The document discusses the chi-square test, a non-parametric statistical method used to assess the difference between observed and expected frequencies in categorical data. It covers essential concepts such as degrees of freedom, conditions for applying the test, and its application in various scenarios, including fabric defect analysis and testing for associations between categorical variables. Examples illustrate how to perform the test, interpret results, and draw conclusions based on statistical significance.

1. DR. B.R. AMBEDKAR, NATIONAL INSTITUTE OF TECHNOLOGY, JALANDHAR
DEPARTMENT OF TEXTILE TECHNOLOGY
STATISTIC ASSIGNMENT ON
CHI SQUARE DISTRIBUTION
Submitted by: Submitted to:
Bhagwan Singh Rathore (20110017) Dr. A. Mukhopadhyay
Divya (20110019)
Jitendra Singh (20110031)
Manpreet Singh(20110041)
Ravi Kumar (20110064)
Sourav Kumar (20110077)

2. IMPORTANT TERMS
 PARAMETRIC TEST: The test in which, the population constants like mean,std deviation, std error,
correlation coefficient, proportion etc. and data tend to follow one assumed or established distribution
such as normal, binomial, poisson etc.
 NON PARAMETRIC TEST: the test in which no constant of a population is used. Data do not follow any
specific distribution and no assumption are made in these tests. E.g. to classify good, better and best we
just allocate arbitrary numbers or marks to each category.
 DEGREE OF FREEDOM: It denotes the extent of independence (freedom) enjoyed by a given set of
observed frequencies Suppose we are given a set of n observed frequencies which are subjected to k
independent constraints(restrictions) then, d.f. = (number of frequencies) -(number of independent
constraints on them)
Or we can say, df = (r - 1)(c - 1)
r = total number of rows &c= total number of columns
 When the table is prepared byenumeration of qualitative data by entering the actual frequencies, and if
that table represents occurance of two sets of events, that table is called the contingency table. (Latin,
con- together, tangere- to touch). It is also called as an association table.In other terms,where

3. INTRODUCTION
 The Chi-Square test is a statistical procedure for determining the difference between observed and expected data.
This test can also be used to determine whether it correlates to the categorical variables in our data. It helps to find
out whether a difference between two categorical variables is due to chance or a relationship between them
 Chi-square test was introduced by Karl Pearson as a test of association. The Greek Letter χ² is used to denote this
test.
 It can be applied when there are few or no assumptions about the population parameter.
 It can be applied on categorical data or qualitative data using a contingency table.
 Used to evaluate unpaired/unrelated samples and proportions.

4. DEFINITION
 A chi-square test is a statistical test that is used to compare observed and expected results. The goal of this test is to
identify whether a disparity between actual and predicted data is due to chance or to a link between the variables
under consideration. As a result, the chi-square test is an ideal choice for aiding in our understanding and
interpretation of the connection between our two categorical variables.
 A chi-square test or comparable nonparametric test is required to test a hypothesis regarding the distribution of a
categorical variable. Categorical variables, which indicate categories such as animals or countries, can be nominal
or ordinal. They cannot have a normal distribution since they can only have a few particular values.

5. CHARACTERISTICS OF CHI-SQUARE
The chi-square test has several key characteristics:
 It is non-parametric, meaning it does not assume a specific probability distribution for the data.
 It is sensitive to sample size; larger samples can result in more significant outcomes.
 It works with categorical data and is used for hypothesis testing and analysing associations.
 The test output provides a p-value, which indicates the level of significance for the observed
relationship between variables.
 It can be used with different levels of significance (e.g., 0.05 or 0.01) to determine statistical
significance

6. CHI SQUARE DISTRIBUTION
If X1, X2 .... Xn are independent normal variates and each is distributed normally with mean zero and
standard deviation unity, then (X1)^2 +(X2)^2+......+(Xn )^2= 1
𝑛
𝑥^2is distributed as chi square (c2)with n
degrees of freedom (d.f.) where n is large. The chi square curve for d.f. N = 1.5 and 9 is as follows.
If degree of freedom > 2: Distribution is
bell shaped.
If degree of freedom = 1 : Distribution is
L shaped with maximum ordinate at zero.
If degree of freedom <2 ( 0 <) :
Distribution L shaped with infinite
ordinate at the origin.

7. GENERAL FORMULA FOR CHI SQUARE TEST
Where,
c = Degrees of freedom
O = Observed Value
E = Expected Value

8. PROPERTIES OF CHI-SQUARE DISTRIBUTION
 The Mean of χ² distribution is equal to the number of degrees of freedom (n)
 The variance is equal to two times the number of degrees of freedom. i.e. the variance of equal to 2n
 The median of χ² distribution divides, the area of the curve into two equal parts.
 The mode of χ² distribution is equal to (n-2)
 Since Chi-square values always positive, the Chi-square curve is always positively skewed.
 Since Chi-square values increase with the increase in the degrees of freedom, there is a new Chi-square
distribution with every increase in the number of degrees of freedom.
 The lowest value of Chi-square is zero and the highest value is infinity.

