product reliability slides

8/10/2019 Product Reliability slides

1/94

Basic Failure Data Analysis

Failure data is basically the time to failure for many items.

From this data we have to calculate:

Failure Density f(t)

Failure Rate (t)

Reliability R(t)

MTTF

Standard deviation

Confidence levels


2/94

Failure data is of various types:

Field Data vs Test Data

Field datais collected from the user sites. This will cover a large number of

samples. But the data from the individual samples will not be precise. Usually

The failure times are grouped into intervals containing a large number of samples.

Test data is obtained from reliability testing labs. The number of samples will be small,

but failure times will be precise.


3/94

Grouped Vs Ungrouped data

In grouped data, the failure times of many items are grouped into intervals

(typical of field data).

Eg: 0-100 hrs, Items failed=20.

101-200hrs, items failed =23

In ungrouped data, the precise failure times of individual items are noted.

They have to be later ranked according to the order of failure

(1stfailure, 2ndfailureetc).


4/94

3. Censored Vs. Uncensored data

(incomplete Vs. complete data)

Censoring is a common problem in failure data.

Some usual reasons are:

a) items are removed before failure (because they failed in a

different way).

b) test is completed before all items fail.

We have to take into account the censored units, otherwise only the

weakest units would be considered and reliability would be underestimated.


5/94

Example of Grouped failure data (Uncensored):

The following field data is assembled, observing the failure of 500 items.

Plot the Probability Density, Reliability, Cumulative failure, Failure rate.

Find the MTTF and Standard deviation.

Time

(hr)

0 100 200 300 400 500 600 700 800 900 1000

Survi

vors

500 480 421 360 343 295 220 175 130 70 0


6/94

The following items have to be tabulated, then plotted as histogram.

Calculation is carried out as finite steps of intervals 0-100, 101-200, 201-300 etc.

Note that t=100hrs.

Time Ns(t) Ns(t+t) Nf(t) h(t) f(t) R(t) F(t)

0-100 - - - - - - -

101-200

(etc)

- - - - - - -

Where:

Ns(t)=Numbers of survivors at time t.Ns(t+t)=Number of survivors at t+ t.

Nf(t)=Number of failures in the interval t.


7/94

No. of failures in an interval( )

(No. of survivors in the beginning of the interval)( t)

( ) =

( ) 100

f

s

h t

N t

N t

h(t) or(t) is the failure rate in the interval t

and given by,

Its unit is fraction of failures per hour (or,probability of failure per hour).

We can multiply the above with 100 and get percentage of failures per hour.


8/94

No. of failures in an interval( )

Initial sample size

( ) =

500

f

tf t

N t

f(t) is the probability density of failure (discrete form)in each

interval:


9/94

Reliability: ( )

Net survivors at the end of an interval( )

Initial sample size

( ) =

500

Cumulative failure: ( )

Net failed items at the end of an interval( )

Initial sample size

=

s

R t

R t

N t t

F t

F t

N

( ) 5001 ( )500

s t t

R t


10/94

Time Ns(t) Ns(t+t) Nf(t) h(t) f(t) R(t) F(t)

0-100 500 480 20 4x10-4 0.04 0.96 0.04

101-200 480 421 59 12.3x10-4 0.118 0.842 0.158

201-300 421 360 61 14.5x10-4 0.122 0.72 0.28

301-400 360 343 17 4.72x10-4 0.034 0.686 0.314

401-500 343 295 48 14.0x10-4 0.096 0.59 0.41

501-600 295 220 75 25.4x10-4 0.150 0.44 0.56

601-700 220 175 45 20.5x10-4 0.09 0.35 0.65

701-800 175 130 45 25.7x10-4 0.09 0.26 0.74

801-900 130 70 60 46.2x10-4 0.12 0.14 0.86

901-1000 70 0 70 100.0x10-4

0.14 0 1.0

f =1.0

(here, n=10i.e, number of time intervals, or number of observations)


11/94

Hazard rate


12/94

Failure density


13/94

Cumulative failure


14/94

Reliability


15/94

Based on the data we can answer questions like these:

1. What is the probability that the item will fail before 300 hours?

i,e cumulative probability of failure upto and including the 201-300 hr step.

