3.Folio-project Screw & Washer

8/7/2019 3.Folio-project Screw & Washer

1/38

1.0 INTRODUCTION

Data analysis is a process of gathering, modeling, and transforming data with the goal of

highlighting useful information, suggesting conclusions, and supporting decision making. Dataanalysis has multiple facets and approaches, encompassing diverse techniques under a variety of

names, in different business, science, and social science domains. Figure below shows the steps

in the engineering problem-solving method.

Note that the engineering method features a strong interplay between the problem, the

factors that may influence its solution, a model of the phenomenon and experimentation to verify

the adequacy of the model and the proposed solution to the problem. In our analysis, we try to

analyze about screw and washer. Now, we bring you to know a little fact of screw and washer.

What we know about screws and washers? Almost know about that.

1.1 NEEDS AND PARAMETER

In real life, screws are very familiar parts either in engineering or manufacturing processthat included of most users. Whereas the washer commonly used together with screw as fastener.

It function is to avoid screw from loosen easily and also to take care about the work piece face in

case of vibration. In manufacturing, washers are dimensioned in BS4320: 1968 and the screw

sizes are in accordance with BS 3692:1967 (ISO 272 equivalent). This standard has now been

1

Draw conclusionsand make

recommendations

Collect data

Develop acleardescription of

the problem

Identify theimportantfactors

Propose orrefine amodel

Manipulatethe model

Confirm thesolution


2/38

superseded by BS 3692:2001. The data has not yet been checked against the latest revision. The

thread form is in accordance with BS 3643:2007.

In our analysis, we have decided to measure the parameter of screw and washer to

manipulate the data in order to achieve the final result. In operation or experimental, we choose

screw in specs of M5 x 50 and washer M5 as selected specimen and vernier caliper as

measurement tools. We measure the length and diameter of thread for screw while for the washer we measure the inner and outer diameter.

1.2 EXPECTED RESULTS AND OUTCOMES

We had chosen the screw and washer from two different sources of manufacturer, but both were

made from the same material and same specification. Initially, we expected that both source

appear the same results when measure because that made from same material and also same

method of manufacturing. However, we try to measure some parameter from both materials to

prove and explain our outcomes with do some statistical methods.

In our objectives, we want to measure the critical parameter of screw and washer that

caused by ineffective machine. Normally, in industry of manufacturing, if any defection were

found the instant maintenance must be run. It done to avoid defect of work piece beside

increasing productivity. From this project, we will learn some minors generic skills such as

ability to make conclusions statistically from data collected, data analyze and so on. We also

ability to apply critical thinking and creatively in using method of statistic that how engineers

solve engineering problems.

2.0 DATA COLLECTION

2


3/38

2.1 SAMPLE AND MEASUREMENT TOOLS

In our analysis, we had taken 2 samples of screw and washer from different manufacturer. One

of them was taken from Machine Laboratory in Mechanical Department UTM, and other one we

taken from maintenance store in College 10, UTM. Figure below shows that our specimen

physically, with two different sources and also there measurement tools:

1) Specimen samples of screw.

2) Specimen samples of washer.

3

Silvercolour

Goldcolour

a) Source from machine labb) Source from maintenance

store

c) From left: source (a), source(b)

d) Source from machine labe) Source from maintenance

store


4/38

3) Measurement tools.

2.2 METHOD OF MEASURING

1) Method of measure samples of screw.

4

a) Vernier caliper

The vernier caliper borrowedfrom machine lab. Thats only ourmeasurement tools to get persistof specific value of measure.


5/38

Figure (a) shows how we take the diameter of thread and (b) how we measure the length of screw.

2) Method of measure samples of washer.

5

(a) (b)

(a)

(b) (c)


6/38

i. 50 calibration of vernier scale = 49 mmmain scale

ii. 1 calibration = 49/50

= 0.98 mm

iii. Difference calibration value = 1 mm 0.98 mm

Figure (a) shows that how to measure outer diameter and both of (b) and (c) are to determine that

value of inner diameter.

Table below shows the method to determine one of calibration value at vernier scale:

Example:

6


7/38

a) Main scale reading

52 x 1 = 52.00 mm

b) Vernier scale reading at 15 th calibration thatparallel with other one line at main scale

15 x 0.02 = 0.30 mm

c) Final reading caliper

52.00 mm + 0.30 mm = 52.30 mm

2.3 TABLE

All of table below shows the data which we had been measure.

