1 1 Slide © 2005 Thomson/South-Western Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population Goodness of.

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© 2005 Thomson/South-Western© 2005 Thomson/South-Western

Chapter 12Chapter 12 Tests of Goodness of Fit and Tests of Goodness of Fit and

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

Goodness of Fit Test: PoissonGoodness of Fit Test: Poisson and Normal Distributionsand Normal Distributions

Test of IndependenceTest of Independence

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Hypothesis (Goodness of Fit) TestHypothesis (Goodness of Fit) Testfor Proportions of a Multinomial for Proportions of a Multinomial

PopulationPopulation1.1. Set up the null and alternative hypotheses. Set up the null and alternative hypotheses.

2.2. Select a random sample and record the observed Select a random sample and record the observed

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

3.3. Assuming 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.

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PopulationPopulation

22

1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

4.4. Compute the value of the test statistic. Compute the value of the test statistic.

Note: The test statistic has a chi-square distributionNote: The test statistic has a chi-square distributionwith with kk – 1 df provided that the expected frequencies – 1 df provided that the expected frequenciesare 5 or more for all categories.are 5 or more for all categories.

ffii = observed frequency for category = observed frequency for category iieeii = expected frequency for category = expected frequency for category ii

kk = number of categories = number of categories

where:where:

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PopulationPopulation

where where is the significance is the significance level andlevel and

there are there are kk - 1 degrees of - 1 degrees of freedomfreedom

pp-value approach:-value approach:

Critical value approach:Critical value approach:

Reject Reject HH00 if if pp-value -value <<

5.5. Rejection rule: Rejection rule:

2 2 2 2 Reject Reject HH00 if if

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Multinomial Distribution Goodness of Fit Multinomial Distribution Goodness of Fit TestTest

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

Finger Lakes Homes manufacturesFinger Lakes Homes manufactures

four models of prefabricated homes,four models of prefabricated homes,

a two-story colonial, a log cabin, aa two-story colonial, a log cabin, a

split-level, and an A-frame. To helpsplit-level, and an A-frame. To help

in production planning, managementin production planning, management

would like to determine if previous would like to determine if previous

customer purchases indicate that therecustomer purchases indicate that there

is a preference in the style selected.is a preference in the style selected.

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Split- A-Split- A-Model Colonial Log Level FrameModel Colonial Log Level Frame

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

The number of homes sold of eachThe number of homes sold of each

model for 100 sales over the past twomodel for 100 sales over the past two

years is shown below.years is shown below.


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

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HypothesesHypotheses


where:where:

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

ppL L = population proportion that purchase a log cabin = population proportion that purchase a log cabin

ppS S = 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

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

HHaa: The population proportions are : The population proportions are notnot

ppCC = .25, = .25, ppLL = .25, = .25, ppSS = .25, and = .25, and ppAA = .25 = .25

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Rejection RuleRejection Rule

22

7.815 7.815

Do Not Reject H0Do Not Reject H0 Reject H0Reject H0


With With = .05 and = .05 and

kk - 1 = 4 - 1 = 3 - 1 = 4 - 1 = 3

degrees of freedomdegrees of freedom

Reject H0 if if pp-value -value << .05 or .05 or 22 > 7.815. > 7.815.

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Expected FrequenciesExpected Frequencies

Test StatisticTest Statistic

22 2 2 230 25

2520 25

2535 25

2515 25

25

( ) ( ) ( ) ( )22 2 2 230 25

2520 25

2535 25

2515 25

25

( ) ( ) ( ) ( )


ee1 1 = .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

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

= 10= 10

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Conclusion Using the Conclusion Using the pp-Value Approach-Value Approach

The The pp-value -value << . We can reject the null hypothesis. . We can reject the null hypothesis.

Because Because 22 = 10 is between 9.348 and 11.345, = 10 is between 9.348 and 11.345, thethe area in the upper tail of the distribution is area in the upper tail of the distribution is betweenbetween .025 and .01..025 and .01.

Area in Upper Tail .10 .05 .025 .01 .005Area in Upper Tail .10 .05 .025 .01 .005

22 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

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Conclusion Using the Critical Value ApproachConclusion Using the Critical Value Approach


We reject, at the .05 level of significance,We reject, at the .05 level of significance,

the assumption that there is no home stylethe assumption that there is no home style

preference.preference.

