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Midterm Review

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Econ 240ADescriptive StatisticsProbability InferenceDifferences between populationsRegression

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I. Descriptive StatisticsTelling stories with Tables and Graphs

That are self-explanatory and esthetically appealing

Exploratory Data Analysis for random variables that are not normally distributed Stem and Leaf diagrams Box and Whisker Plots

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Stem and Leaf DiagtamExample: Problem 2.24Prices in thousands of $ of houses sold

in a Los Angeles suburb in a given year

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Subsample

Prices289208255215270222206221210224209250222213220250209

Problem 2.24Prices in thousands $Houses sold in a Los Angeles suburb

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Sorted Data

Prices192195198200202205206206208208209209209209209210211

Problem 2.24Prices in thousands $Houses sold in a Los Angeles suburb

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Prices

Mean 237.9882Standard Error 3.314365Median 230Mode 222Standard Deviation 30.55693Sample Variance 933.7261Kurtosis 1.620493Skewness 1.164885Range 149Minimum 192Maximum 341Sum 20229Count 85

Summary StatisticsProblem 2.24Prices in thousands $Houses sold in a Los Angeles suburb

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Stem & Leaf Display

Stems Leaves19 ->25820 ->02566889999921 ->0123345778922 ->001222222334669923 ->0033624 ->01224446778825 ->0000225526 ->056927 ->0023568928 ->692930 ->6831323334 ->01

Problem 2.24Prices in thousands $Houses sold in a Los Angeles suburb

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Box and Whiskers PlotsExample: Problem 4.30Starting salaries by degree

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SubsampleBA BSc BBA Other

26819 28930 38968 3455025797 36602 35187 3024529115 35098 29452 3152032877 36793 30943 2668030015 36171 31610 2904725090 28396 39738 3503723163 26204 37444 2655028225 37280 38403 3570425103 37660 36459 3226229742 24539 37963 3420624587 27222 34138 2691720780 39536 42062 2672330353 32653 32700 36297

Problem 4.50Starting salariesBy degree

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BASmallest = 18719Q1 = 25730Median = 27765Q3 = 29835.5Largest = 37025IQR = 4105.5Outliers: 37025, 36345, 18719,

BScSmallest = 23451Q1 = 29927Median = 33396.5Q3 = 36745.25Largest = 40105IQR = 6818.25Outliers:

BBASmallest = 23401Q1 = 31316Median = 34284Q3 = 39551Largest = 47639IQR = 8235Outliers:

OtherSmallest = 21994Q1 = 28253.5Median = 29950.5Q3 = 32905.25Largest = 38812IQR = 4651.75Outliers:

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

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BASmallest = 18719Q1 = 25730Median = 27765Q3 = 29835.5Largest = 37025IQR = 4105.5Outliers: 37025, 36345, 18719,

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

BScSmallest = 23451Q1 = 29927Median = 33396.5Q3 = 36745.25Largest = 40105IQR = 6818.25Outliers:

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

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BBASmallest = 23401Q1 = 31316Median = 34284Q3 = 39551Largest = 47639IQR = 8235Outliers:

BoxPlot

0 10000 20000 30000 40000 50000

BoxPlot

0 10000 20000 30000 40000 50000

OtherSmallest = 21994Q1 = 28253.5Median = 29950.5Q3 = 32905.25Largest = 38812IQR = 4651.75Outliers:

BoxPlot

0 10000 20000 30000 40000 50000

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II. ProbabilityConcepts

Elementary outcomes Bernoulli trials Random experiments events

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Probability (Cont.)Rules or axioms:Addition rule

P(AUB) = P(A) + P(B) – P(A^B)Conditional probability

P(A/B) = P(A^B)/P(B) Independence

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Probability ( Cont.)Conditional probability

P(A/B) = P(A^B)/P(B) Independence

P(A)*P(B) = P(A^B) So P(A/B) = P(A)

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Probability (Cont.)Discrete Binomial Distribution

P(k) = Cn(k) pk (1-p)n-k

n repeated independent Bernoulli trials k successes and n-k failures

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Binomial Random Number Generator

Take 50 statesSuppose each state was a battleground

state, with probability 0.5 of winning that state

What would the distribution of states look like? How few could you win? How many could you win?

