AN APPLICATION OF THE χ2 [CHI – SQUARE] TOOL IN COTTON SPINNING

INSIDE

Warming Up

AN APPLICATION OF THE χ² [CHI – SQUARE] TOOL IN SPINNING

Textile Applications

Example 1

Table 1 : Neps per Gram of Card Slivers on a set of 36 cards

Table 2: Observed and Expected Frequencies for Example 1

Table 3 : Probability Points of the

Chi Square

Distribution

Simplified Procedure

Example 2

Table 4 : Number of Absentees from Work

Table 5 : Calculation of Chi square for Example 2

Going Beyond to Identify and Plan Action

Going Beyond to Perform Product and Process Analysis

Example 3 : Case – Health Checks at Spinning Preparation Process

Table  6 : Mill Studies : Nep Levels at Spinning Preparation and Process Analysis

Fine Print

Annexure I : Sampling Distribution of the Variance

Further Readings

End Note

Warming Up: Three in the Morning……

A Simple Sum [3+4=7*] For Chief Executive Officers

WHAT IS THIS THREE IN THE MORNING?

IT IS ABOUT A MONKEY TRAINER
WHO WENT TO HIS MONKEYS AND TOLD THEM:
"AS REGARDS YOUR CHESTNUTS,
YOU ARE GOING TO HAVE THREE MEASURES IN THE MORNING,
AND FOUR IN THE AFTERNOON."

ON HEARING THIS ALL THE MONKEYS BECAME ANGRY.
SO THE KEEPER SAID:
"ALL RIGHT THEN,
I WILL CHANGE IT
TO FOUR MEASURES IN THE MORNING,
AND THREE IN THE AFTERNOON."
THE ANIMALS WERE SATISFIED WITH THIS ARRANGEMENT.

THE TWO ARRANGEMENTS WERE THE SAME --
THE NUMBER OF CHESTNUTS DID NOT CHANGE,
BUT IN ONE CASE THE MONKEYS WERE DISPLEASED,
AND IN THE OTHER CASE THEY WERE SATISFIED.

THE KEEPER WAS WILLING
TO CHANGE HIS PERSONAL ARRANGEMENT
IN ORDER TO MEET OBJECTIVE CONDITIONS.
HE LOST NOTHING BY IT.

THE TRULY WISE MAN,
CONSIDERING BOTH SIDES OF THE QUESTION
WITHOUT PARTIALITY,
SEES THEM BOTH IN THE LIGHT OF TAO.
THIS IS CALLED FOLLOWING TWO COURSES AT ONCE.

   -adopted from osho's vetruppadagu

* In calculating the expected frequencies in

Chi Square

distribution, we make sure that their sum was equal to the total observed frequencies. The values  are not, in fact, all independent.

AN APPLICATION OF THE χ² [CHI – SQUARE] TOOL IN SPINNING

TEXTILE APPLICATIONS

One of the most widely used analytical methods for dealing with data obtained by counting, is based on the χ2 distribution. 

The simplest way of obtaining numerical data is to count the number of times an event happens. The value obtained  by this procedure can only be whole numbers, or integers, and the variables whose values are found in this way are called discrete or discontinuous.

Examples of discrete variables are:

(a)    neps  per gram in raw cotton; blow room output; card sliver; comb feed and comb sliver;

(b)   sliver breaks in draw frame; roving breaks in speed frame and end breaks in spinning;

(c)    clearer cuts at auto-coners; clearer cuts at open end rotor spinning;

(d)   yarn imperfections – number of thin places; thick places and neps per kilometer of yarn;

(e)    number of yarn faults per hundred kilometer of yarn;

(f)     breaks at warping; breaks at sizing and loom breaks;

(g)    fabric defects in 100 m length of fabric;

(h)    the number of absentees per day in a mill.

The concept of sample variance and the capitulation of chi-square distribution is given in Annexure-I.

Example 1

Neps per gram of card slivers as tested using AFIS on a set of 36 carding machines processing S-6 cotton is given in Table-1.

An examination of the data reveals that the lowest  recorded values is 82 on card no 17 and the highest 142 on card no 3. The question that arises is whether the data really imply that there are significant differences in the nep levels of 36 cards or not.