2 2 = 10 = 10 >> 7.815 7.815

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Test of Independence: Contingency Test of Independence: Contingency TablesTables

ei j

ij (Row Total )(Column Total )

Sample Sizee

i jij

(Row Total )(Column Total ) Sample Size

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

2.2. Select a random sample and record the observed 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.

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

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Test of Independence: Contingency Test of Independence: Contingency TablesTables

22

( )f e

eij ij

ijji2

2

( )f e

eij ij

ijji

5.5. Determine the rejection rule. Determine the rejection rule.

Reject Reject HH00 if if p p -value -value << or or . .

2 2 2 2

4.4. Compute the test statistic. Compute the test statistic.

where where is the significance level and, is the significance level and,with 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.

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Each home sold by Finger LakesEach home sold by Finger Lakes

Homes can be classified according toHomes can be classified according to

price and to style. Finger Lakes’price and to style. Finger Lakes’

manager would like to determine ifmanager would like to determine if

the price of the home and the style ofthe price of the home and the style of

the home are independent variables.the home are independent variables.

Contingency Table (Independence) TestContingency Table (Independence) Test

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

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

The number of homes sold forThe number of homes sold for

each model and price for the past twoeach model and price for the past two

years is shown below. For convenience,years is shown below. For convenience,

the price of the home is listed as eitherthe price of the home is listed as either

$99,000 or less $99,000 or less or or more than $99,000more than $99,000..

> $99,000 12 14 > $99,000 12 14 16 316 3<< $99,000 18 $99,000 18 6 19 12 6 19 12


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

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HH00: Price of the home : Price of the home isis independent of the independent of the

style of the home that is purchasedstyle of the home that is purchasedHHaa: Price of the home : Price of the home is notis not independent of the independent of the

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

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

<< $99K $99K

> $99K> $99K

TotalTotal 30 20 35 15 10030 20 35 15 100

12 12 14 16 3 45 14 16 3 45

18 6 19 12 5518 6 19 12 55

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2.05 7.815 2.05 7.815 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 pp-value -value << .05 or .05 or 22 >> 7.8157.815

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

( . ).

( ). .

( . ).

.

= .1364 + 2.2727 + . . . + 2.0833 = 9.149= .1364 + 2.2727 + . . . + 2.0833 = 9.149


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The The pp-value -value << . We can reject the null hypothesis. . We can reject the null hypothesis.

Because Because 22 = 9.145 is between 7.815 and = 9.145 is between 7.815 and 9.348, the9.348, the area in the upper tail of the distribution is area in the upper tail of the distribution is betweenbetween .05 and .025..05 and .025.

Area in Upper Tail .10 .05 .025 .01 .005Area in Upper Tail .10 .05 .025 .01 .005

22 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Value (df = 3) 6.251 7.815 9.348 11.345 12.838


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Conclusion Using the Critical Value ApproachConclusion Using the Critical Value Approach


We reject, at the .05 level of We reject, at the .05 level of significance,significance,the assumption that the price of the the assumption that the price of the home ishome isindependent of the style of home that independent of the style of home that isispurchased.purchased.

2 2 = 9.145 = 9.145 >> 7.815 7.815

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


HH00: Population has a Poisson probability distribution: Population has a Poisson probability distribution

HHaa: Population does not have a Poisson distribution: Population does not have a Poisson distribution

3.3. Compute the expected frequency of occurrences Compute the expected frequency of occurrences eeii for each value of the Poisson random variable.for each value of the Poisson random variable.

2.2. Select a random sample and Select a random sample and

a.a. Record the observed frequency Record the observed frequency ffii for each value of for each value of

the Poisson random variable.the Poisson random variable.

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

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22

1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k


ffii = observed frequency for category = observed frequency for category iieeii = expected frequency for category = expected frequency for category ii

kk = number of categories = number of categories

where:where:

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where where is the significance is the significance level andlevel and there are there are kk - 2 degrees of - 2 degrees of freedomfreedom

pp-value approach:-value approach:

Critical value approach:Critical value approach:

Reject Reject HH00 if if pp-value -value <<

5.5. Rejection rule: Rejection rule:

2 2 2 2 Reject Reject HH00 if if


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Example: Troy Parking GarageExample: Troy Parking Garage

In studying the need for anIn studying the need for an

additional entrance to a city additional entrance to a city

parking garage, a consultant parking garage, a consultant

has recommended an analysishas recommended an analysis

approach that is applicable approach that is applicable

only in situations where the number of carsonly in situations where the number of cars

entering during a specified time period follows aentering during a specified time period follows a

Poisson distribution.Poisson distribution.


