Outline Chapter 6

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# Outline Chapter 6 - PowerPoint PPT Presentation

Schaum’s Outline PROBABILTY and STATISTICS Chapter 6 ESTIMATION THEORY Presented by Professor Carol Dahl Examples from D. Salvitti J. Mazumdar C. Valencia. Outline Chapter 6. Trader in energy stocks random variable Y = value of share want estimates µ y , σ Y

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Schaum’s OutlinePROBABILTY and STATISTICS Chapter 6 ESTIMATION THEORYPresented by Professor Carol DahlExamples from D. Salvitti J. Mazumdar C. Valencia
Outline Chapter 6
• random variable Y = value of share
• want estimates µy, σY
• Y = ß0 + ß1X+ 
• want estimates Ŷ, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates
Outline Chapter 6
• Types of estimators
• Point estimates
• µ = 7
• Interval estimates
• µ = 7+/-2
• confidence interval
Outline Chapter 6
• Population parameters and confidence intervals Means

Large sample sizes

Small sample sizes

• Proportions
• Differences and Sums
• Variances
• Variances ratios
Properties of Estimators - Unbiased

Unbiased Estimator of Population Parameter

estimator expected value = to population parameter

Unbiased Estimates

Population Parameters:

Sample Parameters:

are unbiased estimates

Expected value of standard deviation not unbiased

Properties of Estimators - Efficient
• Efficient Estimator –

if distributions of two statistics same

more efficient estimator = smaller variance

efficient = smallest variance of all unbiased estimators

Unbiased and Efficient Estimates

Target

Estimates which are efficient and unbiased

Not always possible

often us biased and inefficient

easy to obtain

Types of Estimates for Population Parameter

Point Estimate

single number

Interval Estimate

between two numbers.

Estimates of Mean – Known VarianceLarge Sample or Finite with Replacement
• X = value of share
• sample mean is \$32
• volatility is known σ2 = \$4.00
• confidence interval for share value
• Need
• estimator for mean
• need statistic with
• mean of population
• estimator
Estimates of Mean- Sampling Statistic
• P(-1.96 < <1.96) = 95%

2.5%

Confidence Interval for Mean
• P(-1.96 < <1.96) = 95%
• P(-1.96 < <1.96 ) = 95%
• P(-1.96 -X < -µ <1.96 - X ) = 95%
• Change direction of inequality
• P(+1.96 +X > µ > -1.96 + X ) = 95%
Confidence Interval for Mean
• P(+1.96 +X > µ > -1.96 + X ) = 95%
• Rearrange
• P(X - 1.96 < µ <X + 1.96 ) = 95%
• Plug in sample values and drop probabilities
• X = value of share, sample = 64
• sample mean is \$32
• volatility is σ2 = \$4
• {32 – 1.96*2/64, 32 + 1.96*2/64} = {31.51,32.49}
Estimates for Mean for Normal
• Take a sample
• point estimate
• compute sample mean
• interval estimate – 0.95 (95%+) = (1 - 0.05)
• X +/-1.96
• X +/-Zc
• (Z
• 95% of intervals contain
• 5% of intervals do not contain
Estimates for Mean for Normal
• interval estimate – 0.95 (95%+) = (1 - 0.05)
• X +/-Zc
• (Z
• interval estimate – (1-) %
•  X +/-Zc
• (Z
• % of intervals don’t contain
• (1- )% of intervals do contain

(Z

Confidence Interval Estimatesof Population Parameters

Common values for corresponding to various confidence levels used in practice are:

Confidence Interval Estimatesof Population Parameters

Functions in EXCEL

Menu Click on Insert Function or

=confidence(,stdev,n)

=confidence(0.05,2,64)= 0.49

X+/-confidence(0.05,2,64)

=normsinv(1-/2) gives Zc value

X+/-normsinv(1-/2)

32 +/- 1.96*2/64

Confidence interval

Confidence level

Example: Finite Population without replacement

Evaluate density of oil in new reservoir

81 samples of oil (n)

from population of 500 different wells

samples density average is 29°API

standard deviation is known to be 9 °API

 = 0.05

Confidence Intervals for MeansFinite Population (N) no ReplacementKnown Variance

X = 29 , N= 500, n = 81 , σ = 9 ,  = 0.05

Zc = 1.96

But don’t know Variance

t-Distribution

=N(0.1)

