you have to do 75 calculations (which would take forever by hand).\nLess than 75 means it landed tails 0, 1, 2,..., or 74 times.  The probability of tails is .5 and the probability of not tails is also .5\nSay you want tails to come up k times where k = 0,1, 2,...,74\nThe probability of x = k is\nP(X=k) = C(n,k)(.5)^k(.5)^(n-k) Add the exponents on .5 to get\n= C(n,k)(.5)^n where n = 200 (because that is the amount of trials)\nk goes from 0 to 74, so we need to take the sum from 0 to 74\nThe sum (k=0 to k=74) of (C(200,k)(.5)^200\n=.000144637632\nI see no other way of doing this except with a calculator or computer.  (I'm sure your teacher does not expect you to do 75 calculations by hand and then add them all up.)
923495
You can use a z-test for a population proportion.  The test statistic is\n\n(phat-p)/sqrt(p(1-p)/n).\n\nHere phat = 34/66, p = 0.5 under the null hypothesis that the coin is fair, and n = 66.  Compare this test statistic to a standard normal to determine if you would reject the null or not.  I would not suspect hat you would reject though.\n\nIt is appropriate to use this test when the number of successes and the number of failures are greater than or equal to 5, which is the case here.