At a young and impressionable age students are taught how to find the variance (and hence the standard deviation, its square root). The method is to work out the sum of the squares of the deviations of the mean, and then divide by n, the number of objects, to get an average squared deviation from the mean. This is the variance. Students remember this. Then, when they come to University, unkind lecturers tell them that sometimes they should divide by n-1 instead of n. This tends to make the students unhappy. They thought that they knew what they were doing with variances, and now they don't. Most of them don't understand why they need to do anything different from before, which makes it difficult to remember when it is that they are supposed to do it.
Thus I am often asked to explain what is going on with the n-1. I seldom have time to do it properly, so I'm writing down here why we need to use it. It's called Bessel's correction, after Friedrich Bessel, who worked it out.[1] I shall also go on to explain why it isn't, really, the right thing to do in many practical cases, and some alternative approaches. I finish with a problem to which I don't know the answer. But if you just want to know why, how, and when to apply the correction, just read the first two sections.
Why We need to Do Something
The first thing to say is that the formula taught at GCSE is correct. If you want to find the variance of a set of numbers, then you should work out the sum of the squared deviations from the mean, and divide it by n. That is the definition of the variance. The problem is that you don't always have all of the numbers. And if you don't have all of the numbers, then you can't work out the exact, true variance. You can only make an estimate of it. The challenge is to make this estimate as good as it can be!
Suppose that we want to find the variance of lengths of tails on rats in the United Kingdom. To work out the exact, true variance, we would need to catch all of the rats in the Kingdom and measure their tails. This is impracticable.[2] Instead, what we'd do is catch a few rats at random,[3] measure their tails, and try to extrapolate. The numbers that we measure on the rats we've caught are the "sample". If it was the true mean tail length that we wanted to find, the right thing to do[4] would be to take the mean of the sample and declare this to be the best estimate. This is just what you might expect. But if it's the true variance that we want to estimate, we shouldn't just find the variance of the sample and declare this to be our best estimate. On average, it will be too small.
Why is this? The problem is that we don't know the true mean: we only have an estimate of it, and this estimate is biased to the numbers that we have in our sample. The deviations that we work out are therefore deviations from the sample mean, not from the true mean. For example, if the rats that we caught had a slightly longer mean tail length than the true population mean, then we'd produce an estimate of the mean that was a bit bigger than the true value. Our deviations would then be smaller (in absolute value) than the true deviations, because the true deviations are those from the true mean, which is a smaller number than the one that we're using and therefore further away on average from the numbers that we have. Likewise, if we've caught shorter-tailed rats than average, we will underestimate the mean. But again this causes us to get deviations smaller than the true deviations, because this time the tails are atypically short, so they're closer to the sample mean than they would be to the population mean. Hence it doesn't matter which sort of rats we've got: the sample variance is always going to tend to be[5] an underestimate of the true variance.
Therefore we need some sort of recipe for taking the sample variance and making it a bit bigger so that we somehow get a better estimate of the true variance. This is what Bessel's correction does.
Why Correcting the Variance by n/(n-1) Might be the Right Thing to Do
An intuitive way of thinking about Bessel's correction goes as follows. We started off with n pieces of information, the n pieces of sample data. We have used them to find an estimate of the mean. This effectively uses up one degree of freedom: instead of thinking of it as n random (free) measurements and their (dependent) sample mean, we can think of the sample mean as the actual mean of some distribution (with the same variance as the true one) but with only n-1 measurements' worth of free variation amongst the n readings. Thus we need to correct our estimate by n/(n-1).
The trouble with this argument is that it appeals to a particular sort of intuition that is difficult to make rigorous. A better approach goes as follows. (This is more-or-less just writing out in words the "Alternate 1" from the Wikipedia page on Bessel's Correction.)
Lemma
A lemma is just a theorem, a mathematical result, whose main purpose is that it's a stepping stone to a more important result. In this case the more important result is Bessel's correction, and the lemma is as follows: for any set of data, the expected value of the squared difference between two randomly chosen values is exactly twice the variance of the data.
To prove this, note that the difference between two values is the same as the difference between their deviations from the mean. Thus instead of finding the expected value of (x1 - x2)2, which is what we want, we can find the expected value of (d1 - d2)2, where d1 = x1 - mean, and d2 = x2 - mean.[6] Hence what we want, multiplying out, is d12 - d1d2 + d22. The expectations of the first and third terms are each the variance of the data, by definition. The expectation of the middle term must be zero, because the expected values of the deviations d1 and d2 are each zero.[7] Thus the total expected value is twice the variance.
It's important to note that the lemma applies to any set of data, whether it's what we've been calling the sample, or what we've been calling the population.