=0.04+0.118+0.122=0.28

2. What is the reliability of the item at 300 hours?

R(300) is 0.72.

3. What is the probability of failure between 300-700hours of duty?

Add up failure probabilities f(t)for 301-400, 401-600 and 601-700.

=0.034+0.096+0.15+0.09=0.37


16/94

Failure density

Calculation of MTTF

50150

i+1 i

( )t +t

Here is the mean time of each interval (i.,e )2

(because many items fail in an interval; their precise failure times are not known)

50 0.04 150 0.118 ........ 950 0.14 548.8

i i

i

MTTF t f t

t

hrs


17/94

2 22 2 2 2

2

Standard Deviation (s):

( )

(50 0.04 150 0.118 .... 950 0.14) 548.8

381060 548.8 79879

79879 283

i iVariance t f t MTTF

s hrs


18/94

Calculating the confidence levels of the MTTF

As per Central Limit Theorem, the distribution of means are Gaussian.If we know the exact value of 0, we can work out the confidence levels.

However, we have only s, the standard deviation of sample with nobservations.

Hence we use Students t-distribution, which approximately resembles the

Gaussian distribution, but the spread or uncertainty is more than Gaussian.

The Students t-distribution also realistically represents the distribution for small

values of n(i.e no of observations).


19/94

The shape of the t-distributionchanges depending on the value of n.

i.e., a parameter called degree-of-freedom (n-1) is used to specify the

shape of the t-distribution.

Higher the value of n, the closer the Student distribution matches the Gaussian

distribution.


20/94

Students-t distribution, random variable t is defined as:

/

where:

is the mean calculated from the data table.

is standard deviation calculated from observations.

is the random variable, representing the means of obser

x MTTFts n

MTTF

s n

x n

vations.


21/94

Calculating the 90% confidence level for MTTF prediction using T-Distribution?

1

1

MTTF

( corresponds to =0.05)

is defined as (1-Confidence Level)/2

st

n

t


22/94

For our specific problem:

n=10 (no. of observations)

DOF= n-1= 9;

Confidence level = 0.9Hence

0 05

Look at the chart to find value of t1corresponding to DOF=9, 0 05


23/94

Look at the chart to find value of tcorresponding to DOF=9,0 05


24/94

t1=1.833

Lower limit is : 548.81.833 x 238/sqrt(10) = 410.8

Upper limit is: 548.8 + 1.833 x 238/sqrt(10) =686.8

We can say with 90% certainty that the MTTF in this case will be

will be between 410.8 and 686.8.


25/94

Time Ns(t)

0-100 500

101-200 480

201-300 421

301-400 360

401-500 343

501-600 295

601-700 220

701-800 175

801-900 130

901-1000 70

Note:

In this example we have N=500samples, divided

into n=10classes or observations.

What is the ideal number of classes to divide 500

samples into? If the number of classes are too small

the histogram will be too approximate.

If the number of classes are too large, each class will

contain only a few samples (not representative).

The optimum number of classes is approximately

given by Sturges rule:

n=integer(1+3.3 log10N)

Here n=integer [1+3.3 log 500]

=integer[9.9]

=9 classes.


26/94

Ungrouped data (without censoring):

These are data typically obtained by Lab testing for reliability.

There will be only a few samples, but the time to failure is accurately noted.

The ungrouped data has to be ordered first, and then we have to plot the various

parameters.

Example:

Six machines A,B,C,D,E, F are tested until their first failure occurs. The failure

times in hours are 14.0, 12.2, 14.6, 14.1, 13.1 and 14.7 hrs.

Let us plot the cumulative failure frequency (F(ti)) and study the curve.


