STATISTICAL INFERENCEPART V

CONFIDENCE INTERVALS

1

2

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.

3

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

5

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

6

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 .

7

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

8

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

22

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 .

9

2

11

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.

10

n

1iiXS

11

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.

12

EXAMPLE

• Let X1,X2,…,Xn be a r.s. of Exp(), >0. Find a 100 % CI for . Interpret the result.

13

EXAMPLE

• Let X1,X2,…,Xn be a r.s. of N(,2). Find a 100 % CI for and 2 . Interpret the results.

14

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

15

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

16

x

nz2 2

nzx 2

n

zx 2

Lower confidence limit Upper confidence limit

1 -

Confidence level

Graphical Demonstration of the Confidence Interval for

17

Inference About the Population Mean when is Unknown

• The Student t Distribution

Standard Normal

Student t

0

18

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.

19

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

20

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

21

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

22

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