Proof of Bessel's Correction
Consider the squared difference between two randomly chosen values in the sample. Most of the time it's the squared difference between two distinct values, which means that it has the same expected value as that equivalent quantity in the population. However, sometimes (1/n of the time, in fact) it's just zero, because we happen to have chosen the same value for x1 and x2. Thus the expected value of the squared difference between values is smaller in the sample than in the population, by a factor, on average, of 1 - 1/n.
Now according to the lemma, the variance is always exactly half of the expected value of the squared difference between values. So the variance of the sample is, on average, smaller than the variance of the population by this same factor 1 - 1/n, which is (n-1)/n. So if what we have is the sample variance, we should multiply by n/(n-1) to get a better estimate of the population variance.
A Note on Notation
Some of the confusion surrounding this subject is due to the notation used. People sometimes call the standard deviation of a set of numbers the "population standard deviation", and write it as σn. (It is often called σn on calculators. The notation makes sense because, when finding the variance, we divide by n.) They then sometimes refer to the corrected quantity, √(n/(n-1)) times this, the estimate of the true population standard deviation if what we had was a sample, as the "sample standard deviation", and write it as σn-1. The σn-1 notation at least reminds us what we're doing, but calling it the sample standard deviation is potentially confusing, because it's not the standard deviation of the sample. Instead, it's an estimate of the population standard deviation if what we had was a sample.
In more serious work, the true, unknown variance of the population is usually denoted σ2, and the actual variance of the sample is usually denoted s2. Thus what Bessel's correction tells us is that an unbiased estimator for σ2 is given by (n/(n-1))s2. I'll use this notation in subsequent sections. I'll also use μ for the true, unknown mean of the population, and x for the actual mean of the sample.
In Just What Sense it is the Right Thing To Do
What we find using Bessel's correction is what is known as an "unbiased estimator". This means that, if we used the same underlying population and took lots of samples, using the estimation technique on each one, then the average of all of the estimates would approach the true value, σ. In other words, the expectation value of the estimate is the true value. This is why we used (effectively) the expectation value of the sample variance in the proof above.
Unbiased estimators sound difficult to argue with. How could bias be a good thing, after all?
Why it is, in Practice, Probably Not the Right Thing to Do
There are several problems with using the unbiased estimator for variance. For a start, what we really want to know is probably the standard deviation, not the variance. This doesn't sound like much of a problem: if we know the variance, we can just take the square root to get the standard deviation. But we don't know the variance; we only have an (unbiased) estimate of it. And the square root of that is not, in fact, an unbiased estimate of standard deviation. I've briefly outlined what has gone wrong, and how it can be fixed, in the next section.
There's a more fundamental problem, though: lack of bias doesn't really mean that an estimate is much use. To take an extreme example, suppose that a certain quantity has a true value of 11. Then an estimator that is always 1, except for one time in every million when it is 10000001, would technically be unbiased, because the expected value of the estimator would equal the true value. But it would almost always be an order of magnitude too small. This is a particularly silly example, since we know the true value, so we can give the best possible estimate according to all possible criteria as just being 11, every time. But what is shows is that being unbiased isn't enough to make an estimator useful. A more practical example of this sort of thing is to be found on the Wikipedia page about bias of estimators in statistics. Although requiring estimators to be unbiased seems like a good thing in principle, it doesn't really capture the properties that we really need. I'll address this later on.
Finding an Unbiased Estimator of the Population Standard Deviation
Even if we do want an unbiased estimator, the trouble with just taking the square root of the unbiased estimate of the variance is that the square root of an expected value is not the expected value of the square roots. (After all, the mean of 9 and 25 is 17, but the mean of 3 and 5 is 4, which isn't the square root of 17.) Correcting for this problem is inconvenient. The correct approach depends on what the probability distribution actually is, a thing that we didn't need to know in the case of the variance. If we make the assumption that the distribution is Gaussian[8], then we can find the equivalent thing to Bessel's correction, but it's a nasty formula involving the gamma function. It can be approximated, very roughly, by using n-1.5 instead of n-1 in the denominator of the correction. This is all explained in the "bias correction" section of an informative Wikipedia page. [9]
However, as I mentioned above, finding the unbiased estimator needn't be the right way to summarize what we know about the standard deviation.
Describing the Distribution of Possible Population Variances
The most complete statement of what we know about an uncertain quantity is not an estimator, nor is it a selection of estimators flawed in different ways. It isn't even an essay about how well or badly different estimators work in a particular case. The most complete statement of what we know about an uncertain quantity is a probability distribution. We start off with a probability distribution, the prior, that represents the information we have before the experiment. (This will be some sort of uniform prior if we have negligible information before the experiment.) We then update this using the data from the experiment, together with Bayes' theorem, to get the posterior probability distribution.[10] Finding an estimator is really just a way to find a value that summarizes that distribution.
