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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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    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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    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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    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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    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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    58Answer 7 Completed.

    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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