Binomial Distribution Standard Error Of The Mean
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Binomial Error
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Standard Error Binary Distribution
Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Standard error for the mean of a sample of binomial random variables up vote 21 down vote favorite 8 Suppose I'm running an experiment that can have 2 outcomes, and I'm assuming that the underlying "true" distribution of the 2 binomial error bars outcomes is a binomial distribution with parameters $n$ and $p$: ${\rm Binomial}(n, p)$. I can compute the standard error, $SE_X = \frac{\sigma_X}{\sqrt{n}}$, from the form of the variance of ${\rm Binomial}(n, p)$: $$ \sigma^{2}_{X} = npq$$ where $q = 1-p$. So, $\sigma_X=\sqrt{npq}$. For the standard error I get: $SE_X=\sqrt{pq}$, but I've seen somewhere that $SE_X = \sqrt{\frac{pq}{n}}$. What did I do wrong? binomial standard-error share|improve this question edited Jun 1 '12 at 17:56 Macro 24.1k496130 asked Jun 1 '12 at 16:18 Frank 3561210 add a comment| 4 Answers 4 active oldest votes up vote 25 down vote accepted It seems like you're using $n$ twice in two different ways - both as the sample size and as the number of bernoulli trials that comprise the Binomial random variable; to eliminate any ambiguity, I'm going to use $k$ to refer to the latter. If you have $n$ independent samples from a ${\rm Binomial}(k,p)$ distribution, the variance of their sample mean is $$ {\rm var} \left( \frac{1}{n} \sum_{i=1}^{n} X_{i} \right) = \frac{1}{n^2} \sum_{i=1}^{n} {\rm var}( X_{i} ) = \frac{ n {\rm var}(X_{i}) }{ n^2 } = \frac{ {\rm var}(X_{i})}{n} = \f
be calculated for a binary variable? In a graph showing the progress over time of the probability to find a pathogen within plant tissues, I'm wondering if standard deviation
Sample Variance Bernoulli
or standard error bars can be added. The probability to find the binomial sampling plan pathogen, is obtained dividing the number of findings (positive events) by the total number of attempts (total events). The binomial sample size probability in the graph is a mean of several replicates. If so, standard deviation should be square root of N*P*Q. How can the standard error be calculated? Topics Standard Error × 119 http://stats.stackexchange.com/questions/29641/standard-error-for-the-mean-of-a-sample-of-binomial-random-variables Questions 11 Followers Follow Standard Deviation × 237 Questions 19 Followers Follow Statistics × 2,242 Questions 89,810 Followers Follow Feb 8, 2013·Modified Feb 8, 2013 by the commenter. Share Facebook Twitter LinkedIn Google+ 1 / 0 Popular Answers Todd Mackenzie · Dartmouth College If one is estimating a proportion, x/n, e.g., the number of "successes", x, in a number of trials, n, using https://www.researchgate.net/post/Can_standard_deviation_and_standard_error_be_calculated_for_a_binary_variable the estimate, p.est=x/n, one formula for an estimate of the standard error is sqrt(p.est*(1-p.est)/n). This "behaves well" in large enough samples but for small samples may be unsatisfying. For instance, it equals zero if the proportion is zero. There are a number of alternatives which resolve this problem, such as using SE=sqrt(p.h*(1-p.h)/(n+1)) where p.h=(x+1/2)/(n+1). You might gain some insights by looking at http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval Feb 8, 2013 Genelyn Ma. Sarte · University of the Philippines Diliman in a binomial experiment, the variable of interest is number of successes or positive results. this will be in the form of a sum of Bernoulli experiments which are assumed to be independent and identical. a Bernoulli random variable has variance=pq, hence a binomial random variable will have variance=npq because the variances of the Bernoulli experiments will just be additive. i wasn't able to follow all discussions in the thread, but i think your interest is not the sum of the successes but the mean or average success (which is sum of independent and identical Bernoulli experiments divided by the number of trials, say n). In which case, the variance of this
distribution of p State the relationship between the sampling distribution of p and the normal distribution Assume that in an election race between Candidate A and Candidate B, 0.60 of the voters prefer Candidate http://onlinestatbook.com/2/sampling_distributions/samp_dist_p.html A. If a random sample of 10 voters were polled, it is unlikely that exactly 60% of them (6) would prefer Candidate A. By chance the proportion in the sample preferring Candidate A could easily be a little lower than 0.60 or a little higher than 0.60. The sampling distribution of p is the distribution that would result if you repeatedly sampled 10 voters and standard error determined the proportion (p) that favored Candidate A. The sampling distribution of p is a special case of the sampling distribution of the mean. Table 1 shows a hypothetical random sample of 10 voters. Those who prefer Candidate A are given scores of 1 and those who prefer Candidate B are given scores of 0. Note that seven of the voters prefer candidate A standard error of so the sample proportion (p) is p = 7/10 = 0.70 As you can see, p is the mean of the 10 preference scores. Table 1. Sample of voters. Voter Preference 1 1 2 0 3 1 4 1 5 1 6 0 7 1 8 0 9 1 10 1 The distribution of p is closely related to the binomial distribution. The binomial distribution is the distribution of the total number of successes (favoring Candidate A, for example) whereas the distribution of p is the distribution of the mean number of successes. The mean, of course, is the total divided by the sample size, N. Therefore, the sampling distribution of p and the binomial distribution differ in that p is the mean of the scores (0.70) and the binomial distribution is dealing with the total number of successes (7). The binomial distribution has a mean of μ = Nπ Dividing by N to adjust for the fact that the sampling distribution of p is dealing with means instead of totals, we find that the mean of the sampling distribution of p is: μp = π The standard deviation of the bino
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