27/94

Failure order (i) Time of failure (hr) Cumulative failure

order; F(t)=i/n

1 12.2 1/6=0.167

2 13.1 2/6=0.333

3 14.0 3/6=0.500

4 14.1 4/6=0.667

5 14.6 5/6=0.833

6 15.0 6/6=1.000

Time of failure is now ordered according to increasing number of failures

This means, there is 1/6th

probability of failure at 12.2hrs,2/6that 13.1 hrs.etc


28/94

This cumulative curve has to be improved because:

F=1 is not really achievable in a finite time.

Hence a corrected cumulative distribution has to be used.


29/94

1.Mean rank formula:

We assume there are 1 samples, one of which doesnt fail at the end of the test.

Hence cumulative failure ( ) is given by .1

This gives F


30/94

Failure order

(i)

Time of

failure (hr)

F(i)=i/n F(i)=

i/(n+1)

F(i)=

(i-0.3)/(n+0.4)

1 12.2 1/6=0.167 0.143 =1/7 0.1094

2 13.1 2/6=0.333 0.286=2/7 0.2656

3 14.0 3/6=0.500 0.4286 0.4219

4 14.1 4/6=0.667 0.5714 0.5781

5 14.6 5/6=0.833 0.7183 0.7344

6 15.0 6/6=1.000 0.8571=6/7 0.8906

The two new estimates of F(t) are shown in red


31/94

It is seen that Median fitting is more parallel to the original plot, and hence

captures the trend better than the Mean fitting.


32/94

Detailed Example:

The unordered data of failure times (hr) of 10 machines are as follows

24.5, 18.9, 54.7, 48.2, 20.1, 29.3, 15.4, 33.9, 72.0, 86.1.

Plot Cumulative failure, Reliability, Failure Density, Hazard rate, MTTF.

And 90% confidence interval for MTTF.

Procedure:1. Find cumulative failure F(ti) using Median formula = (i-0.3)/(n+0.4)

2. Reliability is 1-F(ti) or R = (n-i+0.7)/(n+0.4) .

3. Failure density f(t)= -d R/dt = dF/dt

It can be calculated approximately by: f(ti)= -[R(t i+1)-R(ti)] / (ti+1-ti)

4. Hazard rate h(t)= f(t)/R(t)

It can be calculated approximately by : h(ti)= f(ti)/R(ti)


33/94

Density

f(t)

Hazard

h(t)

0.0044 0.0044

0.0275 0.0295

0.0801 0.0958

0.0219 0.0295

0.0200 0.0311

0.0209 0.0381

0.0067 0.0149

0.0148 0.0416

0.0056 0.0214

0.0068 0.0417

- -

Failure

Order(i)

Time

(hr)R(ti)

0 1.0

i=1 15.4 0.933

i=2 18.9 0.837

. 20.1 0.740

. 24.5 0.644

. 29.3 0.548

. 33.9 0.452

. 48.2 0.356

. 54.7 0.260

i=9 72.0 0.164

i=10 86.1 0.0673

T= 403.1

MTTF=403.1/10 =40.31 (each failure time pertains to one item, hence we

can find MTTF by calculating the simple average)

s=23; The 90% confidence intervals are 26.9 and 53.64

(0.933 0.837)

18.9 15.4

0.0275

0.933


34/94


35/94


36/94


37/94

Censored Data


38/94

How to deal with Incomplete or Censored Data:

These are cases where some samples are prematurely withdrawn beforefailure; or failure occurs due to some other causes.

a) Singly censored data:

Herenitems are tested for failure in a fixed test time.

b) Multiply censored data:

Here the test times differ among then items (as shown previous examples).

We will deal with only Multiply censored data as it is more widely used.

There are two cases of multiply censored data: Grouped and Ungrouped.


39/94

Life Table method for Grouped data which is censored.

(originally used in medical trials)

Let the failure data be organized into i=1,2,3..ntime intervals or bands;

say 0-100hrs, 101-200 hrs etc.

N= Total number of samples.

Nf.i = Number of samples failing in the interval i.

Nc.i= Number of censored items in interval i.

