design of experiment Experiments of Evaluation

1. 67.
One of the important tasks in designing engineering cxpctiments is to draw
conclusions about a population from a sample. An engineer seldom knows
the characteristics of the population. In most cases such information just
does not exist. Therefore, be chooses II. sample of a limited number of items.
Such a sample mayor may not be truly representative of the whole population,
and therefore it is essential to predict the characteristics of the population
from the sample. .
Experiments can be designed around samples which would evaluate
the following characteristics: the quality of the product, the uniformity of
the product, and the probability of occurrence of a particular event associated
witr the product. The first two are the basic parameters of the normal
and Wcibull distributions. In the case of the normal distribution, the quality
is expressed by mean 1.1 and the uniformity by standard deviation (1 or variance
(12. For the Weibull distribution, the corresponding two parameters' are
the characteristic value 0 and the Wcibull slope b. The: probab.lity 0:-
occurrence of a particular event p is 1.11cbasic pzramete: -:;f the bi';~::n;G.j
distribution.
!
/3I . ,;. .:.~,!.. . . .
Experiments of Evaluation-
Mood. A. M., and F. A. GraybiJ(:·"JnttOduction·to the Theory or Statistics," 2d ed.,
McGraw·Hill Book Company;:New York,'J9Q3.: ' ':
Moroney, M. 1.: ..Facts from 'FigureS;""Pengliin Books; Inc., Baltimore, 1965.
Natrella, M. G.: £,cpcnmc:,ta.1 Statistics, Nat, Dur. Stand. Hand.-91, Aug. I, 196J.
Weibull, W.: A Statistical Representation of Fatigue 'Failures in Solids. A~I" Po;yuc;'.,
Mrch, Eng. Ser., vol, I, no. 9, 1949.
_._; A SlallsttcaJ Distribution Function of Wide Appljc.~hility. 1. App'. Mech., vel, IS.
pp. 293-297, September, J9SJ.
56 STAr:STIC/,L O.sSIGN AND ANALYSIS OF ENGINEEfW~G EXPERIMENTS

2. Graphically, this means that the probability that X2 > X;" is given h~ the
. h d f r d b···..een v1 ~-.' ecarea a: under the curve, Wit v egrets 0 rcc: onl. '. ~ - ' .,.: T Q",~ .
r'
a=.j21'
Hence. the mean of a i distribution is I' and the standard deviation is j:;;.
Tbe probability that 'r ;:::X;;. is
PC':/'> r.;,,) = r luI) dx1
1.;..
or
fIx' ) (3.4)
G'~, = J'" (r!)lj(xl) d(J.l)
, 0
0" J (xl)l( 2.. /2 - I
J
' . _'_!_!__ (,-~"2 d(x2)
... 0 2,/2 f(v/2) I,
'= 2'
(3.1)
TI1<.:vm iunce ofax.2 distribution is
5. If the, variable is distributed with mean II and variance ,;l, then
Hence
£(J.2) = JI = v
I. The experiment consists of v trials,
2. The variables XI' x 2 .... ,X, are obtained by the first, second, ... , nh
trial, respectively,
3. These variables arc independent and. normally distributed.
4. If the normal distribution has mean J.I ee 0 and variance a1 :;:: I, then
The chi-squo re (x') distribution Consider an experiment which. has
the following characteristics:
where v - number of trials = degrees of frcedolO, and r represents tbc gamma
function (Table A-J 0). Figure 3.1 shows such .a distribution. Th~re arc
many physical phenomena which have a probability distribution approximated
b 2' , ' "
y X • . 2 • . , •
111C expected value £(:/) of ax distribution IS
E(:r) = ( (xl)j(J.2) dX1
...I( l),/l
=f X C-x"ld(xl)
• () 2V/2r(v/2)
l. f(I'J2 + I)
'" 2·/lr(v/2) +<.,,/2+ 1)
2,/2+1 (1'/2)r(vj2)
._-- ='
- 2"1 r('/2)
x2 = sum of squares of random variable
(3.2)
!
