12.1

12 P."obabilityand Statistics "

L COI'\ffiINATIONS AND PERlI.fUTATIONS

There are a finite number of ways in which n elements can be combined into distinctly different groups ofr items For example, suppos e a fanner has a chicken, a rooster, a duck, and a cage that holds only two birds. The possible combiMtiom of w e. birds taken two at a time are (chicken, rooster), (chicken, duck), and (rooster, duck). The birds in the cage will not remain stationary, so the combination (rooster, chicken) is not distinctly different from (chicken, rooster) That is, the combinations are not armr conxiou,

Etpmtion 12.1: Combinations

P(n,r) C(n ,r ) = --

,! n!

r !(n - r )! 12.1

The number of combiMtiom ofn items taken r at a time is written C(n , .. ), ~, ne" "C" or (:.') (pronounced "n choos e r"). It is sometimes r eferred to as the binomial cO

12.2

An order-conscious subset ofr items taken from a set ofn items is thep

12.3

Solution The marble colors represent different typ es of objects The number of permutations of the marbles taken 13 at a time is

P(U; 4, 2, 7)

The tmSWeT is (D).

n! 13! n , !n2! .. . nk! 4!2!7! 13 X 12 X 11 X 10 X 9 X 8 X 7

x 6 x 5x4x3x2xl 4 x 3 x 2xlx2xl

x 7 x 6 x 5 x4 x3 x 2 X I = 25,740 (26,OOO)

2. LAWS OF PROBABILITY

Probability tixlory determines the relative likdihood that a particular event will occur An ovont. E. is one of the possible outcomes of a trial Theprobability of E occurring is denoted as P(E)

Probabilities are real numbers in the range of zero to one Ifan event E is certain to occur. then the probability P(E) of the event is equal to one If the event is certain not to occur. then the probability P(E) of the event is equal to zero. The probability of any other event is between zero and on,

The probability ofan event occurring is equal to one minus the probability of the event not occurring This is known as a complw",ntary probability

PtE) = 1 - P(not E)

Complementary probability can be used to simplifY some probability calculations. For example. calculation of the probability of numerical events being "greater than or "less than or quantities being "at least"" a certain number can often be simplified by calculating the probability of the complementary event

Probabilities of multiple events can be calculated from the probabilities of individual events using a variety of methods \Vhen multiple events are considered. thos e events can either be independent or dependent The probability of an i"'*p""'*nt ovont do es not affect (and is not affected by) other events. The assumption of independence is appropriate when sampling from infinite or very large populations. when sampling from finite populations with replacement. or when sampling from different populations (univers es). For example. the outcome of a second coin toss is generally not affected by the outcome of the first coin toss. The probability of a '*'p""'*nt ovont is affected by what has previously happened For example. drawing a second card from a deck of cards without replacement is affected by what was drawn as the first card

Events can be combined in two basic ways. according to the way the combination is described. Events can be connected by the words "and' and "or For example. the question. "\Vhat is the probability of eventA and event B occurring?' is different than the question. "\Vhat is the probability of eventA or event B occurring?' The combinatorial "and' is designated in various ways AB. A B. A x B . An B. and A, B. among others. In this book. the probability of A and B both occurring is designated as PtA, B)

The combinatorial "or" is designated as A + B and A U B In this book. the probability of A or B occurring is designated as PtA + B) Eqllation 12.-1: La"", of TIH4l Probnbilily

P(A + B) = P(A ) + P(B) - P(A,B) 124

EqualIon 124 gIveS the probabililytlm either evtlll A or B will occur P(A. B) i, \heprobability tim both A and B will occur

A deck often children's cards CotltaJru three fish cards, two dog cards, and five cal cards \\!haI lS \he probabilily of drawmg ather a cat card or a do g card from a full dec\(/

12.4

A deck of len children's cards contains w e. fish cards, Iwo dog cards, and five cal cards. \VIla1 is the probabilily of drawing either a cal card or a dog card from a full deck?

(A) 1110

(B) 2/10

(C) 5/10

(D) 7110

Solution The Iwo evenls are mUlually exclusive, so the probability of both happening, A;A, B), is zero. The lotal probability of drawing either a cat card or a dog card is

PtA + 8) = P(A) + P{B) - PtA, B) = }50 + }20 - 0 = 1/ 10

The tmSWeT is (D).