Let's assume that we know nothing about σ before the experiment. But what does it really mean to know nothing? We could say that we think any value equally likely, leading to a uniform prior. But this isn't the right thing to do if we don't even know what order of magnitude a number will have, because according to a uniform distribution a number between 10 and 90 is ten times more likely to occur than a number between 1 and 9. In this case we probably want a log-uniform prior. I'll quote results below for both choices of prior distribution for the standard deviation.[11]
The Distribution of Possible Standard Deviations
If we assume that the underlying distribution is Gaussian,[12] and use Bayesian analysis to work out the probability distribution for the possible values of σ, we find a distribution which is a transformation of the well-known Chi-squared distribution. I don't want to write out the formula, but the distribution looks something like this.
The horizontal scale represents values of σ and the vertical scale is probability density.
This is the probability density function for σ, given that a sample of n=7 values has s=1, and assuming a uniform prior for the standard deviation. Notice that it isn't symmetrical, but is positively skewed: there is a heavy tail to the right of the maximum, compared to the left. This is what we'd expect, because it isn't possible that σ is less than zero, and it's unlikely that it's very small if s is 1; on the other hand, it is possible that σ is fairly big but that all of the readings in the sample just happen to be clustered together.
The trouble with an asymmetric distribution is that it's difficult to summarize. We could quote the mode, defined as the most likely value, i.e. the maximum of the distribution. This turns out to be, curiously, at the place that Bessel's approximation tells us to look, i.e. at s√(n/(n-1)). On the graph above, it occurs at √(7/6). (The result needs modifying slightly if we assume a log-uniform prior: in this case it turns out that the mode is actually at s.)
The trouble with quoting the mode is that, for this distribution, it's well to the left of the distribution, because of the positive skew. (In the example plotted above, the probability of σ being larger than the mode is about 70%.) What should we use instead? One obvious candidate is the median. This is by definition the value which is in the middle of the distribution in the sense that the probability that it is exceeded is exactly 50%. Unfortunately, there isn't a nice formula for the median of this distribution as there is for the mode. It does, though, imply using a bigger correction factor than Bessel recommends. In the example plotted above, the median is at 1.268, implying a correction factor of 1.268, whereas Bessel's approximation, and the mode, occur at 1.080 = √(7/6).
Another option is to find the mean of the distribution. Because of the positive skew, this is even bigger than the mode. I think that there's a nice formula for it, but since (as explained below) it isn't really the best thing to do, I haven't worked it out. Numerically, for the example plotted above, it occurs at 1.981 or so.
At first sight it might seem surprising that the mean of the posterior distribution isn't the same as the unbiased estimator, as they are both in some sense an expectation value for σ. (For the example above, n=7 and s=1, the unbiased estimate of σ is 1.042.) The reason that they're different is that they are averaging over different sorts of repeats. The mean of the posterior distribution, the one referred to in the previous paragraph, averages over all repeats of the experiment, with all possible values of σ that happen to give the same data. It gives the average value of σ that will produce the data. In contrast, the unbiased estimator gives a quantity which, if averaged over many repeats of the experiment with the same σ, will give the true value of that σ.
Which mean is the right mean? Neither. We're not repeating the experiment[13] in any way. The most complete way to state what we know about σ is to give the whole distribution, and any scheme to summarize it ought to be chosen based on what we happen to want the summary for.
Why this Discussion feels Frustrating
Let's take a step back for a moment: why do we want an estimate of σ, the population standard deviation? In many cases, we're not that interested in it for its own sake, because the variation between numbers is due to experimental error, and the thing that we're actually trying to measure is μ, the mean of the distribution. But we need to find it out because we want to give an estimate of the error in the result that we give. Our best estimate of the actual result, μ is given by the mean of the measurements, x, and then our estimate of the error in that mean is given by our estimate of σ (an estimate of the root-mean-square error on each measurement) divided by the square root of n. (We're allowed to divide by the square root of n because taking several readings tends to make the errors partially cancel out on average. This is a well-known result; I don't want to explain why it works here.)
Thus the thing that we're eventually going to use our estimate of σ to try to find out is, itself, a measure of the width of a distribution: the distribution of possible true means. We shouldn't be worrying about how to summarize the distribution of possible σ. The problem that we should really be tacking, using a Bayesian approach, is the distribution of possible μ.
Describing the Distribution of Possible Population Means
As last time, let's assume that the data are Gaussian distributed, and use Bayesian methods, this time to find the distribution of possible μ. The probability density function for μ turns out to be proportional to (1 + z2)-(n-1)/2, where z = (μ - x)/s. With a few changes of variable name, this turns out to be a well-known statistical distribution called Student's t-distribution.[14] If we assume a log-uniform prior for σ rather than a uniform one (which, as mentioned above, is probably a better expression of what we think that we know in most cases, and also meshes better with the way that Student's t distribution is used in classical statistics) then the n-1 in the formula is replaced by n.
