Download - 1117 Final
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SITRA
Interpretation of Test Data
Part - I
1
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SITRA 2
INTERPRETATION OF FIBRE
AND YARN DATA
- With the advent of large no. of testing
instruments, it has become possible to test
various aspects of the quality of products
and a large volume of data is presented to
the manager for decision making
- Interpretation of test data is made complex in
view of the instrumental & sampling errors
associated with them
- These variabilities have been expressed in
terms of critical difference
- With the help of such values the test data can
be interpreted meaningfully
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SITRA 3
The definition of
some statistical methods
1. The Critical Difference
To Compare Two Mean
Values where a Prior
Knowledge of CV% of the
Property is Known
2. Co-efficient of Variation (CV)
To Compare two Mean
Values where the C.D. is
not Known
3. 2Test
To Compare Frequencies
where the Nature of
Distribution is not Known
4. FTest To Compare Variances
5. Analysis of Variance
To Compare Mean Values
where Individual data is
available
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SITRA 4
The Values of CD for Various Fibre and Yarn
Properties are given in the following Tables
No. of Tests and Critical Difference (%) forVarious Fibre Properties
Fibre Property No. of TestsCritical Difference
(% of Mean)
2.5% Span Length 4 combs / sample 4
Uniformity ratio 4 combs / sample 5
Micronaire value 4 plugs / sample 6
Fibre Strength at 3mm gauge Length
10 breaks/sample 5
Maturity coefficient 600 fibres/sample 7
Trash Content 8 test / sample 7
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SITRA 5
Sample Size and Critical DifferenceFor Yarn Properties
Yarn Property No. of TestsCritical Difference
(% of Mean)
Lea Count 40 2.0
Strength 40 4.0
S.Y.S.Uster 100 2.8
EvennessU% 5 7.0
Twist Henry Baer
(Single Yarn)
(Double Yarn)
50
50
3.4
2.0
Yarn Appearance5 Boards
10 Readings
Half a Grade or 5
Grade Index
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SITRA 6
New Critical Difference
N1New CD% = CD% (From Table) x -----
N2
Where, N1 = No. of tests recommended in
Tables
N2 = No. of tests actually conducted
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A Mill received a basic sample of Varalakshmi
Cotton, whose Mic. value on testing was found to be
3.4. The delivered sample gave a Mic. Value of 3.6.
The management is interested to know whether they
could accept the delivered sample?
(4 tests were carried out both on basic &
delivered samples to determine the Mic. value)
Diff. in Mic. value bet 2 samples = 3.6 - 3.4 = 0.2
3.6 + 3.4Ave. of Mic. value of 2 samples = ----------- = 3.5
2
0.2Diff. Expr. as a % of the average = ---- x 100 = 5.7
3.5
The CD for Mic. value as per table = 6%
Consistency bet basic and delivery samples:
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A Mill wanted to purchase a Cotton of 3.7 Mic.
value to spin 50s count. The sample cotton recd.
from a party was tested for Mic. & it was foundto be 3.9 (on the basis of 4 tests). The mill is
interested to know whether the sample cotton
confirms to the mills requirement ?
Diff. in Mic. value bet specificvalue(3.7) & actual value(3.9) = 3.9 - 3.7 = 0.2
Diff. Expr. as a % of specific 0.2value. = ---- x 100 = 5.3%
3.8
The CD for Mic. value as per the table = 6%
Since we are comparing a mean value with a
specific value, here CD is to be calculated on the
basis of the specific value
Comparison of test data with a specific value:
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A Mill received MCU5 cotton samples from Station A
& Station B. Their strength values were foundto be 22 g/tex & 24 g/tex resp. (based on 5 tests).
The mill would like to know whether the two
Cottons are similar in their strength values?
