Chi Distributions

8/9/2019 Chi Distributions

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Chi Square Distributions

Prof G.R.C.Nair


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Applications

To test if, more than two populationproportions can be considered equal.

To determine if two attributes areindependent of each other.

To test goodness of fit .

To test the discrepancies betweenobserved and expected frequencies.


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1. Continuous distribution.

2. Assumes non negative values only .

3. Chi square distribution curve starts at the originand lies entirely to the right of Y axis.

4.Parameter degrees of freedom (d f) also figure.

5.Shape of Chi square distribution curve is skewed

for very small d.f and changes drastically as d.fincreases . For large d.f, Chi-square distribution

looks like a normal distribution curve.

Characteristics


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Conditions for use

1.Used for large samples.

2.All expected frequency > 10 for good

accuracy, and should be minimum 5.

3.Take Ho: There is no significant

difference between the sample

proportions or between the observedand the corresponding expected values.


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4. If the degrees of freedom and area

required in the right tail (ie, significance

level of test ) are given, the criticalvalues of chi-square can be found from

the table.

5. If the Chi sq value got is > criticalvalue, H0is rejected.


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Goodness Of Fit Test

This is a test to see if a particular

theoretical probability distribution is

appropriate in a specific case.Frequencies obtained from the actual

performance of an experiment /

observation in a survey are calledobserved frequencies: fO


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We test the null hypothesis that these observed

frequencies follow a certain expected frequency

as per a specific theoretical distribution fE. The

hypothesis tested is how good the observed

frequencies fit a given probability distribution.

H0: The observed values follow the selectedprobability distribution.

H1 : They do not follow.


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Test statistic for a Chi square test is calculated

as G =7 ( fO- fE) 2/ fE

Degrees of freedom = n-1, where n denotes thenumber of possible outcomes/samples.

The expected frequency for each categoryshould not be < 5. If there is a category with anexpected frequency of less than 5 either

increase the sample size or combine two ormore categories to make expected frequencyequal to 5.


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Contingency table

For testing association of attributes.

Expected frequency in a contingencytable to be calculated by the formula,

fE =T R x T C / n. and d.f = (c-1)(r-1)

H0:Attributes have no association, ie

independent

H1: they have association ie, dependent


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Example -1

The number of car accidents in a city

during the last 10 months was as follows

20,17,12,6,7,15,8,5,16, and 14. Does it

agree with the general belief that the

number of accidents are more or less the

same every month? Test at 10%Sig level.


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H0 : No of accidents are same every month

H1: They are not the same

Expected number of accidents everymonth, if they are the same =120/10=12

G!7_fO-fE)2/fEa!_

a_aetc !

This is " the table value for d.fat Sign Level, which is 14.684.

So reject H0


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Example - 2

The following table shows the goals made in the world cup

matches. Test if they are as per the Poissons distribution

at 20%S.L

No of goals 0 1 2 3 4 5 6 7

No of match 95 158 108 63 40 9 5 2

H0: Poisson Distribution can be fitted well, H1: Do not fit well

Mean no of goals Q=

(95*0+1

58*1+10

8*2+

.) / (95+1

58+10

8+

)=1

.7

Expected frequency as per Poisson fE= n* e-Q Qx/x!


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No of goals f O fE (fO-fE)2/fE

0 95 88 49/88=0.56

1 158 150 64/150=0.43

2 108 126 324/126=2.57

3 63 72 =1.12

4 40 30 =3.33

5 9* 10* =0.29

6 5* 3*

7 2* 1*

7 * Combine 8.3d.f= 5, S.L=20% G critical= 7.289

calculated value is > critical value. So reject H0.

Poisson distribution do not fit well.


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Example 3

100 students selected randomly from 10,000 doing MBA,were cross classified by their age at entry and final grade

obtained. Test if there is any association between age at

entry and final grade obtained at 5%.

Grade under 21yrs 21 to24 Over24 Total

Up to 3 6 9 5 20

3.1 to 3.5 18 14 8 40

3.6 to 4.0 11 12 17 40

Total 35 35 30 100


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Expected frequency Table

Grade under 21yrs 2

1to24 Over24

Total

Up to 3 7 7 6 20

3.1 to 3.5 14 14 12 40

3.6 to 4.0 14 14 12 40

Total 35 35 30 100

Gcalculated =6.37 d.f= 4, S.L=5% G critical = 9.49

calculated value is < critical value. So cannot reject H0.

Age and Grade points are independent. ( ie no association)


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Example 4

In a survey of 200 boys, of the 75 found to

be intelligent,40 had skilled fathers; while

85 of the unintelligent boys had unskilledfathers. Do these figures support the

hypothesis that skilled fathers get intelligent

boys? Use G2 test at 5% significance.


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Example 5 / HW

A number of managers from various levels were randomlyselected and interviewed for their concern for environmentalissues. The response of each persons was tallied into one of thethree categories as given below:

Use 0.01 significance level to determine whether there is anyrelation ship between the level ofmanagement andenvironmental concern.

No concern Some concern Great Concern

Top level 15 13 12

Middle level 20 19 21

Supervisor 7 7 6

Group leader 28 21 13


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FurtherExamples / HW

2009 mid term make up Q. 6

2007Terminal Part B. Q. 3

2009 terminal part B Q.3

2007 make up Terminal, 7

2007 mid term 2?,4

Mid term 2009 Part C . Q. 6

2007 Mid term Make up Q. 5

Chi Distributions

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