Calculate P Value Average Standard Error
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Calculate P Value From Mean And Standard Deviation In Excel
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How To Calculate Standard Error Of The Mean Difference
to obtain the P value from a confidence interval Research Methods & Reporting Statistics Notes How to obtain the P value from a confidence interval BMJ 2011; 343 doi: http://dx.doi.org/10.1136/bmj.d2304 (Published 08 August 2011) Cite this as: BMJ 2011;343:d2304 Article Related content Metrics Responses Peer review Douglas G Altman, professor of statistics in medicine 1, J Martin Bland, professor of health statistics21Centre for Statistics in Medicine, University of Oxford, Oxford OX2 6UD2Department of Health Sciences, University how to calculate standard error of the mean in r of York, Heslington, York YO10 5DDCorrespondence to: D G Altman doug.altman{at}csm.ox.ac.ukWe have shown in a previous Statistics Note1 how we can calculate a confidence interval (CI) from a P value. Some published articles report confidence intervals, but do not give corresponding P values. Here we show how a confidence interval can be used to calculate a P value, should this be required. This might also be useful when the P value is given only imprecisely (eg, as P<0.05). Wherever they can be calculated, we are advocates of confidence intervals as much more useful than P values, but we like to be helpful. The method is outlined in the box below in which we have distinguished two cases.Steps to obtain the P value from the CI for an estimate of effect (Est) (a) P from CI for a differenceIf the upper and lower limits of a 95% CI are u and l respectively: 1 calculate the standard error: SE = (u − l)/(2×1.96) 2 calculate the test statistic: z = Est/SE 3 calculate the P value2: P = exp(−0.717×z − 0.416×z2). (b) P from CI for a ratioFor a ratio measure, such as a risk ratio, the above formulas should be used with the estimate Est and the confidence limits on the log scale (eg, the log risk ratio and its CI).NotesAll P values are two sided.A
Comments The p-value, while it is one how to calculate standard error of the mean in minitab of the most widely-used and important concepts in statistics, is actually widely misunderstood. http://www.bmj.com/content/343/bmj.d2304 Today we'll talk about what it is, and how to obtain it. (If you're in a statistics class, or using this stuff out there in the real http://trendingsideways.com/index.php/the-p-value-formula-testing-your-hypothesis/ world, consider ordering "Statistics in Plain English" by Timothy Urdan. It's got the readability of the Idiot's Guide on the same subject, and (thank God) a non-textbook price, but without the glaring mistakes.) Update: As I've been doing with my newer posts, I decided to put together a spreadsheet you can use to skip the dirty work. This spreadsheet calculates the p-value for Z-tests, t-tests (both single and two-sample), and proportion tests. Email Address © 2016 Trending Sideways
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site http://math.stackexchange.com/questions/80848/calculate-p-value About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody standard error can ask a question Anybody can answer The best answers are voted up and rise to the top Calculate p value up vote 1 down vote favorite 1 taking stat 101, I was wondering how I could figure out the p-value, with the hypothesis mean being equal to -4 given the data below. Could someone explain the p-value? statistics share|cite|improve this question asked Nov how to calculate 10 '11 at 13:44 Mark 1,28931534 add a comment| 1 Answer 1 active oldest votes up vote 9 down vote accepted The $p$ value of a test is the probability of seeing a result at least as extreme as the one that you actually saw, assuming the null hypothesis is true. In your example the null hypothesis is that $\mu=-4$. The standard test here is a two-sided $t$ test, where we first compute the $t$-statistic: $$t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$$ where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation and $n$ is the number of observations in your sample. In your data $\bar{x}=-5.033$, $s=3.567$ and $n=90$, so $$t = \frac{-5.033 + 4}{3.567/\sqrt{90}} = -2.747$$ This is then compared to a $t$ distribution with $n-1$ degrees of freedom to calculate a $p$ value. We want the probability that the result is at least as extreme as the one we saw, so we use a two-sided $t$ test, since $t<-2.747$ and $t>2.747$ are both considered equally extreme. Let $T_{89}$ be a $t$-distributed random variable with 89 degrees of freedom. We have $$P(T_{89} \leq -2.747) = 0.00364$$ and $P(