E'llifwon 11. 5: Lnw ,!!ColHJ'OflHd (Joint) Probnbility

P(A, B) = P(A )P(BIA) = P{B)P(AIB)

[independent I

events

12.5

Equation 12.5. the !.:;w of compouM ('Oint) probabll!ry, gives the probab:lily thaI evenls A and B will both occur. A;B~) is the COMItIOOO! prcbability thai B wll occur given tha A has already occurred. Likewis e, A;AIB) is the ccm titional probabilty thai A will occur given thai B h

12.5

Solution There is a total of 17 balls There are 2 white balls. The probability of picking a white ball as the first ball is

p eA) 2 17

After picking a white ball first, there are 16 balls remaining, 7 of which are orange The probability ofpicki"lg an orange ball second givm that a white ball was chosm first is

P(BIA) 7

"

The probability of picking a white ball first and an (range ball second is

p eA, H) = P(A)P(BIA )

~ C'7)C'6) = 0.05147 (0.052)

The tmSWeT is (B). E'I'mtion 1Z.': BIJJ_' Thell"m

( ) ~P-"(B-,,;)--,P(,,,A!=:IB,!.C) :-c P HilA = r "'" P(AI8; )P(B;) L....~ l

p(BAA) = PCB and A) peA)

12. 6

G.vm two d"prruItnt ,ets of ""rm:;. A and B, the probability thaI l:Vm1 B will occur glvm the facl\hat the dep mdtlll l:Vm1 A has already occurred i, wmtma, p(BjIA) and is givm by ~,' 1""0,""",. Eq 12.6

A medical pabrm ~bits a ,ymptom that oc curs nalurally 10% of!he tune in all people. The symptom IS also ~hlI.ed by all patimls who have a particular disease. The incidtnee of that particular dI, ease among all ptop1e i, 0.0002% \VIlat i, !he probability of !he patimt havtng \hat particular eli,eas""

(A) 0. 002%

(8) 0. 01%

(e) 0.3%

(Dj 4%

12.6

""iu/ion ~his problem is asking for a conditional probability: the probability that a ,erson has a diseas e, D, given that the person has a symptom, S. Us e Bayes' theorem to calculate the probability: The probability that a person has the symptom S. given that they have the diseas e D is F\.51D) and is 100%. Multiply by 100% to get the answer as a percentage

P{DIS)

The tmSWeT is (A).

P(D JP(SID) ~ ="",,,,~~=m P{SI D)P(D) + P{Slnot D)P(not D) (0.OOOO(2)( 1.00)

(1.00)(0.000Cl02) + (0. 10)(0.999998) = 0.00002 (0.002%)

3. MEASURES OF CENTRAL TENDENCY It i, often unnec essary to present =erimental:tata in their entirety, 6ther in tabular or graj:hic fonn. In such ca,"s, the data and distritution can be repres",ted by various parameters One typ e of parameter is a measure of contrai to,1

12.7

45 measurements were between 0.859 and 0.900

0.901 was observed once

0.902 was observed three times

0.903 was observed twic e

0.9[14 was observed four times

45 measurements were between 0.905 and 0.958

The smallest value was 0.859, and the largest value was 0.958 The sum ofall100 measurements was 91.170. Except thos e noted, no measurements occurred more than twic e

\VIlat are the (a) mean, (b) mode, and (c) median of the measurements, respectively?

(A) 0.908; 0.902; 0.902

(B) 0.908; 0.9[14; 0.903

(e) 0.912; 0.902; 0.902

(D) 0.912; 0.9[14; 0.903

Solution (a) From Eq 127. the arithmetic mean is

.. ( I ) X = (l In) LX; = 100 (91.170) = 0.9117 ....

(0.912)

(b) Tte mode is the value that occurs most frequer.Uy The value of 0.9[1.1 occurred four times, and no other measurements repeated more ttan four tines. 0.9[14 is the mode

(e) Tte median is the value

12.8

s{wiem score 80%

95%

71%

95%

\VIlat is most nearly the student's final grade in the course?

(A) 82%

(B) 85%

(e) 87%

(D) 89%

Solution The student's final grade is the weighted arithmetic mean of the individual exam scores

x . LW;X; 2: wi

(1)(80%) + (2)(95%) + (2)(72%) + (5)(95%) 1 + 2 + 2 + 5

= 88.9% (89%)

The tmSWeT is (D).