Diff. in Strength value bet 2 cottons = 24 - 22 = 2
24 + 22Ave. of Sth. values of 2 cottons = --------- = 23
2
2
Diff. Expr. as a % of the aver.=----- x 100 = 8.7%23
The CD for Sth. as per table=5% based on 10 tests
Homogeneity of cottons from different stations:
10The new CD% =5 x ----- = 7.1%
5
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The 2.5% S.L. of mixing and lap were found to be
25 mm and 24.2 mm respectively. To ascertainwhether the B.R. is creating any fibre rupture?23
Diff. in 2.5% S.L. of mixing&lap = 2524.2 = 0.8
25 + 24.2Ave. of 2.5% S.L. of mixing&lap = ----------- = 24.6
2
0.8Diff. Expr. as a % of the aver.= ------- x 100 = 3.25
24.6
The CD for 2.5% S.L. as per table = 4%
Fibre rupture in Blowroom:
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A Mill produces 40s yarn. While testing two
samples from two different spinning frames forevenness, the U% values are found to be 13.8 &
15.0 resp. on the basis of 10 observations in each
case. It is required to assess whether the yarn
produced on the frame are equally even?
Diff. in U% Expr. as a % of 1.2average = ------ x 100 = 8.3%
14.4
The CD for U% as per the table = 7% (for 5 tests)
5Find the new CD% = 7 x ----- = 5%
10
Yarn evenness between samples:
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Co-efficient of Variation (CV) : To compare two mean values
where the C.D. is not Known
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Application of CV Method
A mill is manufacturing 16s OE Yarn using drawing
sliver of nominal hank 0.14. The actual hank of
Drawing sliver is found to be 0.142 The mill wants to
know whether the DF sliver hank is to be corrected?
In this problem, the std CV of drawing sliver hank is
taken as 1%
SD of drawingsliver hank:
mean x CV 0.14 x 1= ------------- = ----------- = 0.0014100 100
The hank of drawing sliver for the dept.
3 SD 3 x 0.0014
0.14 + ------ = 0.14 + -----------n 8
= 0.14 + 0.0015
The nominal hank of the department
Lies between 0.1385 & 0.1415
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SITRA 14
A mill produces 50s P/V for export market.The mill intends to
produce yarn with mini.CSP of 2500 & CV of Sth. 6% The mill
would like to decide the average CSP it has to achieve so that the
minimum CSP is 2500?
The relation between Average & Minimum CSP is
given by the formula,
Mini. CSP x 100Average CSP = -------------------(100 - 3 CV)
2500 x 100Average CSP = --------------- = 3049
(100 - 18)
Average and minimum CSP:
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SITRA 15
Special Tests
1. Ttest : to compare two mean values
2. Ftest : to compare two variances (square
of standard deviation)
3. 2test : This method is to be used when
there is no prior knowledge of the
distribution of the test values.
4. Analysis of variance (ANOVA)
: A mill engages 10 operatives in conewinding and their production for five days are known.
In order to ascertain whether there is a significant
difference in production between operatives or days, a
special test namely, analysis of variance is to be
adopted.
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SITRA 16
Where, X1=Ave. Hairi. of Sample1X2=Ave. Hairi. of Sample2
n=no.of tests for Sa.1 & 2
S1 & S2 = Std. deviation for sample 1 & 2
From CV & Mean, the SD could be deduced.Thus, S1 = 2400 & S2 = 4000
The value of t for 38 d.o.f. is 2(100008000) 20
t = ------------------------ = 1.9
(24002
+ 40002
)
Hairiness between samples:
A mill produces 80s P/C yarn. On testing two yarn
samples one each from G5/1 & DJ/5 frame for hairiness,
the no. of hairs/1000 m are found to be 8000 & 10000 &CV of hairiness as 30% & 40% resp.(on the basis of 20
test for each sample). The mill wantsto know whether
DJ/5 R.F. is prod. More hairiness?
(X1X2) nt = ---------------
(S12 + S2
2)
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SITRA 17
Due to incorporating auto levellers in the high prod.
cards in a mill the CV% of card sliver decreased from
4 to 3.5. Is the difference statistically significant?