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24

24

28

25

18

29

25

24

24

23

25

24

29

32

28

30

23

27

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Subsample

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Histogram of States Won

0

2

4

6

8

10

Bin

Fre

qu

ency

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Discrete Probability Density, p=0.5

0

0.02

0.04

0.06

0.08

0.1

0.12

15 20 25 30 35 40

States Won

Pro

ba

bil

ity

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Discrete Cumulative Distribution, p=0.5

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35 40

States Won

Pro

ba

bil

ty

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Discrete Cumulative Distribution

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35 40

States Won

Pro

ba

bil

ity

p=0.5

p=0.48

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Probability (Cont.)Continuous normal distribution as an

approximation to the binomial n*p>5, n(1-p)>5 f(z) = (1/2½ exp[-½*z2] z=(x- f(x) = (1/ (1/2½ exp[-½*{(x-

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III. InferenceRates and ProportionsPopulation Means and Sample MeansPopulation Variances and Sample

VariancesDecision Theory

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Decision Theory In inference, I.e. hypothesis testing, and

confidence interval estimation, we can make mistakes because we are making guesses about unknown parameters

The objective is to minimize the expected cost of making errors

E(C) = C(I) + C(II)

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Sample Proportions from Polls

Where n is sample size and k is number of successes

nkp /ˆ

)1(,(~ pnpnpBk

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Sample Proportions

)1()/1(,(~ˆ

/)1()1()/1()/1(ˆ

)/1()/1(ˆ22

ppnpNp

npppnpnVarknpVAR

pnpnEknpE

So estimated p-hat is approximately normal for large sample sizes

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Sample ProportionsWhere the sample size is large

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Problem 9.38A commercial for a household

appliances manufacturer claims that less than 5% of all of its products require a service call in the first year. A consumer protection association wants to check the claim by surveying 400 households that recently purchased one of the company’s appliances

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Problem 9.38 (Cont.)What is the probability that more than

10% require a service call in the first year?

What would you say about the commercial’s honesty if in a random sample of 400 households, 10% report at least one service call?

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Problem 9.38 Answer Null Hypothesis: H0: p=0.05

Alternative Hypothesis: p>0.05 Statistic:

59.4

95.0)05.0)(400/1(/)05.010.0(

)ˆ1(ˆ)/1(/)ˆ(/)ˆˆ( ˆ

z

z

ppnpppEpz p

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0.0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4

Z

NO

RM

DE

NS

Continuous Density of the Standardized Normal Variate, Z

4.59

Z .

Z critical

1.645

5%

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Sample means and population means where the population variance is known

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Problem 9.26, Sample MeansThe dean of a business school claims

that the average MBA graduate is offered a starting salary of $55,000. The standard deviation of the offers is $4600. What is the probability that in a sample of 38 MBA graduates , the mean starting salary is less than $53,000?

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Problem 9.26 (Cont.) Null Hypothesis: H0: Alternative Hypothesis: HA: Statistic:

68.23.746/2000

38/4600/()5300055000(

)//()(/)(

z

z

nxxExz x

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0.0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4

Z

NO

RM

DE

NS

Continuous Density of the Standardized Normal Variate, Z

Zcrit(1%)= -2.33

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Sample means and population means when the population variance is unknown

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Problems 12.33A federal agency responsible for

enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the package. A random sample of 18 containers whose packaging states that the contents weighs 8 ounces was drawn.

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Problems 12.33 (Cont.) Can we conclude that on average the

containers are mislabeled? Use

)//()(/)( nsxxExt x

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0.0

0.1

0.2

0.3

0.4

-2 -1 0 1 2

RANDT

TD

EN

S

Density Function for Student's t-distribution, 17 Degrees of Freedom

t crit 5%

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Problems 12.33 (Cont.)

7.8 7.97 7.92

7.91 7.95 7.87

7.93 7.79 7.92

7.99 8.06 7.98

7.94 7.82 8.05

7.75 7.89 7.91

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Mean 7.913888889

Standard Error 0.019969567

Median 7.92

Mode 7.91

Standard Deviation 0.084723695

Sample Variance 0.007178105

Kurtosis -0.24366084

Skewness -0.22739254

Range 0.31

Minimum 7.75

Maximum 8.06

Sum 142.45

Count 18

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Problems 12.33 (Cont.) Can we conclude that on average the

containers are mislabeled? Use

3.4

020.0/086.0)18/0847.0/()8914.7(

)//()(/)(

t

t

nsxxExt x

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Confidence Intervals for Variances

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Problems 12.33 &12.55A federal agency responsible for

enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the package. A random sample of 18 containers whose packaging states that the contents weighs 8 ounces was drawn.

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Problems 12.33 &12.55 (Cont.)

Estimate with 95% confidence the variance in contents’ weight.

variable with n-1 degrees of freedom is (n-1)s2 /

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0.00

0.02

0.04

0.06

0.08

5 10 15 20 25

RANDCHI

CH

IDE

NS

Chi Square Density for 17 Degrees of Freedom

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Problems 12.33 &12.55(Cont.)

