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PARAMETER
ESTIMATION
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Parameters Populations are described by their
probability distributions and/or parameters.
For quantitative populations, the locationand shape are described by mand s.
Binomial populations are determined bya single parameter,p.
If the values of parameters are unknown,we make inferences about them usingsample information.
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Statistical inference is the process bywhich we acquire information aboutpopulations from samples.
There are two types of inference:
Estimation
Hypotheses testing
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Types of Inference
Estimation
Estimating or predicting the value of theparameter
What is (are) the most likely values of
m orp? Hypothesis Testing
Deciding about the value of a parameter
based on some preconceived idea. Did the sample come from a population
with m = 5 orp= .2?
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Types of Inference
Examples:
A consumer wants to estimate the averageprice of similar homes in her city beforeputting her home on the market.
Estimation: Estimate m, the average home price.
Hypothesis test: Is the new average resistance, mN greater thanthe old average resistance, mO?
A manufacturer wants to know if a new typeof steel is more resistant to high temperatures
than an old type was.
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Types of Inference
Whether you are estimating parameters ortesting hypotheses, statistical methods
are important because they provide: Methods for making the inference
A numerical measure of the goodness or
reliability of the inference
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Concepts of Estimation
The objective of estimation is to determinethe value of a population parameter on thebasis of a sample statistic.
There are two types of estimators: Point Estimator Interval estimator
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Definitions
An estimatoris a rule, usually a formula, thattells you how to calculate the estimate based onthe sample.
Point estimation:A single number iscalculated to estimate the parameter.
Interval estimation/Confidence Interval:Two numbers are calculated to create an
interval within which the parameter isexpected to lie. It is constructed so that, witha chosen degree of confidence, the trueunknown parameter will be captured inside
the interval.
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Point Estimator
A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single value
or point.
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Point Estimator of Population Mean
A sample of weights of 34 male freshman students was obtained.
185 161 174 175 202 178 202 139 177170 151 176 197 214 283 184 189 168
188 170 207 180 167 177 166 231 176
184 179 155 148 180 194 176If one wanted to estimate the true mean of all male freshman students, you might
use the sample mean as a point estimate for the true mean.
sample mean x 182.44= =
n
xxi=
An point estimate of population mean, m, is the
sample mean
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Point Estimation of Population
Proportion
A sample of 200 students at a large university isselected to estimate the proportion of studentsthat wear contact lens. In this sample 47 wearcontact lens.
An point estimate of population mean, p, is thesample proportion where x is the numberof successes in the sample.
nxp /=
235.200/47 ==p
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Population distribution
Point Estimator
Parameter
?
Sampling distribution
A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single value
or point.
Point estimator
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Properties of
Point Estimators
Since an estimator is calculated from sample values, itvaries from sample to sample according to itssampling distribution.
An estimatoris unbiased if the mean of its sampling
distribution equals the parameter of interest. It doesnot systematically overestimate or underestimate thetarget parameter.
Both sample mean and sample proportion areunbiased estimators of population mean andproportion. The following sample variance is anunbiased estimator of population variance.
1
)( 22
=
n
xxs
i
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Properties of
Point EstimatorsOf all the unbiased estimators, we preferthe estimator whose sampling distributionhas the smallest spread orvariability.
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An interval estimator draws inferencesabout a population by estimating the valueof an unknown parameter using an
interval.
Interval estimator
Population distribution
Sample distribution
Parameter
Interval Estimator
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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 and
the parameter grows smaller as the sample sizeincreases.
Relative eff iciency :For two unbiased estimators, the onewith a smaller variance is said to be relatively efficient.
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Uncertainty Analysis The estimate of the error is called the uncertainty.
It includes both bias and precision errors. We need to identify all the potential significant errors for
the instrument(s). All measurements should be given in three parts
Mean value
Uncertainty Confidence interval on which that uncertainty is based
(typically 95% C.I.) Uncertainty can be expressed in either
absolute terms (i.e., 5 Volts 0.5 Volts)or in percentage terms (i.e., 5 Volts 10%) We will use a confidence interval throughout this course.
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Estimating the Population Mean
when the Population Variance is
Known How is an interval estimator produced froma sampling distribution?
