1 1 Slide Mátgæði Kafli 11 í Newbold Snjólfur Ólafsson + Slides Prepared by John Loucks ©...

1 1 Slide

Slide

MátgæðiMátgæði

Kafli 11 í Newbold Kafli 11 í Newbold

Snjólfur ÓlafssonSnjólfur Ólafsson++

Slides Prepared bySlides Prepared by John LoucksJohn Loucks

© 1999 ITP/South-Western College Publishing© 1999 ITP/South-Western College Publishing

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Tests of Goodness of Fit and Tests of Goodness of Fit and IndependenceIndependence

Goodness of Fit Test: A Multinomial Goodness of Fit Test: A Multinomial Population Population

Tests of Independence: Contingency TablesTests of Independence: Contingency Tables Goodness of Fit Test: Poisson and Normal Goodness of Fit Test: Poisson and Normal

DistributionsDistributions

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Goodness of Fit Test:Goodness of Fit Test:A Multinomial PopulationA Multinomial Population

1. Set up the null and alternative hypotheses.1. Set up the null and alternative hypotheses.

2. Select a random sample and record the observed2. Select a random sample and record the observed

frequency, frequency, ffi i , for each of the , for each of the kk categories. categories.

3. Assuming 3. Assuming HH00 is true, compute the expected is true, compute the expected frequency, frequency, eei i , in each category by multiplying the , in each category by multiplying the category probability by the sample size.category probability by the sample size.

4. Compute the value of the test statistic.4. Compute the value of the test statistic.

5. Reject 5. Reject HH00 if if (where (where is the significance is the significance level and there are level and there are kk - 1 degrees of freedom). - 1 degrees of freedom).

22

1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

2 2 2 2

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Example: Finger Lakes Homes (A)Example: Finger Lakes Homes (A)

Multinomial Distribution Goodness of Fit TestMultinomial Distribution Goodness of Fit Test

Finger Lakes Homes manufactures four models of Finger Lakes Homes manufactures four models of

prefabricated homes, a two-story colonial, a ranch, aprefabricated homes, a two-story colonial, a ranch, a

split-level, and an A-frame. To help in productionsplit-level, and an A-frame. To help in production

planning, management would like to determine ifplanning, management would like to determine if

previous customer purchases indicate that there is aprevious customer purchases indicate that there is a

preference in the style selected.preference in the style selected.

The number of homes sold of each model for 100The number of homes sold of each model for 100

sales over the past two years is shown below.sales over the past two years is shown below.

ModelModel Colonial Ranch Split-Level A-Frame Colonial Ranch Split-Level A-Frame

# Sold # Sold 30 30 20 35 15 20 35 15

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LetLet

ppCC = population proportion that purchase a colonial = population proportion that purchase a colonial

ppRR = population proportion that purchase a ranch = population proportion that purchase a ranch

ppSS = population proportion that purchase a split-level = population proportion that purchase a split-level

ppAA = population proportion that purchase an A-frame = population proportion that purchase an A-frame

HypothesesHypotheses

HH00: : ppCC = = ppRR = = ppSS = = ppAA = .25 = .25

HHaa: The population proportions are not : The population proportions are not ppCC = .25, = .25,

ppRR = .25, = .25, ppSS = .25, and = .25, and ppAA = .25 = .25

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Expected FrequenciesExpected Frequencies

ee11 = .25(100) = 25 = .25(100) = 25 ee22 = .25(100) = 25 = .25(100) = 25

ee33 = .25(100) = 25 = .25(100) = 25 ee44 = .25(100) = 25 = .25(100) = 25

Test StatisticTest Statistic

= 1 + 1 + 4 + 4 = 1 + 1 + 4 + 4

= 10= 10

22 2 2 230 25

2520 25

2535 25

2515 25

25 ( ) ( ) ( ) ( )2

2 2 2 230 2525

20 2525

35 2525

15 2525

( ) ( ) ( ) ( )

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Rejection RuleRejection Rule

With With = .05 and = .05 and

kk - 1 = 4 - 1 = 3 degrees of - 1 = 4 - 1 = 3 degrees of

freedomfreedom

ConclusionConclusion

We reject the assumption there is no home style We reject the assumption there is no home style

preference, at the .05 level of significance.preference, at the .05 level of significance.

22

7.81 7.81

Do Not Reject H0Do Not Reject H0 Reject H0Reject H0


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Tests of Independence: Contingency Tests of Independence: Contingency TablesTables


2. Select a random sample and record the observed2. Select a random sample and record the observed

frequency, frequency, ffij ij , for each cell of the contingency table., for each cell of the contingency table.