Etpllliion 12. J : ~o_tI'ic ~

129

The goo'""tric ,""an ofn nonnegative values is defined by Eq 12.9. The geometric mean is the number that. when raised to the power of the sample size. produces the same result as the product ofall samples. It is appropriate to us e the geometric mean when the values being averaged are used as consecutive multipliers in other calculations. For example. the total revenue earned on an investment of C earning an effective interest rate of it in year k is calculated as R - 0:)]i2i3 ... i0. The interest rate. i. is a multiplicative element. Ifa $100 investment earns 10% in year 1 (resulting in $110 at the end of the y ear) . then the $110 earns 30% in year 2 (resulting in $143). and the $143 earns 50% in year 3 (resulting in $215). the average interest earned each year would not be the arithmetic mean of (1 0% + 30%+ 50%)/3 - 30% The average would be calculated as a geometric mean (24.66%)

\VIlat is most nearly the geometric mean of the following data set?

0.820, 1.96,2.22,0.190,1.00

12.9

(A) 079

(B) 0.81

(C) 0.93

(D) 0.96 Solution T:1e geometric m= of the data ,et is

The tmSWeT is (C).

(0.820)(1.96)(2.22) x(O. I90)(J.OO)

= 0.925 (0.93)

Etpmtion 12.10: Root-ldmn-StpllU

sample root-wea&8Quare value = J( l jnlLJ4 12. 1 0

T:1e root-m

12.10

4. :MEASURES OF DISPERSION

Mwwr., oj di'p"r5ion describe the variability in observed data

EtpUJtion 12.11 TlU"OI'6" Eq. 12.15: SttmiIanJ Devintion

1211

1212

1213

U-. =

1214

1215

One measu-e of dispersion is:he ,tandard deviation, defined in Eq. 12.11 . Nis the total. population size, not the sample size, n. This implies that the entire j:opulation is measured

Equation 12.11 can be used te calculate the stancard deviation only when the entire poj:ulation can be included in the calculation \Vben only a small subs

12.11

A cat colony living in a small town has a total population of seven cats The ages of the cats are as shown

nwnber

\VIlat is most nearly the standard deviation of the age of the cat population?

(A) 1.7 yr

(B) 2.0 yr

(e) 2.2 yr

(D) 2.4 yr

S:l)ution Using Eq 12.7. the arithmetic mean of the ages is the population mean. If-

~ (') (1)(7 YT) + (1)(8 YT) + (2)(10 yr ) 1 +(1)(12 yr) + (2)( 13 yr)

= lOA YT

From Eq. 12.11 . the standard deviation of the ages is

12.12

EtpUJtion 12.16: Sampk StJUuJanl Devintion

" ~ = [l/(n- l )] L (Xi- X )l ; - 1

12. 16

The standard deviation oj a wmplo (particularly a small sample) ofn items calculated from Eq 12.11 is a bia",d .stimaior of (i. e., on the average, it is not equal to) the population standard deviation. A different measure of dispersion called the wmplo standard deviation, s (not the same as the standard deviation of a sample), is an unbiased estimator of the population standard deviation. The sample standard deviation can be found using Eq 12. 16

Samples of aluminum-alloy channels were tested for stiffness The following distribution of results was obtained

s tiffness

2480

2440

2400

2360

2320

frequency

If the mean of the samples is 2402, what is the approximate standard deviation of the population from which the samples are taken?

(A) 48.2

(B) 49.7

(e) 50.6

(D) 50.8

Solution The number of samples is

n = 23 + 35+40+33 + 21 = 152

The sample standard deviation, s, is the unbiased estimator of the population standard deviation, (5 .

~ [1/(n-II]f:(X, - XI' .-,

I 152 1

= 50.82 (50.8)

(23)(2480 - 2402)2 + (35)(2440 - 2402)1 + (40)(2400 - 2402)1 + (33)(2360 - 2402)2 + (21 )(2320 _ 2402)2

12.13

The tmSWeT is (D).

EtpUJtion 1Z.1 7 TlUOf'6" Eq. 1Z.IJ: Vtuitutct!o tmiI Sampk Vtuitutct!o

.;: = (l / N) [(X I _ 1')2 + (X , _ p)l + ... + (XJV _ 1')2] 12.17

N

~ = (1 / N) L (X; - p)2 12. 18

" 8 2 = [l / (n - 1)1 L (X ; - X )!