Student's t Distribution
If we have x=0 and s=1[15], and assume a log-uniform prior for σ[16], then the probability density function becomes proportional to (1 + μ2)-n/2, which, for n=7, looks like this.
The horizontal scale represents values of μ and the vertical scale is probability density.
It looks fairly similar to the Gaussian distribution, which is not a surprise. In the limit as n goes to infinity it must approach the Gaussian distribution, because of the central limit theorem. But it's slightly different: it has heavier tails, i.e. a larger probability of extreme values relative to its own standard deviation than the Gaussian. (The technical term for this sort of distribution is leptokurtic.) The problem of finding a measure of the expected error in the measurement of μ becomes the problem of characterizing the width of this distribution.
The Standard Deviation of Student's t Distribution
Perhaps the first thing that springs to mind, when looking for a measure of the width of a distribution, is to find its standard deviation. For the distribution above, the standard deviation of μ is 1/√(n-3). (Thus in the specific case n=7 illustrated above, it's exactly 0.5.) If we then work back to what we would have to estimate for σ in order to derive this error in the mean when dividing by √(n), it's equivalent to replacing Bessel's correction by a factor of √(n/(n-3)).
It's nice that the result is so simple and exact, but I think that it's worth asking whether its standard deviation is really the best way to characterize the distribution. When people are given a standard error, they tend to assume that it obeys Gaussian statistics, and this doesn't. So it's perhaps worth looking at how to characterize the width in a way that preserves probabilities compared to the Gaussian distribution.
Matching Percentage Points of Student's t Distribution, and an Unsolved Problem
We know that, for the Gaussian distribution, 68%[17] of the data lie within one standard deviation of the mean. So perhaps one way to quote the standard error in μ would be to quote the +/- value such that μ lies within that range (i.e this distance either side of x) with 68% probability. The problem can be solved numerically. It turns out that the answer (as a +/- value for μ given x=0 and s=1) is approximately 1/√(n-2); it is well approximated by that formula even for n as small as 5, and the approximation gets better as n increases. This is equivalent to replacing Bessel's correction by a factor of √(n/(n-2)).
Somewhat to my annoyance, I can't prove this result, though the numerical evidence for it is very strong. What I want to prove is that if k is defined as satisfying
then k-2 = n - 2 + O(1/n). Yes, I can evaluate both of the denominators. I've just left them in integral form to make it clear where the calculation is coming from.
Matching the t-distribution to the Gaussian at the one standard deviation (of the Gaussian) point may not be the right thing to do, though. When interpreting a range of possible results, people perhaps usually make heavier use of the assumption that the probability that the value lies within two standard errors of the mean is 95%[17]. Similar numerical work to that of the last section suggests that Bessel's correction should then be replaced by √(n/(n-3.5)). Again I don't have a proof. (And this time it doesn't become a good approximation until n=10 or so.)
If we need to bring the probability of being wrong down further, say to beyond the 5σ level that CERN insists on before admitting that it's discovered a new particle, we'll need more formulae. At this point the numerical methods that I've been using start to lack sufficient precision!
Conclusion
The result of the last section suggests that we should use n-2, or possibly n-3.5, instead of the n-1 of Bessel's correction. Under the assumptions made, we should. But I've been assuming for several sections that the underlying distribution is Gaussian, and it might not be. (Bessel's correction, on the other hand, is correct on its own, albeit artificial, terms whatever the underlying distribution.) If we really want to use a proper Bayesian approach, which will make the answer as right as it can be, we'll need to incorporate whatever information we have about the underlying distribution. And then, as always, the fullest statement of what we think we know is a probability distribution.
But there is a quick, practical answer to the question of whether we should use n, n-1, n-1.5, n-2, n-3, n-3.5, or something else in simple, everyday cases where working out the full posterior distribution for μ would be excessive. If these numbers are very different from each other, then n is small, so you should take more readings. And then it won't matter very much which you use. Moreover, you'll have a smaller error, and perhaps, even, more idea as to what's going on in your experiment.
Matthew Smith, 2.viii.2014
[1] ^ It appears that he wasn't the first to do so, and that it was first used by Gauss. But Gauss invented practically everything, and we can't name it all after him, or it gets too confusing.
[2] ^ It's at moments like these that the excellent "What If?" articles insert something like [citation needed].
[3] ^ The major challenge would be ensuring that the rats really are randomly chosen, i.e. that the catching process doesn't somehow favour larger or smaller rats. In practice this is likely to be a bigger problem than whether we do or don't use Bessel's correction. But it's one that I'm happy to leave to the biologists.