3. Compute the expected frequency, 3. Compute the expected frequency, eeij ij , for each cell. , for each cell.


5. Reject 5. Reject HH00 if (where if (where is the significance level is the significance level and with and with nn rows and rows and mm columns there are columns there are

((nn - 1)( - 1)(mm - 1) degrees of freedom). - 1) degrees of freedom).

ei j

ij (Row Total )(Column Total ) Sample Size

ei j

ij (Row Total )(Column Total ) Sample Size

22

( )f e

eij ij

ijji2

2

( )f e

eij ij

ijji

2 2 2 2

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Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B) Contingency Table TestContingency Table Test

Each home sold can be classified according to price and Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the determine if the price of the home and the style of the home are independent variables.home are independent variables.

The number of homes sold for each model and price for The number of homes sold for each model and price for the past two years is shown below. For convenience, the the past two years is shown below. For convenience, the price of the home is listed as either price of the home is listed as either $65,000 or less $65,000 or less or or more than $65,000more than $65,000..

Price Colonial Ranch Split-Level A-FramePrice Colonial Ranch Split-Level A-Frame

<< $65,000 18 $65,000 18 6 19 6 19 1212

> $65,000 12 14 16 3> $65,000 12 14 16 3

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Contingency Table TestContingency Table Test


HH00: Price of the home is independent of the style of: Price of the home is independent of the style of

the home that is purchasedthe home that is purchased

HHaa: Price of the home is : Price of the home is notnot independent of the independent of the

style of the home that is purchasedstyle of the home that is purchased


PricePrice Colonial Ranch Split-Level A-Frame Total Colonial Ranch Split-Level A-Frame Total

<< $65K 16,5 $65K 16,5 11 19,25 8,25 55 11 19,25 8,25 55

> $65K 13,5 9 15,75 6,75 > $65K 13,5 9 15,75 6,75 4545

Total 30 20 35 15 Total 30 20 35 15 100100

Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B)

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Contingency Table TestContingency Table Test


= .14 + 2.27 + . . . + 2.08 = 8.00= .14 + 2.27 + . . . + 2.08 = 8.00


With With = .05 and (2 - 1)(4 - 1) = 3 d.f., = .05 and (2 - 1)(4 - 1) = 3 d.f.,

Reject Reject HH00 if if 22 > 7.81 > 7.81


We reject We reject HH00. We reject the assumption that the. We reject the assumption that the

price of the home is independent of the style of theprice of the home is independent of the style of the

home that is purchased.home that is purchased.

22 2 218 16 5

16 56 11

113 6 75

6 75 ( . )

.( )

. .( . )

. . 2

2 2 218 16 516 5

6 1111

3 6 756 75

( . ).

( ). .

( . ).

.

. .052 7 81 . .052 7 81

Example: Finger Lakes Homes (B)Example: Finger Lakes Homes (B)

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Goodness of Fit Test: Poisson DistributionGoodness of Fit Test: Poisson Distribution


2. Select a random sample and2. Select a random sample and

a. Record the observed frequency, a. Record the observed frequency, ffi i , for each of the, for each of the

k k values of the Poisson random variable.values of the Poisson random variable.

b. Compute the mean number of occurrences, b. Compute the mean number of occurrences, ..

3. Compute the expected frequency of occurrences, 3. Compute the expected frequency of occurrences, eei i , , for each value of the Poisson random variable.for each value of the Poisson random variable.


5. Reject 5. Reject HH00 if if (where (where is the significance level is the significance level

and there are and there are kk - 2 degrees of freedom). - 2 degrees of freedom).

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1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

2 2 2 2

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Example: Troy Parking GarageExample: Troy Parking Garage Poisson Distribution Goodness of Fit TestPoisson Distribution Goodness of Fit Test

In studying the need for an additional entrance to a city In studying the need for an additional entrance to a city parking garage, a consultant has recommended an parking garage, a consultant has recommended an approach that is applicable only in situations where the approach that is applicable only in situations where the number of cars entering during a specified time period number of cars entering during a specified time period follows a Poisson distribution.follows a Poisson distribution.

A random sample of 100 one-minute time intervals A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a test must be conducted to see if the assumption of a Poisson distribution is reasonable.Poisson distribution is reasonable.

# Arrivals # Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1212

FrequencyFrequency 0 1 4 10 14 20 12 12 9 8 6 3 0 1 4 10 14 20 12 12 9 8 6 3 11

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Example: Troy Parking GarageExample: Troy Parking Garage

Poisson Distribution Goodness of Fit TestPoisson Distribution Goodness of Fit Test


HH00: The number of cars entering the garage during a : The number of cars entering the garage during a one-minute interval is Poisson distributed. one-minute interval is Poisson distributed.

HHaa: The number of cars entering the garage during a : The number of cars entering the garage during a one-minute interval is one-minute interval is notnot Poisson distributed. Poisson distributed.

Estimate of Poisson Probability FunctionEstimate of Poisson Probability Function

otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600

Total Time Periods = 100Total Time Periods = 100

Estimate of Estimate of = 600/100 = 6 = 600/100 = 6

f xex

x

( )!

6 6

f xex

x

( )!