1) Screws table

SOURCE 1 (MACHINE LABORATORY)

7


8/38

SpecimenLength

AverageDiameter of

Thread Average1 2 3 1 2 3

1 51.940 51.100 51.900 51.647 4.860 4.840 4.880 4.860

2 51.640 51.800 51.500 51.647 4.860 4.880 4.860 4.8673 51.840 51.910 51.800 51.850 4.860 4.820 4.860 4.8474 52.010 51.900 51.910 51.940 4.880 4.860 4.840 4.8605 51.880 51.920 51.880 51.893 4.840 4.840 4.860 4.8476 51.800 51.680 51.900 51.793 4.860 4.880 4.820 4.8537 51.980 52.000 51.900 51.960 4.860 4.860 4.820 4.8478 51.960 51.880 51.860 51.900 4.860 4.880 4.860 4.8679 51.780 51.920 51.780 51.827 4.860 4.860 4.860 4.860

10 52.000 51.960 51.980 51.980 4.880 4.860 4.880 4.87311 51.960 51.900 51.920 51.927 4.820 4.840 4.860 4.840

12 51.300 51.780 51.700 51.593 4.860 4.840 4.860 4.85313 52.000 51.980 51.780 51.920 4.820 4.860 4.840 4.84014 51.660 51.680 51.700 51.680 4.880 4.840 4.860 4.86015 51.900 51.940 51.920 51.920 4.820 4.840 4.860 4.84016 51.700 51.740 51.760 51.733 4.840 4.860 4.880 4.86017 51.920 51.980 51.940 51.947 4.880 4.820 4.840 4.84718 51.940 51.900 51.980 51.940 4.860 4.880 4.820 4.85319 51.840 51.900 51.920 51.887 4.820 4.850 4.820 4.83020 51.920 51.860 51.900 51.893 4.840 4.880 4.860 4.86021 51.980 51.780 51.940 51.900 4.860 4.840 4.820 4.84022 51.980 52.000 51.920 51.967 4.880 4.820 4.880 4.86023 51.900 51.900 51.980 51.927 4.880 4.840 4.860 4.86024 52.000 51.820 51.880 51.900 4.820 4.860 4.880 4.85325 51.980 51.920 51.900 51.933 4.860 4.820 4.840 4.84026 51.820 51.880 51.780 51.827 4.860 4.860 4.820 4.84727 51.900 51.880 51.980 51.920 4.820 4.840 4.860 4.84028 52.000 51.940 51.920 51.953 4.840 4.860 4.880 4.86029 51.980 51.940 51.880 51.933 4.840 4.840 4.880 4.85330 51.900 51.840 51.960 51.900 4.880 4.860 4.820 4.853

SOURCE 2 (MAINTENANCE STORE)

SpecimenLength

AverageDiameter of

Thread Average1 2 3 1 2 3

1 51.900 51.900 51.100 51.633 4.860 4.840 4.840 4.8472 51.500 51.740 51.800 51.680 4.860 4.880 4.880 4.8733 51.800 51.800 51.910 51.837 4.860 4.840 4.820 4.8404 51.910 51.880 51.900 51.897 4.880 4.860 4.880 4.873

8


9/38

5 51.880 52.000 51.920 51.933 4.840 4.840 4.840 4.8406 51.900 51.860 51.680 51.813 4.860 4.860 4.880 4.8677 51.900 52.000 52.000 51.967 4.840 4.860 4.860 4.8538 51.860 51.880 51.880 51.873 4.860 4.840 4.880 4.8609 51.780 51.880 51.920 51.860 4.860 4.880 4.860 4.867

10 51.980 51.970 51.960 51.970 4.880 4.840 4.860 4.86011 51.920 51.900 51.900 51.907 4.820 4.840 4.840 4.83312 51.700 51.760 51.780 51.747 4.860 4.860 4.840 4.85313 51.780 51.880 51.980 51.880 4.820 4.860 4.840 4.84014 51.700 51.680 51.680 51.687 4.880 4.840 4.840 4.85315 51.920 51.920 51.940 51.927 4.840 4.820 4.840 4.83316 51.760 51.740 51.740 51.747 4.840 4.860 4.850 4.85017 51.940 51.960 51.980 51.960 4.880 4.820 4.820 4.84018 51.980 51.960 51.920 51.953 4.860 4.860 4.880 4.86719 51.920 51.900 51.900 51.907 4.820 4.850 4.850 4.840