=

= tdf

2/df

df

t distribution

Normal

X +/-Zc

X +/-tc

Confidence Interval Unknown Variance

Example:

Eight independent measurements diameter of drill bit

3.236, 3.223, 3.242, 3.244, 3.228, 3.253, 3.253, 3.230

99% confidence interval for diameter of drill bit

X +/-tc

Confidence Intervals for MeansUnknown Variance

X +/-tc

X = ΣXi/n

3.236+3.223+3.242+3.244+3.228+3.253+3.253+3.230

8

X = 3.239

ŝ2 = Σ(Xi - X) = (3.236- X)2 + . . .(3.230 - X)2

(n-1) (8-1)

ŝ = 0.0113

Confidence Intervals for MeansUnknown Variance

X +/-tc

• X = 3.239, n = 8, ŝ= 0.0113, =0.01,
• 1- /2=0.995
• From the t-table with 7 degrees of freedom, we find tc= t7,0.995=3.50

1-/2=.975

.005%

-tc

tc

Find tc from Table of Excel

Confidence Intervals for MeansUnknown Variance

1-/2=.975

/2= 0.005%

Depends on Table

-tc

tc

GHJ /2 = 0.005  tc = 2. 499

Schaums 1- /2 = 0.995  tc = 2.35

Excel =tinv(0.01,7) = 3.499483

Confidence Intervals for MeansUnknown Variance

X +/-tc

X = 3.239, n = 8, ŝ= 0.0113, =0.01,

Confidence Intervals for Proportions

Example

600 engineers surveyed

250 in favor of drilling a second exploratory well

95% confidence interval for

proportion in favor of drilling the second well

Approximate by Normal in large samples

Solution: n=600, X=250 (successes), = 0.05

zc = 1.96 and

Confidence Intervals for Proportions

Example

600 engineers surveyed

250 in favor of drilling a second exploratory well.

95% confidence interval for

proportion in favor of drilling the second well

Approximate by Normal in large samples

Solution: n=600, X=250 (successes), = 0.05

zc = 1.96 and

Confidence Intervals for Proportions

sampling from large population

or finite onewith replacement

Example

sample of 200 steel milling balls

average life of 350 days - standard deviation 25 days

new model strengthened with molybdenum

sample of 150 steel balls

average life of 250 days - standard deviation 50 days

samples independent

Find 95% confidence interval for difference μ1-μ2

Confidence Intervals for Differences and Sums

Example

Solution: X1=350, σ1=25, n1=200, X2=250, σ2=50, n2=150

Where:

P1, P2 two sample proportions,

n1, n2 sizes of two samples

Confidence Intervals for Differences and Sums

Example

random samples

200 drilled holes in mine 1, 150 found minerals

300 drilled holes in mine 2, 100 found minerals c

Construct 95% confidence interval difference in proportions

Solution: P1=150/200=0.75, n1=200, P2=100/300=0.33,n2=300

With 95% of confidence the difference of proportions {0.42, 0.08}

Confidence Intervals Differences and Sums

Example

Solution: P1=150/200=0.75, n1=200,

P2=100/300=0.33, n2=300

95% of confidence the difference of proportions

[0.08,0.42]

Confidence Intervals for Variances

Need statistic with

population parameter 2

estimate for population parameter ŝ2

its distribution - 2

Confidence Intervals for Variances

has a chi-squared distribution

n-1 degrees of freedom.

