chi distributions
TRANSCRIPT
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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