N s.i = Number of survivors in the beginning of an interval = N-Nf. i-1 - Nc. i-1

N s.i = Adjusted N s.i = N s.i - Nc.i /2


40/94

1

th

1. 1'

( . ., Reliability of the i band is the product of the local Reliabilities of the

previous band from 1 to i).

2. Hazard rate, h' .

3. Failure de

i

fii

si

fi

i

si

NRN

i e

N

N t

i-1 i

i

(R -R )nsity is approximately found out by f = - .

t


41/94

Example:

A total of N=20 samples are tested in intervals of time 50hrs.

5 are censored. 12 are failed. Remaining are left untested.

Calculate the Reliability and Failure Density distribution.Time Nf.i Nc.i

0-50 3 0

51-100 1 1101-150 2 1

151-200 2 2

201-250 1 0

251-300 2 1301-350 1 0

Ri hi

0.850 0.003

0.85x0.94=0.8 0.0012

0.689 0.0028

0.560 0.00364

0.489 0.0025

0.339 0.0062

0.254 0.005

Ns.i Ns.i 1-Nf.i/Ns.i(local

reliability)

20 20 0.850

17 16.5 0.940

15 14.5 0.862

12 11 0.812

8 8 0.875

7 6.5 0.692

4 4 0.750


42/94

Ungrouped Censored data.

The method preferred is the Rank adjustment method, by Johnson [1959].

Consider the table below.

Rank (no

censoring)

Rank with censoring

i=1 1

i=2 2

i=3 Censored.

Rank is skipped.(Rank Increment =1.2,say)

i=4 2+1.2=3.2 (instead of 4 without censoring)(previous rank +R.I)

i=5 3.2+1.2=4.4(instead of 5 without censoring)(previous rank +R.I)

i=6 Censored.Rank is skipped.

(Rank Increment =1.3, say)

i=7 4.4+1.3=5.7(instead of 7 without censoring)

(previous rank +R.I)

We give a Rank Increment

for the missing (censored)

data.

Rank is skipped for the

censored data.

The RI is added to previously

given rank, to obtain the

next rank.

A new RI is calculated for

the next censored data.

The later a component is

censored, the higher is the

RI given to it.


43/94

.

How to find rank increment (RI), Reliability etc.

Let be the highest rank of failure. . ., it is the no. of items being tested.

is the original rank without censoring.

is the corrected rank.t i

n i e

i

i

t.i-1

t.i

This is obtained by recursively adding Reliability

Index to the previous ranks.

1 previous corrected rank( )RI=

1 number of units beyond the censored unit

0.3

Reliability=1- 0.4

n i

i

n


44/94

Rank (i) Time(hr) Rank Index Corrected

Rank (iti)

Reliability

1 150 1 (start) 1 0.933

2 340 C (11-1)/(1+8)=1.11 - -

3 560 1+1.11=2.11

0.826

4 800 2.11+1.11=3.22

0.719

5 1130 C (11-3.22)/(1+5)=1.2963 - -

6 1720 3.22+1.2963=4.518

0.594

7 2470 C - -

8 4210 C (11-4.518)/(1+2)=2.16 - -

9 5230 4.518+2.16=6.679

0.387

10 6890 6.679+2.16=8.839

0.179


45/94


46/94

By plotting the Hazard rate - h(t) or(t)


47/94

If it is a constant hazard rate, the distribution could be Exponential.

If it is linearly increasing hazard rate, the distribution could be Rayleigh.

If it is a non-linearly increasing or decreasing it could indicate a Weibull

Distribution which require further investigation.


48/94

Fitting the data into standard probability distributions

(Weibull, Exponential..etc)

using Curve fitting.

This is a very convenient and general purpose method to identify a

repair distribution.

The plot of ti versus Fi(cumulative probability) is drawn on a linearized

scale.

A line is drawn with the best fit through all plotted points.

The graph should be a straight line if it matches the appropriate

probability distribution.

From the straight line, noting the slope and intercepts we can calculatethe failure rate and other distribution parameters.