3.1.1 ESTIMATE OF THE lJNIFORMrrY OF A PRODUCT
In many engineering problems the proper functioniog of the product is
determined by the uniformity of the characteristics of (he product. A test
therefore is rUII, sample standard deviation s is determined. and from this
the population standard deviation (J is predicted. In these cases, therefore.
it is desirable to know the relation between sand a. This relation Ciln be
derived from the Xl distribution. -
Tbe ./ is another random variable which has a '/ distribution, and its density
function is3.1 NORMAL DISTRIBUTION
69u:~,[R1,....1("'1:; 0 F EVALU, no«STp. TIS~ICAL :; t 51Gl AN :)' ANALYSIS Of ENG:~,E oRING EXFEnIV." '~TS6S
/

3. SO/III;OIl
x = 8.254 X )0-' in s=» 1.94 x 10-l in [from Eq. {1.I4»)
Y~-1l-1 v=IO-I~9(3.5)with v = n -- I
with degrees of freedom ,. equal to n - 1.
The two quantities s· and r? are related by the equation
Dctc~m;nc the standard deviation of the ......ear of the population from whi:h this
sample was llke!l, at ?O. 95, and 99 percent confidence. .
when mean ::: J.I and variance = 0'2, Therefore,
8.05 5.0
9.1 8.S3
6.2 6.8
7.S 9.6
10.3 11.46
Wear, 0.001 ;rt
r I0 h I removed after n cert ain U!'C, were
ExamQll' 3,1 Friction liningS rom W ces. -
- measured fOi wear with the following results:
(3.6)( .
Prediction of the population variance from the sample variance
The following derivation is based on a general case of 1 random variable
when distributed normally with mean J.l. and variance 0'2. From Eq, (1.14),
when x is assumed to be equal to p ;
2 I(x - 11)2
S = =-"----'-
n,- I
where n is the sample size and n - I arc the degrees of freedom (sec Sec. 1.12)
for unbiased estimate of variance. From Eq. (3.1).
From Table A.2, IX is given as 0.62, er 2 percent.
From fig. 3.3:
P{~2 ..- 2.r ,,2 ) - 1 - 0:
A'(I-ct'l);v'::'): ~""l;. - .
p[xt1-.,1IIY s (n - 1) ;::S1.;/2;.] '" 1 - 0:
or the confidence interval for the population standard deviation a is
This is shown in Fig. 3.2. The distribution starts at zero and therefore is
skewed to the right. These probabilities are given in Table A-2 for tlirrcrcnt
values of 0: and I'. For example, if I = 2 and ,;;. => 7.824, then
, 3 3 'I'he (I - ox)cor.fl(.ko~ interval fc~rn X' distributioll.F,g. .
Fig. 3.2 The mc:Irung.r X:,•.
x' 0;'o
I,
Ij(X' )
"drytt1 of freedom
7~
'...<?ERI:....'.r.N;S 0;: F...',...i..~.I/...TlCN
STATISTICAL C~SIGt; AND ANALYSIS OF ENGIN(CflING o(PEfUMCNTS70

4. t .
(
nJl
=-=.:/l
n
_.
I
':;'- (p + II + ... + II)
II
J I 1
= - £(x,) + - £(Xl) + ... + -E(x,.)
n n' II
(I I I )
E(i)=E -Xl+-X.+···+-x.
Vf If n
Hence, x can be considered a new variable which is the linear combination
of independent random variables x,, X2' •• , x•. The expected value of this
variable is
X, X2 X.
o:! -+-+ ...+-
11 II n
... r.
_ :<:+X2+X3+···+X•
x,. .cc . ..,.';'_--"---"--
Relationship between the distributions of x and i Consider a
lest conducted with the following sample of II items, XII Xl •••. ' X.' Then
In order to predict the population mean J.I from the sample mean X. for
both cases it is necessary first 10 establish rhe relationship between the dis.
tribution of the variable x and of x.
l~ yr the standard deviation of th'c population o is known, in which case the
~. sample size 'chosen may be small: or if a is not known, in which case
sample size must be large (» 30) so that the standard deviation of the
sample s is approximately equal to the standard deviation of U1C
population rJ-in these cases use the z -distribuuon (Subsec. 2.1.1).
yt: If the standard deviation of the population (J is-not known and the sample
~ 'J size is small (< 30), use the t distribution (discussed later in this section).