20 51.900 51.880 51.860 51.880 4.840 4.860 4.880 4.86021 51.940 51.880 51.780 51.867 4.820 4.840 4.840 4.83322 51.920 51.980 52.000 51.967 4.860 4.820 4.820 4.83323 51.980 52.000 51.900 51.960 4.860 4.880 4.840 4.86024 51.880 51.900 51.820 51.867 4.840 4.820 4.860 4.84025 51.900 51.940 51.920 51.920 4.860 4.840 4.820 4.84026 51.780 51.800 51.880 51.820 4.860 4.860 4.860 4.86027 51.980 52.000 51.880 51.953 4.820 4.840 4.820 4.82728 51.920 51.900 51.940 51.920 4.840 4.880 4.860 4.86029 51.880 51.900 51.940 51.907 4.880 4.860 4.860 4.867

30 51.960 51.880 51.840 51.893 4.850 4.860 4.860 4.857

2) Washer table

SOURCE 1 (MACHINE

LABORATORY)

Specimen

Outer Diameter Average

Inner Diameter Average1 2 3 1 2 3

1 12.27412.28

012.27

812.27

7 5.668 5.670 5.668 5.669

2 12.27212.27

912.28

412.27

8 5.670 5.668 5.670 5.669

9


10/38

3 12.27812.27

812.27

612.27

7 5.668 5.668 5.668 5.668

4 12.28212.28

212.27

812.28

1 5.674 5.674 5.668 5.672

5 12.28012.27

412.27

612.27

7 5.676 5.664 5.668 5.669

6 12.27812.27

612.27

812.27

7 5.674 5.678 5.668 5.673

7 12.27812.27

412.28

012.27

7 5.668 5.674 5.674 5.672

8 12.27612.27

612.28

012.27

7 5.670 5.672 5.672 5.671

9 12.27412.26

812.26

812.27

0 5.668 5.672 5.668 5.669

10 12.28412.27

812.28

012.28

1 5.674 5.664 5.670 5.669

11 12.28

0

12.27

8

12.27

6

12.27

85.666 5.674 5.668

5.66912 12.284

12.274

12.278

12.279 5.670 5.664 5.674 5.669

13 12.27812.27

412.28

012.27

7 5.674 5.664 5.668 5.669

14 12.27212.27

012.27

612.27

3 5.662 5.668 5.674 5.668

15 12.28412.27

812.27

812.28

0 5.674 5.678 5.672 5.675

16 12.26812.27

212.27

412.27

1 5.682 5.672 5.668 5.674

17 12.27

4

12.28

8

12.28

4

12.28

25.670 5.678 5.684

5.67718 12.282

12.278

12.274

12.278 5.680 5.672 5.688 5.680

19 12.28812.28

412.27

012.28

1 5.660 5.662 5.668 5.663

20 12.27412.27

612.28

212.27

7 5.688 5.670 5.672 5.677

21 12.27012.27

812.29

012.27

9 5.674 5.682 5.680 5.679

22 12.27412.26

812.27

812.27

3 5.678 5.672 5.670 5.673

2312.27

812.28

812.28

012.28

2 5.684 5.688 5.690 5.687

24 12.27212.26

812.27

612.27

2 5.672 5.674 5.682 5.676

25 12.26812.27

812.27

212.27

3 5.674 5.688 5.700 5.687

26 12.27012.28

212.27

812.27

7 5.660 5.660 5.660 5.66027 12.27 12.28 12.29 12.28 5.672 5.668 5.670 5.670

10


11/38


12/38

16 12.27412.28

012.26

812.27

4 5.678 5.670 5.660 5.669

17 12.28412.27

812.27

412.27

9 5.670 5.668 5.678 5.672

18 12.27412.28

412.28

212.28

0 5.668 5.674 5.672 5.671

19 12.27012.27

612.28

812.27

8 5.668 5.676 5.678 5.674

20 12.28212.27

812.27

412.27

8 5.674 5.674 5.672 5.673

21 12.29012.27

612.27

012.27

9 5.664 5.668 5.662 5.665

22 12.27812.27

812.27

412.27

7 5.678 5.670 5.670 5.673

23 12.28012.28

012.27

812.27

9 5.674 5.668 5.682 5.675

24 12.27

6

12.28

0

12.27

2

12.27

65.672 5.674 5.672

5.67325 12.272

12.268

12.268

12.269 5.672 5.666 5.688 5.675

26 12.27812.28

012.27

012.27

6 5.664 5.670 5.674 5.669

27 12.29212.27

412.27

012.27

9 5.674 5.674 5.688 5.679

28 12.28412.27

212.28

812.28

1 5.664 5.662 5.660 5.662

29 12.28212.27

