STATISTICAL INFERENCEPART V
CONFIDENCE INTERVALS
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INTERVAL ESTIMATION
• Point estimation of : The inference is a guess of a single value as the value of . No accuracy associated with it.
• Interval estimation for : Specify an interval in which the unknown parameter, , is likely to lie. It contains measure of accuracy through variance.
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INTERVAL ESTIMATION
• An interval with random end points is called a random interval. E.g.,
5 5,
8 3
X X
is a random interval that contains the true value of with probability 0.95.
5 5Pr 0.95
8 3
X X
Interpretation ( )
( )( )
( ) ( ) ( ) ( )
( ) ( )
( ) μ (unknown, but true value)
490% CI Expect 9 out of 10 intervals to cover the true μ
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INTERVAL ESTIMATION• An interval (l(x1,x2,…,xn), u(x1,x2,…,xn)) is
called a 100 % confidence interval (CI) for if
where 0<<1.• The observed values l(x1,x2,…,xn) is a lower
confidence limit and u(x1,x2,…,xn) is an upper confidence limit. The probability is called the confidence coefficient or the confidence level.
1 2 1 2Pr , , , , , ,n nl x x x u x x x
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INTERVAL ESTIMATION
• If Pr(l(x1,x2,…,xn))= , then l(x1,x2,…,xn) is called a one-sided lower 100 % confidence limit for .
• If Pr( u(x1,x2,…,xn))= , then u(x1,x2,…,xn) is called a one-sided upper 100 % confidence limit for .
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METHODS OF FINDING PIVOTAL QUANTITIES
• PIVOTAL QUANTITY METHOD: If Q=q(x1,x2,…,xn) is a r.v. that is a function of
only X1,…,Xn and , and if its distribution does not depend on or any other unknown parameter (nuisance parameters), then Q is called a pivotal quantity.
nuisance parameters: parameters that are not of direct interest
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PIVOTAL QUANTITY METHODTheorem: Let X1,X2,…,Xn be a r.s. from a
distribution with pdf f(x;) for and assume that an MLE (or ss) of exists:
• If is a location parameter, then Q= is a pivotal quantity.
• If is a scale parameter, then Q= / is a pivotal quantity.
• If 1 and 2 are location and scale parameters respectively, then
1 1 2
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ˆ ˆ and
ˆ
are PQs for 1 and 2.
Note
• Example: If 1 and 2 are location and scale parameters respectively, then
is NOT a pivotal quantity for 1
because it is a function of 2
A pivotal quantity for 1 should be a function of only 1 and X’s, and its distribution should be free of 1 and 2 .
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2
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Example
• X1,…,Xn be a r.s. from Exp(θ). Then,is SS for θ, and θ is a scale
parameter.S/θ is a pivotal quantity. So is 2S/θ, and using this might be more
convenient since this has a distribution of χ²(2n) which has tabulated percentiles.
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n
1iiXS
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CONSTRUCTION OF CI USING PIVOTAL QUANTITIES
• If Q is a PQ for a parameter and if percentiles of Q say q1 and q2 are available such that
Pr{q1 Q q2}=,
Then for an observed sample x1,x2,…,xn; a 100% confidence region for is the set of that satisfy q1 q(x1,x2,…,xn;)q2.
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EXAMPLE
• Let X1,X2,…,Xn be a r.s. of Exp(), >0. Find a 100 % CI for . Interpret the result.
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EXAMPLE
• Let X1,X2,…,Xn be a r.s. of N(,2). Find a 100 % CI for and 2 . Interpret the results.
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APPROXIMATE CI USING CLT• Let X1,X2,…,Xn be a r.s.
• By CLT,
0,1/
dX E X X
NnV X
The approximate 100(1−)% random interval for μ:
/2 /2 1P X z X zn n
The approximate 100(1 −)% CI for μ:
/2 /2x z x zn n
Non-normal random sample
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APPROXIMATE CI USING CLT
• Usually, is unknown. So, the approximate 100(1)% CI for :
/2, 1 /2, 1n n
s sx t x t
n n
•When the sample size n ≥ 30, t/2,n-1~N(0,1).
/2 /2
s sx z x z
n n
Non-normal random sample
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x
nz2 2
nzx 2
n
zx 2
Lower confidence limit Upper confidence limit
1 -
Confidence level
Graphical Demonstration of the Confidence Interval for
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Inference About the Population Mean when 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 DFStudent t with 30 DF
The “degrees of freedom” (a function of the sample size) determines how spread the distribution is compared to the normal distribution.
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Finding t-scores Under a t-Distribution (t-tables)
t.100 t.05 t.025 t.01 t.005
Degrees of Freedom
123456789101112
3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.356
6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.782
12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.179
31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.681
63.6579.9255.8414.6044.0323.7073.4993.3553.2503.1693.1063.055
t0.05, 10 = 1.812
1.812
.05
t0
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EXAMPLE• A new breakfast cereal is test-marked for 1
month at stores of a large supermarket chain. The result for a sample of 16 stores indicate average sales of $1200 with a sample standard deviation of $180. Set up 99% confidence interval estimate of the true 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 for :
(1067.3985, 1332.6015)With 99% confidence, the limits 1067.3985 and
1332.6015 cover the true average sales 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 is normally distributed, or at least not extremely nonnormal.
• Look at the sample histograms, Q-Q plots …• There are statistical methods to test for
normality