(The mean hank of card sliver is 0.2 and 40
readings were taken to measure CV%)
As 2 CV values are to be comp. F test can be used
CV1 x mean 4 x 0.2SD1
= ------------- = -------- = 0.008100 100
CV2 x mean 3.5 x 0.2SD2
= ------------- = ---------- = 0.007
100 100
SD12 (0.008)2
F = ----- = ---------- = 1.31SD2
2 (0.007)2
The value of F (for n1 = 39 & n2 = 39) = 1.53
Auto levellers in Cards:
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SITRA 18
A mill produces 40s yarn. On testing two samples from
two spindles for twist, the std deviation of twist were
found to be 1.31 & 2.85 based on 50 and 60 tests resp.
Can it be concluded that the two samples differ in
terms of their twist variation ?
Here, S1 = 2.85 & S2 = 1.31
n1 = 60 & n2 = 50
(2.85)2F = --------- = 4.7
(1.31)2
The value of F (for (df)1 = 59 & (df)2 = 49) = 1.6
Twist variability among samples:
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SITRA 19
A mill maintains an average end breakage rateof 150 per 1000 spl. hrs.
in 60s Ne. Due to change of mixing, the breakage rate increasedto the
level of 220 per 1000 spl. hrs. Has thechange of mixing increased the
breakage rate? (Breaks were observed for 1000 spl. hrs.)(OE)2
2 is defined as = ----------E
Where, O & E are observed & expected values
Here, O = 220 & E = 150(220150)2 4900
2 = ---------------- = ------- = 32.7150 150
2 value for 1 degree of freedom is 3.84
Comparison of end breaks:
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SITRA 20
A Mill processes DCH-32 cotton through tandem card &
high prod. card. Neps in card web were assessed by taking10 readings in both the cards. The neps were found to be
120 & 80 per 10 gms. Does the tandem card generate more
neps?
An assumption is made in the formula(OE)2 (AB)2
2 = --------- will reduce to --------E A + B
Where, A & B are the two observed values
(12080)2
2 = --------------- = 8120 + 80
2 value for 1 degree of freedom is 3.84
Comparison of neps during carding:
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SITRA 21
Application of Analysis of Variance Method
Cone Winding Production
A mill engages 10 operatives in cone winding and their
production for five days are given in table below:
Days
Production per winder per Shift (in Kg)
1 2 3 4 5 6 7 8 9 10 Total
weigh
t
1 62 67 63 66 64 68 59 63 60 69 641
2 59 63 66 60 64 59 60 62 66 63 622
3 65 67 59 60 62 66 59 62 66 60 626
4 69 66 59 68 64 60 63 67 64 65 645
5 66 62 60 59 68 64 63 62 61 67 632
Total 321 32
5
307 313 322 317 304 316 317 324 3166
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SITRA 22
In order to ascertain whether there is a significant difference in
production between operatives or days, a special test namely,
Analysis of Variance is to be adopted. The method of
computation is given below:
1. Find the sumT of all 50 readings T = 3166
2. Calculate the correction factor (CF) = T2/50 = 200471
3. Calculate the sum of the squares
of the individual readings A = 200938
4. The corrected sum of the squares
S = ACF = 200938200471 = 467
5. Between days sum of squares =
(641)2 + (622)2 + (626)2 + (645)2 + (632)2 - CF
---------------------------------------------------------
10
= 200509200471 = 386. Between operatives sum of squares =
(321)2 + (325)2+ + (324)2 - CF
---------------------------------------------------------
5
= 200559200471 =88
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SITRA 23
07.Error sum of squares = 467 38 88 = 341
The above information are then summarized in thefollowing Analysis of Variance table
Source ofVariation
Degree of* Freedom
Sum ofsquares
MeanSquares
F
Between days 4 38 9.50 1.00
Between
Operatives
9 88 9.78 1.03
Error 36 341 9.47 -
*Degree of freedom for between days = No. of days 1
Degree of freedom for between operatives = No. of operatives 1
Degree of freedom for error = Total No. of readings
No. of days
No. of operatives + 1
Mean Square
F = ----------------------
Error
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SITRA 24
Since the actual values of F i.e., 1.00 for between
days and 1.03 for between operatives are lower than 2.63
and 2.15, which are given in statistical tables for the
corresponding degrees of freedom, it can be concluded that
there is no real difference in production rates either
between operatives or between days.