[4] ^ This is the right thing to do, anyway, if we assume a uniform prior for the underlying mean of the population. Everything becomes much more horrible if we don't.
[5] ^ It's not always going to be an underestimate: sometimes, by chance, the individuals chosen will be significantly more scattered than the general population, and it will be an overestimate. But the point is that it is biased towards being an underestimate on average. In what sense I mean "on average" is to be explored later on.
[6] ^ I'm using x1 and x2 as the two randomly-chosen values. Hopefully this is clear.
[7] ^ I'm using, implicitly, some technical results about combining expectation values here: firstly that we can add them up, and secondly that, provided that the variables are uncorrelated, we can multiply them. These results may make intuitive sense. At any rate they can be proved.
[8] ^ The Gaussian distribution and the normal distribution are two names for the same thing. In statistics generally the name normal distribution is more common. But physicists of a Bayesian persuasion tend to call it the Gaussian distribution.
[9] ^ The other half of that page is taken up by the effect of serial correlation. I'm assuming throughout these notes that the sample data are all purely random selections from the population. (As with the rats.)[3] But in real life serial correlation is an effect to watch out for: it may well be much more important than whether we use Bessel's correction or not!
[10] ^ This is the method of Bayesian inference. I'm rather keen on it, though not everyone agrees.
[11] ^ Whichever prior we assume for the underlying standard deviation, I'll assume a uniform prior for the underlying mean. There are sometimes good theoretical reasons for doing this. In principle we oughtn't to use improper priors (prior distributions like the uniform distribution that can't be normalized), but should instead incorporate such information as we have, however limited it may be, in the form of a cut-off. Otherwise we get all kinds of problems. However, using an un-normalized uniform prior still leads to a normalizable posterior distribution, and the answer is fairly insensitive to where we put the cut-offs provided that they really are well away from the data.
[12] ^ This is a key assumption, and it may not be a good one. We should look at the problem at hand and see what we know about the underlying distribution rather than just assume that it's Gaussian. However, if we want to look at how Bessel's approximation might have to be modified, we will need to focus on a specific example, and the Gaussian distribution is the best example to choose, because it's one that often occurs in the analysis of real experimental errors, it's relatively simple to analyse, and it allows us to make comparisons with standard results from classical statistics.
[13] ^ If we were repeating the experiments, we'd just combine our results and have a larger sample size. That larger sample is then what we should analyse, rather than artificially chopping our results into pieces based on when we happened to take them.
[14] ^ The variable t that appears in the usual form of Student's t distribution differs from my z by a factor of √(ν), where ν is the number of degrees of freedom, equal to n-1.
[15] ^ If we have any other values for x and s, we can just shift and re-scale the answers. And I shall.
[16] ^ I'll do this for the rest of this section. I know that it's inconsistent with the graphical example in the previous section, but I don't want to go back and change it now!
[17] ^ It's really 68% and a bit for one standard deviation, and 95% and a bit for two standard deviations. I'm just writing them to the nearest whole percentage point for brevity.
FAQs
Standard Deviations: Division by n or n-1? ›
The standard deviation calculated with a divisor of n−1 is a standard deviation calculated from the sample as an estimate of the standard deviation of the population from which the sample was drawn.
Do you divide standard deviation by n or n-1? ›The n-1 equation is used in the common situation where you are analyzing a sample of data and wish to make more general conclusions. The SD computed this way (with n-1 in the denominator) is your best guess for the value of the SD in the overall population.
Why standard deviation is divided by n-1 or n? ›measures the squared deviations from x rather than μ . The xi's tend to be closer to their average x rather than μ , so we compensate for this by using the divisor (n-1) rather than n.
How do you know when to use N or N-1 in standard deviation? ›When you collect data from a sample, the sample standard deviation is used to make estimates or inferences about the population standard deviation. With samples, we use n – 1 in the formula because using n would give us a biased estimate that consistently underestimates variability.
Why we use n-1 in variance instead of n? ›WHY DOES THE SAMPLE VARIANCE HAVE N-1 IN THE DENOMINATOR? The reason we use n-1 rather than n is so that the sample variance will be what is called an unbiased estimator of the population variance ��2.
What is the sum of the squared deviations about the mean divided by n-1? ›Divide the sum of the squares of the deviations by n-1. This is the Variance! Take the square root of the variance to obtain the standard deviation, which has the same units as the original data.
Does Excel use N or N 1 for standard deviation? ›The standard deviation is calculated using the "n-1" method. Arguments can either be numbers or names, arrays, or references that contain numbers.
Why is standard deviation divided by n 2? ›the reason why we use n-2 df instead n-1 in estimating error variance is there are two parameters estimated in each equation, we deduct 2 from the number of observations to obtain the df.