7.8 7.97 7.92

7.91 7.95 7.87

7.93 7.79 7.92

7.99 8.06 7.98

7.94 7.82 8.05

7.75 7.89 7.91

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Mean 7.913888889

Standard Error 0.019969567

Median 7.92

Mode 7.91

Standard Deviation 0.084723695

Sample Variance 0.007178105

Kurtosis -0.24366084

Skewness -0.22739254

Range 0.31

Minimum 7.75

Maximum 8.06

Sum 142.45

Count 18

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Problems 12.33 &12.55(Cont.)7.564<(n-1)s2 /<30.1917.564<17*0.00718/<30.191 (1/7.564)*17*0.00718>>(1/30.191)*17*0

.007180.0161>>0.0040

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IV. Differences in Populations Null Hypothesis: H0: or =0

Alternative Hypothesis: HA: ≠ 0

212121

22112

222

1121

2221121

2212121

2121

2][

)])((2)()[(][

)]()[(][

)]()[(][

/)]()[(21

xxCovxVarxVarxxVar

xxxxExxVar

xxExxVar

xxExxVar

xxt xx

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IV. Differences in Populations

)]/()/[(/)]()[(

][/)]()[(

2][

/)]()[(

2221

212121

212121

212121

2121 21

nnxxt

xVarxVarxxt

xxCovxVarxVarxxVar

xxt xx

Reference Ch. 9 & Ch. 13

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V. Regression Model: yi = a + b*xi + ei

n

i

n

iii

n

ii

n

ii

iii

n

i

n

iiii

ii

exxbTSS

USSSumlainedUnESSSumExplainedTSS

yyTSSSquaresofSumTotalANOVA

esidualsSquaredofSum

yyeerrorestimated

xbyaestimate

xxxxyybestimate

xbayFitted

1 1

222

2

1

1

2

1 1

2

ˆ][ˆ

)(_exp)(_

][)(___:

ˆ:Re___

)ˆ(ˆ:_

*ˆˆ:

][/]][[ˆ:

*ˆˆˆ:

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Fortune 500, 1999: Assets Vs. Revenue, In Logs

General Motors

Exxon Mobil

Wal-Mart

Kroger

Ingram Micro

Costco Wholesale

McKesson HBOC

General Electric

CitigroupBank of AmericaFannie May

Chase ManhattenMorgan Stanley

Merrill LynchPrudential

Bank One American InternationalTIAA-CREF

State Farm

Allstate

1000

10000

100000

1000000

10000 100000 1000000

Log Revenue

Lo

g A

ss

ets

Lab Five

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rank firm industry revenue M$ profits M$ assets M$5 General Electric Diversified Financials 111630 10717 4052007 Citigroup Diversified Financials 82005 9867 716900

11 Bank of America Corp. Commercial banks 51392 7882 63257426 Fannie Mae Diversified Financials 36968.6 3911.9 575167.431 Chase Manhatten Corp. Commercial Banks 33710 5446 40610548 Prudential Ins.Co. of America Insurance: Life, Health(stock) 26618 813 28509450 Bank One Corp. Commercial Banks 25986 3479 26942530 Morgan Stanley Dean Witter Securities 33928 4791 36696729 Merrill Lynch Securities 34879 2618 32807119 TIAA-CREF Insurance: Life, Health(mutual) 39410.2 1024.07 289247.9917 American International Group Insurance; P&C(stock) 40656.08 5055.44 268238

The Financials

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The Financials: Eleven Firms

y = 0.4335x + 8.2535

R2 = 0.3039

12.4

12.6

12.8

13

13.2

13.4

13.6

10 10.2 10.4 10.6 10.8 11 11.2 11.4 11.6 11.8

ln Revenue M$

ln A

ss

ets

M$

Excel Chart

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SUMMARY OUTPUT

Regression StatisticsMultiple R 0.5512779R Square 0.3039073Adjusted R Square 0.2265636Standard Error 0.3117374Observations 11

ANOVAdf SS MS F Significance F

Regression 1 0.381851405 0.381851 3.929312 0.078773838Residual 9 0.874622016 0.09718Total 10 1.256473421

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Lower 95.0%Upper 95.0%Intercept 8.2534951 2.33138973 3.540161 0.006313 2.979521108 13.52747 2.979521 13.52747X Variable 1 0.4335105 0.218696259 1.982249 0.078774 -0.061215204 0.928236 -0.06122 0.928236

Excel Regression

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12.4

12.6

12.8

13.0

13.2

13.4

13.6

10.0 10.5 11.0 11.5 12.0

LNSALES

LNA

SS

ET

S

Eleven Financial Firms

Eviews Chart

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Eviews Regression

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-0.4

-0.2

0.0

0.2

0.4

0.6

12.4

12.6

12.8

13.0

13.2

13.4

13.6

1 2 3 4 5 6 7 8 9 10 11

Residual Actual Fitted

Eviews: Actual, Fitted & residual