A sample of size n is drawn from thepopulation, and its mean is calculated.
By the central limit theorem, isnormally distributed (or approximatelynormally distributed.), thus
X
X
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n
x
Z s
m
=
We have established before that
=s
ms
m 1)n
zxn
z(P 22
Estimating the Population Mean when
the Population Variance is Known
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=sms 1)n
zxn
zx(P 22
This leads to the following equivalent statement
The Confidence Interval form( s is known)
The confidence interval
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CONFIDENCE INTERVAL FOR
KNOWN
Confidence Level :1 - = The probabilitythat a stated interval contains the unknownparameter.
m
s
/ 2X zn
s
/ 2X z
n
s
/ 2X z
n
s
Upper Confidence Limit
(UCL)
Lower Confidence Limit
(LCL)
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Interpreting the Confidence
Interval form1 of all the values of obtained in repeated
sampling from a given distribution, construct an interval
that includes (covers) the expected value of the
population.
x
s
s
n
zx,
n
zx 22
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x
nz2 2
s
n
zx 2s
n
zx 2s
Lower confidence limit Upper confidence limit
1 - Confidence level
Graphical Demonstration of the
Confidence Interval form
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The Confidence Interval form( s is known)
Four commonly used confidence levels
Confidence
level /20.90 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
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Example: Estimate the mean value of the distribution resulting from
the throw of a fair die. It is known that s = 1.71. Use a 90%
confidence level, and 100 repeated throws of the die.
Solution: The confidence interval is
The Confidence Interval form( s is known)
=s
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.
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The Confidence Interval form( s is known)
Recalculate the confidence interval for 95% confidence level.
Solution: =s
n
zx 2 34.x
100
71.196.1x =
34.x 34.x
.95
.90
28.x 28.x
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The Confidence Interval form( s is known)
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 ofm.
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Example Doll Computer Company delivers computers
directly to its customers who order via theInternet.
To reduce inventory costs in its warehouses Doll,employs an inventory model, that requires theestimate of the mean demand during lead time.
It is found that lead time demand is normallydistributed with a standard deviation of 75computers per lead time.
Estimate the lead time demand with 95%
confidence.
The Confidence Interval form( s is known)
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Example Solution
The parameter to be estimated is m, the meandemand during lead time.
We need to compute the interval estimation form.
From the data provided in the data file, thesample mean is
The Confidence Interval form( s is known)
.16.370=x
56.399,76.34040.2916.37025
7596.116.370
25
75z16.370
nzx
025.2
===
=s
Since 1 - =.95, = .05.
Thus /2 = .025. Z.025 = 1.96
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Example (MINITAB Output)
Stat > Basic Statistics > 1-Sample Z
Z Confidence Intervals
The assumed sigma = 75.0
Variable N Mean StDev SE Mean 95.0 % CI
Demand 25 370.2 80.8 15.0 (340.8, 399.6)
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Example
To help make a decision about expansionplans, the president of a music companyneeds to know how many CDs teenagersbuy annually. 250 teenagers are selected;
each reports the number of CDs boughtwithin the last 12 months. The responseshave a sample mean of 4.26. The (population)standard deviation is 3. Construct 99, 95, 80
percent confidence intervals for the trueaverage number of CDs bought by teenagersin a year.
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ANSWER
99% CI:
95% CI: 80% CI:
. .z z
X Xn n
. .. .
. . %
005 005
2 57 3 2 57 34 26 4 26
250 250
3 771 4 749 99
s s m = m = m =
= 95%3.89 4.63m = 80%4.015 4.501m
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INTERPRETATION
3.8 as a guess regarding the average
number of CDs bought by teenagers is:
a) reasonable corresponding to 99%confidence
b) unreasonable corresponding to 95%confidence, and
c) unreasonable corresponding to 80%confidence.
I f ti d th Width f th
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Wide interval estimator provides littleinformation.
Where is m ????????????????
Information and the Width of the
Interval
I f ti d th Width f th
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Here is a much narrower interval.
If the confidence level remains
unchanged, the narrower interval
provides more meaningfulinformation.
Wide interval estimator provides littleinformation.Where is m ?
Ahaaa!
Information and the Width of the
Interval
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The width of the confidence interval isaffected by
the population standard deviation (s) the confidence level (1-) the sample size (n).