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A random sample of 100 one-A random sample of 100 one-

minute time intervals resultedminute time intervals resulted

in the customer arrivals listedin the customer arrivals listed

below. A statistical test mustbelow. A statistical test must

be conducted to see if thebe conducted to see if the

assumption of a Poisson distribution is assumption of a Poisson distribution is reasonable.reasonable.


Example: Troy Parking GarageExample: Troy Parking Garage

# Arrivals# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 120 1 2 3 4 5 6 7 8 9 10 11 12

Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1

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

HH00: Number of cars entering the garage during: Number of cars entering the garage during a one-minute interval is Poisson distributeda one-minute interval is Poisson distributed

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Estimate of Poisson Probability FunctionEstimate of Poisson Probability Function

f xex

x

( )!

6 6

f xex

x

( )!

6 6

f xex

x

( )!

6 6

f xex

x

( )!

6 6


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

Hence,Hence,

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

Total Time Periods = 100Total Time Periods = 100

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x x f f ((x x ) ) nf nf ((x x ))

00

11

22

33

44

55

66

13.7713.77

10.3310.33

6.886.88

4.134.13

2.252.25

2.012.01

100.0100.000

.137.13777

.103.10333

.068.06888

.041.04133

.022.02255

.020.02011

1.0001.00000

77

88

99

1010

1111

12+12+

TotalTotal

.002.00255

.014.01499

.044.04466

.089.08922

.133.13399

.160.16066

.160.16066

.2.255

1.491.49

4.464.46

8.928.92

13.313.399

16.016.066

16.016.066

xx f f ((x x )) nf nf ((x x ))

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Observed and Expected FrequenciesObserved and Expected Frequencies


ii ffii eeii ffii - - eeii

-1.20-1.20 1.081.08 0.610.61 3.943.94-4.06-4.06-1.77-1.77-1.33-1.33 1.121.12 1.611.61

6.206.20 8.928.9213.313.39916.016.06616.016.06613.713.77710.310.333 6.886.88 8.398.39

5510101414202012121212 99 881010

0 or 1 or 0 or 1 or 22 33 44 55 66 77 88 9910 or 10 or moremore

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2 2 22 ( 1.20) (1.08) (1.61)

. . . 3.2686.20 8.92 8.39

2 2 22 ( 1.20) (1.08) (1.61)

. . . 3.2686.20 8.92 8.39


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), 2.05 14.067 2.05 14.067

Reject Reject HH00 if if pp-value -value << .05 or .05 or 22 >> 14.067. 14.067.


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The The pp-value > -value > . We cannot reject the null . We cannot reject the null hypothesis. There is no reason to doubt the hypothesis. There is no reason to doubt the assumption of a Poisson distribution.assumption of a Poisson distribution.

Because Because 22 = 3.268 is between 2.833 and = 3.268 is between 2.833 and 12.017 in the Chi-Square Distribution Table, the 12.017 in the Chi-Square Distribution Table, the area in the upper tailarea in the upper tailof the distribution is between .90 and .10. of the distribution is between .90 and .10.

Area in Upper Tail .90 .10 .05 .025 .01 Area in Upper Tail .90 .10 .05 .025 .01

22 Value (df = 7) 2.833 12.017 14.067 16.013 18.475 Value (df = 7) 2.833 12.017 14.067 16.013 18.475


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Goodness of Fit Test: Normal DistributionGoodness of Fit Test: Normal Distribution


3.3. Compute the expected frequency, Compute the expected frequency, eei i , for each interval., for each interval.

2.2. Select a random sample and Select a random sample and

a.a. Compute the mean and standard deviation. Compute the mean and standard deviation.

b.b. Define intervals of values so that the expected Define intervals of values so that the expected

frequency is at least 5 for each interval. frequency is at least 5 for each interval.

c.c. For each interval record the observed frequencies For each interval record the observed frequencies

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Goodness of Fit Test: Normal DistributionGoodness of Fit Test: Normal Distribution

22

1

( )f ee

i i

ii

k2

2

1

( )f ee

i i

ii

k

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

and there are and there are kk - 3 degrees of freedom). - 3 degrees of freedom).

