CHI SQUARE THEORY AND PRACTICAL PROBLEMS

1. CHI SQUARE
BY:
M RS.KEERTHI SAM UEL
ASST.PROFESSOR
VIJAY M ARIE CON

2. INTRODUCTION TO TERMS
1. PARAMETRIC TEST: The test in which the population constants like mean,
standard deviation, standard error ,correlation coefficient etc. and data tends to follow
one assumed or established distribution such as normal, Poisson etc.
2. NON PARAMETRIC: The test in which no constant of a population is used. Sdata do
not follow any specific distribution and no assumptions are made in these tests. Ex:
classifying good, better best.
3. HYPOTHESIS: it is a definite statement about the population parameters.

3. INTRODUCTION TO TERMS
4. NULL HYPOTHESIS: states there is no association between two cross tabulated variables in the population
and therefore the variables are statistically independent.
Ex: if we want to compare 2 methods A and B and if the assumption is that both the methods are
equally good is null hypothesis.
5. ALTERNATIVE /RESEARCH HYPOTHESIS: proposes that the two variables are related in the population. If
we assume that from 2 methods , method A is superior than method B , then this assumption is called Alternative
hypothesis.
6.CONTINGENCY TABLE: when the table is prepared by enumeration of qualitative data by entering the actual
frequencies , an if the table represents occurrence of two sets of events , the table is called contingency table. It
is also called association table.

4. INTRODUCTION TO TERMS
7. DEGREES OF FREEDOM: it denotes to the extent of independence (freedom) enjoyed by a
given set od observed frequencies . Suppose we are given a set of n observed frequencies which
are subjected to k independent constraints ( restrictions) then,
df=(number of frequencies)-(number of independent constraints on them)
In other terms df=(r-1)(C-1)
WHERE r= no of rows
c=no of columns

5. INTRODUCTION TO CHISQUARE
•The chi-square test is an important test amongst the several tests of significance developed by
statisticians.
•It was developed by Karl Pearson in 1990.
•It is a non parametric test not based on any assumption or distribution of any variable.
•This statistical test follows a specific distribution known as chi-square distribution.
•In general the test we use to measure the differences between what is observed and what is
expected according to an assumed hypothesis is called chi-square test.

6. CHARACTERISTICS
•The test is based on frequencies and not on the parameters like mean and SD.
•The test is used for testing the hypothesis and is not useful for estimation
•Applied to a complex contingency table with several classes and such a very useful test in
research work.
•This test is an important non parametric test as no rigid assumptions are necessary in regard to
the type of population , no need of parameter values and relatively less mathematical details are
involved.

7. Chi square distribution
•If 𝑥1 , 𝑥2 …. Are independent normal variates and each is
distributed normally with a mean zero and SD as unity , then
X12+X22+…….Xn2 = ∑Xi2 is distributed as chi-square with n
degrees of freedom when n is large. The chi-square curve for df
N=1,5 and 9 is as follows
• if DF >2: distribution is bell shaped
•if DF =2: distribution is L shaped with maximum ordinate 0
•if DF >2: distribution is L shaped with infinite ordinate at the origin

8. APPLICATIONS
•Goodness of fit of distributions
•Test of independence attributes
•Test of homogeneity

9. TEST OF GOODNESS OF FIT
•This test enables us to see how well does the assumed theoretical distribution ( such as
binomial or normal distribution) fit to the observed data.
•Formula 𝑥2
=
𝑜−𝑒 2
𝑒
•Where o= observed frequency and e=expected frequency
•If 𝑥2caluculated is > than table value then null hypothesis is rejected

10. TEST OF independence of attributes
•Test enables us to explain whether or not two attributes are associated.
•For instance, we may be interested in knowing whether a new medicine is effective in controlling fever or not, 𝑥2
test
is useful.
•In such situation we proceed with the null hypothesis that the 2 attributes ( new medicine and control of fever) are
independent which means the new medicine is not effective in controlling fever.
•The calculated value is greater than table value at a certain level of significance for a given df the H0 is rejected and
viceversa.
•When H0 is rejected it can be concluded that there is a significant association between 2 attributes.

