# 1117 final

Post on 04-Apr-2018

217 views

Embed Size (px)

TRANSCRIPT

7/30/2019 1117 Final

1/78

SITRA

Interpretation of Test Data

Part - I

1

7/30/2019 1117 Final

2/78

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

7/30/2019 1117 Final

3/78

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

7/30/2019 1117 Final

4/78

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

7/30/2019 1117 Final

5/78

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

7/30/2019 1117 Final

6/78

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

7/30/2019 1117 Final

7/78SITRA 7

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:

7/30/2019 1117 Final

8/78SITRA 8

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:

7/30/2019 1117 Final

9/78SITRA 9

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

7/30/2019 1117 Final

10/78SITRA 10

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:

7/30/2019 1117 Final

11/78SITRA 11

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:

7/30/2019 1117 Final

12/78SITRA 12

Co-efficient of Variation (CV) : To compare two mean values

where the C.D. is not Known

7/30/2019 1117 Final

13/78SITRA 13

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

7/30/2019 1117 Final

14/78

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:

7/30/2019 1117 Final

15/78

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.

7/30/2019 1117 Final

16/78

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)

7/30/2019 1117 Final

17/78

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:

7/30/2019 1117 Final

18/78

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:

7/30/2019 1117 Final

19/78

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:

7/30/2019 1117 Final

20/78

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:

7/30/2019 1117 Final

21/78

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

7/30/2019 1117 Final

22/78

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

7/30/2019 1117 Final

23/78

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

7/30/2019 1117 Final

24/78

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.

7/30/2019 1117 Final

25/78

SITRA 25

Interpretation of Test Data

Part - II

7/30/2019 1117 Final

26/78

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

7/30/2019 1117 Final

27/78

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

7/30/2019 1117 Final

28/78

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

7/30/2019 1117 Final

29/78

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.

7/30/2019 1117 Final

30/78

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

7/30/2019 1117 Final

31/78

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):

7/30/2019 1117 Final

32/78

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.)

7/30/2019 1117 Final

33/78

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

7/30/2019 1117 Final

34/78

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.)

7/30/2019 1117 Final

35/78

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.

7/30/2019 1117 Final

36/78

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

7/30/2019 1117 Final

37/78

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 ( )

7/30/2019 1117 Final

38/78

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.

7/30/2019 1117 Final

39/78

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

7/30/2019 1117 Final

40/78

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.

7/30/2019 1117 Final

41/78

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.

7/30/2019 1117 Final

42/78

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.

7/30/2019 1117 Final

43/78

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.

7/30/2019 1117 Final

44/78

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.

7/30/2019 1117 Final

45/78

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 )

7/30/2019 1117 Final

46/78

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 )

7/30/2019 1117 Final

47/78

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.

7/30/2019 1117 Final

48/78

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 )

7/30/2019 1117 Final

49/78

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.)

7/30/2019 1117 Final

50/78

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.)

7/30/2019 1117 Final

51/78

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.

7/30/2019 1117 Final

52/78

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 )

7/30/2019 1117 Final

53/78

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 )

7/30/2019 1117 Final

54/78

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.)

7/30/2019 1117 Final

55/78

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.)

7/30/2019 1117 Final

56/78

SITRA

56

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.)

7/30/2019 1117 Final

57/78

SITRA

57

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.)

7/30/2019 1117 Final

58/78

SITRA

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

7/30/2019 1117 Final

59/78

SITRA

59

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

7/30/2019 1117 Final

60/78

SITRA

60

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

7/30/2019 1117 Final

61/78

SITRA

61

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.)

7/30/2019 1117 Final

62/78

SITRA

62

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.)

7/30/2019 1117 Final

63/78

SITRA

63

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.)

7/30/2019 1117 Final

64/78

SITRA

64

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

7/30/2019 1117 Final

65/78

SITRA

65

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.)

7/30/2019 1117 Final

66/78

SITRA

66

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.)

7/30/2019 1117 Final

67/78

SITRA

67

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.)

7/30/2019 1117 Final

68/78

SITRA

68

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

7/30/2019 1117 Final

69/78

SITRA

69

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)

7/30/2019 1117 Final

70/78

SITRA

70

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

7/30/2019 1117 Final

71/78

SITRA

71

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

7/30/2019 1117 Final

72/78

SITRA

72

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

7/30/2019 1117 Final

73/78

SITRA

73

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

7/30/2019 1117 Final

74/78

SITRA

74

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

7/30/2019 1117 Final

75/78

SITRA

75

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

7/30/2019 1117 Final

76/78

SITRA

76

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

7/30/2019 1117 Final

77/78

SITRA

77

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

7/30/2019 1117 Final

78/78