2 2 2 2

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Normal Distribution Goodness of Fit TestNormal Distribution Goodness of Fit Test

Example: IQ ComputersExample: IQ Computers

IQIQIQIQ

IQ Computers (one better than HP?)IQ Computers (one better than HP?)

manufactures and sells a generalmanufactures and sells a general

purpose microcomputer. As part ofpurpose microcomputer. As part of

a study to evaluate sales personnel, a study to evaluate sales personnel, managementmanagement

wants to determine, at a .05 significance level, wants to determine, at a .05 significance level, if theif the

annual sales volume (number of units sold by aannual sales volume (number of units sold by a

salesperson) follows a normal probability salesperson) follows a normal probability distribution.distribution.

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A simple random sample of 30 ofA simple random sample of 30 of

the salespeople was taken and theirthe salespeople was taken and their

numbers of units sold are below.numbers of units sold are below.


Example: IQ ComputersExample: IQ Computers

(mean = 71, standard deviation = 18.54)(mean = 71, standard deviation = 18.54)

33 43 44 45 52 52 56 58 63 6433 43 44 45 52 52 56 58 63 6464 65 66 68 70 72 73 73 74 7564 65 66 68 70 72 73 73 74 7583 84 85 86 91 92 94 98 102 10583 84 85 86 91 92 94 98 102 105

IQIQIQIQ

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HHaa: The population of number of units sold: The population of number of units sold does does notnot have a normal distribution with have a normal distribution with

mean 71 and standard deviation 18.54.mean 71 and standard deviation 18.54.

HH00: The population of number of units sold: The population of number of units sold has a normal distribution with mean 71has a normal distribution with mean 71 and standard deviation 18.54.and standard deviation 18.54.

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Interval DefinitionInterval Definition


To satisfy the requirement of an To satisfy the requirement of an expectedexpectedfrequency of at least 5 in each interval frequency of at least 5 in each interval we willwe willdivide the normal distribution into 30/5 = divide the normal distribution into 30/5 = 66equal probability intervals.equal probability intervals.

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Interval DefinitionInterval Definition

Areas = 1.00/6 = .1667

Areas = 1.00/6 = .1667

717153.0253.02

71 .43(18.54) = 63.0371 .43(18.54) = 63.0378.9778.9788.98 = 71 + .97(18.54)88.98 = 71 + .97(18.54)


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Observed and Expected FrequenciesObserved and Expected Frequencies


11

-2-2

11

00

-1-1

11

55

55

55

55

55

55

3030

66

33

66

55

44

66

3030

Less than 53.02Less than 53.02

53.02 to 63.0353.02 to 63.03

63.03 to 71.0063.03 to 71.00

71.00 to 78.9771.00 to 78.97

78.97 to 88.9878.97 to 88.98

More than 88.98More than 88.98

ii ffii eeii ffii - - eeii

TotalTotal

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2 2 2 2 2 22 (1) ( 2) (1) (0) ( 1) (1)

1.6005 5 5 5 5 5

2 2 2 2 2 22 (1) ( 2) (1) (0) ( 1) (1)

1.6005 5 5 5 5 5


With With = .05 and = .05 and kk - - pp - 1 = 6 - 2 - 1 = 3 d.f. - 1 = 6 - 2 - 1 = 3 d.f.

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

of population parameters estimated), of population parameters estimated), 2.05 7.815 2.05 7.815

Reject Reject HH00 if if pp-value -value << .05 or .05 or 22 >> 7.815. 7.815.



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The The pp-value > -value > . We cannot reject the null . We cannot reject the null hypothesis. There is little evidence to support hypothesis. There is little evidence to support rejecting the assumption the population is rejecting the assumption the population is normally distributed with normally distributed with = 71 and = 71 and = 18.54. = 18.54.

Because Because 22 = 1.600 is between .584 and = 1.600 is between .584 and 6.251 in the Chi-Square Distribution Table, the 6.251 in the Chi-Square Distribution Table, the area in the upper tailarea in the upper tailof the distribution is between .90 and .10. of the distribution is between .90 and .10.

Area in Upper Tail .90 .10 .05 .025 .01 Area in Upper Tail .90 .10 .05 .025 .01

22 Value (df = 3) .584 6.251 7.815 9.348 11.345 Value (df = 3) .584 6.251 7.815 9.348 11.345

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End of Chapter 12End of Chapter 12

1 1 Slide © 2005 Thomson/South-Western Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population Goodness of.

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