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From Wikipedia

Point estimation

In statistics, point estimation involves the use of sampledata to calculate a single value (known as a statistic) which is to serve as a "best guess" for an unknown (fixed or random) population parameter.

More formally, it is the application of a point estimator to the data.

In general, point estimation should be contrasted with interval estimation.

Point estimation should be contrasted with general Bayesian methods of estimation, where the goal is usually to compute (perhaps to an approximation) the posterior distributions of parameters and other quantities of interest. The contrast here is between estimating a single point (point estimation), versus estimating a weighted set of points (a probability density function). However, where appropriate, Bayesian methodology can include the calculation of point estimates, either as the expectation or median of the posterior distribution or as the mode of this distribution.

In a purely frequentist context (as opposed to Bayesian), point estimation should be contrasted with the specific interval estimation calculation of confidence intervals.

Routes to deriving point estimates directly

Routes to deriving point estimates via Bayesian Analysis

Properties of Point estimates

Mean difference

The mean difference is a measure of statistical dispersion equal to the average absolute difference of two independent values drawn from a probability distribution. A related statistic is the relative mean difference, which is the mean difference divided by the arithmetic mean. An important relationship is that the relative mean difference is equal to twice the Gini coefficient, which is defined in terms of the Lorenz curve.

The mean difference is also known as the absolute mean difference and the Gini mean difference. The mean difference is sometimes denoted by Δ or as MD. Themean deviation is a different measure of dispersion.


For a population of size n, with a sequence of values yi, i = 1 to n:

MD = \frac{1}{n(n-1)} \Sigma_{i=1}^n \Sigma_{j=1}^n | y_i - y_j | .

For a discrete probability functionf(y), where yi, i = 1 to n, are the values with nonzero probabilities:

MD = \Sigma_{i=1}^n \Sigma_{j=1}^n f(y_i) f(y_j) | y_i - y_j | .

For a probability density functionf(x):

MD = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x)\,f(y)\,|x-y|\,dx\,dy .

For a cumulative distribution function F(x) with quantile function x(F):

MD = \int_0^1 \int_0^1 |x(F_1)-x(F_2)|\,dF_1\,dF_2 .

Relative mean difference

When the probability distribution has a finite and nonzero arithmetic mean, the relative mean difference, sometimes denoted by ∇ or RMD, is defined by

RMD = \frac{MD}{\mbox{arithmetic mean}}.

The relative mean difference quantifies the mean difference in comparison to the size of the mean and is a dimensionless quantity. The relative mean difference is equal to twice the Gini coefficient which is defined in terms of the Lorenz curve. This relationship gives complementary perspectives to both the relative mean difference and the Gini coefficient, including alternative ways of calculating their values.


The mean difference is invariant to translations and negation, and varies proportionally to positive scaling. That is to say, if X is a random variable and c is a constant:

  • MD(X + c) = MD(X),
  • MD(-X) = MD(X), and
  • MD(cX) = |c| MD(X).

The relative mean difference is invariant to positive scaling, commutes with negation, and varies under translation in proportion to the ratio of the original and translated arithmetic means. That is to say, if X is a random variable and c is a constant:

  • RMD(X + c) = RMD(X) · mean(X)/(mean(X) + c) = RMD(X) / (1+c / mean(X)) for c≠ -mean(X),
  • RMD(-X) = −RMD(X), and
  • RMD(cX) = RMD(X) for c> 0.

If a random variable has a positive mean, then its relative mean difference will always be greater than or equal to zero. If, additionally, the random variable can only take on values that are greater than or equal to zero, then its relative mean difference will be less than 2.

Compared to standard deviation

Both the standard deviation and the mean difference measure dispersion—how spread out are the values of a population or the probabilities of a distribution. The mean difference is not defined in terms of a specific measure of central tendency, whereas the standard deviation is defined in terms of the deviation from the arithmetic mean. Because the standard deviation squares its differences, it tends to give more weight to larger differences and less weight to smaller differences compared to the mean difference. When the arithmetic mean is finite, the mean difference will also be finite, even when the standard deviation is infinite. See the examples for some specific comparisons. The recently introduced distance standard deviation plays similar role than the mean difference but the distance standard deviation works with centered distances. See also E-statistics.

