# Outline Chapter 6 - PowerPoint PPT Presentation

1 / 70

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Outline Chapter 6

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

Schaum’s OutlinePROBABILTY and STATISTICS Chapter 6 ESTIMATION THEORYPresented by Professor Carol DahlExamples from D. Salvitti J. Mazumdar C. Valencia

### Outline Chapter 6

• Trader in energy stocks

• 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<Zc) = 0.975 = (1 – 0.05/2)

• 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<Zc) = 0.975 = (1 – 0.05/2)

• interval estimate – (1-) %

• X +/-Zc

• (Z<Zc) = (1 – /2)

• % of intervals don’t contain

• (1- )% of intervals do contain

(Z<Zc) = 0.975 = (1 – 0.05/2)

### 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 Intervals for MeansFinite Population (N) no Replacement

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

Confidence Intervals of Meanst- distribution

=

=

=

=

Confidence Intervals of MeansNormal compared to t- distribution

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

### Confidence Intervals Differences and SumsKnown Variances

Samples are independent

### Confidence Interval for Differences and Sums – Known Variance

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

### Confidence Intervals for Differences and Sums – Large Samples

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

F-Distribution

df1

df2

F-Distribution

### Confidence Intervals for Ratio of Variances

Need statistic with

population parameter 21/ 22

estimate for population parameter ŝ21/ ŝ22

its distribution - F

Rearrange

### 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

• Variances

### Sum Up Chapter 6

• Variances ratios

### Sum Up Chapter 6

• Maximum Likelihood Estimators

• Pick  which maximizes the function