9. CONDITIONS FOR APPLYING Χ² TEST:
following conditions should be satisfied before χ2 test can be applied.
 The data must be in the form of frequencies
 The frequency data must have a precise numerical value andmust be organised into categories or
groups.
 Observations recorded and used are collected on a randombasis.
 All the itmes in the sample must be independent.
 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. (Some statisticians take this number as 5, but 10 is regarded as better by most
of the statisticians.)
 The overall number of items must also be reasonably large. It should normally be at least 50.

10. APPLICATIONS OF A CHI SQUARE TEST
This test can be used in
 Goodness of fit of distributions
 test of independence of attributes
 test of homogeneity.

11. CHI-SQUARE TEST FOR DISCRETE DATA:
For discrete data, the chi-square test assesses the difference between observed and expected frequencies
within one categorical variable.
Where,
χ2 = chi-square statistic.
Oi = observed frequency for each category.
Ei = expected frequency for each category

12. Fabric Defects
Cotton 45
Polyester 30
Silk 25
A textile company is testing the quality of three different types of fabrics: cotton, polyester, and silk. They
collected data on the number of defects found in a sample of each fabric type. The observed frequencies
are as follows:
QUESTION -1
The company believes that the expected distribution of defects should be uniform, with each fabric type
having an equal chance of defects. Use the chi-square test to determine whether the observed frequencies
differ significantly from the expected frequencies at a 5% significance level.

13. SOLUTION:
Step 1: State the hypothesis.
Null hypothesis (H0): The observed frequencies do not differ significantly from the expected frequencies.
Alternative hypothesis (H1): The observed frequencies differ significantly from the expected frequencies.
Step 2: Calculate the expected frequencies.
Since the expected distribution is uniform, each fabric type should ideally have 100/3 = 33.33 defects.
Step 3: Calculate the chi-square statistic.
Using the formula:
χ2 = Σ ((O - E)2 / E)
where,
O = Observed frequency
E = Expected frequency

14. For the given data, the calculations are as follows:
χ2 = 4.08 + 1.33 + 2.08 = 7.49
Fabric Defects (O - E)2 χ2 = Σ ((O - E)2 / E)
Cotton 45 136.18 4.08
Polyester 40 44.4 1.33
Silk 25 69.38 2.08
Total 7.49

15. Step 4: Find the degrees of freedom (df).
Since there are 3 fabric types, df = n - 1 = 3 - 1 = 2.
Step 5: Find the critical value.
With a 5% significance level and 2 degrees of freedom, the critical value of chi-square from the table is approximately
5.99.
Step 6: Compare the chi-square statistic with the critical value.
Since 7.49 > 5.99, we reject the null hypothesis.
Step 7: conclusion.
Based on the chi-square test, there is significant evidence to suggest that the observed frequencies of defects in the
fabrics differ significantly from the expected frequencies.

16. QUESTION - 2
A textile company is analyzing the fineness of fibers used in their production process. They collected data on the
fiber diameters from a sample of 100 fibers. The observed frequencies for different ranges of fiber diameters are as
follows:
Diameter(micrometer) Fiber
10-15 25
15-20 30
20-25 20
25-30 15
30-35 10

17. SOLUTION
Step 1: State the hypothesis.
Null hypothesis (H0): The observed distribution of fiber diameters follows the hypothesized distribution.
Alternative hypothesis (H1): The observed distribution of fiber diameters significantly differs from the hypothesized distribution.
Step 2: Determine the expected frequencies.
If the distribution is assumed to be uniform, then each category should ideally contain 100/5 = 20 fibers.
Step 3: Calculate the chi-square statistic.
Using the formula:
χ2 = Σ ((O - E)2 / E)
where,
O = Observed frequency
E = Expected frequency

18. For the given data, the calculations are as follows:
For 10-15 micrometers: ((25 - 20)2 / 20) = 1.25
For 15-20 micrometers: ((30 - 20)2 / 20) = 2.50
For 20-25 micrometers: ((20 - 20)2 / 20) = 0
For 25-30 micrometers: ((15 - 20)2 / 20) = 1.25
For 30-35 micrometers: ((10 - 20)2 / 20) = 5.00
Diameter(micrometer) Fiber O-E (O-E)^2 Σ =(O-E)^2/E
10-15 25 5 25 1.25
15-20 30 10 100 5
20-25 20 0 0 0
25-30 15 -5 25 1.25
30-35 10 10 100 5
Total Σ =12.50
χ2 = 1.25 + 5 + 0 + 1.25 + 5.00 = 10.00

19. Step 4: Find the degrees of freedom (df).
Degrees of freedom (df) = number of categories - 1 = 5 - 1 = 4.
Step 5: Find the critical value.
With a 5% significance level and 4 degrees of freedom, the critical value from the chi-square table is
approximately 9.49.
Step 6: Compare the chi-square statistic with the critical value.
Since 12.50 > 9.49, we reject the null hypothesis.
Step 7: conclusion.
Based on the chi-square test, there is significant evidence to suggest that the observed distribution of fiber
diameters differs significantly from the hypothesized distribution.