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SITRA 25
Interpretation of Test Data
Part - II
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SITRA 26
Let N be the number of tests and
P the percentage accuracy
1.96 V
N = ----------
P
2
Where V is the Coefficient of Variation
So, No. of Tests required to bring down the
Error to 1.0%
1.96 x 10.89N = -----------------
1.0
2
= 455.58
456 tests (app.)
Answer 1 Completed
Answer for Question No. 1
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SITRA27
Answer for Question No. 2Required Percentage Error P = 0.5
----- x 10040
= 1.25
So, No. of Tests Required
1.96 V
N = ----------
P
2
Where,
V = Coefficient of Variation
1.96 x 1.5
= ----------
1.25
2
= 5.53 6 tests
Answer 2 Continued
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SITRA 28
CV% = 3
Answer for Question No. 2 (Contd)
Therefore, the No. of Tests required
1.96 x 3= -------------
1.25
2
= 22.12 23 tests
Answer 2 Completed
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SITRA 29
Answer for Question no. 3 (a)
Yarn Imperfections Follow Poisson Distribution
For Poisson Distribution
SD = Mean
SD of Imperfections = 45
6.7
2 SD 13.0
The Mill can Conclude that the Imperfection
have Increased Statistically when the Incidence of
Yarn Imperfections (Checked Between Diff. Frames
on a Particular Day)
Exceed 45 + 13 = 58.0
Answer 3 (a) Completed.
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SITRA 30
Answer for Question no. 3 (b)
Answer 3 (b) Contd.
Maximum Imperfections = 45.0/km
Mean + 2SD = 45.0
Mean = 45.02SD
= 45.013.0
= 32.0/km
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SITRA 31
Answer for Question No. 3 (b) Contd.
Therefore, Only if the Mill Maintains,
the Imperfections at or Below 32.0/km,
then the Maximum Imperfections (as
Requested by the Buyer) can be Maintained
Below 45.0/km in Most of the Occasions.
Answer 3 (b) Completed.
Answer for Question no 4 (a):
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SITRA 32
Answer for Question no. 4 (a):
Rewinding Breaks
No. of cones and number of hours of observation
depends on the level of accuracy required. End
breaks in winding follows Poisson Distribution.
In Poisson Distribution,
SD = Mean
If C is the cone winding breaks/cone hr,
Error of estimate = 2 C
n
n
i.e. e = 2 C
Answer 4 (a) (contd.)
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SITRA 33
Answer for Question No. 4 (a) (contd.)
Answer 4 (a) (contd.)
n = 2 C
e
Suppose we assume that
C = 1.5 and
e = 20% of C
i.e. 0.2 C,
Then n = 2 x 1.5
0.2 C
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SITRA 34
Answer for Question No. 4 (a) (contd.)
n = 20 x 1.52C
n = 400 x 1.5
4C2
= 400 x 1.5
4 x 1.5 x 1.5
= 267 cone66.75 = 70 cone Hrs.
4
Answer 4 (a) (contd.)
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SITRA 35
Hence, breakage rate during rewinding shouldbe assessed for 70 cone hours for an accuracy
of 20%.
Suppose one lot (for disposal contains) 140Cones,
Take 20% 28 cones
Observe the breakage for 2.5 hoursfor each cone.
Answer for Question No. 4 (a) (contd).
Answer 4 (a) Completed.
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SITRA 36Answer 4(b) contd.
Answer for Question No. 4 (b)
Rkm CV of 9% is Subjected to Statistical
Fluctuation on Day To Day Basis.
95% of the Rkm CV Readings will Lie at
9 + 2 x 9
2 x 100
= 9 +18
14.14
= 9 + 1.3
10.3 & 7.7
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SITRA 37
Answer for Question No. 4(b) (contd.)99% of the Rkm CV Readings will Lie at
9 + 3 x 9
2 x 100
= 9 +27
14.14
= 9 + 1.91,
10.91 & 7.09
This Aspect has to be Kept in Mind when
Rkm CV Readings Obtained Between Days
are considered.
Answer 4 (b) Completed.