812.27

812.27

9 5.664 5.672 5.668 5.668

30 12.27

2

12.28

2

12.28

0

12.27

85.668 5.672 5.662

5.667

3.0 ANALYSIS & DISCUSSION

3.1 DATA SUMMARY AND PRESENTATION

Well-constructed data summaries and displays are essential to good statiscal thinking because

they focus the analyst on important features of the data or provide insight about the type of

12


13/38

model that should be used in a problem situation. In case, we have done some of data display that

more useful for our analysis.

3.1.1 STEM-AND-LEAF DIAGRAM

A stem-and-leaf diagram is a good way to obtain an informative visual display of a data

set x 1, x2,, x n, where each number xi consists of at least two digits.

a) Screw

Stem-and-leaf of Length N = 30Leaf Unit = 0.010

1 515 93 516 444 516 85 517 36 517 98 518 2212 518 5899

(14) 519 000022222334444 519 5668

Stem-and-leaf of Diameter Of Thread N = 30Leaf Unit = 0.0010

1 483 01 4837 484 00000012 484 77777

(6) 485 33333312 48512 486 0000000003 486 771 487 3

b) Washer

Stem-and-leaf of Outer Diameter N = 30Leaf Unit = 0.0010

2 1227 016 1227 23337 1227 5

(9) 1227 77777777714 1227 8889998 1228 011113 1228 22

13


14/38

52.051.951.851.751.6

Median

Mean

51.9451.9251.9051.8851.8651.84

Anderson-Darling Normality Test

Vari ance 0 .011Skewness -1.44669Kurtosis 1.06723N 30

Minimum 51.593

A-Squared

1st Quartile 51.827Median 51.9003rd Quartile 51.935Maximum 51.980

90% C onfidence Interv al for Mean

51.835

2.44

51.900

90% C onfidence Interv al for Median

51 .893 51. 927

90% Confidence Interval for StDev

0. 087 0.134

P-Value < 0 .005

M ean 51. 868S tDev 0.105

90% Confidence Intervals

Length for Source 1

1 1228 5

Stem-and-leaf of Inner diameter N = 30Leaf Unit = 0.0010

1 566 02 566 32 5663 566 713 566 8899999999

(3) 567 00114 567 223310 567 458 567 6775 567 893 568 02 5682 5682 568 77

3.1.2 GRAPH

a) Screw

14


15/38


16/38

4.8724.8604.8484.836

Median

Mean

4.8604.8554.8504.8454.840


Variance 0.0002Skewness -0.03228Kurtosis -1.24437N 30

Minimum 4.8270

A-Squared

1st Quartile 4.8400Median 4 .85303rd Quartile 4.8600Maximum 4.8730

90% C onfidence I nterval for Mean

4.8467

0.83

4.8550


4. 8400 4 .86 00


0. 0111 0 .01 72

P -Val ue 0 .02 9

M ean 4.8509S tD ev 0. 0135


Diameter Of Thread for Source 2

12.28412.28012.27612.272

Median

Mean

12.279012.278512.278012.277512.277012.276512.2760


Var ian ce 0. 000Skewness -0.269346Kurtosis -0.062932N 30

Minimum 12.270

A-Squared

1st Quartile 12.277M edi an 12. 2773rd Quartile 12.280Maximum 12.285


12.276

0.68

12.279


12 .277 12. 279

90% C onfidence Interv al for StDev

0.003 0.004

P -Value 0. 068

M ean 12. 277StDev 0.004


Outer Diameter for Source 1

This is unimodal histogram it is have only one single peak. In the histogramdiagram you can see this unimodal histogram is not symmetric diagram and is saidto be skewed. From the graph histogram of diameter of thread for source 1 can seethe lower tail is much longer than the upper tail, the histogram is negativelyskewed. This situation happens because number of frequency is less at earlyreading data before mean value.

