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find the point estimate of the population mean
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From Wikipedia
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
 maximum likelihood (ML)
 method of moments, generalized method of moments
 minimum mean squared error (MMSE)
 minimum variance unbiased estimator (MVUE)
 best linear unbiased estimator (BLUE)
Routes to deriving point estimates via Bayesian Analysis
Properties of Point estimates
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 Î” or as MD. Themean deviation is a different measure of dispersion.
Calculation
For a population of size n, with a sequence of values y_{i}, i = 1 to n:
 MD = \frac{1}{n(n1)} \Sigma_{i=1}^n \Sigma_{j=1}^n  y_i  y_j  .
For a discrete probability functionf(y), where y_{i}, 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)\,xy\,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 âˆ‡ 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.
Properties
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) Â· mean(X)/(mean(X) + c) = RMD(X) / (1+c / mean(X)) for câ‰ mean(X),
 RMD(X) = âˆ’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 dispersionâ€”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 Estatistics.
Sample estimators
For a random sample S from a random variable X, consisting of n values y_{i}, the statistic
 MD(S) = \frac{\sum_{i=1}^n \sum_{j=1}^n  y_i  y_j }{n(n1)}
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 }{(n1)\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
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
Answers:If you're asking me, I can see the answer to this question within the text you gave.
Answers:98
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 2tailed tdistribution with n1 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.
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