The Chi Square Test.ppt

7/28/2019 The Chi Square Test.ppt

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The Chi Square Test

2 By SDK, AIM


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Chi sq: The test of

the goodness of fit


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The Chi Square Test

The Chi Square Test (2) is used to

determine how well theoretical

distributions (Normal, Binomial, Poisson,

etc) fit empirical distributions (Those

obtained from samples)

Pearson developed this test in 1990 to

check the goodness of fit of distributions


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Consider A Particular Sample : A set of possible events

E1, E2,, Ek, that are observed to occur withfrequencies o1, o2,, ok (called observed

frequencies). As per the rules of probability,

these events are expected to occur with

frequencies e1, e2,, ek

Event E1 E2 Ek

Observed frequency o1 o2 okExpected frequency e1 e2 ek


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Example

If we toss a fair coin 100 times, we may

expect 50 heads and 50 tails. However the

results may not be obtained exactly


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The 2

VariableThe 2 Variable gives a measure of thedisparity existing between theobserved and the expected frequencies

2=i=1k (oi-ei)2/eiN = total frequency =i=1

koi =i=1kei


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Thus

2 =i=1n[(oi2-2oiei+ei2)/ei]=i=1

n[(oi2/ei) 2N + N]

=i=1n

(oi2

/ei)-N


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Example

I assume that themarks of a class aredistributed

normally. Howeverwhen theexamination takesplace I realize that

the class has had abetter performance

Marks Observedfrequency

Expectedfrequency

0-20 2 5

21-40 5 10

41-60 18 30

61-80 23 10

81-100 12 5

Total 60 60


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Marks Observedfrequency

Expectedfrequency

(oi-ei)2/ei

0-20 2 5 1.8

21-40 5 10 2.5

41-60 18 30 4.861-80 23 10 16.9

81-100 12 5 9.8

Total 60 60 35.8 =2


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Also, 20 as it is the sum of squares& the larger the 2 the greater thedifference in the two distributions


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The probability function of2


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Calculation of

Populationparameters are notknown and have to

be estimated fromsample statistics

= k 1 m,

m = no of populationparametersestimated

Populationparameters areknown m = 0

= k - 1


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The 2

Curve

Table value of2

AcceptanceArea Rejection Area

o X

Y


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Example at = 0.05 and df = 5-1 = 4

Table value of2


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Steps of the Chi Square Test

Define H0 and H1 List the observedfrequencies

Calculate the expectedfrequencies if the datafollows a theoreticaldistribution

Compute2


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Accept H0 if computed 2 Chi sq comp Accept H0

H0: Data follow the Binomialdistribution


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Chi sq as a test of

Independence


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Note that :

The tests of significance are allbased on the assumption that the

population is normally distributed.However it is not always possibleto assume the underlying

distribution pattern for thesampling done


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If we classify a population into

several categories with respectto two attributes (e.g.: age, jobpreference), we can use the Chi

Sq Test to determine if the twoattributes are independent of

each other


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Example

In 4 regions National Health Company

samples its employees attitudes towards

job performance reviews. Respondents

are given a choice: between the presentmethod of 2 reviews a year and the

proposed method of quarterly reviews.


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Also,

1. pN is the proportion of employees from the northwho prefer the present plan

2. pE is the proportion of employees from the eastwho prefer the present plan

3. pS is the proportion of employees from the southwho prefer the present plan

4. pW is the proportion of employees from the westwho prefer the present plan

H0: pN = pE = pS = pW


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Contingency table

North South East West TotalNumber

who

prefer

present

method

68 75 57 79 279

Number

who

prefer

new

method

32 45 33 31 141

Total 100 120 90 110 420


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Thus combined proportion of employees preferringthe new method = 1 0.6643 = 0.3357

Thus,

1. 0.6643 = Estimate of population proportion who prefer thecurrent method

2. 0.3357 = Estimate of population proportion who prefer thenew method

Multiply the estimate with the total number of employeessampled in each region to get the expected number


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Contingency table

North South East West Total

Number

who

prefer

present

method

68 75 57 79 279

Number

who

prefer

new

method

32 45 33 31 141

Total 100 120 90 110 420

Observed values

North South East West Total

Number

who

prefer

present

method

66 80 60 73 279

Number

who

prefer

new

method

34 40 30 37 141

Total 100 120 90 110 420

Expected Values


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Thus H0 isaccepted

Degrees of

freedom=(4-1)(2-1)

= 3


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Criteria C1 C2 C3 Total

R1 O11 O12 O13R1

R2 O21 O22 O23 R1

Total C1 C2 C3 n

Consider a contingency table


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H0: Ri is independent of Cj

Or P(Ri

Cj

) = P(Ri

) P(Cj

)

But P(Ri Cj) = Eij/ n

Also P(Ri) = Ri/ nP(Cj) = Cj/ n

Thus =P(Ri Cj) = P(Ri) P(Cj)

= (Ri/ n)(Cj/ n)Thus Eij = Ri Cj/ n2


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Example

In order to study the profits and losses of

firms by industry, a random sample of 100

firms is selected, and for each form in the

sample, we record whether the company

made ,money or lost money, and whetherthe firm is a service company. The data

are summarized in a 2x2 contingency

table. Is the possibility of making a profitindependent of type of industry?


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An insurance companys data regarding

claims gathered by studying three

different age groups of sample size 100each is given below

25 and

under

Over25 and

under

50

50 and

over

Claim 40 35 60No

claim60 65 40

Age group H0: Claim is

not related toage


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25 and

under

Over

25 and

under

50

50 and

over

Claim 40 35 60 135

No

claim 60 65 40 165100 100 100 300

Age group

Contingency Table(Observed Values)

25 and

under

Over

25 and

under

50

50 and

over

Claim 45 45 45 135

No

claim 55 55 55 165100 100 100 300

Age group

Contingency Table(Expected Values)


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fo fe (fo-fe)2/fe

40 45 0.56

35 45 2.22

60 45 5.00

60 55 0.45

65 55 1.82

40 55 4.09

Chi sq comp 14.14df 2

Level of

significance0.05

Chi sq

tabulated5.99 Thus reject H0


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THE MEDIAN TEST


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Example

An economist wants to

testy the nullhypothesis that medianfamily incomes in threerural areas areapproximately equal.

For simplicity, an equalsample size of 10 ineach population waschosen. The familyincomes are shownalongside

Region A Region B Region C

22 31 28

29 37 42

36 26 21

40 25 4735 20 18

50 43 23

38 27 51

25 41 16

62 57 30

16 32 48

Family incomes $1000 per year


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22 31 28

29 37 42

36 26 2140 25 47

35 20 18

50 43 23

38 27 51

25 41 16

62 57 30

16 32 48

Median 31.5



No of incomes less

than median4 5 6

No of incomes

greater than than

median

6 5 4



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Region A Region B Region C Total

No of incomes less

than median 4 5 6 15

No of incomes

greater than than

median

6 5 4 15

Total 10 10 10 30


Contingency Table(Observed Values)


No of incomes less

than median5 5 5 15

No of incomes

greater than than

median

5 5 5 15

Total 10 10 10 30

Contingency Table(Expected Values)Expected value =(10x15)/30


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fo fe (fo-fe)2/fe

4 5 0.2

5 5 06 5 0.2

6 5 0.2

5 5 0

4 5 0.2

Chi sq comp 0.8

df 2

Level of

significance0.05

Chi sq tabulated 5.99Thus accept H0

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The Chi Square Test.ppt