Why does dividing by standard deviation make standard deviation 1? ›as @Silverfish already pointed out in a comment, the standard deviation has the same unit as the measurements. Thus, dividing by standard deviation as opposed to variance, you end up with a plain number that tells you where your case is relative to average and spread as measured by mean and standard deviation.
Why is standard deviation always 1? ›When we convert our data into z scores, the mean will always end up being zero (it is, after all, zero steps away from itself) and the standard deviation will always be one.
Why is standard deviation 1 in normal distribution? ›
A Standard Normal Distribution is a type of normal distribution with a mean of 0 and a standard deviation of 1. This means that the normal distribution has its center at 0 and intervals that increase by 1. It gives the actual weights of the students above the x-axis.
How do you know which standard deviation to use? ›The population standard deviation is relevant where the numbers that you have in hand are the entire population, and the sample standard deviation is relevant where the numbers are a sample of a much larger population.
What are the two formulas for standard deviation? ›| Population Standard Deviation Formula | σ = ∑ ( X − μ ) 2 n |
|---|---|
| Sample Standard Deviation Formula | s = ∑ ( X − X ¯ ) 2 n − 1 |
The sum of n natural numbers is represented as [n(n+1)]/2. Natural numbers are the numbers that start from 1 and end at infinity. Natural numbers include whole numbers in them except the number 0.
Do you divide standard deviation? ›Step 1: Find the mean. Step 2: For each data point, find the square of its distance to the mean. Step 3: Sum the values from Step 2. Step 4: Divide by the number of data points.
What happens to standard deviation when divided? ›Common areas of testing: (a) If you multiply or divide every term in the set by the same number, the SD will change. SD will change by that same number. The mean will also change by the same number.
Can you divide standard deviation by mean? ›The coefficient of variation(CV%) is the intrasubject standard deviation divided by the mean, expressed as a percentage.
What does N-1 mean in variance? ›Here since Variance is dependent on the calculation of the sample means, therefore we have one constraint, hence the degree of freedom is N-1. Therefore, if you divide by N-1 then the Standard deviation becomes an unbiased estimate.
How does N affect standard deviation? ›The larger n gets, the smaller the standard deviation of the sampling distribution gets. (Remember that the standard deviation for the sampling distribution of ¯X is σ√n.) This means that the sample mean ¯x must be closer to the population mean μ as n increases.
What is the use of N vs N in statistics? ›N usually refers to a population size, while n refers to a sample size. Can also consider n to be the within-cell size, while N is the entire-sample size.
What is standard deviation squared divided by n? ›
In statistics, you will encounter the formula s/√n in different scenarios. This formula is used to calculate the standard error of a sample mean. In the formula, s represents the sample standard deviation and n represents the sample size.
What is standard deviation divided by square root of n called? ›SEM is calculated simply by taking the standard deviation and dividing it by the square root of the sample size. Standard error gives the accuracy of a sample mean by measuring the sample-to-sample variability of the sample means.
What is sum of the squared deviations divided by the number of values? ›The variance is the average of the sum of squares (i.e., the sum of squares divided by the number of observations). The standard deviation is the square root of the variance.
What is n 1 called in standard deviation? ›In statistics, Bessel's correction is the use of n − 1 instead of n in the formula for the sample variance and sample standard deviation, where n is the number of observations in a sample. This method corrects the bias in the estimation of the population variance.
What is the difference between STDEV and STDEV s? ›The STDEV. P Function calculates the standard deviation of a dataset if the dataset is considered to be from a population. The STDEV. S Function calculates the standard deviation of a dataset if the dataset is considered to be from a sample.
What is the correct standard deviation formula in Excel? ›Say there's a dataset for a range of weights from a sample of a population. Using the numbers listed in column A, the formula will look like this when applied: =STDEV.S(A2:A10). In return, Excel will provide the standard deviation of the applied data, as well as the average.
What is the 2 standard deviation rule in statistics? ›According to this rule, 68% of the data falls within one standard deviation, 95% within two standard deviations, and 99.7% within three standard deviations from the mean.
What is the difference between N and N 1 in statistics? ›Basically, you should use N-1 when you estimate a variance, and N when you compute it exactly.
What is the 2 3 rule standard deviation? ›The Empirical Rule states that 99.7% of data observed following a normal distribution lies within 3 standard deviations of the mean. Under this rule, 68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean.
Why do we divide by n 1 when calculating variance? ›The variance estimator makes use of the sample mean and as a consequence underestimates the true variance of the population. Dividing by n-1 instead of n corrects for that bias. Furthermore, dividing by n-1 make the variance of a one-element sample undefined rather than zero.