This is unimodal histogram it is have only one single peak. In the histogramdiagram you can see this unimodal histogram is not symmetric diagram and is saidto be skewed. From the graph histogram of diameter of thread for source 2 can seethe lower tail is much longer than the upper tail, the histogram is negativelyskewed. This situation happens because number of frequency is less at earlyreading data before mean value.

b) Washer

16


17/38

12.28012.27812.27612.27412.27212.270

Median

Mean

12.279012.278512.278012.277512.277012.2765


Vari ance 0 .000Skewness -1.18615Kurtosis 2.42836N 30

Minimum 12.269

A-Squared

1st Quartile 12.276Median 12.2783rd Quartile 12.279Maximum 12.281


12.277

0.73

12.278


12 .277 1 2. 279

90% C onfidence Interv al for StDev

0. 002 0.003

P -Val ue 0 .05 0

M ean 12. 277S tDev 0.003


Outer Diameter for Source 2

5.6885.6805.6725.664

Median

Mean

5.6745.6735.6725.6715.6705.669


Variance 0.0000Skewness 0.73803Kurtosis 1.08157N 30

Minimum 5.6600

A-Squared

1st Quartile 5.6690Median 5 .67053rd Quartile 5.6763Maximum 5.6870


5.6704

0.94

5.6742

90% C onfidence Interval for Median

5 .6690 5 .6730


0 .0049 0 .0077

P -Val ue 0. 01 5M ean 5. 6723S tD ev 0. 00 60


Inner Diameter for Source 1

This is unimodal histogram it is have only one single peak. In the histogramdiagram you can see this unimodal histogram is not symmetric diagram and is saidto be skewed. From the graph histogram of outer diameter for source 1 can see thelower tail is much longer than the upper tail, the histogram is negatively skewed.This situation happens because number of frequency is less at early reading databefore mean value.

This is also the unimodal histogram but it has some part different from the

histogram of outer diameter for source 2.This is not the symmetric and you can see

the upper tail of the histogram stretches out much farther than the distribution of

value is positively skewed. This situation happens because number of frequency is

less at early reading data before mean value.

17


18/38

Value51.9651.9051.8451.7851.7251.6651.60

Length

5.6805.6755.6705.665

Median

Mean

5.67405.67355.67305.67255.67205.67155.6710


Vari ance 0 .0000Skewness -0.196456

Kurtosis -0.281616N 30

Minimum 5 .6620

A-Squared

1st Quartile 5.6690M edi an 5. 673 03rd Quartile 5.6753Maximum 5.6810


5.6709

0.26

5.6739

90% C onfidence Interval for Median

5. 6710 5. 674 0


0. 0040 0. 006 1

P -Value 0. 697

M ean 5. 6724S tDev 0. 0048


Inner Diameter for Source 2

This is the unimodal histogram but it has some part different from thehistogram of inner diameter for source 1.This is not the symmetric and you can see

the upper tail of the histogram stretches out much farther than the distribution of

value is positively skewed. This situation happens because number of frequency is

less at early reading data before mean value.

This is unimodal histogram it is have only one single peak. In the histogramdiagram you can see this unimodal histogram is not symmetric diagram and is saidto be skewed. From the graph histogram of inner diameter for source 2 can see thelower tail is much longer than the upper tail, the histogram is negatively skewed.This situation happens because number of frequency is less at early reading databefore mean value.

3.1.3 DOT PLOT

a) Screw

18


19/38


20/38

Value5.6885.6845.6805.6765.6725.6685.6645.660

Dotplot of Inner Diameter

N

Value

302520151050

52.0

51.9

51.8

51.7

51.6

Length

Dot plot of Outer and Inner Diameter

The dot plot is a very useful plot for displaying a small body of data, up toabout 50 observation. This dot plot is allows us to easily see two important featuresof data: the location or the middle and the scatter or variability. When the numberof observation is small, it is usually difficult to see any specific pattern in variability.

3.1.4 MULTIVARIATE DATA

a) Screw

20


21/38


22/38

N

Value

302520151050

12.286

12.284

12.282

12.280

12.278

12.276

12.274

12.272

12.270

Outer Diameter

Probability plot of washer

Figure Probability plot of Washer is a lognormal probability plot. The data fallmuch closer to the straight line this plot, particularly the observations in the tail.The lognormal distribution is more likely to provide a reasonable model for diameterof washer.

