Schaum

1
Schaums OutlinePROBABILTY and STATISTICS
Chapter 6 ESTIMATION THEORYPresented by
Professor Carol DahlExamples from D. Salvitti
J. Mazumdar C. Valencia
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Outline Chapter 6

• Trader in energy stocks
• random variable Y value of share
• want estimates µy, sY
  
• Y ß0 ß1X ?
• want estimates Y, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates

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Outline Chapter 6

• Types of estimators
• Point estimates
• µ 7
• Interval estimates
• µ 7/-2
• confidence interval

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Outline Chapter 6

• Population parameters and confidence intervals
  Means
• Large sample sizes
• Small sample sizes
• Proportions
• Differences and Sums
• Variances
• Variances ratios

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Properties of Estimators - Unbiased
Unbiased Estimator of Population Parameter
estimator expected value to population
parameter
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Unbiased Estimates

• Population Parameters
• Sample Parameters
• are unbiased estimates
• Expected value of standard deviation not unbiased

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Properties of Estimators - Efficient

• Efficient Estimator
• if distributions of two statistics same
• more efficient estimator smaller variance
• efficient smallest variance of all unbiased
  estimators

8
Unbiased and Efficient Estimates
Target Estimates which are efficient and
unbiased Not always possible often us biased and
inefficient easy to obtain
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Types of Estimates for Population Parameter

• Point Estimate
• single number
• Interval Estimate
• between two numbers.

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Estimates of Mean Known VarianceLarge Sample
or Finite with Replacement

• X value of share
• sample mean is 32
• volatility is known s2 4.00
• confidence interval for share value
• Need
• estimator for mean
• need statistic with
• mean of population
• estimator

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Estimates of Mean- Sampling Statistic

• P(-1.96 lt lt1.96) 95

2.5
12
Confidence Interval for Mean

• P(-1.96 lt lt1.96) 95
• P(-1.96 lt lt1.96 )
  95
• P(-1.96 -?X lt -µ lt1.96 -
  ?X ) 95
• Change direction of inequality
• P(1.96 ?X gt µ gt -1.96
  ?X ) 95

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Confidence Interval for Mean

• P(1.96 ?X gt µ gt -1.96
  ?X ) 95
• Rearrange
• P(?X - 1.96 lt µ lt?X 1.96
  ) 95
• Plug in sample values and drop probabilities
• X value of share, sample 64
• sample mean is 32
• volatility is s2 4
• 32 1.962/?64, 32 1.962/?64 31.51,32.49

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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
• (ZltZc) 0.975 (1 0.05/2)
• 95 of intervals contain
• 5 of intervals do not contain

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Estimates for Mean for Normal

• interval estimate 0.95 (95) (1 - 0.05)
• ?X /-Zc
• (ZltZc) 0.975 (1 0.05/2)
• interval estimate (1-?)
• ? X /-Zc
• (ZltZc) (1 ?/2)
• ? of intervals dont contain
• (1- ?) of intervals do contain

(ZltZc) 0.975 (1 0.05/2)
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Confidence Interval Estimatesof Population
Parameters
Common values for corresponding to various
confidence levels used in practice are
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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 /-
1.962/?64
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Confidence Intervals for MeansFinite Population
(N) no Replacement

Confidence interval
Confidence level
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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
29API standard deviation is known to be 9
API ? 0.05
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Confidence Intervals for MeansFinite Population
(N) no ReplacementKnown Variance

• X 29 , N 500, n 81 , s 9 , ? 0.05
• Zc 1.96

21
But dont know Variance
t-Distribution
N(0.1)

tdf
??2/df
df
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Confidence Intervals of Meanst- distribution




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Confidence Intervals of MeansNormal compared to
t- distribution
t distribution
Normal
?X /-Zc
?X /-tc
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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
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Confidence Intervals for MeansUnknown Variance
?X /-tc
?X SXi/n 3.2363.2233.2423.2443.2283.2533.
2533.230 8 ?X 3.239 s2 S(Xi - ?X)
(3.236- ?X)2 . . .(3.230 - ?X)2
(n-1) (8-1) s 0.0113
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Confidence Intervals for MeansUnknown Variance
?X /-tc

• ?X 3.239, n 8, s 0.0113, ?0.01,
• 1- ?/20.995
• From the t-table with 7 degrees of freedom, we
  find tc t7,0.9953.50

1-?/2.975
.005
tc
-tc
Find tc from Table of Excel
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Confidence Intervals for MeansUnknown Variance
3.499483
1-?/2.975
?/2 0.005
Depends on Table
tc
-tc
?
3.499483
GHJ ?/2 0.005 ? tc 2. 499 Schaums 1- ?/2
0.995 ? tc 2.35 Excel tinv(0.01,7) 3.499483
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Confidence Intervals for MeansUnknown Variance
?X /-tc
?X 3.239, n 8, s 0.0113, ?0.01,
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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 n600, X250 (successes), ? 0.05
• zc 1.96 and