The Width of the Confidence Interval
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90%
Confidence level
To maintain a certain level of confidence, a larger
standard deviation requires a larger confidence interval.
n)645.1(2
nz2 05.
s=
s
/2 = .05/2 = .05
n
5.1
)645.1(2n
5.1
z2 05.s
=
s
Suppose the standard
deviation has increased
by 50%.
The Affects ofs on the interval width
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n)96.1(2
nz2
025.
s=
s
/2 = 2.5%/2 = 2.5%
/2 = 5%/2 = 5%
n
)645.1(2
n
z2 05.s
=s
Confidence level90%95%
Let us increase the
confidence level
from 90% to 95%.
Larger confidence level produces a wider confidence interval
The Affects of Changing the
Confidence Level
Th Aff t f Ch i th S l
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90%
Confidence level
n)645.1(2
nz2 05.
s=
s
Increasing the sample size decreases the width of the
confidence interval while the confidence level can remain
unchanged.
The Affects of Changing the Sample
Size
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The required sample size to estimate themean is
22
w
zn
s=
Selecting the Sample size
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Example To estimate the amount of lumber that
can be harvested in a tract of land, the
mean 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 s = 6 inches.
Selecting the Sample size
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Solution The estimate accuracy is +/-1 inch. That is w = 1.
The confidence level 99% leads to = .01, thusz
/2= z
.005= 2.575.
We compute239
1
)6(575.2
w
zn
22
2 =
=
s=
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
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EXAMPLE
A sample survey of kindergarten children in NYcity is being planned to estimate, among theother things, the mean number of older siblingsof such children. It is desired to estimate this
mean within , with a 90% confidencelevel. A reasonable value for is 0.6.
a) What sample size is needed to estimate themean number of older siblings of such children?
w=0.08, s = 0.6,
0.08 s
0.05z 1.645=2
1.645* 0.6n 152.2 153
0.08
= =
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EXAMPLE(contd.)
b) If w=0.06, how will n change?
2
1.645* 0.6n 270.6 2710.06
= =
So, when w=0.08 0.06, the required sample
size increases substantially.
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Inference about the Population
Mean when the s is unknown When s is unknown, we need to estimate it to
construct confidence intervals for the population
mean s is estimated by s.
When s is unknown, the sample mean of a smallsample is no longer normally distributed. That is,we use t-scores (i.e., scores from a tdistribution) in place of z-scores.
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Inference About the PopulationMean when s is Unknown
The Student t Distribution
Standard Normal
Student t
0
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Effect of the Degrees of Freedom
on the t Density Function
Student t with 10 DF
0
Student t with 2 DF
Student t with 30 DF
The degrees of freedom, (a function of the sample size)
determine how spread the distribution is compared to the
normal distribution.
Finding t scores Under a
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Finding t-scores Under at-Distribution (t-tables)
t.100 t.05 t.025 t.01 t.005Degrees of
Freedom
1
2
3
45
6
7
8
910
11
12
3.078
1.886
1.638
1.5331.476
1.440
1.415
1.397
1.3831.372
1.363
1.356
6.314
2.920
2.353
2.1322.015
1.943
1.895
1.860
1.8331.812
1.796
1.782
12.706
4.303
3.182
2.7762.571
2.447
2.365
2.306
2.2622.228
2.201
2.179
31.821
6.965
4.541
3.7473.365
3.143
2.998
2.896
2.8212.764
2.718
2.681
63.657
9.925
5.841
4.6044.032
3.707
3.499
3.355
3.2503.169
3.106
3.055
t0.05, 10=1.812
1.812
.05
t0
C fid I t l
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Confidence Intervals on m(s unknown)
A (1100% confidence interval on m(when s is unknown) based on
samples of size n drawn from aNormal Population is given by:
/2, n-1 /2x t , (t with n-1 d.f.)
s
n
EXAMPLE
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EXAMPLE A new breakfast cereal is test-marked for
1 month at stores of a large supermarketchain. The result for a sample of16 storesindicate average sales of $1200 with a
sample standard deviation of $180. Set up99% confidence interval estimate of thetrue average sales of this new breakfast
cereal. Assume normality.
