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Point Estimation of Parameters and Sampling Distributions

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Point Estimation of Parameters and Sampling Distributions Outlines: Sampling Distributions and the central limit theorem Point estimation Methods of point estimation – PowerPoint PPT presentation

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Title: Point Estimation of Parameters and Sampling Distributions


1
Point Estimation of Parameters and Sampling
Distributions
  • Outlines
  • Sampling Distributions and the central limit
    theorem
  • Point estimation
  • Methods of point estimation
  • Moments
  • Maximum Likelihood

2
Sampling Distributions and the central limit
theorem
  • Random Sample
  • Sampling distribution the probability
    distribution of a statistic.
  • Ex. The probability distribution of is called
    distribution of the mean.

3
Sampling Distributions of sample mean
  • Consider the sampling distribution of the sample
    mean.
  • Xi is a normal and independent probability, then

4
Central limit theorem
  • n gt 30, sampling from an unknown population gt
    the sampling distribution of will be
    approximated as normal with mean µ and ?2/n.

5
Central limit theorem
6
Central limit theorem
  • Ex. Suppose that a random variable X has a
    continuous uniform distribution
  • Find the distribution of the sample mean of a
    random sample of size n40
  • Method 1.Calculate the value of mean and
    variance of x
  • 2.

7
Sampling Distribution of a Difference
  • Two independent populations.
  • Suppose that both populations are normally
    distributed.
  • Then, the sampling distribution of is
    normal with

µ1 ,?12
µ2 ,?22
8
Sampling Distribution of a Difference
  • Definition

9
Sampling Distribution of a Difference
  • Ex. The effective life of a jet-turbine aircraft
    engine is a random variable with mean 5000 hr.
    and sd. 40 hr. The distribution of effective life
    is fairly close to a normal distribution. The
    engine manufacturer introduces an improvement
    into the manufacturing process for the engine
    that increases the mean life to 5050 hr. and
    decrease sd. to 30 hr.
  • 16 components are sampling from the old process.
  • 25 components are sampling from the improve
    process.
  • What is the probability that the difference in
    the two sample means is at least 25 hr?

10
Point estimation
  • Parameter Estimation calculation of a reasonable
    number that can explains the characteristic of
    population.
  • Ex. X is normally distributed with unknown mean
    µ.
  • The Sample mean( ) is a point estimator of
    population mean (µ) gt
  • After selecting the sample, is the point
    estimate of µ.

11
Point estimation
  • Unbiased Estimators

12
Point estimation
  • Ex. Suppose that X is a random variable with mean
    µ and variance s2 . Let X1,X2 ,..., Xn be a
    random sample of size n from the population. Show
    that the sample mean and sample variance S2
    are unbiased estimators of µ and s2 ,
    respectively.
  • proof,
  • proof,

13
Point estimation
14
Point estimation
  • Sometimes, there are several unbiased estimators
    of the sample population parameter.
  • Ex. Suppose we take a random sample of size n
    from a normal population and obtain the data x1
    12.8, x2 9.4, x3 8.7, x4 11.6, x5 13.1,
    x6 9.8, x7 14.1,x8 8.5, x9 12.1, x10
    10.3.

all of them are unbiased estimator of µ
15
Point estimation
  • Minimum Variance Unbiased Estimator (MVUE)

16
Point estimation
  • MVUE for µ

17
Method of Point Estimation
  • Method of Moment
  • Method of Maximum Likelihood
  • Bayesian Estimation of Parameter

18
Method of Moments
  • The general idea of the method of moments is to
    equate the population moments to the
    corresponding sample moments.
  • The first population moment is E(X)µ........(1)
  • The first sample moment is
    ........(2)
  • Equating (1) and (2),
  • The sample mean is the moment estimator of the
    population mean

19
Method of Moments
  • Moment Estimators
  • Ex. Suppose that X1,X2 ,..., Xn be a random
    sample from an exponential distribution with
    parameter ?. Find the moment estimator of ?
  • There is one parameter to estimate, so we must
    equate first population moment to first sample
    moment.
  • first population moment E(X)1/?, first sample
    moment

20
Method of Moments
  • Ex. Suppose that X1,X2 ,..., Xn be a random
    sample from a normal distribution with parameter
    µ and s2. Find the moment estimators of µ and s2.
  • For µ k1 The first population moment is E(X)µ
    ........(1)
  • The first sample moment is
    ........(2)
  • Equating (1) and (2),
  • For s2 k2 The second population moment is
    E(X2)µ2s2 ......(3)
  • The second sample moment is
    ......(4)
  • Equating (3) and (4),

21
Method of Maximum Likelihood
  • Concept the estimator will be the value of the
    parameter that maximizes the likelihood function.

22
Method of Maximum Likelihood
  • Ex. Let X be a Bernoulli random variable. The
    probability mass function is

23
Method of Maximum Likelihood
  • Ex. Let X be normally distributed with unknown µ
    and known s2. Find the maximum likelihood
    estimator of µ

24
Method of Maximum Likelihood
  • Ex. Let X be exponentially distributed with
    parameter ?. Find the maximum likelihood
    estimator of ?.

25
Method of Maximum Likelihood
  • Ex. Let X be normally distributed with unknown µ
    and unknown s2. Find the maximum likelihood
    estimator of µ, and s2.

26
Method of Maximum Likelihood
  • The method of maximum likelihood is often the
    estimation method that mathematical statisticians
    prefer, because it is usually easy to use and
    produces estimators with good statistical
    properties.
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