3.1.Z 'ESTIMATE OF THE QUALITY OF A 'PRODUCT
. 'Probably the most important characteristic of It product is its quality, which
statistically can be best expressed by the mean value. Since the mean of
tr;e population J1 is seldom known, a test is run on a sample taken from this
population, and l~~ sample mean :x: is determined. From x the value of J.I
is then predicted. To make this prediction. it is essential to establish the
relation between X. and /1. This relation is established here for the following
cases:j
EXPERIMENT:; OF llALUATlON
1.22 X 10-' in :$ C1 ~ 4 . 52 x 10-' in
and f01' 99 percent confidence:
Hence, it an be concluded with 90 percent confidence that the standard deviation
of-wear of the population of the lining i.~.o.minimum of 1.43 x 10-> in and a maxi-
mum of 3.24 x IO-J in. Similarly. {or.9S percent confidence:
1.43 x 10-> in :S 0' :S J.24 )( 10-) in
or
Therefore
/(10 - 1)(1.94') )( 10·< sC1 sJ(IO - 1)(I.'14'} x 10--
'" 16.919 3.325
I 33.8 '" C' J3l.810·' x -- ;:,a;;:, -- X 10-'
". 16.919 3 .32.5
For W pc",cnt conrideoce:
,--- ,---
J (n - 1)1' '(It - I}s'
-,-- ~a$ I-,-,-;__
X./ll" "v' X(I-"1)I.
From Eq. (3.6):
xl" I' = ,f.,OO'" - 23.589
ex
1- 2 =0.995
ex
2 - 0.005
For 99 percent confidence:
X!1l I" cr X~,OJ':o .. 19,023
(I
1 - ~ -A 0.975
-' .
ex
'2 ~ 0.025
For 95 percent confidence:
x;.-.",:. = ,f..9J,. - ~.325
;,
ex
1 ~ 2 = 0.95
..'2 = O.OS
For 90 r<:rccnt confidence:
STI:-ISTlCAl C~SlGN ANO ,NALysrs 0;:' (NC;N~Ef!lNG [XPERI.1tNr~72
j

5. a: = 0.05J - IX - 0.95
IX
2 ~ 0.02.5n = 36
Soluuon
With 95 percent confidence. predict the true (population) mean value of the resistance.
s ~ 4.2 ill ;;;(1..1'=20 )(11-£:xample 3.2 Asample of 36 resistors was tested. and the following data were: obtained:
.,f ••
!
Prediction of tho poputatlon mean from the sample mean by
using z distribution As pointed out before, z distribution applies to the
situations when the standard deviation of the population a is known, in which
case the sample .size rnay .bc small; or if a is oot known, ·the sample size
must be over ~O.
Substituting Eq. (3.12) into Eq. (3.10),
p( -Z0/2 ~ :,In.$Z."l) ""I - a (3.13)
II will be recalled that 'this is applicable when the sample size is large (over 30).
For smaller sample sizes, I distribution •. discussed later in this section,
should be used, .
(l,-1~)
. X-I-'
z = <J/.J~
This equation is used for the prediction of the population mean from the
sample mean.
With the aid or [q. (3.'7) this becomes,
.(3.11)
In the present case, the variable under consideration.is x instead of x; there-
fore z can be defined as .
X-Jl
z=--.-.
(3.10)·
Of', in general,
and the probability that x is larger than the predetermined value b was
defined as
PCr >b) ~ l'(z > b : II) 7a
I
7"
J
I
I,.~
.~
j
I
I
)
Fig, 3.4 Relation between the distributions of SAmple mean x and toe
variable x,
Oi",lbullon or.f "
/ (SI"nd",dd••I.,lon. -::-1
. V"
Diudbution of .t
(Standard d.·bliun ~ 0)
.~.~
i ~..~ r.