A f Q ti N 4 ( )
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SITRA 38
Answer for Question No. 4 (c)
This Exercise is to Estimate the Extent of
Influence of Splices Made During AutoWinding on Warping Breaks
In 60s Yarn, Let us Assume that,
Cop Content
Cone Weight
= 50 gms.
= 1.8 kg.
No. of Splices/Cone Made Entirely
Due to Cop Changes
1800
50= 36
Answer 4(c) contd.
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SITRA39
Answer for Question No. 4(c) (contd.)
Answer 4(c) contd.
Assuming One Break/Cop Due to Yarn Fault (Itis Assumed that One Cop Contains 5000 mts. in
60s), Splices made on Account of Fault Removal
= 36
Total No. of Splices = 36 + 36 = 72
Total No. of Splices in a 1.8 kg Cone = 72
No. of Splices/1.8 Lakh Metre = 72
No. of Splices/Million Metre = 400
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SITRA 40
Answer for Question No. 4 (c) (contd.)
Standard for Warp Breaks
No. of Breaks/Million Metre = 0.4 (for very good
Performance)
It is Evident that the Breaks during Warping
are Just 0.1% of the Splices in the Yarn.
This leads to the Conclusion that Generally
(Under Normal Working Conditions) Splices
in Yarn are not the Reason for Excessive
End Breaks During High Speed Warping.
Answer 4(c) contd.
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SITRA 41
Answer for Question No. 4 (c) (contd.)
It can be further Explained that in a
Yarn if the Splices Made During Auto
Coner Winding is 65/Lakh metre under
Normal Circumstances and on a
Particular Occasion, the Splices Have
Increased, Say, to a Level of 85/Lakh
Metre, This will not Make Very High
Difference in the Warping Breakage
Rate, Say, from 0.4 Breaks/Million
metre to 1.2 Breaks/Million metre.
Answer 4(c) Completed.
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SITRA 42
Answer for Question no. 5
Snap Study Technique falls under
Binomial DistributionHere SD = pq
n
p = efficiency of the Loom Shed
(in a fraction)
q = loss in efficiency (as a fraction)
q = 1 - p
n = total no. of Looms observed
Where,
Answer 5 contd.
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SITRA 43
Answer for Question No. 5 (contd.)
For 99% confi. interval, proportion of idle
looms will be at + 3SD
= + 3 x .0072
= + .0216
= 0.0072
Therefore SD = 0.93 x .07
1250
Answer 5 contd.
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SITRA 44
The actual efficiency will vary between
0.93 + 0.0216
= 0.9516, 0.9084
i.e. 95.16% and 90.84%.
Hence, it is concluded that the actual
Loom shed eff. is not significantly lower
than the mill Norm of 95%.
Answer for Question No. 5 (contd.)
Answer 5 Com leted.
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SITRA 45
Answer for Question no. 6
DaysWarp Breaks/3000 Ends/1,00,000 Picks
Loom Number
1 2 3 4 5 6 7 8 9 10 Total
1 12 11 10 13 14 9 8 10 12 15 114
2 9 13 10 11 8 12 11 13 10 12 109
3 13 10 9 12 10 13 12 14 11 9 113
4 11 12 10 13 9 10 12 10 14 11 112
5 10 14 13 12 10 14 10 9 13 12 117
Total 55 60 52 61 51 58 53 56 60 59 565
Answer 6 (contd.)
Answer for Question No 6 (contd )
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SITRA 46
Answer for Question No. 6 (contd.)
In order to ascertain whether there is
any significant difference in warp
breakage rate between looms or between
days, a special test viz.,
Analysis of Varianceis to be carried out.
The method of computation is given below:
1. Find the sum (total) of
all 50 readings T= 565
2. Calculate the Correction
Factor CFT2
50
=
Answer 6 (contd.)
= 6384.5
Answer for Question No 6 (contd )
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SITRA 47
Answer for Question No. 6 (contd.)
3. Calculate the sum of the Squares
of the individual reading A = 6537
4. The corrected sum of Squares S = A - CF
= 65376384.5
= 152.5
5. Between days sum of Squares = BD
(114)2+(109)2+(113)2+(112)2+(117)2
10- CF
= 6387.96384.5 = 3.4
BD =
Answer 6 contd.