The scatter plot of the screw vs. washer

The scatter plot of the screw vs. washer indicates no strong relationshipbetween screw and washer. There is no tendency for screw either to increase ordecrease as washer increases.

3.2 TOOLS AND CALCULATION

In this chapter, we illustrated how a parameter can be estimated from a sample data. The field of

statistical inference consists of those methods used to make decisions or to draw conclusions

about a population. Statistical methods are important tools in these activities that could assist

analyst with both descriptive and analytical methods in handling with the variability in theobserved data. Statistical inference may be divided into 2 major areas, parameter estimation and

hypothesis testing.

3.2.1 NORMAL DISTRIBUTION

22


23/38

Value

N

52.152.051.951.851.751.6

7

6

5

4

3

2

1

0

Mean StDev N

51.87 0.1051 30

51.87 0.09088 30

Variable

Source 1

Source 2

NormalLength

Value

N

4.8724.8604.8484.8364.824

8

7

6

5

4

3

2

1

0

Mean StDev N

4.852 0.01007 30

4.851 0.01347 30

Variable

Source 1

Source 2

NormalDiameter Of Thread

Normal distribution is the most important distribution in statistics. A statistical table

typically provides two types of tables associated to a standard normal distribution.

a) Screw

23


24/38

Value

N

12.28412.28212.28012.27812.27612.27412.27212.270

9

8

7

6

5

4

3

2

1

0

Mean S tDev N

12.28 0.003511 30

12.28 0.002609 30

Variable

Source 1

Source 2

NormalOuter Diameter

Value

N

5.6885.6845.6805.6765.6725.6685.6645.660

10

8

6

4

2

0

Mean S tDev N

5.672 0.005989 30

5.672 0.004789 30

Variable

Source 1

Source 2

NormalInner Diameter

b) Washer

24


25/38

3.2.2 SAMPLING DISTRIBUTION

SAMPLING DISTRIBUTION OF X 1 X 2

SCREW

i) LENGTH

From Analysis :, X 1 = 51.871, X 2 = 51.868, 1 = 0.105054, 2 = 0.090888

X1 ~ N ( 51.871, )

X2 ~ N ( 51.868, )

X1 X2 ~ N ( 3 x10 -3 , 6.432 x10 -4 )

P (X 1 > X 2) = P (X 1 X2 > 0 )

25


26/38

= P ( Z > )

= P ( Z > -0.118 )

= 0.54697

ii ) DIAMETER OF THREAD

From Analysis :, X 1 = 4.852, X 2 = 4.851, 1 = 0.1010, 2 = 0.013417

X1 ~ N ( 4.852, )

X2 ~ N ( 4.851, )

X1 X2 ~ N ( 1 x10 -3 , 3.4603 x10 -4 )

P (X 1 > X 2) = P (X 1 X2 > 0 )

= P ( Z > )

= P ( Z > -0.0375 )

= 0.52144

WASHER

iii) OUTER DIAMETER

From Analysis :, X 1 = 12.278, X 2 = 12.277, 1 = 0.003499, 2 = 0.002579

X1 ~ N ( 12.278, )

X2 ~ N ( 12.277, )

X1 X2 ~ N ( 1 x10 -3 , 6.29 x10 -7 )

26


27/38


28/38

Assume :

= = 1.96

90 CI Difference Of The Two Population Means.

1 - 2 = 51.871 51.868 1.96

= ( 3 x 10 -3 ) 0.0497

0 1 - 2 0.0527

ii) DIAMETER OF THREAD

From Analysis :, X 1 = 4.852, X 2 = 4.851, 1 = 0.1010, 2 = 0.013417

Assume :

= = 1.96

90 CI Difference Of The Two Population Means.

1 - 2 = 4.852 4.851 1.96

= ( 1 x 10 -3 ) 0.06

0 1 - 2 0.061

WASHER

iii) OUTER DIAMETER

From Analysis :, X 1 = 12.278, X 2 = 12.277, 1 = 0.003499, 2 = 0.002579

28


29/38


30/38

Ho : 1 = 2

H1 : 1 2

Z test =

=

= - 0.11829

= Z 0.025

= 1.96

Z test > -

-0.11829 -1.96

Accept H o

There is no sufficient evidence to reject null hypothesis.