-e ~
'" ...
2
z=x-p
a
The z distribution In Eqs. (2.2) and (2.4), the variable was).' and the
standardized normal variate }. was defined as
Figure 3.4 shows the relation between the distribution of x and the.
dist~ibution of x,
(3.9)
(3.8)
s
$;=7
....;n
£(x) r.: J.I.
Similarly,
0"" therefore. is t hc standard deviation of the variable X. It is frequently
referred to as the standard error of the mean. Thus
(3.7)
or
I 2
'" - (11(1 )
III
(
<7) 1 (v J (<J)2
= -: + -) + ... + -
h n n
In a silr.ilJJ[ ::)311,IC[ the variance of x is given by
~ 2 _ 2 2! 2 2 '1 '1 1
v s n. G1 (71 -. (/1 <71 + Q,) 17J + ... + 0. <7"
74

6. '_"
Fig. 3.6 Meaning of c.:
Prediction of the population mean from the sarnpte mean by
usi~g I distribu~.ion It has been shown before that when X,. Xl, •.•• :".
arc Independent and normally distributed with mean J1 and varia nee (Jl. the
variable
!
I
.[
I,
I
I
I
!
Values of c ' arc given for VIti ious values of I. in Table A-J. It is. . .f;. ,
mtcrcsung to no re that the values of :x for I.;. when ' = CO are the same as
the values' of :x given in Table A-I for the z distribution. This follows from
the fact that as I' _, eo the I distribution asymptotically approaches a normal
distribution. The tabulation of :x is made only up to ..... 500. This is
because :x does not change appreciably with changes in ' above 500. For
example. to find the probability that a variable with t distribution has a
value greater than or equal to. 2.787. given that v = 25. Ptt > I.:.) = «.
Then, frOJT; Table 1·3 for . ee 25 and to;. "" 2.787. r.J. = 0.005 = required
probability.
o
Fig. 3.5 Density plOI of I distribution.
I
I
I
.I.
·1
I
((II
I
77EXPEfUM~NT$ OF EViLUATION
,
I !
for ' = 2
and the variance of a t distribution is
.-"
for v >£(1) = (' (f(l) d: = 0
. i z distribution extends symmetrically from - c.o to + oc whereas a
X2 distribution extends from 0 to + cc. Hence." I distribution also extends
symmetrically from - 00 to + 00. Figure 3.5 shows a plot of j(l} versus I
From this figure it is seen thai as v - co , a I distribution tends to be" normal
distribution.
The expected value of a I distribution is
-OO<I<cr. (3.14)
X
jl.~'.
has a I distr ibution wit h v degrees of !reedom.
Such a distriburion has II frequency distributio« function
1 f[(I'+I)!2J( (!)-C''''1:2
J(I) = -:-= I +-
J;rl' r(I'J2} v,
The t distribution /distribution is used for the prediction of the popula-
tion mean II from the sample mean x when the population standard deviation
a is not known and the sample size is small. The I distribution is mathc-
rnatically defined as follows:
l . If Xl' Xl' " ', x. arc normally distributed random variables with zero
mean and unit variance, and
2. Xl is a random variable' independent of x and having 2 -; distribution
with ,. degrees of freedom. then the random variable
Therefore. with a 95 percent confidence levcl.jhc true rncan value j.J. or the rtsis'~:1ce
is between J 8.63 and 21.37 kO.
'I'[(X-:"l ;~)$}L$(X~='(l J~)]=l-~
(20 - 1.96 ~/':) $ ".~(20 _ 1.96 4;~) ~ 0.95
., " 36; .jJ6 .
P(l8.63 sIis:21.37),., 0.95
STATISTICAL CcSIGN AND ANALYSIS OF ENGil,EEflII'iG EXPfn:r ....i·,-S76
(
.. fT ,.