49/94

For more accurate fitting we have to use specific tests for

specific distributions:

Eg. Bartletts test for Exponential Distribution

Manns test for Weibull Distribution

Kolmogorov Smirnov Test for Normal and Lognormal distributions.


50/94

In these exercises we use only uncensored data for simple illustration.

We can equally well used censored data and plot the tiversus Fiusing the

censoring methods previously studied.


51/94


52/94

Example 1. Manually fit an Exponential Distribution with the data:

Time to first failure times of 10 machines are given in hours

80,134,148,186,238,450,581,890.

i ti Fi(median rank) loge( 1/(1-Fi) )

1 80 0.083 0.0870

2 134 0.2024 0.2261

3 148 0.3214 0.3878

4 186 0.4405 0.5807

5 238 0.5595 0.8199

6 450 0.6786 1.1350

7 581 0.7976 1.5976

8 890 0.9167 2.4849


53/94


54/94

Instead of manually drawing a line, a more accurate option is the

linear least square fit, by minimizing the square of the y deviations.

Assuming the tivalues are taken as xi, and the loge() values are taken as

yi,

If y =a+bx is the fitted line, a and bare calculated as follows:


55/94

The same can also be drawn on standard Exponential Distribution

Chart paper.

Here we can directly plot Fi (y axis) against ti (x axis) on the ready-made

logarthmic scale.

The value of ti corresponding to Fi =0.632 gives the MTTF.

The reciprocal of MTTF gives the failure rate.


56/94

F(t)

Time (t)

Standard

Exponential Chart

Paper.


57/94

Exponential plot,

drawn on chart paper.


58/94


59/94

We can plot z versust ,with orwithout standard chart paper.

Without chart paper:

The failure times of 20 machines are observed to be 68,69.6,71.1,71.4,74.3,74.6,75.5,

77.6,77.8,78,78.2,80.2,,80.3,81.9,83.0,85.6,87.4,87.7,88.4,98.3 hrs.

Check the fit of Gaussian distribution and find MTTF and Standard deviation.

i t i Fi(median rank) zi (from table)

1 68.0 0.0343 -1.8212 69.6 0.0833 -1.3832

3 71.1 0.1324 -1.1151

4 71.4 0.1814 -0.9100

5 74.3 0.2304 -0.7375

..

20 98.3 0.9657 1.8211


60/94

Approximately:

Slope =0.124

Y intercept=-10

Standard deviation=8.09

MTTF=80hrs

zi


61/94

0

50 100 hrs

MTTF=80

=10

Data is plotted

on chart paper.

MTTF and S.D are

calculated from

F=0.5

And F=0.84.

Plotting Log-normal distributions


62/94

g g

2

2

1log

2

1

We have to use the alternate form of Log-Normal distribution;

1( )

2( . ; .)

1( ) log .

:

( )

emed

t

tt

med

e

med

f t e

twhere is the std dev of t t is the median time to failure

tF t

t

This is written as

z F

1 1

log log log .

is calculated from charts - using the same procedure as Normal

Distribution plotting discussed previously.

e e e med

med

tt t

t

z


63/94

2

e

2

1. ., log log

( . . x-cordinate is log (time).

y intercept is log from which we can find .

1

Slope is )

or MTTF is calculated by a separate formula:

MTTF

e e med

e med med

mean

med

i e z t t

bx a

i e

t t

t

t e


64/94

loge ti

zi1

-1

-2

0

0 10 1001000

Illustration of a Log-normal plot

Slope is equal to 1/standard_deviation;

Y intercept is equal to -loge tmed


65/94

0.1 10.0

Log-normal probability

chart paper.

Note the log-scale on the

x-axis for time.

Here tmed is the timecorresponding to

F=0.5

1/ is the slope.

We can then find MTTF.

100.01.0


66/94

How to fit a Weibull Distribution:

Since Weibull distribution can model a wide range of failure rates, it ishighly important to learn to model Weibull distributions.

In the two parameter Weibull distribution, we can find the shape factor () and

scale factor ().