Find interval such that σ lies in the interval for

95% of samples

95% confidence interval

Confidence Intervals for Variances

Rearrange

Take square root if want confidence interval for

standard deviation

Confidence Intervals for Variances and Standard Deviations

Drop probabilities when substitute in sample values

1 -  confidence interval for variance

1 -  confidence interval for standard deviation

Confidence Intervals for Variance

Example

Variance of amount of copper reserves

16 estimates chosen at random

ŝ2 = 2.4 thousand million tons

Find 99% confidence interval variance

Solution: ŝ2=2.4, n=16,

degrees of freedom = 16-1= 15

How to get 2Critical Values

Not symmetric

/2

/2

2 lower 2 upper

How to get 2Critical Values

1-/2

Not symmetric

1-/2

/2

/2

GHJ area above 20.995, 20.005 4.60092, 32.8013

Schaums area below 20.005, 20.995 4.60, 32.8

Excel = chiinv(0.995,15) = 4.60091559877155

Excel = chiinv(0.005,15) = 32.8013206461633

Confidence Intervals for Variances and Standard Deviations

Example

99% confidence interval variance of reserves

Solution: ŝ=2.4 (n-1)=15

2lower = 4.60, 2upper = 32.8

Confidence Intervals for Ratio of Variances

Two independent random samples

size m and n

population variances

estimated variances ŝ21, ŝ22

• interested in whether variances are the same
• 21/ 22
Confidence Intervals for Ratio of Variances

Need statistic with

population parameter 21/ 22

estimate for population parameter ŝ21/ ŝ22

its distribution - F

Confidence Intervals for Ratio of Variances

Need statistic with

population parameter 21/ 22

estimate for population parameter ŝ21/ ŝ22

its distribution - F

Confidence Intervals for Ratio of Variances

Put smallest first, largest second

When substitute in values drop probabilities

1- confidence interval for 21/ 22

Confidence Intervals for Variances

Example

Two nickel ore samples

of sizes 16 and 10

unbiased estimates of variances 24 and 18

Find 90% confidence limits for ratio of variances

Solution: ŝ21 = 24, n1 = 16, ŝ22 = 18, n2 = 10,

Confidence Intervals for Ratio of Variances

/2

/2

Tablesdf1df2 

F upper

F lower

GHJ area above F0.95,15,9, F0.05,15,9 ?3.01 Schaums area below F0.05,15,9, F0.95,15,9 ?3.01

Area above

Excel = Finv(0.95,15,9) = 0.386454546279388

Excel = Finv(0.05,15,9) = 3.00610197251669

Confidence Intervals for Ratio of Variances

GHJ area above F0.95,15,9

P(F15,9>Fc) = 0.95

P(1/F15,9<1/Fc) = 0.95

But 1/F15,9 = F9,15

P(F9,15<1/Fc) = 0.95

P(F9,15<1/Fc) = 0.05

1/Fc = 2.59 Fc = 0.3861

/2

/2

F upper

F lower

Confidence Intervals for Variances

Example

Two nickel ore samples

Solution: ŝ21 = 24, n1 = 16, ŝ22 = 18, n2 = 10,

Maximum Likelihood Estimates

Point Estimates

x is population with density function f(x,)

if know  - know the density function

2 where  = degrees of freedom

Poisson λxe-λ/x!  = λ (the mean)

If sample independently from f n times

x1, x2, . . .xn

a sample

if consider all possible samples of n

a sampling distribution

Maximum Likelihood Estimates

If sample independently from f n times

x1, x2, . . .xn

a sample

if consider all possible samples of n

a sampling distribution

called likelihood function

Maximum Likelihood Estimates

 which maximizes the likelihood function

Derivative of L with respect to  and setting it to 0

Solve for 

Usually easier to take logs first

log(L) = log(f(x1,) + log(f(x2,)+ . . .+ log(f(xn,)

Maximum Likelihood Estimates

log(L) = log(f(x1,) + log(f(x2,) +. . .+ log(f(xn,)

   

Solution of this equation is maximum likelihood estimator

work out example 6.25

work out example 6.26

Sum Up Chapter 6
• Y = ß0 + ß1X
• Ŷ, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates
• Types of estimators
• Point estimates
• Interval estimates
Sum Up Chapter 6
• Y- µY, Y, Y, ŝ2
• In 590-690
• Y = ß0 + ß1X
• Ŷ, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates
• Types of estimators
• Point estimates
• Interval estimates
Sum Up Chapter 6
• Need statistic with
• population parameter
• estimate for population parameter
• its distribution
Sum Up Chapter 6
• Population parameters and confidence intervals
• Mean – Normal

Know variance and population normal

Large sample size can use estimated variance

Sum Up Chapter 6

Proportions

• large sample approximate by normal
• Differences of means (known variance)
Sum Up Chapter 6
• Mean
• population normal - unknown variance
Sum Up Chapter 6
• Variances ratios
Sum Up Chapter 6
• Maximum Likelihood Estimators
• Pick  which maximizes the function