; _ 1 12. 19

The varicmc. is the square of the standard deviation Since there are two standard deviations. there are two varianc es. The varicmc. ojt"" population (i. e .. thepopulation varicmc.) is,r. and the wmplo varicmc. is ,2 The population varianc e can be found using either Eq 12.17 or Eq. 12.18. both derived from Eq 12.11 . and the sample varianc e can be found using Eq. 12.19. derived from Eq. 12.16

Most nearly. what is the sample varianc e of the following data set?

2, 4,6, 8,10, 12, 14

(A) 4.3

(B) 5.2

(e) 8.0

(D) 19

Solution Find the mean using Eq 12.7

X =(I/nl tX,= n ) (Z+4+6+8+1O +1Z+14) .-,

~ 8

From Eq. 12.19. the sample varianc e is

8 2 = p/(n - l )l t (X, - X)2

12.14

(A) 4.3

(B) 5.2

(e) 8.0

(D) 19

Solution Find the mean using Eq 12.7

x =(1/ n) tX,= n)(2+4+6+8+1O +12+14) .-,

From Eq. 12.19, the sample varianc e is

$% = [l/(n - l)l t (X, - X)2

~ (7 ~ 1) ((2 :(:)~:)~4:(:~1:8~~ :~:22 _ 8)2 ) +(14 _ 8)2

= 18.67 (19)

The tmSWeT is (D).

EtpUJlion 12.20: Sampk CHfflCWnJ ofVarintion

cv = $jX 12. 20

The rdativ. di'p"rolon is defined as a measure of dispersion divided by a measure of central tendency The wmplo coojjici.nt o/variation, CV, is a relative dispersion calculated from the sample standard deviation and the mean

The following data were recorded from a laboratory experiment

20,25,30, 32,27, 22

12.15

The mean of the data is 26 \VIlat is most nearly the sample coefficient of variation of the data?

(A) 0.18

(B) 1.1

(e) 2.4

(D) 4.6

Solution Find the sample standard deviation of the data using Eq 12. 16

[I/ (n - IlJ t (Xi - X )2 ; _ 1

(

(20 - 26)1 + (25 _ 26)1 ) = (6~I) +(30 _ 26)1 + (32 _ 26)1

+ (27 - 26)2 + (22 _ 26)2 = 4.6

From Eq. 12.20, the sample co efficient of variation is

cv =.rx = ~: = 0.177 (0.18)

The tmSWeT is (A). 5. NUMERICAL EVENTS A dixroto nwmrical ovont is an occurrence that can be described by an integer. For example, 27 cars passing through a bridge toll booth in an hour is a discrete numerical event Most numerical events are continoou,ly di,triDutod and are not constrained to discrete or integer values For example, the resistance ofa 10% 1 II resistor may be any value between 0.9 II and 1.1 II

6. EXPECTED VALVES

Etpmtion lZ.Z1: Exp~d Vah"" of ... Disc"~ VtJrinbk

" /J = E[X] = L Z",(Z~ )

.-, 12. 21

The oxpoctod val"", E, ofa discrete random variable, X, is given by Eq. 12.21. !(Zk) is the probability mass function as defined in Eq. 12.28

The probability distribution of the number of calls, X, that a customer service agent receives each hour is shown

12.16

, M c 000 , eM , 0.05

0 0.10

0 0.35

W 0.46 \VIla! is most nearly the average number of phone calls that a customer servic e agent expects to receive in an hour?

CAl 5

(B) 7

(e) 8

CD) 9

Solution The expected number of received calls is

p. = E[X] = L,,*,(Xk) ~,

= (0)(0.00) + (2)(0.04) + (4)(0.05) + (6)(0.10) + (8)(0.35) + (10)(0.46)

= 8.28 (8)

The tmS>W!T is (0. Etpmtion 12.22: VIJI'ianuo of /J Disc"" VaritJbk

rr = V[Xl = L (::r~ - 1')' f(x.) .-,

1222

Equation 12.22 gives the varianc e, d2, ofa discrete function of variable X To us e Eq. 12.22, the population mean, /1-, must be known, having bem calculated from the total population ofn values The name "discrete" requires only that n be a finite number and all values ofx be known It do es not limit the values ofx to integers