What is the difference between the two means divided by the standard deviation? ›
In statistics, the strictly standardized mean difference (SSMD) is a measure of effect size. It is the mean divided by the standard deviation of a difference between two random values each from one of two groups.
Will standard deviation always be 1? ›The standard deviation of the z-scores is always 1. The graph of the z-score distribution always has the same shape as the original distribution of sample values. The sum of the squared z-scores is always equal to the number of z-score values.
Why is standard deviation always squared? ›The variance is defined as the average squared difference of the scores from the mean. We square the deviation scores because, as we saw in the Sum of Squares table, the sum of raw deviations is always 0, and there's nothing we can do mathematically without changing that.
Is standard deviation always normally distributed? ›No. The use of standard deviation does not assume normality. The variance of a random variable is defined as Var(X)=E[(X−E[X])2].
Does standard normal distribution mean 1? ›The standard normal distribution is a normal distribution with mean μ = 0 and standard deviation σ = 1. The letter Z is often used to denote a random variable that follows this standard normal distribution.
What is the standard deviation in a normal distribution? ›In a normal distribution the mean is zero and the standard deviation is 1. It has zero skew and a kurtosis of 3. Normal distributions are symmetrical, but not all symmetrical distributions are normal.
What is the standard deviation for dummies? ›What is standard deviation? Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean.
What is the best standard deviation to have? ›Statisticians have determined that values no greater than plus or minus 2 SD represent measurements that are are closer to the true value than those that fall in the area greater than ± 2SD.
Which standard deviation is better? ›Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out. A standard deviation close to zero indicates that data points are close to the mean, whereas a high or low standard deviation indicates data points are respectively above or below the mean.
What does N 1 represent? ›At its most basic definition, N+1 simply means that there is a power backup in place should any single system component fail. The 'N' in this equation stands for the number of components necessary to run your system. The '+1' means there is one independent backup should a component of that system fail.
Why are there two different formulas for standard deviation? ›
The main difference is that when your goal is estimating the variance rather than the standard deviation, the formula with 1/n is the maximum likelihood estimate but is biased, whereas the formula with 1/(n−1) is unbiased.
How do you find the standard deviation of two distributions? ›We can find the standard deviation of the combined distributions by taking the square root of the combined variances.
What is n 1 divided by 2 used for? ›If the number of observations is odd, the number in the middle of the list is the median. This can be found by taking the value of the (n+1)/2 -th term, where n is the number of observations. Else, if the number of observations is even, then the median is the simple average of the middle two numbers.
What is the use of n & n 1? ›n & (n-1) helps in identifying the value of the last bit. Since the least significant bit for n and n-1 are either (0 and 1) or (1 and 0) . Refer above table. (n & (n-1)) == 0 only checks if n is a power of 2 or 0.
What is the difference between n 1 and n == 1? ›n=1 means assigning 1 to n. 2. n==1 means comparing n with 1. n value is 1 condition is true and if condition is also true condition will execute i.e output is 1.
What is the best way to combine standard deviations? ›We'll do this in the simplest way possible: Namely, after we square the SDs, we divide each by its own sample size. Then add square root to get the combined standard deviation.
What is standard deviation divided by return? ›The coefficient of variation (COV) is the ratio of the standard deviation of a data set to the expected mean. Investors use it to determine whether the expected return of the investment is worth the degree of volatility, or the downside risk, that it may experience over time.
What is standard deviation divided by sample size? ›The smaller the standard error, the more representative the sample will be of the overall population. The relationship between the standard error and the standard deviation is such that, for a given sample size, the standard error equals the standard deviation divided by the square root of the sample size.
What is a deviation score divided by the standard deviation? ›Standard Score (cont...)
As the formula shows, the standard score is simply the score, minus the mean score, divided by the standard deviation.
...
Effect of Changing Units.
| When multiplying or dividing by a constant | Effect on the Measure |
|---|---|
| Variance | Multiply or divide by the square of that constant |
Is standard deviation range divided by 6? ›
In current practice, the median is often substituted for the mean, and the Range/4 or Range/6 for the standard deviation.
Does standard deviation divide observations into equal parts? ›So it divides a series of observations into two equal parts where 50% of the observations are below the median value and other 50% are above the median value.
What is subtracting its mean and dividing by its standard deviation? ›Normalization is subtracting the mean then dividing by the standard deviation.
Why divide by n-1 instead of n in variance? ›The reason dividing by n-1 corrects the bias is because we are using the sample mean, instead of the population mean, to calculate the variance. Since the sample mean is based on the data, it will get drawn toward the center of mass for the data.
Why do we use n-1 instead of n in sample variance? ›The reason we use n-1 rather than n is so that the sample variance will be what is called an unbiased estimator of the population variance 2.
