Date post: | 14-Apr-2018 |
Category: |
Documents |
Upload: | andy-william |
View: | 230 times |
Download: | 0 times |
of 15
7/27/2019 061 Intro to Prediction
1/151
Introduction toPredictionIntroduction toPrediction
Introduction
Statistical inference is the process bywhich we acquire information aboutpopulations from samples.
There are two types of inference:
Estimation
Hypotheses testing
7/27/2019 061 Intro to Prediction
2/152
Concepts of Estimation
The objective of estimation is todetermine the value of a populationparameter on the basis of a samplestatistic.
There are two types of estimators:Point EstimatorInterval estimator
Point Estimator
A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single valueor point.
7/27/2019 061 Intro to Prediction
3/153
Population distribution
Point Estimator
Parameter
?
Sampling distribution
A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single valueor point.
Point estimator
An interval estimator draws inferencesabout a population by estimating the valueof an unknown parameter using aninterval.
Interval estimator
Population distribution
Sample distribution
Parameter
Interval Estimator
7/27/2019 061 Intro to Prediction
4/154
Selecting the right sample statistic to estimatea parameter value depends on thecharacteristics of the statistic.
Estimators Characteristics
Estimators desirable characteristics:Unbiasedness: An unbiased estimator is one whose
expected value is equal to the parameter it estimates.Consistency: An unbiased estimator is said to be
consistent if the difference between the estimator andthe parameter grows smaller as the sample sizeincreases.
Relative efficiency:For two unbiased estimators, the onewith a smaller variance is said to be relatively efficient.
Estimating the Population Mean whenthe Population Variance is Known
How is an interval estimator producedfrom a sampling distribution?
A sample of size n is drawn from thepopulation, and its mean is calculated.
By the central limit theorem is normallydistributed (or approximately normallydistributed.), thus
x
x
7/27/2019 061 Intro to Prediction
5/155
n
xZ
=
We have established beforethat
=
+
1)nzxnz(P22
Estimating the Population Mean whenthe Population Variance is Known
=
+
1)n
zxn
zx(P 22
This leads to the followingequivalent statement
The Confidence Interval for ( isknown)
The confidence interval
7/27/2019 061 Intro to Prediction
6/156
Interpreting the Confidence Interval for
1 of all the values of obtained in repeated
sampling from a given distribution, construct an interval
that includes (covers) the expected value of thepopulation.
1 of all the values of obtained in repeated
sampling from a given distribution, construct an interval
that includes (covers) the expected value of thepopulation.
x
+
nzx,
nzx 22
x
nz2 2
nzx 2
nzx 2
+
Lower confidence limit Upper confidence limit
1 -
Confidence level
Graphical Demonstration of the
Confidence Interval for
7/27/2019 061 Intro to Prediction
7/157
The Confidence Interval for ( isknown)
Four commonly used confidence levels
Confidence
level /2/ 2/ 2/ 2
0.90 0.10 0.05 1.645
0.95 0.05 0.025 1.96
0.98 0.02 0.01 2.33
0.99 0.01 0.005 2.575
Confidence
level /2/ 2/ 2/ 2
0.90 0.10 0.05 1.645
0.95 0.05 0.025 1.96
0.98 0.02 0.01 2.33
0.99 0.01 0.005 2.575
z/2/2/2/2
Example: Estimate the mean value of the distribution resulting from
the throw of a fair die. It is known that = 1.71. Use a 90%
confidence level, and 100 repeated throws of the die
Solution: The confidence interval is
The Confidence Interval for ( isknown)
=
n
zx 2 28.x100
71.1645.1x =
The mean values obtained in repeated draws of samples of size100 result in interval estimators of the form
[sample mean - .28, Sample mean + .28],
90% of which cover the real mean of the distribution.
7/27/2019 061 Intro to Prediction
8/158
The Confidence Interval for ( isknown)
Recalculate the confidence interval for 95% confidence level.
Solution: =
n
zx 2 34.x100
71.196.1x =
34.x +34.x
.95
.90
28.x +28.x
The Confidence Interval for ( isknown)
The width of the 90% confidence interval = 2(.28) = .56
The width of the 95% confidence interval = 2(.34) = .68 The width of the 90% confidence interval = 2(.28) = .56
The width of the 95% confidence interval = 2(.34) = .68
Because the 95% confidence interval is wider, it is
more likely to include the value of. Because the 95% confidence interval is wider, it is
more likely to include the value of.