20. CHI-SQUARE TEST FOR CONTINGENCY TABLES (INDEPENDENCE TEST):
For contingency tables (cross-tabulation) or when dealing with associations between two
categorical variables:
Where,
χ2 = chi-square statistic.
Oij = observed frequency in the cell of the contingency table.
Eij = expected frequency in the cell assuming no association or independence between the
variables.

21. QUESTION-3
A textile company produces and sells three types of fabric (Cotton, Silk, and Wool) in four different colours (Red,
Blue, Green, and Yellow). The company is interested in determining if there is a significant association between the
choice of fabric type and the colour of the fabric purchased by the customers. They collected data from 600
customers. The observed frequencies are as follows:
Red Blue Green Yellow Total
Cotton 100 90 70 40 300
Silk 60 80 50 30 220
Wool 40 60 30 20 150
Total 200 230 150 90 600

22. SOLUTION
First condition: Making the hypothesis
Null Hypothesis (H0) : There is no association between the choice of fabric type and the colour of the fabric is
purchased by the customers.
Alternative Hypothesis (H1) : There is an association between the choice of fabric type and the colour of the fabric is
purchased by the customers.
Now creating Contingency Table with Observed Frequencies:
Material Red Blue Green Yellow Total
Cotton 100 90 70 40 300
Silk 60 80 50 30 220
Wool 40 60 30 20 150
Total 200 230 150 90 600

23. Now Calculate Expected Frequencies:
Expected Frequency = (Row Total * Column Total) / Grand Total
Red Blue Green Yellow
Cotton 100 115 75 45
Silk 60 69 45 27
Wool 40 46 30 18
Calculate the Chi-Squared Statistic:
Chi-Square(χ2 ) = Σ [(Observed Frequency - Expected Frequency )^2 / Expected Frequency]
χ2 = [(100 – 100 )^2 / 100] + [(90 - 115)^2 / 115] + [(70 - 75)^2 / 75] + [(40 - 45)^2 / 45] + [(60 – 60 )^2 / 60] +
[(80 - 69)^2 / 69] + [(50 - 45)^2 / 45] + [(30 - 27)^2 / 27] + [(40 – 40 )^2 / 40] + [(60 - 46)^2 / 46] + [(30 - 30)^2 / 30] +
[(20 - 18)^2 / 18]
χ2 ≈ 15.76

24. Degrees of freedom (df) = (R - 1) * (C - 1),
where R is the number of rows and C is the number of columns in the contingency table.
In this case, df = (3 - 1) * (4 - 1) = 6.
Find the Critical Value:
To determine if the chi-squared statistic is statistically significant, we need to compare it to a critical value from
the chi-squared distribution table.
Assuming a significance level (alpha) of 0.05.
For df = 6, the critical value is approximately 12.591.
The calculated chi-squared statistic (15.76) is greater than the critical value (12.591) for a significance level of
0.05. Therefore, we reject the null hypothesis (H0).
Conclusion:
There is a significant association between the choice of fabric type and the color of the fabric purchased. The data
suggests that the choice of fabric type is not independent of the color of the fabric purchased.

25. QUESTION-4
Is there a significant association between the choice of manufacturing process (New Textile Manufacturing
Method or Conventional Textile Manufacturing Method) and the quality of textile products (Favourable Quality
or Non-Favourable Quality) in a sample of 250 textile products?
Favourable Quality Non-Favourable Quality Total
New Manufacturing
Method
140 30 170
Conventional
Manufacturing Method
60 20 80
Total 200 50 250
Observed Data:

26. SOLUTION
Favourable Quality Non-Favourable Quality Total
New Manufacturing (170*200)/250 (170*50)/250 170
Conventional
Method
(80*200)/250 (80*50)/250 80
Total 200 50 250
First condition: Making the hypothesis
Null Hypothesis (H0): There is no significant association between the choice of manufacturing
process and the quality of textile products.
Alternative Hypothesis (H1): There is an association between the choice of manufacturing
process and the quality of textile products.
Create a contingency table with the expected values (assuming no association between
treatment and outcome).