6 6

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xx f f ((x x )) xf xf ((x x )) xx f f ((x x )) xf xf ((x x ))

00 .0025.0025 .25 .25 7 7 .1389.1389 13.8913.89

11 .0149.0149 1.49 1.49 8 8 .1041.1041 10.4110.41

22 .0446.0446 4.46 4.46 9 9 .0694.0694 6.946.94

33 .0892.0892 8.92 8.92 1010 .0417.0417 4.174.17

44 .1339.1339 13.3913.39 1111 .0227.0227 2.272.27

55 .1620.1620 16.2016.20 1212 .0155.0155 1.551.55

66 .1606.1606 16.0616.06 Total Total 1.00001.0000 100.00100.00

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Example: Troy Parking GarageExample: Troy Parking Garage Poisson Distribution Goodness of Fit TestPoisson Distribution Goodness of Fit Test

Observed and Expected FrequenciesObserved and Expected Frequencies

ii ffii eeii ffii - - eeii

0 or 1 or 20 or 1 or 2 55 6.206.20 -1.20-1.20

33 1010 8.928.92 1.081.08

44 1414 13.3913.39 .61.61

55 2020 16.2016.20 3.803.80

66 1212 16.0616.06 -4.06-4.06

77 1212 13.8913.89 -1.89-1.89

88 99 10.4110.41 -1.41-1.41

99 88 6.946.94 1.061.06

10 or more10 or more 1010 7.997.99 2.012.01

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With With = .05 and = .05 and kk - - pp - 1 = 9 - 1 - 1 = 7 d.f. - 1 = 9 - 1 - 1 = 7 d.f.

(where (where kk = number of categories and = number of categories and pp = number = number

of population parameters estimated), of population parameters estimated),

Reject Reject HH00 if if 22 > 14.07 > 14.07


We cannot reject We cannot reject HH00. There is no reason to question. There is no reason to question

the assumption of a Poisson distribution.the assumption of a Poisson distribution.

22 2 21 20

6 201 088 92

2 017 99

3 42

( . )

.( . )

.. . .

( . ).

. 22 2 21 20

6 201 088 92

2 017 99

3 42

( . )

.( . )

.. . .

( . ).

.

. .052 14 07 . .052 14 07


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Mátgæði fyrir normaldreifinguMátgæði fyrir normaldreifingu

Tvær ólíkar leiðir.Tvær ólíkar leiðir.

A. Nota A. Nota 22 - próf eins og hér á undan. - próf eins og hér á undan.

B. Reikna B. Reikna skekkinguskekkingu og og ferilrisferilris og nota og nota Bowman-Shelton próf. Bowman-Shelton próf.

Þið þurfið að vita um þessar leiðir og vita hvað Þið þurfið að vita um þessar leiðir og vita hvað skekkingskekking og og ferilrisferilris segja, en þurfið ekki að segja, en þurfið ekki að reikna þetta.reikna þetta.

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A. A. 22 - próf - próf

Meðaltal og staðalfrávik úrtaks notað í Meðaltal og staðalfrávik úrtaks notað í tilgátuprófi.tilgátuprófi.

HH00: Þýðið er normaldreift með meðalgildi og : Þýðið er normaldreift með meðalgildi og staðalfrávik eins og í úrtaki.staðalfrávik eins og í úrtaki.

HHaa: Svo er ekki.: Svo er ekki.

Rauntöluásnum skipt í hafilega mörg bil, þ.a. Rauntöluásnum skipt í hafilega mörg bil, þ.a. væntanlegur fjöldi á bili, samkvæmt H0, sé meiri væntanlegur fjöldi á bili, samkvæmt H0, sé meiri en 5. en 5.

ReiknaðReiknað o.s.frv.o.s.frv.

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1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

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B. Skekking og ferilrisB. Skekking og ferilris

Skekking (skewness) mælir hve samhverf Skekking (skewness) mælir hve samhverf tíðnidreifing er tíðnidreifing er

(assymmetry of a frequency distribution).(assymmetry of a frequency distribution).

Ferilris (kurtosis) mælir hve tíðnidreifing rís hátt Ferilris (kurtosis) mælir hve tíðnidreifing rís hátt

(flatness or peakedness of a frequency distribution)(flatness or peakedness of a frequency distribution)

31

3 /)(

s

nxxSkekking

n

ii

41

4 /)(

s

nxxFerilris

n

ii

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Í Í Bowman-Shelton prófi er skekking og ferilris Bowman-Shelton prófi er skekking og ferilris reiknað fyrir úrtakið og athugað hvort það sé í reiknað fyrir úrtakið og athugað hvort það sé í nægilega góðu samræmi við skekkingu og nægilega góðu samræmi við skekkingu og ferilris fyrir normaldreifingu.ferilris fyrir normaldreifingu.

Skekking og ferilris - framhaldSkekking og ferilris - framhald

1 1 Slide Mátgæði Kafli 11 í Newbold Snjólfur Ólafsson + Slides Prepared by John Loucks ©...

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