Questions of this kind are answered by tests of significance. The appropriate null hypothesis in this case is

H₀: There are no difference in nep levels among the cards.

The alternate hypothesis is

H₁:There are differences in nep levels among the cards.

Table 1 : Neps per Gram of Card Slivers on a set of 36 cards
Card No	Neps/Gram	Card No	Neps/Gram	Card No	Neps/Gram
1	82	13	106	25	102
2	98	14	88	26	100
3	142	15	110	27	116
4	90	16	86	28	90
5	118	17	82	29	140
6	104	18	84	30	136
7	126	19	120	31	86
8	109	20	94	32	102
9	84	21	106	33	128
10	92	22	107	34	98
11	118	23	130	35	112
12	120	24	110	36	116

Table 2: Observed and Expected Frequencies for Example 1
Card No	O	E	O-E	(O-E)2/E
1	82	106.4	-24.4	5.6
2	98	106.4	-8.4	0.7
3	142	106.4	35.6	11.9
4	90	106.4	-16.4	2.5
5	118	106.4	11.6	1.3
6	104	106.4	-2.4	0.1
7	126	106.4	19.6	3.6
8	109	106.4	2.6	0.1
9	84	106.4	-22.4	4.7
10	92	106.4	-14.4	2.0
11	118	106.4	11.6	1.3
12	120	106.4	13.6	1.7
13	106	106.4	-0.4	0.0
14	88	106.4	-18.4	3.2
15	110	106.4	3.6	0.1
16	86	106.4	-20.4	3.9
17	82	106.4	-24.4	5.6
18	84	106.4	-22.4	4.7
19	120	106.4	13.6	1.7
20	94	106.4	-12.4	1.5
21	106	106.4	-0.4	0.0
22	107	106.4	0.6	0.0
23	130	106.4	23.6	5.2
24	110	106.4	3.6	0.1
25	102	106.4	-4.4	0.2
26	100	106.4	-6.4	0.4
27	116	106.4	9.6	0.9
28	90	106.4	-16.4	2.5
29	140	106.4	33.6	10.6
30	136	106.4	29.6	8.2
31	86	106.4	-20.4	3.9
32	102	106.4	-4.4	0.2
33	128	106.4	21.6	4.4
34	98	106.4	-8.4	0.7
35	112	106.4	5.6	0.3
36	116	106.4	9.6	0.9
Totals	3832	3832	0.0	94.5

We begin, by assuming the null hypothesis is true. If it is, we should expect the average nep level of each card at 106.4. i.e the total of 3832 would be equally divided among the 36 cards [3832/36=106.4].  What is now of primary interest is how the observed [O] frequencies shown in Table 1 deviate from the expected [E] frequencies just calculated. The deviations are shown in Table 2 Col 4[O-E]; obviously the greater the deviations, the less credence we would give to the null hypothesis. We therefore need to develop a method for concisely summarizing the deviations that will form the basis of a test of significance.

Imagine any one of the cards being observed over many days, the neps  per gram in each day being tested. The neps levels will vary from day to day and, we would expect that if it is at random, this variation could be described by a poisson distribution. Consequently, the observed frequencies [O] during the single day can be thought of as values of a poisson variable. Our best estimate of the mean of this distribution (if H0 is true) is the expected frequency [E], and, because for a poisson variable the standard deviation is equal to the square root of the mean, an estimate of is standard deviation is √E.

μ = E,     σ =√E                                   [equation 1]

Further that, provided that the mean is large enough, the normal distribution can be used as an approximation to the poisson distribution. Thus, under these circumstances,  the observed frequencies [O] can be regarded as values of a normal distribution with mean and standard deviation given by equation 1. The linear transformation of any normal distribution into the standard normal distribution with zero mean and unit standard deviation  is

U = (O- μ) /σ  = (O-E)/√E

The values of

U² = (O-E)²/E

are given in Column 5 of Table 2.