/ 2 ,n 1 0.005,15
n 16 ,x $1200,s $180, 0.01
t t 2.947
= = = =
= =
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ANSWER
99% CI form:
(1067.3985, 1332.6015)
With 99% confidence, the limits 1067.3985
and 1332.6015 cover the true averagesales of the new breakfast cereal.
/ 2 ,n 1
s 180x t 1200 2.947 1200 132.6015
n 16
= =
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Checking the required conditions
We need to check that the population isnormally distributed, or at least not extremelynonnormal.
There are statistical methods to test fornormality
From the sample histograms we see
0
5
10
15
20
25
30
-4 2 8 14 22 30 More
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Example
A random sample of n=8 E-glass fiber testspecimens of a certain type yielded asample mean interfacial shear yield
stress of 30.2 and a sample standarddeviation of 3.1. Assuming that interfacialshear yield stress is normally distributed,compute a 95% CI for true average stress.
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One Sided Confidence Intervals
An upper confidence bound form:
, 1n
s
x t nm
A lower confidence bound form:
, 1n
sx t
n
m
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Example
A sample of 14 joint specimens of aparticular type gave a sample meanproportional limit stress of 8.48 MPaand a sample standard deviation of 0.79MPa. Calculate and interpret a 95% lowerconfidence bound for the true averageproportional limit stress of all such joints.
What, if any, assumptions did you makeabout the distribution of proportional limitstress?
Estimation of a Population
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Estimation of a PopulationProportion
When the population consists of nominaldata, the only inference we can make isabout the proportion of occurrence of a
certain value. The parameter p was used before to
calculate these probabilities under the
binomial distribution.
Inference About a Population
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Inference About a PopulationProportion
Statistic and sampling distribution
the statistic used when making inference about p is:
Under certain conditions, [np > 5 and n(1-p) > 5],is approximately normally distributed, withm = p and s2= p(1 - p)/n.
.
.
xp where
nx the number of successes
n sample size
=
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Estimating the Proportion
Interval estimator for p (1- confidence level)
100(1-% Confidence Interval for p:
5)p1(nand5pnprov ided
n/)p1(pzp2/
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Nielsen Ratings
In a survey of2000 TV viewers at 11.40 p.m. on acertain night, 226indicated they watched TheTonight Show.
Estimate the number of TVs tuned to the Tonight
Show in a typical night, if there are 100 millionpotential television sets. Use a 95% confidencelevel.
Solution:
x=226, n=2000
014.113.
2000/)113.1(113.96.1113./)1( 2/
= nppzp
226 0.1132000
xpn = = =
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Solution
z - Estimate: ProportionViewers
Sample Proportion 0.113
Observations 2000
LCL 0.099
UCL 0.127
A confidence interval estimate of the
number of viewers who watched the
Tonight Show:
LCL = .099(100 million)= 9.9 millionUCL = .127(100 million)=12.7 million
Selecting the Sample Size to
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Selecting the Sample Size toEstimate the Proportion
Recall: The confidence interval for theproportion is
Thus, to estimate the proportion to within W,we can write
nppzp /)1( 2/
nppzW /)1(2/ =
Selecting the Sample Size to
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Selecting the Sample Size toEstimate the Proportion
The required sample size is
2
2/ )1(
=W
ppzn
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Example
Suppose we want to estimate the proportion ofcustomers who prefer our companys brand to
within .03 with 95% confidence. Find the sample size.
Solution:W = .03; 1 - = .95,
therefore /2 = .025,
so z.025 = 1.96
2
03.
)p1(p96.1n
=
Since the sample has not yetbeen taken, the sample proportion
is still unknown.
We proceed using either one of the
following two methods:
Sample Size to Estimate the
Proportion
Sample Size to Estimate the
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Method 1: There is no knowledge about the value of
Let . This results in the largest possible n needed fora 1- confidence interval of the form .
If the sample proportion does not equal .5, the actual W will
be narrower than .03 with the n obtained by the formulabelow.
5.p =03.p
p
068,103.
)5.1(5.96.1n
2
=
=
68303.
)2.1(2.96.1n
2
=
=
Sample Size to Estimate theProportion
Method 2: There is some idea about the value of
Use the value of to calculate the sample sizep
p