STAT 211
Handout 6 (Chapter 6): Point Estimation
A point estimate of a parameter ( is a single number that can be regarded as the most plausible value of (.
Unbiased Estimator: A point estimator, [pic] = ( + error of estimation, is an unbiased estimator of ( if E([pic])= ( for every possible value of (. Otherwise, it is biased and Bias = E([pic])- (.
Read the example 6.2 (your textbook).
Example 1: When X is a binomial r.v. with parameters, n and p, the sample proportion X/n is an unbiased estimator of p.
To prove this, you need to show E(X/n)=p where [pic]=X/n.
E(X/n) = E(X)/n, Using the rules of the expected value.
= np / n =p If X~Binomial(n,p) then E(X)=np (Chapter 3)
Example 2: A sample of 15 students who had taken calculus class yielded the following information on brand of calculator owned: T H C T H H C T T C C H S S S (T: Texas Instruments, H: Hewlett Packard, C=Casio, S=Sharp).
a) Estimate the true proportion of all such students who own a Texas Instruments calculator.
Answer=0.2667
b) Three out of four calculators made by only Hewlett Packard utilize reverse Polish logic. Estimate the true proportion of all such students who own a calculator that does not use reverse Polish logic.
Answer=0.80
Example 3 (Exercise 6.8) : In a random sample of 80 components of a certain type, 12 are found to be defective.
a) A point estimate of the proportion of all such components which are not defective.
Answer=0.85
b) Randomly select 5 of these components and connect them in series for the system. Estimate the proportion of all such systems that work properly.
Answer=0.4437
Example 4 (Exercise 6.12) :
X: yield of 1st type of fertilizer. [pic], E(X)= [pic] Var(X)= [pic]
Y: yield of 2nd type of fertilizer. [pic], E(Y)=[pic] Var(Y)= [pic]
Show [pic] is an unbiased estimator for [pic]
It means that you need to show[pic]
[pic]
[pic]
Example 5 (Exercise 6.13) : X1,X2,….,Xn be a random sample from the pdf f(x)=0.5(1+(x), -1(x(1, -1(((1. Show that [pic] is an unbiased estimator for [pic].
It means that you need show [pic].
[pic]=[pic] where
[pic], [pic]
The standard error: The standard error of an estimator [pic] is its standard deviation [pic].
The estimated standard error: The estimated standard error of an estimator is its estimated standard deviation [pic]=[pic].
The minimum variance unbiased estimator (MVUE): The best point estimate. Among all estimators of ( that are unbiased choose the one that has minimum variance. The resulting [pic] is MVUE.
Example 6: If we go back to example 1, the standard error of [pic]is [pic] where
[pic][pic]
Example 7: If we go back to example 5, the standard error of [pic] is
[pic]
where Var(X)=[pic]
E(X2)= [pic]
Example 8: For normal distribution, [pic] is the MVUE for (. Proof is as follows.
The following graphs are generated by creating 500 samples with size 5 from N(0,1) and calculating the sample mean and the sample median for each sample.
Example 9 (Exercise 6.3): Given normally distributed data yield the following summary statistics.
Variable n Mean Median TrMean StDev SE Mean
thickness 16 1.3481 1.3950 1.3507 0.3385 0.0846
Variable Minimum Maximum Q1 Q3
thickness 0.8300 1.8300 1.0525 1.6425
a) A point estimate of the mean value of coating thickness.
b) A point estimate of the median value of coating thickness.
c) A point estimate of the value that separates the largest 10% of all values in the coating thickness distribution from the remaining 90%.
Answer=1.78138
d) Estimate P(X-1.
A random sample of 10 students yield the data: 0.92, 0.79, 0.90, 0.65, 0.86, 0.47, 0.73, 0.97, 0.94, 0.77.
(a) Obtain the MME of ( and compute the estimate using the data. [pic]
Set E(X)= [pic]and then solve for (.
The given data yield [pic]= 0.80 then the method of moment estimator for ( is [pic]=3
(b) Obtain the MLE of ( and compute the estimate using the data.
L=Likelihood=[pic]
ln(L)= [pic]
[pic]=0 then solve for (.
The given data yield [pic]-2.4295 then the maximum likelihood estimator for ( is [pic]=3.1161
Proposition: Under very general conditions on the joint distribution of the sample when the sample size is large, the MLE of any parameter ( is approximately unbiased and has a variance that is nearly as small as can be achieved by an estimator.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.