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SITRA 48
Answer for Question No. 6 (contd.)
6. Between Looms Sum of Squares = BL
7. Error of Sum of Squares = e
e = SBD - BL
= 152.53.423.7
= 125.4
(55)2+(60)2+(52)2+ + (59)2
5
- CF
= 6408.26384.5 = 23.7
BL =
Answer 6 contd.
A f Q ti N 6 ( td )
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SITRA 49
Answer for Question No. 6 (contd.)
The above information are then
Summarised in the following
ANALYSIS OF VARIANCE TABLE
Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean
SquaresF
Between
Days4 3.4 0.85 0.24
Between
Looms9 23.7 2.63 0.76
Error 36 125.4 3.48 -
Answer 6 (contd.)
Answer for Question No. 6 (contd.)
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SITRA 50
Degrees of Freedom
of Error
= Total No. of Readings
- No. of Days
- No. of Looms + 1
Answer for Question No. 6 (contd.)
Degrees of Freedom for
Between Days= No. of Days - 1
Degrees of Freedom for
Between Looms= No. of Looms - 1
F =Mean Square
Error
Answer 6 (contd.)
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SITRA
51
Answer 6 Completed.
Answer for Question No. 6 (contd.)
8. Since the actual values ofF 0.24for between days and 0.76 forbetween looms are lower than2.61 and 2.18, which are givenin
Statistical tables for thecorresponding degrees of freedom,it can be concluded that there isno real difference in warp breakage
rate either between days orbetween looms.
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SITRA 52
Answer for Question no. 7
Answer 7 (contd.)
DaysEnd Breaks/100 Spindle Hours
Ring Frame Numbers
1 2 3 4 5 6 7 8 9 10 Total
1 4 6 5 5 3 4 6 5 6 7 512 3 7 5 4 5 4 7 7 6 8 56
3 4 9 4 3 6 3 4 5 4 6 48
4 5 6 6 5 4 2 5 6 5 7 51
5 5 5 4 5 4 6 4 6 5 5 49
Total 21 33 24 22 22 19 26 29 26 33 255
A f Q ti N 7 ( td )
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SITRA53
Answer for Question No. 7 (contd.)
T2
In order to ascertain whether there is
any significant difference in end breakage
rate between ring frames or between
days, a special test viz.,
Analysis ofVarianceis to be carried out.
The method of computation is given below:
1. Find the sum (total) of
all 50 readings T= 255
2. Calculate the Correction
Factor CF50
=
Answer 7 (contd.)
= 1300.50
Answer for Question No 7 (contd )
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SITRA 54
Answer for Question No. 7 (contd.)
3. Calculate the sum of the Squaresof the individual reading A = 1393.00
4. The corrected sum of Squares S = A - CF
= 393.00 1300.50
= 92.50
5. Between days sum of Squares = BD
(51)
2
+(56)
2
+(48)
2
+(51)
2
+(49)
2
10
- CF
= 1304.30 1300.50 = 3.8
BD =
Answer 7 (contd.)
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SITRA
55
Answer for Question No. 7 (contd.)
6. Between Ring Frames Sum
of Squares = BR
7. Error of Sum of Squares = e
e = S BD - BR
= 92.50 3.8 42.9
= 45.8
(21)2+(33)2+(24)2 +(22)2+ (33)2
5
- CF
= 1343.40 1300.50 = 42.9
BR =
Answer 7 (contd.)
Answer for Question No. 7 (contd.)
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The above information are thenSummarised in the following
ANALYSIS OF VARIANCE TABLE
Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean
SquaresF
Between
Days4 3.8 0.95 0.75
Between
Ring Frames9 42.9 4.77 3.76
Error 36 45.8 1.27 -
Answer 7 (contd.)
Answer for Question No. 7 (contd.)
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Degrees of Freedomof Error
= Total No. of Readings
- No. of Days
- No. of Ring Frames + 1
Degrees of Freedom forBetween Days = No. of Days - 1
Degrees of Freedom forBetween Ring Frames
= No. of Ring Frame - 1
F =
Mean Square
Error
Answer 7 (contd.)