In the three parameter Weibull distribution, we also have to find the startingtime (t0or ).


67/94

-

-

Two Parameter Weibull Distribution ( shape parameter, =scale parameter);

( ) 1-

1 ( )

1

log 1 ( )

1log log log log

1 ( )

. , ( is

t

t

e

e e e e

F t e

F t e

t

F t

tF t

i e y bx c b

the slope equal to , and y intercept is equal to - log ).

: coordinates are in the form of log (t).

ec

Note x

Note: The value of time t corresponding to F=0 632 gives the value of


68/94

Note: The value of time tcorresponding to F=0.632 gives the value of .

Usually it is preferred to use standard Weibull chart paper.

loget

loge[

loge(1

/(1-F(t)))]

y

x

Slope= y/ x=

Manual plotting of Weibull distribution


69/94

Plot the following data and check for Weibull fit:

i ti(hr) Median

rank (Fi)

loge(ti) Loge [loge(1/(1-Fi)) ]

1 32 0.13 3.4657 -1.9714

2 51 0.31 3.9318 -0.9914

3 74 0.50 4.3041 -0.3665

4 90 0.69 4.4998 0.1580

5 120 0.87 4.7875 0.7131

First let us plot without chart paper.


70/94

Slope=2.2/1 =2.2 =

Y intercept is found out by extending the line to cut the y axis at x=0;

c= -9.8

Therefore is calculated as 85.


71/94

1 100010 100

F=0.632

t

Weibull chartpaper.

F(t)

t


72/94

The same example is solved

Using Weibull chart paper.

1.9

88

Time (t) hrs

F(t)

What is the failure rate ?


73/94

1 0.9

What is the failure rate ?

1.9 88.

( ) 0.0216 (increasing failure rate).88

1MTTF= (1 ) 88 (1.53) 88 0.8876 78

( : Gamma function data is obtained from tables)

med

use and

t t

t

hrs

note

t

1.9

-

-88

1

? Put ( )=0.5 where ( )=

We get 0.5

( 0.69315 , this is a standard formula derived from the above)

72.56 .

med

t

med

t

med

med

R t R t e

e

or t

t hrs


74/94

0.9

( ) 0.021688

tt


75/94


76/94

Three parameter Weibull Distribution

When we plot the 2 parameter Gamma distribution, sometimes we get a curve

which deviates from the straight line, towards the bottom right.

This indicates the need for adding the location parameter or failure free time

which is represented by t0 or g. Failures start occuring only after the time t0.

F(t)

t


77/94

0

0

-

1

0

0

1

0

The Three parameter Weibull equation is obtained by replacing

with ( - ) in the function.

( ) 1 ( )

( )

1(1 )

0.69315

t t

med

t t t

R t F t e

t tt

MTTF t

t t


78/94

Procedure for 3 parameter Weibull plot:

1. Once we obtain a downward deviating Weibull plot (as discussed earlier),it means we need to insert the location parameter t0.

2. This is done by trial and error.

3. Assume a reasonable value of t0. On the x axis, (t- t0) is taken instead

of t. If the guess t0 is correct we get a straight line. and are calculatedas usual.

4. If the estimate of t0 is too high we get a upward deviating curve.


79/94

F(t)

Log Scale (t-t0)

t0

t0

Original curve

Corrected curve

Goodness of Fit and Hypothesis Testing


80/94

using Chi Square (2) distribution.

This is a statistical method for testing wether a given failure data fits into any

predicted distribution.

It can be applied only to Grouped and ordered data. There must be at least

5 samples in each class. If the data is ungrouped, it has to be grouped into

appropriate classes using Sturges Rule.

A hypothesis is made such as The given failure data fits the Exponential

Distribution and it is proved either true or false.

The deviations between the given data and predicted model are expressed

as the Chi-Square parameter 2.

For the hypothesis to be true, the cumulative probability of 2 must be

0.1 or less. This corresponds to 90% confidence level.


