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The weighted mean is similar to an arithmetic mean (the most common type of average), where instead of each of the data points contributing equally to the final average, some data points contribute more than others. The notion of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics.## Example

## Mathematical definition

## Length-weighted mean

## Convex combination

## Dealing with variance

### Correcting for over/under dispersion

## Weighted sample variance

Typically when you calculate a mean it is important to know the variance and standard deviation of that mean. When a weighted mean $mu^*$ is used, the variance of the weighted sample is different from the variance of the unweighted sample. The biased weighted sample variance is defined similarly to the normal biased sample variance:## Accounting for correlations

## See also

## References

## External links

If all the weights are equal, then the weighted mean is the same as the arithmetic mean. While weighted means generally behave in a similar fashion to arithmetic means, they do have a few counter-intuitive properties, as captured for instance in Simpson's paradox.

The term weighted average usually refers to a weighted arithmetic mean, but weighted versions of other means can also be calculated, such as the weighted geometric mean and the weighted harmonic mean.

Given two school classes, one with 20 students, and one with 30 students, the grades in each class on a test were:

- Morning class = 62, 67, 71, 74, 76, 77, 78, 79, 79, 80, 80, 81, 81, 82, 83, 84, 86, 89, 93, 98

- Afternoon class = 81, 82, 83, 84, 85, 86, 87, 87, 88, 88, 89, 89, 89, 90, 90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95, 96, 97, 98, 99

The straight average for the morning class is 80 and the straight average of the afternoon class is 90. The straight average of 80 and 90 is 85, the mean of the two class means. However, this does not account for the difference in number of students in each class, and the value of 85 does not reflect the average student grade (independent of class). The average student grade can be obtained by either averaging all the numbers without regard to classes, or weighting the class means by the number of students in each class:

- $$

Or, using a weighted mean of the class means:

- $$

The weighted mean makes it possible to find the average student grade also in the case where only the class means and the number of students in each class are available.

Formally, the weighted mean of a non-empty set of data

- $[x\_1,\; x\_2,\; dots\; ,\; x\_n],,$

with non-negative weights

- $[w\_1,\; w\_2,\; dots,\; w\_n],,$

is the quantity calculated by

- $bar\{x\}\; =\; frac\{\; sum\_\{i=1\}^n\; w\_i\; x\_i\}\{sum\_\{i=1\}^n\; w\_i\},$

which means:

- $$

So data elements with a high weight contribute more to the weighted mean than do elements with a low weight. The weights must not be negative. They may be zero, but not all of them (because division by zero is not allowed). In the special case, often encountered in practice, where the weights are normalized (i.e. are nonnegative and sum up to 1), the denominator of the fraction simplifies to 1.

For weighting a response variable based upon its dependency on x, a distance variable.

- $$

Since only the relative weights are relevant, any weighted mean can be expressed using coefficients that sum to one. Such a linear combination is called a convex combination.

Using the previous example, we would get the following:

- $$

- $$

- $$

This simplifies to:

- $$

For the weighted mean of a list of data for which each element $x\_i,!$ comes from a different probability distribution with known variance $\{sigma\_i\}^2,$, one possible choice for the weights is given by:

- $$

The weighted mean in this case is:

- $$

and the variance of the weighted mean is:

- $$

which reduces to $sigma\_\{bar\{x\}\}^2\; =\; frac\{\; \{sigma\_0\}^2\; \}\{n\}$, when all $sigma\_i\; =\; sigma\_0.,$

The significance of this choice is that this weighted mean is the maximum likelihood estimator of the mean of the probability distributions under the assumption that they are independent and normally distributed with the same mean.

Weighted means are typically used to find the weighted mean of experimental data, rather than theoretically generated data. In this case, there will be some error in the variance of each data point. Typically experimentally errors are underestimated, because the experimenter does not know all sources of error in calculating the variance of each data point. In this event, the variance in the weighted mean must be corrected to account for the fact that $chi^2$ is too large. The correction that must be made is

- $sigma\_\{bar\{x\}\}^2\; rightarrow\; sigma\_\{bar\{x\}\}^2\; chi^2\_nu$

where $chi^2\_nu,$ is $chi^2,$ divided by the number of degrees of freedom, in this case $n-1,$. This gives the variance in the weighted mean as:

- $sigma\_\{bar\{x\}\}^2\; =\; frac\{\; 1\; \}\{sum\_\{i=1\}^n\; 1/\{sigma\_i\}^2\}\; times\; frac\{1\}\{(n-1)\}\; sum\_\{i=1\}^n\; frac\{\; (x\_i\; -\; bar\{x\}\; )^2\}\{\; sigma\_i^2\; \}\; ,$

- $$

N} ;;; sigma_text{weighted}^2 = frac{ sum_{i=1}^N{{w_i}left(x_i - mu^*right)^2} }{ sum_{i=1}^N{w_i} }.

For small sample of populations, it is customary to use an unbiased estimator for the population variance. In normal unweighted samples, the N in the denominator (corresponding to the sample size) is changed to N − 1. While this is simple in unweighted samples, it becomes tedious for weighted samples. Thus, the unbiased estimator of weighted population variance is given by :

- $$

Which can also be written in terms of running sums for programming as:

- $$

The standard deviation is simply the square root of the variance above.

In the general case, suppose that $mathbf\{X\}=[x\_1,dots,x\_n]$, $mathbf\{C\}$ is the covariance matrix relating the quantities $x\_i$, $bar\{x\}$ is the common mean to be estimated, and $mathbf\{W\}$ is the design matrix [1, ..., 1] (of length n). The Gauss–Markov theorem says that the estimate of the mean having minimum variance is given by:

- $sigma^2\_bar\{x\}=(mathbf\{W\}^T\; mathbf\{C\}^\{-1\}\; mathbf\{W\})^\{-1\},$

and

- $bar\{x\}\; =\; sigma^2\_bar\{x\}\; (mathbf\{W\}^T\; mathbf\{C\}^\{-1\}\; mathbf\{X\}).$

- Bevington, Philip. Data Reduction and Error Analysis for the Physical Sciences.

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Last updated on Saturday October 11, 2008 at 03:00:29 PDT (GMT -0700)

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Last updated on Saturday October 11, 2008 at 03:00:29 PDT (GMT -0700)

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