Etpmtion lZ.Z1 (JIuJ Eq. 12.24: Exp"tl Vah"" (M~an) of tJ CO"IiIUIUfIS Varinbk

00

p. = S[X) = / z/(z)dz - 00

1223

00

E [Y] = E[g(X)] = / g(z)f(it)dJ: _ 00

12.24

12.17

Equation 12.23 calculates the population mean, /1-, of a continuous variable, X, from the probability density function, fX). Equation 12.24 calculates the mean of any continuously distributed variable defined by Y- g(x), whos e values are observed according to the probabilities given by the probability density function (PDF)fx). Equation 12.24 is the general fonn of Eq 12.23. where g(x) - x

~

'" = Vlx] = E[IK - pi'] = 1(' -pi' fl' )" ro

12 15

Equabon 12 15 gives the vanance ofa coIl\lIlu.ous random varr.ble, X /I- 's the mean of X, and / (z) II the dmslly tunCllon of X

1216

The standard deviation is always the square root of the varianc e, as shown in the variation equation. Equation 12.26 gives the standard deviation for a continuous random variable, X

Etpmtion 12.2 7: CH.JIicWnI of V"rintion of tJ COnJimlOJIS V"ritJbk

CV = (T I l' 1227

The co efficient of variation of a continuous variable is calculated from Eq 12.27

12.18

7. PROBABll.JTY DENSITY FUNCTIONS (DISCRETE) Etpmtion 1Z.ZB: ProbabiJiJy ldtJss Frmction

12. 28

A dixr.t. rmilim variaN., X, can take on values from a set of discrete values, Xi. The set of values can be finite or infinite, as long as each value can be expressed as an integer. The probability ma" junction, defined by Eq. 12.28. gives the probability that a discrete random variable, X, is equal to each of the set's possible values, xk. The probabilities ofall possible outcomes add up to unity

Etpmtion 1Z.ZJ: ProbabiJiJy IknsiJy Frurction

P(a :O::; X :O::; b) = J J(z )dz

12. 29

A ""mity junction is a nonnegative function whos e integral taken over the entire range of the independent variable is unity A probability ""mity junction (PDF) is a mathematical fonnula that gives the probability of a numerical event

Various mathematical models are used to describe probability density functions. Figure 12. 1 shows a graph of a continuous probability density function. The area under the probability density function is the probability that the variable will assume a value between the limits of evaluation The total probability, or the probability that the variable will assume any value over the interval, is 1. O. The probability of an exact numerical event is zero. That is, there is no chance that a numerical event will be exacUy a It is possible to determine only the probability that a numerical event will 'b e less than a, greater than b, or between the values of a and b

Figure 12.1 Probability DImity Function

pix }

, b x

If a random variable, X, is continuous over an interval, then a nonnegative probability ""mity junction of that variable exists over the interval as defined by Eq. 12.29

8. PROBABILITY DISTRIBUTION FUNCTIONS (CONTINUOUS) A wmulativ. probability di,tribution junction, FIx), gives the probability that a numerical event will occur or the probability that the numerical event will 'b e less than or equal to some value,,,

12.19

Etpmtion 12.10: C'Ulfldnu",~ Disui1nllion F,mcuon: Disc"" RnniIom V"rinbk

F(Zm) = L P(z~) = P(X ~ Zm) [m =1,2, ... ,n] .-,

1230

For a dixr.t. rmilim variaN . X. the probability distribution function is the sum of the individual probabilities of all possible events up to and including event x",. The wmulativ. di,triDution junction (CDF) is a function that calculates the cumulative sum of all values up to and including a particular end point For discrete probability density functions (PDF s). PIx",). the CDF can be calculated as a summation. as shown in Eq. 1230

Because calculating cumulative probabilities can be cumbersome. tables of values are often used. Table 12. I at the end of this chapter gives values for cumulative binomial probabilities. where n is the number of trials. P is the probability of success for a single trial. and x is the maximum number of successful trials

Etpmtion 12.11: C'Ulfldnu",~ Disui1nllion F,mcuon: COntimiOlIS RnniIom VtJrinbk

,

F (z ) = J f (t)dt -.

12. 31

For continuous functions. the CDF is calculated as

12.20

Therefore, the height of the curve at its peak is 2

The equation of the line from x- O up to x = 1/2 is

/ (x ) =4", [o:

12.21

Solution Since the outcomes are 'either -or" in nature, the outcomes (and combinations of outcomes) follow a binomial distribution. A male kittm is defined as a success The probability ofa success is

p = 1 - 0.52 = 0.48 = P (male kitten) q = 0.52 = P (Ccmale k itten) n = 7 t rials '" = 2"uccesset!