Now in Annexure I, the sum of squares of independent standard normal variables was seen to have a χ2 distribution. Hence the sum of the entries in the last column of Table 2 will tend to have such a distribution, and this fact forms the basis of the test of significance. Before we can proceed further, however, we need to know how many degrees of freedom are associated with our value of χ2. It was emphasized above that χ2 is the sum of squares of independent standard normal variables. The values of u in this example are not, in fact, all independent. The  reason for this is that, in calculating the expected frequencies, we made sure that their sum was equal to the total observed frequencies i.e 3832[Last row of Column 2 and 3 in Table 2]. Consequently, the deviations [O-E] always add to zero. [Last row of Column 4] Knowing this, we find that only 35 of the deviations are independent; for, if 35 of them are known, the 36^th is determined, since the sum of the deviations must be zero.

Table 3 : Probability Points of the Chi Square Distribution
Degree of freedom		areain right tail=0.05
1		3.841
2		5.991
3		7.815
4		9.488
5		11.070
6		12.590
7		14.070
8		15.510
9		16.920
10		18.310
12		21.030
14		23.680
16		26.300
18		28.870
20		31.410
25		37.650
30		43.770
35		49.765
40		55.760
45		61.630
50		67.500

For this example, therefore, the observed value of χ2 is

χ² = 94.5 [Last row of column 5 in Table 2]

with 35 degrees of freedom, and this can be used to test the null hypothesis.

Now, large deviations from the expected frequencies are reflected in large values of χ^2; hence the  significance level of the example is the probability of finding χ^{2 Table} values greater than χ^{2 calculated.}

The probability points of the χ²  distribution for significance level α =0.05 is reproduced from a standard statistical book in Table 3 for  ready reference. χ^{2 Table} for our example is 49.77 with 35 degrees of freedom.

Since χ^{2 calculated.}  > χ^{2 Table}  , we reject the null hypotheis, we conclude that there are differences in nep levels among the cards.

SIMPLIFIED PROCEDURE

The procedure described above for calculating χ² has been developed with reference to a specific example, the technique is, in fact, can be used for analyzing data arising in  a number of apparently different situations. Few textile applications are given in Introduction.

What characterizes them all, is that

(a) a set of expected frequencies [E] can be calculated, based on the assumption that a certain null hypothesis is true and

(b) these have to be compared with some observed frequencies [O] to provide a test of the hypothesis.

(c) calculate χ² = (O-E)²/E and

(d) compare the result with values of χ^{2 Table} in Table 3.

Some further examples will make the application perspective clear.

 Example 2

In the month of Oct, the number of absences from work for male and female employees were recorded in a spinning mill with the results shown in Table 4. The number of absences per employee for males and females are shown in the last row of this table and suggest that there may be differences in absentee rate between male and female employees. We therefore ask whether these differences are significant.

Table 4 : Number of Absences from Work
Mill	Male	Female	Totals
Average number of	190	95	285
Employees during Oct
Number of Absences	28	19	47
during Oct
Absences per employee	0.1474	0.2000	0.1649

The null hypothesis is that the absentee rate is the same for both male and female employees. To get an estimate of this overall absentee rate, we combine the data and find

Overall absentee rate

     Total number of absences            47

= ---------------------------------  = -----------

    Total number of employees         285

=0.1649 absences per employee

If the null hypothesis is true, this rate would be expected to apply to both male and female. Thus the expected number of absences for male employees would be equal to the overall absentee rate x number of male employees = 0.1649 x 190 =  31.3 and that for female employees would be equal to 0.1649 x 95 = 15.9. This is shown in Table 5.

The details of the calculation of χ2 are also shown in this table, the result being χ^{2 calculated.}

=1.064. The degrees of freedom are 1, since only one of the deviation is independent, the other must be such that their sum is zero [See EndNote]. Comparison with χ^{2 Table}   Value of 3.841 indicates that it is rather a low probability and  we conclude that there are no difference among absentee rate between male and female employees.

GOING BEYOND TO IDENTIFY AND PLAN ACTION

Taking the Example 1 further, to identify the cards with large deviations, individual card values are examined as per procedure and χ^{2 calculated} is given in Column 5 of Table 2 against individual cards.

 χ^{2 Table} value to compare is 3.84 with 1 degrees of freedom. So far as the number of degrees of freedom is concerned, the only general rule that can be stated is that it is equal to the number of independent deviations, all totals being regarded as fixed. [See Warming Up and End Note for examples]

Examination of Column 5 in Table 2 reveals that χ^{2 calculated.}  is greater than  χ^{2 Table,} in cards 1,3,9,16,23,29,30,31,33 where cards 1,9,16 and 31 have significantly lower nep levels. The cards 3,23,29,30 and 33 have higher level of neps and needs to be attended.