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Answer for Question No. 7 (contd.)
Actual F
Value
StatisticalF Value
Difference
Between Days0.75 2.61 NS
Between
Ring Frames 3.76 2.18 S
* A real difference in end breakage ratebetween ring frames.
Answer for Question no. 8
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Warping Breaks Follow Poisson Distribution.
The difference between actual and expected breaks isgiven by,
Since, this value is lower than the valueof 3.84, (which is given in X2 statistical
tables for 1 degree of freedom at 95%confidence limit), theactual breakage rate do not differsignificantly from the norms.
(O E)2
E
=
(6 4)2
4
O = Observed breakage rate
E = Expected breakage rate
=22 /4 = 1.0
Answer 8 Completed.
Answer for Question no. 9
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Since, this value is higher than the valueof 3.84, (which is given in X2 statistical
tables for 1 degree of freedom at 95%confidence limits), the actual end breakagerate differ significantly from thenorms.
End Breaks Follow Poisson Distribution.
The difference between actual and expected breaks isgiven by,
(O E)2
E
=(7 3)
2
3
O = Observed breakage rate
E = Expected breakage rate
= 5.33
Answer 9 Completed.
Answer for Question no. 10
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Answer for Question no. 10
Total No. of Warp Breaks = 125
Breaks/Loom =125
150= 0.83
The expected no. of looms with 0, 1, 2, 3, 4Warp breaks can be calculated using Lawsof Poisson Distribution
a. Expected No. of Looms
with 0 Warp Breaks = 150 x e-m
x mo
= 150 x e-0.83 x 0.83o = 65.41
Answer 10 (contd.)
m
Answer for Question No.10 (contd.)
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b. Expected No. of Loomswith 1 Warp Breaks =
150 x e-0.83 x 0.831
1!
= 54.29
c. Expected No. of Loomswith 2 Warp Breaks
=
150 x e-0.83 x 0.832
2!
= 22.53
d. Expected No. of Loomswith 3 Warp Breaks
=
150 x e-0.83 x 0.833
3!
= 6.23
e. Expected No. of Loomswith 4 Warp Breaks
=
150 x e-0.83 x 0.834
4!
= 1.29
Answer 10 (contd.)
Answer for Question No. 10 (contd.)
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Looms with Warp Breaks (Numbers)
No. of Warp
Breaks
Actual
(O)
Expected
(E)
(O-E)2
E
0 70 65.41 0.32
1 50 54.29 0.34
2 25 22.53 0.27
3 4 6.23 0.80
4 1 1.29 0.07
Total 1.80
Answer 10 (contd.)
Answer for Question No. 10 (contd.)
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Q ( )
2 Value from 2 tables for 4
Degrees of freedom at 95%Confidence level is 9.49.
Since the calculated value of 2 is
Lower than the value given instatistical tables, it is concluded
that repeated warp breaks do notoccur in looms.
Answer 10 Completed.
Answer for Question no. 11
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Q
Answer 11 (contd.)
Total No. of End Breaks = 129
Breaks/Spindle =129
480
= 0.27
The expected no. of spindles with 0, 1, 2
and 3 end breaks can be calculated usingLaws of Poisson Distribution
a. Expected No. of Spindles
with 0 end Breaks=
480 x e-m
x mo
= 480 x e-0.27 x 0.27o = 366.4
(m)
Answer for Question No. 11 (contd.)
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b. Expected No. of Spindles
with 1 End Breaks=
480 x e-0.27 x 0.271
1!
= 98.93
c. Expected No. of Spindleswith 2 End Breaks
=
480 x e-0.27 x 0.272
2!
= 13.36
d. Expected No. of Spindleswith 3 End Breaks
=480 x e-0.27 x 0.273
3!
= 1.20
Answer 11 (contd.)
Answer for Question No. 11 (contd.)
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Spindles with End Breaks (Numbers)
No. of EndBreaks
Actual(O)
Expected(E)
(O-E)2E
0 370 366.4 0.035
1 95 98.93 0.156
2 11 13.36 0.417
3 4 1.20 6.533
Total 7.141
Answer 11 (contd.)