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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 n600, X250 (successes), ? 0.05
• zc 1.96 and

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Confidence Intervals for Proportions
sampling from large population or finite one
with replacement
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Confidence Intervals Differences and SumsKnown
Variances
Samples are independent
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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
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Confidence Intervals for Differences and Sums
Example
Solution X1350, s125, n1200, X2250, s250,
n2150
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Confidence Intervals for Differences and Sums
Large Samples
Where P1, P2 two sample proportions, n1, n2
sizes of two samples
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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
P1150/2000.75, n1200, P2100/3000.33,n2300
With 95 of confidence the difference of
proportions 0.42, 0.08
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Confidence Intervals Differences and Sums
Example
Solution P1150/2000.75, n1200,
P2100/3000.33, n2300
95 of confidence the difference of proportions
0.08, 0.42
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Confidence Intervals for Variances

• Need statistic with
• population parameter ?2
• estimate for population parameter s2
• its distribution - ?2

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Confidence Intervals for Variances

• has a
  chi-squared distribution
• n-1 degrees of freedom.
• Find interval such that s lies in the interval
  for
• 95 of samples
• 95 confidence interval

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Confidence Intervals for Variances
Rearrange
Take square root if want confidence interval for
standard deviation
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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
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Confidence Intervals for Variance
Example
Variance of amount of copper reserves 16
estimates chosen at random s2 2.4 thousand
million tons Find 99 confidence interval
variance Solution s22.4, n16,
degrees of freedom 16-1 15
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How to get ?2 Critical Values

Not symmetric
?/2
?/2
?2 lower ?2 upper
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How to get ?2 Critical 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
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Confidence Intervals for Variances and Standard
Deviations
Example
99 confidence interval variance of
reserves Solution s2.4 (n-1)15
?2lower 4.60, ?2upper 32.8
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Confidence Intervals for Ratio of Variances

• Two independent random samples
• size m and n
• population variances
• estimated variances s21, s22
• interested in whether variances are the same
• ?21/ ?22

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Confidence Intervals for Ratio of Variances

• Need statistic with
• population parameter ?21/ ?22
• estimate for population parameter s21/ s22
• its distribution - F

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F-Distribution
df1
df2
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F-Distribution
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Confidence Intervals for Ratio of Variances

• Need statistic with
• population parameter ?21/ ?22
• estimate for population parameter s21/ s22
• its distribution - F

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Confidence Intervals for Ratio of Variances
Rearrange
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Confidence Intervals for Ratio of Variances
Put smallest first, largest second
When substitute in values drop probabilities 1-?
confidence interval for ?21/ ?22
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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 s21 24, n1 16, s22 18,
n2 10,
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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
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Confidence Intervals for Ratio of Variances
GHJ area above F0.95,15,9 P(F15,9gtFc)
0.95 P(1/F15,9lt1/Fc) 0.95 But 1/F15,9
F9,15 P(F9,15lt1/Fc) 0.95 P(F9,15lt1/Fc)
0.05 1/Fc 2.59 ?Fc 0.3861
?/2
?/2
F upper
F lower
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Confidence Intervals for Variances
Example
Two nickel ore samples Solution s21 24, n1
16, s22 18, n2 10,
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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 ?xe-?/x! ? ? (the mean) If
sample independently from f n times x1, x2, .
. .xn a sample if consider all possible
samples of n a sampling distribution
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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
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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,?)

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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

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Sum Up Chapter 6

• Y ß0 ß1X
• Y, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates
• Types of estimators
• Point estimates
• Interval estimates

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Sum Up Chapter 6

• Y- µY, ?Y, ?Y, s2
• In 590-690
• Y ß0 ß1X
• Y, b0, b1
• Properties of estimators
• unbiased estimates
• efficient estimates
• Types of estimators
• Point estimates
• Interval estimates

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Sum Up Chapter 6

• Need statistic with
• population parameter
• estimate for population parameter
• its distribution

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Sum Up Chapter 6

• Population parameters and confidence intervals
• Mean Normal
• Know variance and population normal
• Large sample size can use estimated
  variance

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Sum Up Chapter 6

• Proportions
• large sample approximate by normal
• Differences of means (known variance)

66
Sum Up Chapter 6

• Mean
• population normal - unknown variance

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Sum Up Chapter 6

• Variances

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Sum Up Chapter 6

• Variances ratios

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Sum Up Chapter 6

• Maximum Likelihood Estimators
• Pick ? which maximizes the function

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End of Chapter 6!