81/94

2

12 220

2

Chi - square distribution is represented as,

( , )

Here, is the estimator of deviation (or error value),

is the DOF of the data, i.e., No. of classes-1, Number of random variables

f Y e

or

2

2

-1.

In order for the hypothesis to be acceptable, must be small such that the

cumulative frequency ( ) must be 0.1 or less.

This value is obtained from data tables.

F


82/94

f(,5

2

2Distribution for 5 ; the Cumulative probability is the red region.

Some Chi-square Tables give thearea of the blue region (i.e 1-F(t)).

Some Chi-square tables give F(t).

Cumulative probability

must be less than 0.1


83/94

2

2

2

2

Chi Square parameter is calculated as:

i i

i i

i i

i i

Observed Expected

Expected

x E

E


84/94

Example:

A coin is given. The hypothesis is This is a fair coin.

If it is a fair coin, it has equal chances of obtaining Head or Tails.

To test this hypothesis, we conduct a test of 100 coin tossings

and the data is shown in 2 classes.

Class Observed ExpectedHeads 38 50

Tails 62 50


85/94

2 2

2

2

2

2

(38 50) (62 50)5.7650 50

. 1 2 1 1

5.76 1.

( ) 0.1.

(1 ( )) 0.9.

No of Classes

Look in the Chi square tables for and DOF

F must be less than

Or the confidence level F must beat least


86/94

2

2

5.76

1 ( ) 0.015interpolated.P F

P should be at least 0.9 for the hypothesis to be accepted.

Here it is only 0.015.

Hence the hypothesis is definitely rejected.


87/94

Reliability example-2:

Grouped failure data of transistors are shown in 5 classes.

1. Check the hypothesis that there is constant failure rate of 0.012 per hour

2. Check the hypothesis that there is a variable failure rate defined by,

h(t)=0.2670 t -0.4170

Classes (interval in hours) No. Failed

0-999 18

1000-1999 14

2000-2999 10

3000-3999 12

4000-4999 6

Data given

Let us test the first hypothesis (constant failure rate of 0.012/hr).


88/94

Classes (interval in

hours)

Observed Expected

0-999 18 12

1000-1999 14 12

2000-2999 10 12

3000-3999 12 12

4000-4999 6 12

This means we expect 12 failures every 1000 hours.

2 2 2 2 22

2

(18 12) (14 12) (10 12) (12 12) (6 12)..... 6.67

12 12 12 12 12

. 1 5 1 4

6.67 4.

(1 ( )).

No of Classes

Look in the Chi Square tables for and DOF

We can find the confidence level F t


89/94

2

2

6.7

1 ( ) 0.15interpolated.P F


Here it is only 0.15.

Hence the hypothesis is definitely rejected.

The hypothesis that there is a variable failure rate defined by,

h(t)=0.2670 t -0.4170


90/94

h(t) 0.2670 t

Using the above eqn find the failure rates at the centre of the class

intervals i.,e h(500), h(1500), h(2500),h(3500),h(4500). Let us assume

they represent the failure rate across an interval.

The failure rates are 0.02, 0.0126, 0.0102, 0.0089, 0.008.

The expected failures are 20,13,10,9 and 8 as per the above failure

rates.

Classes (interval inhours)

Observed Expected

0-999 18 20

1000-1999 14 13

2000-2999 10 10

3000-3999 12 9

4000-4999 6 8

2 22 (18 12) (14 13) ..... 1.1; 4

12 13


91/94

2

2

1.1

1 ( ) 0.9 (very rough interpolation).P F


Hence the hypothesis is accepted.


92/94

An alternate

Chi-Square table

Showing F(2) instead

of 1-F(2).

Here we can

get a better estimate

of the cumulative

probabilitycorresponding

to .2 1.1


93/94

Goodness of Fit for Ungrouped data:

There are other tests such as Kolmogorov Smirnov tests which canbe used for ungrouped data.

But they have limitations such as being applicable only for Normal,

and Lognormal distributions.


94/94

product reliability slides

Documents