Answer for Question No. 11 (contd.)
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2
Value from
2
table for 3degrees of freedom at 95%Confidence level is 7.81.
Since the calculated value of 2 is
Lower than the value given instatistical tables, it is concluded
that repeated end breaks do notoccur in spindles.
Answer 11 Completed.
Answer for Question No. 12
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POPULATION S.D.
Answer 12 Continued
S = (n1-1) SD12 + (n2-1) SD22----------------------------------------------
n1 +n2 - 2Here, n1 = n2 = 10
SD1 = 4 ; x 1 = 48
SD2 = 5 ; x 2 = 46
S = (10 -1)42 + (10-1)52-------------------------------------- = 4.5310+10-2
Answer for Question No. 12 (Contd)
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Answer 12 Continued
HENCE t VALUE
[X1 - X2 ]
= -------------------S 1/n1 + 1/n2
[48 - 46 ]= -----------------------
4.53 1/10 + 1/10
= 1.00
So, degrees of freedomV = n1 + n2 2
= 10 + 10 - 2 = 18Comparing the same with statistical tabular value,
t = 2.101 at 95% level
t = 2.878 at 99% level
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Here, the calculated t value ,i.e, 1.00 is
less than the corresponding statistical tabular
value at 95% level. So, there is insufficient
evidence to prove that the chemical treatment
has weakened the fabric.
Answer for Question No. 12 (Contd)
Answer 12 Completed
Answer for Question No 13
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Answer for Question No. 13
Answer 13 Continued
Step 1
Calculate the Standard Error of the means, S.E.1 and S.E.2
S.D.1 7.8S.E.1 = -------- = -------- = 1.42
n1 30
S.D.2 8.2
S.E.2 = -------- = -------- = 1.50n2 30
Step 2
Calculate the Standard Error of the difference,between the means :
S.E.diff= (S.E.12 + S.E.2
2)
S.E.diff= (1.422 + 1.52)
= 2.06
Step 3Answer for Question No. 13
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Calculate the ratio
[mean1 - mean2 ] [X1 - X2 ]---------------------- = -------------
S.E.diff S.E.diff
[58 - 65] [7]---------------------- = -------------
2.06 2.06Step 4
Compare the value of this ratio with 1.96 and 2.58
3.4 > 2.58
Conclusion : Since 3.4 is greater than 2.58, the diff.Between mean lea strengths is significant at the 1 percenti.e. a real difference exists.
Answer 13 Completed
= 3.4
Answer for Question No. 14
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Step 1
Q
Answer 14 Continued
Calculate the S.E. of the Standard Deviation :
S.D. of sampleS.E. =------------------
2n
n = 40
S.D. of sample = 8.6
8.6S.E. =------------------ = 0.96(2 x 40)
Answer for Question No. 14
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Step 2
Calculate the ratio :
Answer 14 Completed
Difference between the S.D.s------------------------------------S.E. of the standard deviation
[ 6.4 - 8.6 ]----------------- = 2.3
0.96Step 3
Compare the value of this ratio with the values1.96 and 2.58, the 5 percent and 1 percent levels;
2.3 exceeds 1.96 but less than 2.58
Conclusions :Although there is some evidence of a difference invariability it is significant at the 5 percent level.
Answer for Question No. 15
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Answer 15 Continued
Step 1
Calculate the variances of the sample & population :
Variance = S.D.2
V1, the sample variance is 2.02 = 4.0
V2, the population variance is 1.52 = 2.25
Step 2
Calculate F, the variance ratio :
Variance expected to be greaterF = -------------------------------------
Variance expected to be smaller
Answer for Question No. 15
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V1 4F = ---- = ----- = 1.78
V2 2.25
Answer 15 Completed
Step 3
Degrees of freedomfor the sample, V1 = 9-1 = 8
Population is very large, hence it is taken asInfinity, V2
F ratio for 5 per cent significance limit is 1.94
F ratio for 1 per cent significance limit is 2.51
Therefore, found to be not significant
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