27. Step 2: Calculate the expected values.
Expected value for "Favourable" in the "New Manufacturing Method " group: (170 * 200) / 250 = 136
Expected value for "Non-Favourable" in the "New Manufacturing Method " group: (170 * 50) / 250 = 34
Expected value for "Favourable" in the "Conventional Manufacturing Method " group: (80 * 200) / 250 = 64
Expected value for "Non-Favourable" in the "Conventional Manufacturing Method " group: (80 * 50) / 250 = 16
Step 3: Calculate the chi-square statistic.
χ² = Σ [(O - E)² / E]
Where O is the observed value, and E is the expected value.
χ² = [(140 - 136)² / 136] + [(30 - 34)² / 34] + [(60 - 64)² / 64] + [(20 - 16)² / 16]
Now, calculate χ²:
χ² = (16/136) + (16/34) + (16/64) + (16/16) = 0.1176 + 0.4706 + 0.25 + 1 = 1.8382

28. Observed value(O) Expected value(E) O-E (O - E)² [(O - E)² / E]
140 136 4 16 0.12
60 64 -4 16 0.25
30 34 -4 16 0.45
20 16 4 16 1.00
Total 1.83
Step 4: Determine the degrees of freedom. Df = (number of rows - 1) * (number of columns - 1) = (2 - 1) * (2 - 1) = 1
Step 5: Look up the critical chi-square value in a chi-square table with 1 degree of freedom and desired significance
level (e.g., 0.05). Here the chi square table value=3.84
Step 6: Compare the calculated chi-square value (1.8382) with the critical chi-square value. So here the chi square
table is greater than calculated value.
Conclusion: (H0) is passed and accepted. So there is no significant association between the choice of manufacturing
process (New Textile Manufacturing Method or Conventional Textile Manufacturing Method) and the quality of
textile products (Favourable Quality or Non-Favourable Quality

29. QUESTION -5
A textile company is assessing the consistency in fabric strength across three distinct suppliers. They
have gathered fabric strength data from each supplier, with five samples obtained from each.
The fabric strength data, recorded in pounds-force, is as follows:
Supplier A: 50, 55, 60, 58, 53
Supplier B: 48, 52, 49, 50, 47
Supplier C: 55, 57, 60, 58, 56
The company is interested in conducting a statistical test to evaluate the homogeneity of fabric strength
across these suppliers. examine if the fabric strength measurements differ significantly among the three
suppliers.

30. SOLUTION
Step 1: Formulate the Hypotheses
Null Hypothesis (H0): The fabric strength is the same across all three suppliers (there is homogeneity).
Alternative Hypothesis (H1): The fabric strength is different across at least two suppliers (there is no
homogeneity).
Step 2: Organize the Data
Create a frequency table showing the observed frequencies for each category (supplier) and fabric
strength.
45-50 51-55 56-60
Supplier A 0 3 2
Supplier B 2 2 1
Supplier C 0 4 1

31. Step 3: Calculate Expected Frequencies
Calculate the expected frequencies assuming homogeneity.
The total number of observations is 15 (5 for each supplier).
45-50 51-55 56-60 Total
Supplier A 1 3 1 5
Supplier B 1 3 1 5
Supplier C 1 3 1 5
Step 4: Compute the Chi-Square Statistic
Use the formula:

32. For Supplier A, 51-55 category: (3−3) ^2/3=0
For Supplier A, 56-60 category: (2−1) ^2 /1=1
For Supplier B, 45-50 category: (2−1) ^2/1=1
For Supplier B, 51-55 category: (2−3) ^2/3=0.33
For Supplier B, 56-60 category: (1−1) ^2 /1=0
For Supplier C, 51-55 category: (4−3) ^2/3=0.33
For Supplier C, 56-60 category: (1−1) ^2/1=0
Hence the χ² = 2.67
Step 5: Determine Degrees of Freedom and Critical Value
Calculate the degrees of freedom. In this case, it is
(row-1)*(column-1) = (3-1)*(2-1)= 4
From chi-square distribution table the critical value for a significance level (e.g., α = 0.05) with the
degrees of freedom 4 = 9.488
Step 6: Conclusion:
Hence, the calculated value of χ² is less than the table value. So the Alternate hypothesis is true. The fabric
strength is different across at least two suppliers (there is no homogeneity).

33. LIMITATION OF A CHI SQUARE TEST
 The data is from a random sample.
 This test applied in a four fould table, will not give a reliable result with one degree of freedom if the
expected value in any cell is less than 5.
 In contingency tables larger than 2*2, Yate's correction cannot be applied.
 Interprit this test with caution if sample total or total of values in all the cells is less than 50.
 This test tells the presence or absence of an association between the events but doesn't measure the
strength of association.
 This test doesn't indicate the cause and effect, it only tells the probability of occurance of association by
chance.
 the test is to be applied only when the individual observations of sample are independent which means
that the occurrence of one individual observation (event) has no effect upon the occurrence of any other
observation (event) in the sample under consideration.