Similarly, the analysis of stoppages on a set of ring frames can be extended to identify the machine numbers with poor performance and type of stoppages, such as spindle breaks, multiple breaks and creel breaks / replacements. Here the reference is made to Chapter 7 Means to Improve Productivity in ATIRA Silver Jubilee Monograph “Process Control in Spinning”.

GOING BEYOND TO PERFORM PROCESS AND PRODUCT ANALYSIS

Example 3, adopted from ATIRA publication on Denim details, health checks carried out at spinning preparation in 7 different mills spinning coarse count ring and open end rotor yarns. This is summarized as a case study, so as to provide the reader, the horizon of application to perform the process analysis, product analysis and to derive standards with little imagination.

Example 3

HEALTH CHECKS AT SPINNING PREPARATION PROCESS –MILL STUDIES

Quality of Feed Sliver

Standards for Nep Levels

Particularly in ring spun Denim yarn, a high incidence of nep in the yarn will cause uneven dye uptake during the warp yarn preparation. It is observed that modern blow room lines create neps, upto an increase of 100% over neps in raw cotton,  that will still allow the carding machine to be able to remove most of these objectionable faults. The nep content expressed in neps/gram as tested using AFIS should not exceed 150.

Mill Studies: Health Checks at Spinning Preparation Processes

In addition to level of neps at feed sliver, AFIS [Advanced Fibre Information System] can also be used as a powerful process control tool for scanning the processes that precede yarn manufacturing periodically.-

One such analysis, done Product wise and Process wise in seven different mills, four open end spinning and three ring spinning, is summarized in Table 6

The product wise analysis refers to Level of neps at mixing, Blow room and Card sliver. The  process wise analysis refers to nep increase at Blow room and nep reduction at card.

It is important to arrive the product standards, considering the end use requirement for card sliver. Once the level of card sliver neps required are decided,  the level of neps required at blow room material and target level at mixing can be arrived. This is  by working backwards with the use of   process standards for nep increase in blow room and nep reduction at cards.

The standards used in the analysis are as follows:

Nep/gram at Card Sliver      : 120

                     Card Feed       : 400

                     Mixing            : 200

Nep increase at Blowroom: 200%

Nep reduction at Card         :    70%       

Table  6 : Mill Studies : Nep Levels at Spinning Preparation and Process Analysis
UNIT	PRODUCT ANALYSIS			PROCESS ANAYSIS
	Mixing	Blowroom	Sliver	Blowroom	Carding	Remarks
	Neps/gram			Nep Inc.%	Nep Red %.
OE Mill 1	218	467	221	214%	53%	Cards 2,4 and 5
OE Mill 2	204	607	163	298%	73%	Cards 16,18 and 21
OE Mill 3 Feeder 1	129	278	104	216%	63%	Card 3
OE Mill 3 Feeder 2	129	360	102	279%	72%	Card 11
OE Mill 4 Feeder 1	199	451	165	227%	63%	Card 2
OE Mill 4 Feeder 2	200	670	383	335%	43%	Cards 10,11,15 and 16
OE Mill 4 Feeder 3	183	269	127	147%	53%	Card 24
OE Mill 4 Feeder 4	162	222	164	137%	26%	Card 29 and 31
RS Mill 5 Feeder 1	161	398	161	247%	60%	Cards 2 and 5
RS Mill 5 Feeder 2	168	326	90	194%	72%	No card
RS Mill 6 Feeder 1	149	413	137	277%	67%	Card 3
RS Mill 6 Feeder 2	149	310	129	208%	58%	No card
RS Mill 7 Feeder 1	241	362	96	150%	73%	Card 6
RS Mill 7 Feeder 2	241	340	101	141%	70%	No card
Standards	200	400	120	200%	70%
Tolerance Values	242	460	150	230%	62%

As can be seen from the Table, the blow room nep generation was high at OE Mill1, OE

Mill2 and OE Mill 4 Feeder 2.  It is observed that Long and bent material transport duct lines were the reasons for nep generation in OE Mill 4 Feeder 2.