Pn(z ) n ! rf rf-I< "'!(n xl ! ~ ( 2) = ( 7' ) (O.48)1(O.52f-1

2!(7 2)! = 0.184 (0.18)

The tmSWeT is (B).

EtpUJtion lZ.J411utn'6" Eq. 12.17: Nonnnl Disuilnliion

1 f (",) = -- ' u../2-i [-oo :s '" S 00]

I {:r) = _1_ (J-~/z [-00 S " :S ooJ ff.

F(- xl = 1- F(z)

12.34

12.35

12. 36

12.37

The norma! di,/riDution (G"''''icm di,/riDution) is a symmetrical continuous distribution, commonly referred to as the wll-,hap

12.22

Figure 12.3 Normal Om ... with M.an If. and Standard Ikviaiion (5

2.15% 13.6% 34.1 % 34.1 % 13.6% 2.15%

"f fl - 2

12.23

From Table 12.2, the cumulative distribution function at Z- 06 is F( Z) = 0.7257 The percentage of boys having height greater than 1.23 m "

percentage taller than 1.23 m = 100% - (0.7257)(100%) = 27.43% (27%)

The tmSWeT is (A),

Etpmtion 12.1B tmiI Eq. 12.1J: C~nJTuI LimiJ I1woTPm

12. 38

u Uj =

12. 39

The contral limil Ihcorom states that the distribution of a significantly large number of sample means of n items where all items are drawn from the same (i. e., parent) population will be normal. According to the centra1limit theorem, the mean of sample means, 1',. , is equal to the population mean of the parent distribution, /1-, as shown in Eq. 12.38. The standard deviation of the sample means, Uj, is equal to the standard deviation of the parent population divided by the square root of the sample size, as shown in Eq. 12.39

Etpmtion 12.40 tmiI Eq. 12.41: t-Disuilnllion

r(~) f (t ) = 2 foir(n

_. ( ")-' 1+ -"

12.40

12.4 1

For the I-di,tribution (commonly referred to as Stu

12.24

INscription

The gamma junclion, r{ n ), is an extension of the factorial function and is used to determine values of the factorial for complex numbers greater than zero (i. e .. positive integers)

Etpmtion 12.41: ClIi-StpllUd Disll'i]",uon

:i = 4+z~+ ... +z~ 12.43

IHscriptlon

The sum of the squares of n independent normal random variables will be distributed according to the chHquar.d di,/riDulion and will have n degrees of freedom The chi-squared distribution is often used with hypothesis testing of varianc es Chi-squared values, X~ n ' for selected values of a and n can be found from Table 12.5 at the end of this chapter

10. STUDENT' S T-TE ST

Etpllltion 12.44: Ex~

a = f~ f{t)dt 12.44

Dlscriptlon

The I-to'l is a method of comparing two variables, usually to test th, significanc, of the differ,",c e betw= _ samples F", example, th, I-test can be used to test i

12.25

12. SUl\IS OF RANDOM VARIABLES

EtpUJtion lZ.45: Sruns of RmriIom Varinbks

12.45

LHscrlptlon

The sum of random variables, Y, is found from Eq. 12.45

EtpUJtion lZ.46: Exp~d Vah~ ofdle Srun of RmriIom Varinbks

~ = E(Y) = aIE(X ,) + aJE(Xz) + ... + a"E(X,,) 12.46

~scrlptlon

The expected value of the sum of random variables 11", is calculated using Eq. 12.46

EtpUJtion lZ.47 and Eq. lZ.4B: Variance ofdle Srun of I~mJmJ RnniIom Varinbks

12.47

u: = a~o-i + ~~ + ... + ~,r" 12.48

The varianc e of the sum of indep endent random variables can be calculated from Eq 12.47 and Eq. 12.48

EtpUJtion lZ.4J: StturiIanl Devintion ofJlIe Srun of I~tuklll RmriIom Varinbks

"". = p, 12.49

LHscrlptlon

The standard deviation of the sum of independent random variables (s ee Eq 12.45) is found from Eq. 12.49