iii) DIAMETER OF THREADFrom Analysis : X 1 = 4.852, X 2 = 4.851, 1 = 0.10101, 2 = 0.013417

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

Z test =

=

= 0.05375

30


31/38

= Z 0.025

= 1.96

Z test >

0.05375 1.96

Accept H o


b) WASHER :

i) OUTER DIAMETER

From Analysis : X 1 = 12.278, X 2 = 12.277, 1 = 0.003499, 2 = 0.002579

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

Z test =

=

= 1.2601

= Z 0.025

= 1.96

Z test >

1.2601 1.96

Accept H o


31


32/38

ii) INNER DIAMETER

From Analysis : X 1 = 5.672, X 2 = 5.672, 1 = 0.006, 2 = 0.004693

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

Z test =

=

= 0

= Z 0.025

= 1.96

Z test >

0 1.96

Accept H o


32


33/38

3.2.5 TEST OF HYPOTHESIS FOR THE RATIO OF THE VARIANCES.

a) SCREW :

i) LENGTH

From Analysis : = 0.01104, = 0.008261

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

f test

=

=

= 2.1

f test =

=

= 1.786

Since the value f test = 1.786 is between 0.47619 and 2.1, we are unable to rejectHo : 1 = 2 at the 0.05 level of significance. Therefore, there is no strong evidence

to indicate that two population variances differ.

33


34/38

ii) DIAMETER OF THREAD

From Analysis : = 0.10101, = 0.013417

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

f test

==

= 2.1

f test =

=

= 56.6784

Since the value f test = 56.6784 is not between 0.47619 and 2.1, we are reject H o : 1= 2 at the 0.05 level of significance. Therefore, there is very strong evidence to

indicate that two population variances differ.

34


35/38

b) WASHER :

i) OUTER DIAMETER

From Analysis : = 0.003499, = 0.002579

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

f test

=

=

= 2.1

f test = =

= 1.8407

Since the value f test = 1.8407 is between 0.47619 and 2.1, we are unable to reject

Ho : 1 = 2 at the 0.05 level of significance. Therefore, there is no strong evidence

to indicate that two population variances differ.

35


36/38

ii) INNER DIAMETER

From Analysis : = 0.006, = 0.004693

Assume : = 0.05

Ho : 1 = 2

H1 : 1 2

f test

==

= 2.1

f test =

=

= 1.63

Since the value f test = 1.6346 is between 0.47619 and 2.1, we are unable to rejectHo : 1 = 2 at the 0.05 level of significance. Therefore, there is no strong evidenceto indicate that two population variances differ.

36


37/38

4.0 CONCLUSION

From our analysis that we had done, we found that there are not enough evidence to

reject null hypothesis, H o. Basically we found that there are very small difference in mean andstandard deviation of screw and washer from both source. We also consider some minor

measurement error since we choose two different people for collecting sample data. This two

people might have different observation when taking the data.

In general, engineers develop new products, improve existing designs, build and test

prototype, troubleshoot ongoing manufacturing process and others. In each of these functions,

engineers collect and analyze data as an integral part of their job. Thus, statistical methods are

inseparable part of how engineers solve engineering problems.

From the hypothesis testing, we accept all null hypothesis which can be conclude that

there are no obvious different between the data from this two sources. Basicly, the smallest

difference is better and we found that our data match to this standard assumption. The estimation

analysis showed that the CI for the difference between two population means are very close and

we can say that there are no different from the two source.

37


38/38

REFERENCES

1. Engineering Statistics 4 th Edition; Douglas C. Montgomery, George C. Runger, Norma Faris

Hubele; John Wiley & Sons, Inc.; 2007.

2. Engineering Statistics Edition 2009; Arifah Bahar, Ismail Mohamad, Muhammad Hisyam

Lee, Noraslinda Mohamed Ismail, Norazlina Ismail, Norhaiza Ahmad, Zarina Mohd Khalid;

Department of Mathematics, Faculty of Science, UTM; 2009.

3. Statistics Formulae and Tables; Dr. Muhammad H. Lee; Faculty of Science, UTM.

4. www.mwindustries.com

5. www.barth-landis.com

6. www.wikipedia.com

Documents

3.Folio-project Screw & Washer