Slivers made from OE Mill1, OE Mill2, OE Mill 4 Feeders 1, 2 and , RS Mill 5 feeder1 had excessive neps. It is also observed, that the overall rep removal efficiency of cards in OE  Mill 1 and 4 were very poor due to poor condition of wire points.

In addition, the cards who’s performance was too poor than the rest were identified and given in remark – though individual values could not find place in this summarized report.

Fine Prints

It must be noted that  the formula χ²  = (O-E)²/E was derived by using the assumption that the Poisson distribution can be approximated by the normal distribution. This requires that the expected frequencies should be reasonably large. A general rule is that all the expected frequencies should be greater than 5, though there is some evidence to suggest that in some applications viz seed coat neps  per gram especially in card sliver; comb feed and comb sliver; yarn imperfections especially number of thin places; classimate yarn faults, this is rather conservative.

ANNEXURE I  : SAMPLING DISTRIBUTION OF THE VARIANCE

Here, we shall first introduce the concept of sample variance and then present the chi square distribution which helps us in working out probabilities for the samle variance, when the population is distributed normally.

Sample Variance

                                                                                                         

We use the sample mean to estimate the population mean, when the parameter is unknown. Similarly, we use a sample statistic called the sample variance to estimate the population variance. The sample variance is usually denoted by s² and it again captures some kind of an average of the square  of deviations of the sample values from the sample mean.   Let us put it in an equation form

s² =∑(x-x)²/(n-1)

By comparing this expression with the corresponding expression for the population variance, we notice two differences. The deviations are measured from the sample mean and not from the population mean and secondly, the sum of squared deviations is divided by (n-1) and not by n. Consequently, we can calculate the sample variance based only the sample values without knowing the value of any population parameter. The division by (n-1) is due to a technical reason to make the expected value of s² equal σ², which it is supposed to estimate.

Chi-square Distribution

A chi-square distribution is known by its only parameter viz. the degrees of freedom.

 If x is a random variable having a standard normal distribution, then x² will have a chi-square distribution with one degree of freedom.

If Y₁ and Y₂ are independent random variables having chi-square distributions with v₁ and v₂ degrees of freedom respectively, then (Y₁ + Y₂) will have a chi-square distribution with v₁₊ v₂ degrees of freedom.

Finally, if a normal population with mean μ and variance σ2 and if the sample mean x  and the sample variance s²,  then (n-1) s²/ σ² will have a chi-square distribution with (n-1) degrees of freedom – a result that has a number of important applications.

FURTHER READINGS

Levin,R.I, 1987. Statistics for Management, Prentice-Hall of India: New Delhi.

Shroff.J.J, Bhatt.S.R, Appasamy.N.R.V,2009.Denim, ATIRA.

Garde.A.R, Subramanian.T.A, Third Edition,1987. Process Control in Spinning, ATIRA Silver Jubilee Monographs.

Leaf.G.A.V. Practical Statistics for the Textile Industry, 1987.The Textile Institute Quality Control and Assessment Series.

End Note: Degree of  Freedom#……

There is an anecdote about Prophet Mohammad and his disciple Ali.

Once Ali asked him,” Gurudev, as a human being what is my freedom. Am I slave to my destiny or fate? Am I completely dependent on fate? If its true then what is the use to do efforts as I will get what is destined for me and I can never get things which are not decided for me. And If there is no role of fate in the life of man and there is no control of some power then why so much stress on following right karmas. What is the difference between a right and wrong karma then?”

Prophet smiled and he ordered him,”Ali, lift one of you leg and stand on one foot only”.

Ali was astonished to hear this but he followed the sayings of his Guru and stood on his left foot only and lifted the right foot.

He saw towards prophet who was looking at him.

Prophet said,” Now lift your left foot also”.

Ali,”What are you saying Gurudev. How I can lift my left foot now. I will fall down”.

Prophet,” Ali, that’s the answer to your question. You have freedom to decide which foot you want to lift first but after lifting either of the foot you don’t have freedom to lift another foot as now your Karmas will decide your destiny.

#A chi-square distribution is known by its only parameter viz. the degrees of freedom. Here the degree of freedom for two legs is one [2-1=1].