(1/ = I /2j(l) dl =--
• -.r, r - 2
The evaluation of these integrals is generally not made because 'lhe £(1) and
a/ arc of little use in applications of the I distribution.
The probability a that the value of / exceeds a preset constant value 1,;,
is given by the shaded area in Fig. 3.6. This is denoted by '] = P(I > I'l.)' C

7. This interval for population distribution is larger than the one that
covers a single parameter of the population.
Table 3.1 gives the value of K for the two-sided cases where the estimate
of both the upper limit (x+ Ks) and of the lower limit (x - Ks) are ~:';ce
.nade. Table 3.2 is to be used. for one-sided t.:a~~. whe,- 31" .r.::.llmat·! !.~ "':"
oe made of only one of the two IBT.:ls.
3.1.3 ESTIMATE OF THE POPULATION LIMITS OF A P RODUCT
Sections 3.), I nod 3,1.2 were concerned with the estimation of the uniformity
of a product in lerms of the standard deviation «(1) and of the quality of a
product in terms of its mean value (J.I) respectively. A pair of confidence
limits were determined so that the interval between the two limits would
contain the popularion parameters 0 or J.I with a certain confidence. How.
ever, in many engineering situations it is important to determine an interval
that would cover a fixed proportion of the population distribution. This
interval can be expressed in the form
i±Ks
where x = sample average based on the sample size n
s = sample standard deviation .
and K = factor that depends on sample size n, the desired proportion P of
the population distribution, and the confidence at which this
interval is estimated
93.24707,:;; It $ 96,75307.
Hence. it can be concluded with 90 percent confidence thaI the range 93.247 10
96.753 oz will contain the population mean J.L.
---- --_.._. - "".._ ..' '" ."..._,,- ..._-_ ........_... "'_" -.._._.........
or
Substituting the values in EQ. (3.16),
4 4
95 - /_. x 1.753 ::::IL:::; 95·t J- x 1.753
"V t6 16
',/1,.-10.0.,,, = 1.753 (rom Table A·3
lI=n-'!-15n ... 16s~ 4 O~
I
. v"x - 95 oz
Solution
Example 3.2 A filling machine is set .to fill packages 10 a certain weight x, Six-
teen pa::kag!:S were chosen at random; their average weight was found (0 be 95 OZ,
and th e standard deviation 4.00 oz, Estimate the population mean I~ with 90
percent confidenee,
(3.16)
or (I ._:r) confidence interval for ,Iis
_ s _ s
x ---r:, './2:. '$ II s: X + .r:. 1./2;.
" n In
79
/
././
,
~
I
I
J
I
I
"'1
' '
o
Fig. 3.7 The (I - «) confidence interval (or a I distribulion,
P(-I./2:. S; (~ +!'f2:.) = I - a
where I -;x is the confidence, and 1= x - f.J.
. s/~~
p( -1.12:,'5,X-: '5, +1012:.) = I-:x
sf";fI
/'( - '.12 > I) == :
2'
where . = /I - I
Referring to rig. 3.7:
P(I.s I./l,.) <= ) - ~
. 2
Since. by definition,
has a X2 dis!ribution with (n, - ) degrees of freedom,
xf=_
.Jx2(v
by combining the Above relations, the f~lJowing is obtained:
,[{x - 11)';;1:(01';;;; =t (.~- J1)~
J(n,-I)s2ioj ... s :<...,/ (lIS)
has a 1 distribution witli (.'I, - I) degrees of freedom.
The usefulness of this variable I becomes apparent when • l
that x and rare th . , . one o,lS(!rVCS
ld b fo '. c paramctcr.~ associated 'with :I given sample and as such
:~u e ound ,from a lest,. TIle, relationship between the sample mean x
.d the population mean II IS derived from the following l 1..'1'
mcnts., pro la", uy stare-
qATI"'T' -A' CE" G~ • ~ IL;" ',Ii N ANa /-t'jALYSIS OF CNGIN",I'lr,'"c c 1 Y rxrEf~a"r,··"f!,;
is also normally oistribu(ed with zero mean and unit vari~ncc' _ h
obtained Iron I" , ,x ere was
h
( , ' .1 n.J( samp C Size, It Wil,'> also shown thai for ,:I S·r.'~I~ size
I e variable ' , ," "I"~ n,!