Is variance denominator n or n-1? ›We define s² in a way such that it is an unbiased sample variance. The (n-1) denominator arises from Bessel's correction, which is resulted from the 1/n probability of sampling the same sample (with replacement) in two consecutive trials.
What does N mean for standard deviation? ›Overview of how to calculate standard deviation
where ∑ means "sum of", x is a value in the data set, μ is the mean of the data set, and N is the number of data points in the population.
Standard deviation is a measure of dispersion of data values from the mean. The formula for standard deviation is the square root of the sum of squared differences from the mean divided by the size of the data set. For a Population. σ=√∑ni=1(xi−μ)2n.
What does N mean in statistics standard deviation? ›The symbol 'n,' represents the total number of individuals or observations in the sample.
Why we use n 1 instead of n in the formula of calculating sample standard deviation? ›measures the squared deviations from x rather than μ . The xi's tend to be closer to their average x rather than μ , so we compensate for this by using the divisor (n-1) rather than n.
How does N affect variance? ›
That is, the variance of the sampling distribution of the mean is the population variance divided by N, the sample size (the number of scores used to compute a mean). Thus, the larger the sample size, the smaller the variance of the sampling distribution of the mean.
Is the sum of the squared deviations from the mean divided by n-1? ›Divide the sum of the squares of the deviations by n-1. This is the Variance! Take the square root of the variance to obtain the standard deviation, which has the same units as the original data.
Why do we divide sigma by square root of n? ›Sigma over square root n talks about how variable the population of average is of size n from that population R. So two different statements and they estimate different things. So, for example, if the Xs are IQ measurements, Sigma talks about how variable IQs are.
What is sigma squared divided by n? ›For a population, the variance is calculated as σ² = ( Σ (x-μ)² ) / N. Another equivalent formula is σ² = ( (Σ x²) / N ) - μ². If we need to calculate variance by hand, this alternate formula is easier to work with.
How do you find the sum of two standard deviations? ›Step 1: Name the independent random variables X and Y , and identify the standard deviations σX and σY . Step 2: Calculate the standard deviation of the sum of the random variables using the formula σX+Y=√σ2X+σ2Y σ X + Y = σ X 2 + σ Y 2 . The standard deviation of the sum of the two random variables is 2.06.
What is the formula for the sum of squared deviations? ›The sum of the squared deviations, (X-Xbar)², is also called the sum of squares or more simply SS. SS represents the sum of squared differences from the mean and is an extremely important term in statistics. Variance.
What is the difference between N and N in standard deviation? ›N is the population size and n is the sample size. The question asks why the population variance is the mean squared deviation from the mean rather than (N−1)/N=1−(1/N) times it.
What do you divide standard deviation by? ›Step 1: Find the mean. Step 2: For each data point, find the square of its distance to the mean. Step 3: Sum the values from Step 2. Step 4: Divide by the number of data points.
What happens to standard deviation when dividing? ›Common areas of testing: (a) If you multiply or divide every term in the set by the same number, the SD will change. SD will change by that same number. The mean will also change by the same number.
Is standard deviation always 1 for normal distribution? ›The standard normal distribution always has a mean of zero and a standard deviation of one.
What is the n 1 rule in maths? ›
Step 1: The nth term of an arithmetic sequence is given by an = a + (n – 1)d. So, to find the nth term, substitute the given values a = 2 and d = 3 into the formula. Step 2: Now, to find the fifth term, substitute n = 5 into the equation for the nth term.
Why do we use n 1 in sequences? ›If n is equal to 1, then it refers to the first term of the sequence. The d stands for the difference between all the successive numbers of your sequence. This is called the explicit formula for an arithmetic sequence.
Is n 1 divided by any number between 2 and n? ›Consequently n! + 1, when divided by any number between 2 and 'n' leaves 1 as remainder. Hence, n! + 1 is not divisible by any number between 2 and 'n'.
What is mean divided by standard deviation called? ›Definition. The coefficient of variation (CV) is defined as the ratio of the standard deviation to the mean. , It shows the extent of variability in relation to the mean of the population.
Why is variance sigma squared divided by n? ›The variance of the sum would be σ2 + σ2 + σ2. For N numbers, the variance would be Nσ2. Since the mean is 1/N times the sum, the variance of the sampling distribution of the mean would be 1/N2 times the variance of the sum, which equals σ2/N.
Why are there 2 equations for standard deviation? ›Explanation: The two formulas, as shown below, are equivalent. They are alternate forms and which one is used depends on which is the most efficient method with the given data.
Why is degrees of freedom n 1? ›The use of n-1 instead of n degrees of freedom fixes this because the lower the degrees of freedom of a chi-square distribution the tighter the distribution. This slightly tighter distribution makes up for our under-estimate of the the true population variance.