11. TEST OF HOMOGENITY
•This test can also be used to test whether the occurance of events follow uniformity or not
• Ex: admission of patients in govt hospital in all days of week is uniform or not
• Chisquare less than tabulated value then null hypothesis is accepted , and it can be concluded that there is a uniformity in the
occurance of the events ( uniformity of patients getting admitted throughout the week)

12. calculation
• 𝑥2 =
𝑜−𝑒 2
𝑒
• Where o= observed frequency and e=expected frequency
• If 𝑥2
calculated is > than table value then null hypothesis is rejected
•If two distributions are exactly alike chisquare is zero which is mostly due to sampling error.
• generally it is not equal to zero

13. Steps in calculation
1. Calculate the expected frequencies and the observed frequencies
2. Expected frequencies: the cell frequencies that would be expected in a contingency table if the two variables
were statistically independent.
3. OBSERVED FREQUENCIES: the cell frequencies actually observed in a contingency table.
𝑓𝑒= (column total)(row total)
N
To obtain the expected value for nay cell in any cross tabulation in which the 2 variables are assumed independent ,
multiply the row and column totals for that cell and divide the product by the total number of cases in the table.

14. CONDITIONS FOR APPLICATION
1. The data must be in the form of frequencies
2. The frequency data must have a precise numerical value and must be organized into categories or groups
3. Observations recorded and used are collected on a random basis.
4. All the items in the samples must be independent.
5. No group should contain very few items, say less than 10. in case where the frequencies are less than 10,
regrouping is done by combining the frequencies of adjoining groups so that the new frequencies become
greater than 10.
6. The overall number of items must be reasonably high . It should be generally be at least 50.

15. Yates correction
If the 2x2 contingency table , the expected frequencies are small say less than 5 , the chi-square test cant be used .
In that case the direct formula of the test cant be used . So we can use yates correction formula.

16. LIMITATIONS
◦ The data is from a random sample
◦ This test is applied in four fold table , will not give a reliable result with one degree of freedom if the expected value
in any cell is less than 5. in such case yates correction is necessary.
◦ Even if yates correction, the test may be misleading if any expected frequency is much below than 5. in that case
another appropriate test should be applied
◦ In contingency tables larger than 2X2. yates correction cannot be applied.
◦ Interpret this test with caution if sample total or total of values in all cells is less than 50.

17. Problem-1
◦ In an antimalarial camp ,in a certain area chloroquine was administered to 812 people out of the population of
3248. The number of fever cases shown are:
TREATMENT FEVER NO FEVER
CHLOROQUIN 20 792
NO CHLOROQUIN 220 2216

18. Problem-1
◦ 1. Formulate the null and alternate hypothesis
◦ H0: chloroquine is not effective against malaria
◦ H1: chloroquine is effective against malaria
◦ 2. Make contingency table
Treatment Fever No fever Total
Chloroquin 20 792 812
No chloroquin 220 2216 2436
Total 240 3008 3248

19. Problem-1
◦ Find out the expected frequencies
◦ E=RTXCT/ Grand total
E1=240X812/3248 =60
E2=240X2436/3248 = 180
E3=3008X812/3248 = 752
E4 = 3008 X 2436 / 3248 =2256

20. Problem-1
o e O-e (O-e)2 (O-e)2/e
20 60 -40 1600 1600/60=26.6
220 180 +40 1600 1600/180=8.8
792 752 +40 1600 1600/752=2.12
2216 2256 -40 1600 1600/2256=0.7
38.39
Df=(2-1)(2-1)=1
At 5% level of significance with1 as df the table value of 𝑋2 is 3.84 which is less than the calculated value .
Hence the hypothesis H0 is rejected and H1 is accepted.

21. Problem-II
◦ The table shown is given the data during the epidemic of cholera. Check the effect of inoculation on cholera.
TREATMENT ATTACKED NOT ATTACKED
Inoculated 31 469
Not Inoculated 185 1315

22. Problem-II
◦ 1. Formulate the null and alternate hypothesis
◦ H0: there will be no effect of inoculation in preventing the attack of cholera
◦ H1: there will be some effect of inoculation on cholera
2. Make contingency table
TREATMENT ATTACKED NOT ATTACKED TOTAL
Inoculated 31 469 500
Not Inoculated 185 1315 1500
Total 216 1784 2000

23. Problem-II
◦ Find out the expected frequencies
◦ E=RTXCT/ Grand total
E1=500X216/2000 =54
E2=216X1500 / 2000 = 162
E3=500X1784 / 2000= 446
E4 = 1500 X 1784 / 2000 =1338
TREATMENT ATTACKED NOT
ATTACKED
TOTAL
Inoculated 31 469 500
Not Inoculated 185 1315 1500
Total 216 1784 2000

24. Problem-II
o e O-e (O-e)2 (O-e)2/e
31 54 -23 529 9.79
185 162 23 529 3.26
469 446 23 529 1.18
1315 1338 23 529 0.39
14.62
Df=(2-1)(2-1)=1
At 5% level of significance with1 as df the table value of 𝑋2 is 3.84 which is less than the calculated value .
Hence the hypothesis H0 is rejected and H1 is accepted.