73
I

8. :;, HC'N "'2"" GUscr.-a:;on~~h-:-"" :-: ,..·,ad' ,r- e~~'1rc Y.!;(, 90 r-e.f'XI" ;i;n!i.dc:;'~H:"t
;'"6~;"-;': ,(:,1 ~ 1.7{~":·
Whal is the average interrelcnCe cf :his ')1 of lOO.O<Xlassemblies? Provide the answer at
95 percent conf:dcr.cc level.
q9An engineer teSlS a $3mplc or 1I bearings for hardness. The S3mplc variance is
2.850. The rest data arc norn1a!ly distributct.!. Find the range of the population variance
",ith 80 yerCCIlI confidence.
,~--"__ 3.7. A eenain Iype of lill)ll buib Ita, a variance in burninG time of 10.OOOh'. A sample of
2O'blllbs was picked. presumably from this lot, and ilS variance was four:d 10 be 12.000 h'.
At 9$ percent conl'tlknce level. dctern1ine whether the 20 bulbs were pic.kcd from the righ!
lot.
I 3~8>Fifteen pl:.~tlC lining.s made out of standard material were tested for wear with the
• tc>i:",,,,,,,,fi.re~:.>IL~·:; "" 0.0090 in. s "" 0 (l02! in. Wh.:.l call be SOlid about the poy;;;al:Ci
-,"1 '" :.';," . ~,~ ~1 ';P" ~O) !;IKcr. ~ "~ ...cr ",ith 95 pc~ccn' <:cnlio·!ncc.
0.00060
O.Ov"'()4()
0.00050
0.00035
0.00060
0.00055
0.00050
0.00045
0.00045
0.00060
I
2
3
"5
(,
7
8
9
10
D;omcrral
inltr/t(tnu. in
Pa.lS-fir
assrmbly nllmbrr
.
I
/':i':D The sample mean of nine bOTCi is 1.004 in~and the standard ceviation is 0,005 in.
The distribl!tion of dir:1ension sires is (lorma!.
(1'1) Determine the average of the population with 95 percent confidence.
(b) Determine the average of the population with 99 percent confidence.
(3;2> Determine by Weibuil rl,clhod, with 90 percenl confidence. the mean life and the
characteri~tic life. on the basis of the following lest data (life cycles in 10'): 0.51. 0.97,1.50,
2.20.3.00.
3.3. Firty [terns were picked at random from 1I 101 of two thousand. and ten items were
found defective. Predict with 91) pcrcen~ confidence the minimum 811d Ihe maximum
number of irerns that could be defective in the remaining items of n'iis lot.
(~A_) TCII metallic pieces were chosen at ranool:1 [rom a normal po;>ulation. and their
Iialdnesses ....-ere found to be 66.68.67.69. n. 70. 70, 71,63. and 63 ac, Plollne mean
ha,dnc!.lo(S of the population versus the confidcn~e levels.
3.5. Ten press-lit 3ssen1blies. invol.ing a ShAft and a bore, were picked out or a 101of one
hundred thousand assemblies. and their inler(ercno:s musured. with the following Te$ullS;
!
PROBLEMS
EXpcRIM<.IHS GF EVA,-UATIO:-.t
97
.,
::
.
'ic;n~I ~
''<
VI
::l.
VI
t
EE
"(/)
I
.1
:I.
f
:::
",'
'"
...
":;)
'1J
e0.
..co
tg
..:;J
..>.,
I II I
I I",
c: I"";0'"
'; <: ~
~:;5 I:M::J...;~ -"0'
f 1 ...
,TAiISTICAL CESIGN AND A'lAlYSIS OF ENGIM~RI,G !.XPEP.:MHiTS96
I
r
;
,