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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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Outline chapter 6

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


Outline chapter 6

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 61

Outline Chapter 6

  • Types of estimators

  • Point estimates

  • µ = 7

  • Interval estimates

  • µ = 7+/-2

  • confidence interval


Outline chapter 62

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

Properties of Estimators - Unbiased

Unbiased Estimator of Population Parameter

estimator expected value = to population parameter


Unbiased estimates

Unbiased Estimates

Population Parameters:

Sample Parameters:

are unbiased estimates

Expected value of standard deviation not unbiased


Properties of estimators efficient

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

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

Types of Estimates for Population Parameter

Point Estimate

single number

Interval Estimate

between two numbers.


Estimates of mean known variance large sample or finite with replacement

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

Estimates of Mean- Sampling Statistic

  • P(-1.96 < <1.96) = 95%

2.5%


Confidence interval for mean

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 mean1

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

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 normal1

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 estimates of population parameters

Confidence Interval Estimatesof Population Parameters

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


Confidence interval estimates of population parameters1

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 means finite population n no replacement

Confidence Intervals for MeansFinite Population (N) no Replacement

Confidence interval

Confidence level


Example finite population without replacement

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 means finite population n no replacement known variance

Confidence Intervals for MeansFinite Population (N) no ReplacementKnown Variance

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

Zc = 1.96


Outline chapter 6

But don’t know Variance

t-Distribution

=N(0.1)

=

= tdf

2/df

df


Outline chapter 6

Confidence Intervals of Meanst- distribution

=

=

=

=


Outline chapter 6

Confidence Intervals of MeansNormal compared to t- distribution

t distribution

Normal

X +/-Zc

X +/-tc


Confidence interval unknown variance

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 means unknown variance

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 means unknown variance1

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 means unknown variance2

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 means unknown variance3

Confidence Intervals for MeansUnknown Variance

X +/-tc

X = 3.239, n = 8, ŝ= 0.0113, =0.01,


Confidence intervals for proportions

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 proportions1

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 proportions2

Confidence Intervals for Proportions

sampling from large population

or finite onewith replacement


Confidence intervals differences and sums known variances

Confidence Intervals Differences and SumsKnown Variances

Samples are independent


Confidence interval for differences and sums known variance

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

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

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 sums1

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

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

Confidence Intervals for Variances

Need statistic with

population parameter 2

estimate for population parameter ŝ2

its distribution - 2


Confidence intervals for variances1

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 variances2

Confidence Intervals for Variances

Rearrange

Take square root if want confidence interval for

standard deviation


Confidence intervals for variances and standard deviations

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

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 2 critical values

How to get 2Critical Values

Not symmetric

/2

/2

2 lower 2 upper


How to get 2 critical values1

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 deviations1

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

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 variances1

Confidence Intervals for Ratio of Variances

Need statistic with

population parameter 21/ 22

estimate for population parameter ŝ21/ ŝ22

its distribution - F


Outline chapter 6

F-Distribution

df1

df2


Outline chapter 6

F-Distribution


Confidence intervals for ratio of variances2

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 variances3

Confidence Intervals for Ratio of Variances

Rearrange


Confidence intervals for ratio of variances4

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 variances3

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 variances5

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 variances6

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 variances4

Confidence Intervals for Variances

Example

Two nickel ore samples

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


Maximum likelihood estimates

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 estimates1

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 estimates2

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 estimates3

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

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 61

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 62

Sum Up Chapter 6

  • Need statistic with

  • population parameter

  • estimate for population parameter

  • its distribution


Sum up chapter 63

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 64

Sum Up Chapter 6

Proportions

  • large sample approximate by normal

  • Differences of means (known variance)


Sum up chapter 65

Sum Up Chapter 6

  • Mean

  • population normal - unknown variance


Sum up chapter 66

Sum Up Chapter 6

  • Variances


Sum up chapter 67

Sum Up Chapter 6

  • Variances ratios


Sum up chapter 68

Sum Up Chapter 6

  • Maximum Likelihood Estimators

  • Pick  which maximizes the function


End of chapter 6

End of Chapter 6!


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