Sample estimators

For a random sample S from a random variable X, consisting of n values yi, the statistic

MD(S) = \frac{\sum_{i=1}^n \sum_{j=1}^n | y_i - y_j |}{n(n-1)}

is a consistent and unbiasedestimator of MD(X). The statistic:

RMD(S) = \frac{\sum_{i=1}^n \sum_{j=1}^n | y_i - y_j |}{(n-1)\sum_{i=1}^n y_i}

is a consistentestimator of RMD(X), but is not, in general, unbiased.

Confidence intervals for RMD(X) can be calculated using bootstrap sampling techniques.

There does not exist, in general, an unbiased estimator for RMD(X), in part because of the difficulty of finding an unbiased estimation for multiplying by the inverse of the mean. For example, even where the sample is known to be taken from a random variable X(p) for an unknown p, and X(p) - 1 has the Bernoulli distribution, so that Pr(X(p) = 1) = 1 âˆ’ p and , then

RMD(X(p)) = 2p(1 âˆ’ p)/(1 + p).

But the expected value of any estimator R(S) of RMD(X(p)) will be of the form:

From Yahoo Answers

Question:Hi, need a little help with a stat problem please. A data set has a sample size of 61, sample average of 11.7, an a standard deviation of 3.8 What is the point estimate of the population mean? Compute a 92% confidence interval for the population mean. Thanks!

Answers:hmm..if i'm not mistaken, the point estimate of the population mean is the sample mean = 11.7 92% confidence interval: 11.7 (1.75)(3.8/[ 61]) =11.7 0.851445 =(10.8486,12.5514) Assumptions: 1) Random sample is selected from target population 2) Sample size n is large ----n=61>30

Question:A survey of 47 houses in NY shows that the average cost of heating per month is $62.83 with the standard deviation of $3.36 , find the best point estimate of the mean heating cost for houses per month in NY? A survey of 47 houses in NY shows that the average cost of heating per month is $62.83 with the standard deviation of $3.36 , find the best point estimate of the mean heating cost for houses per month in NY? (The average of $62.83 is only the average of the sample of 47 houses. Based on this I need to find the mean price of the population, which would be all houses in NY).

Answers:If you're asking me, I can see the answer to this question within the text you gave.

Question:A laboratory tested 83 chicken eggs and found that the mean amount of cholesterol was 233 milligrams with a standard deviation of 12.9 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.


Question:Monitoring the Lead in Air: Listed below are measured amounts of lead (in micrograms per cubic meter, or ug/m3) in the air. The Environmental Protection Agency has established an air quality standard for lead of 1.5ug/m3. The measurements shown below were recorded at Building 5 of the WTC site on different days immediately following the destruction caused by the terrorist attacks of 9/11/01. After the collapse of the two WTC buildings, there was considerable concern about the quality of the air. Use the given values to construct a 95% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very good? 5.40 1.10 0.42 0.73 0.48 1.10 (I know the answer to this problem should be -0.471
Answers:To create a confidence interval, you first need to calculate the sample mean (M) and standard deviation (SD). Using Excel, these are: =AVERAGE(5.4, 1.1, 0.42, 0.73, 0.48, 1.1) 1.538 =STDEV(5.4, 1.1, 0.42, 0.73, 0.48, 1.1) 1.914 You then should find the 95% critical value for the 2-tailed t-distribution with n-1 degrees of freedom (n - 1 = 6 - 1 = 5). This value is 2.571, which we'll call t'. Finally, calculate the confidence interval using CI = M +/- t' * SD/sqrt(n) Upper CI = 1.538 + 2.571 * 1.914/sqrt(6) = 3.547 Lower CI = 1.538 - 2.571 * 1.914/sqrt(6) = -0.471 The sample that we used to calculate this confidence interval, however, wasn't ideal. The value 5.40 appears to be an outlier, which would greatly affect the M & SD for such a small sample. For instance, by removing this value, our mean and SD drop to 0.766 & 0.326. Further investigation would be necessary to see if dropping this apparent outlier is warranted.

From Youtube

Calculating Required Sample Size to Estimate Population Mean :statisticslectures.com - where you can find free lectures, videos, and exercises, as well as get your questions answered on our forums!

Estimating the Population Mean, Part 1 of 2 :Produced by Kent Murdick Instructor of Mathematics University of South Alabama