7/27/2019 061 Intro to Prediction
9/159
Example Doll Computer Company delivers computers
directly to its customers who order via theInternet.
To reduce inventory costs in its warehousesDoll employs an inventory model, thatrequires the estimate of the mean demandduring lead time.
It is found that lead time demand is normally
distributed with a standard deviation of 75computers per lead time.
Estimate the lead time demand with 95%confidence.
The Confidence Interval for ( isknown)
Example 10.1 Solution
The parameter to be estimated is , themean demand during lead time.
We need to compute the interval estimation
for .
From the data provided in file Xm10-01, the
sample mean is
The Confidence Interval for ( isknown)
.16.370=x
[ ]56.399,76.34040.2916.37025
7596.116.370
25
75z16.370
nzx
025.2
===
=
Since 1 - =.95, = .05.
Thus /2 = .025. Z.025 = 1.96
7/27/2019 061 Intro to Prediction
10/1510
Using Excel
Tools > Data Analysis Plus > Z Estimate:Mean
The Confidence Interval for ( isknown)
Wide interval estimator provides littleinformation. Where is ? ? ? ?
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???
Information and the Width of the Interval
7/27/2019 061 Intro to Prediction
11/1511
Here is a much narrower interval.
If the confidence level remains
unchanged, the narrower interval
provides more meaningful
information.
Here is a much narrower interval.
If the confidence level remains
unchanged, the narrower interval
provides more meaningful
information.
Wide interval estimator provides littleinformation. Where is ? ? ? ?
Ahaaa!
Information and the Width of the Interval
The width of the confidence interval isaffected by
the population standard deviation () the confidence level (1-) the sample size (n).
The Width of the Confidence Interval
7/27/2019 061 Intro to Prediction
12/1512
90%
Confidence level
To maintain a certain level of confidence, a larger
standard deviation requires a larger confidence interval.To maintain a certain level of confidence, a larger
standard deviation requires a larger confidence interval.
n)645.1(2
nz2 05.
=
/2 = .05/2 = .05
n
5.1)645.1(2
n
5.1z2
05.
=
Suppose the standard
deviation has increased
by 50%.
The Affects of on the interval width
n)96.1(2
nz2
025.
=
/2 = 2.5%/2 = 2.5%
/2 = 5%/2 = 5%
n)645.1(2
nz2 05.
=
Confidence level
90%95%
Let us increase the
confidence level
from 90% to 95%.
Larger confidence level produces a wider confidence intervalLarger confidence level produces a wider confidence interval
The Affects of Changing the ConfidenceLevel
7/27/2019 061 Intro to Prediction
13/1513
90%
Confidence level
n)645.1(2
nz2 05.
=
Increasing the sample size decreases the width of theconfidence interval while the confidence level can remain
unchanged.
Increasing the sample size decreases the width of theconfidence interval while the confidence level can remain
unchanged.
The Affects of Changing the SampleSize
Selecting the Sample size
We can control the width of the confidenceinterval by changing the sample size.
Thus, we determine the interval width first, andderive the required sample size.
The phrase estimate the mean to within Wunits, translates to an interval estimate of the
form
wx
7/27/2019 061 Intro to Prediction
14/1514
The required sample size to estimate themean is
2
2
w
zn
=
Selecting the Sample size
Example
To estimate the amount of lumber thatcan be harvested in a tract of land, themean diameter of trees in the tractmust be estimated to within one inchwith 99% confidence.
What sample size should be taken?Assume that diameters are normallydistributed with = 6 inches.
Selecting the Sample size
7/27/2019 061 Intro to Prediction
15/15
Solution
The estimate accuracy is +/-1 inch. That isw = 1.
The confidence level 99% leads to = .01,thus z/2 = z.005 = 2.575.
We compute
2391
)6(575.2
w
zn
22
2=
=
=
If the standard deviation is really 6 inches,
the interval resulting from the random sampling
will be of the form . If the standard deviationis greater than 6 inches the actual interval will
be wider than +/-1.
If the standard deviation is really 6 inches,
the interval resulting from the random sampling
will be of the form . If the standard deviation
is greater than 6 inches the actual interval will
be wider than +/-1.
1x
Selecting the Sample size