13. SUl\IS AND DIFFERENCES OF :MEANS \Vhen two variables are sampled from two different standard normal variables (i. e , are independent), their sums will be distributed with mean 14-- /1-1 +/1-2 and variance.r ..... = o-i / n t + ~/fl2 The sample size" nl and n2, do not have to be the same. The relationships for confidence intervals and hypothesis testing can be used for a new variabl e, X"",, - Xl +X2, if /I- is replaced by 14- and "is replaced by""""

For the differenc e in two standard normal variables, the mean is the differenc e in two population mean" 14-- /1-1 -/1-2, but the varianc e is the sum, as it was for the sum of two standard normal variables

12.26

14. CONFIDENCE INTERVALS

Population properties such as means and variarr es must usually be estimated from samples The sample mean, X , and sample standard deviation, " are unbiased estimators, but they ace not nec essarily precisely equal to the true populatim properties. For estimated values, it is common to specilY an interval expected to conlain the true population properties. The interval is kno;.vn as a confidenc e interval becaus e a confidenc e level, C(e.g., 99%), is associated with it (There is still a 1 - C chanc e that the true pop1.lation property is outside of the interval.) The interval will be bounded below by its lowor conjicknco limit (LCL) and above by its upp"r conjicknco limit (UCL)

As a consequenc e of the contral limit thorom, means of samples of n items taken from a distribution that is normally distributed with mean If-and standard deviation (5 will be normally distrituted with mean If- and varianc e ,rln Therefore, the j:robability that any given average, X, exceeds some value, L, is

L ~" } '"

L is the conjicknco limit for the confidenc e level 1 - p { X > L } (expressed as a percentage). Values ofp(x) are r.ad direcUy from the unit normal table (s ee Table 12. 2) . As an example, :r = 1.64:) for a 95% confidenc e level sinc e only 5% of the curve is above that x in the upp e:-tail. This is known as a om-tail conjicknco limit becaus e all of the exceedanc e probability is given to one side of the variation With:wo tail conjidonc< Ii>?'it" the probability i, ,olit between the two , ne, of vori:ttion. Ther< will be upper ond 10w,.- confidenc e limit, U:L and LCL, r espectively This is appropriate when it is not specifically known that the calculated parameter is too high 0:- too low Table 12.3 "t the er.d of this chapter lists standard normal variab:es and t values for two-tail confidenc e limits

p{LCL < X < UCL} _ l LCL- ~ VCL -JJ } _ p ~ < :r < "

'" '"

Etpllllion 12.50 tmiI Eq. 12.51: COiifIMlrcr LimiJstmillttUnwfor ~ of u Nornwl Distri1ndion

12.50

12.51

LCL = X - l"j :,n_l ( :n ) UCL = X + l" j:,n_1 ( :n )

12.27

The conjicknc. limil' jar 1m "",an, /1-, of a normal distribution can be calculated from Eq. 12.50 when the standard deviation, (5, is known

If the standard deviation, (5, of the underlying distribution is not known, the confidence limits must be estimated from the sample standard deviation, " using Eq. 12.51 . Accordingly, the standard normal variable is replaced by the I-distribution parameter, lo!:l, with n - 1 degrees of freedom, where n is the sample size. a = 1 - C, and r:JJ2 is the I-distribution parameter since half of the exceedance is allocated to each confidence limit

EtpUJtion 12.52 aniI Eq. 12.51: C91ifuk,~ LimiJsforOw D!ffrlncr Brtwun TW9~

~' .; +Z"/l - + -.. '" 12. 52 (i + ~)[(n1 - l)sf +(n-: - l)~]

n1 +nt 2

(i +~) [(n1 - l )sf + (n-: - l)~] n1 +R:I 2

[unknown '" and ".1 12. 53

The differenc e in two standard normal variables will be distributed with mean 14-- /1-1 -/1-2 Us e Eq. 12.52 to calculate the confidence interval for the differenc e between two means, /1-1 and /1-2, if the standard deviations "l and"'.l are known. If the standard deviations "l and"'.l are unknown, us e Eq 12.53. The I-distribution parameter, lo!:l, has 1 + n-: - 2 degrees of freedom

100 resistors produced by company A and 150 resistors produced by company B are tested to find their limits before burning out The test results show that the company A resistors have a mean rating of 2 W before burning out, with a standard deviation of 0.25 W; and the company B resistors have a 3 W mean rating before burning out, with a standard deviation of 0.30 W. \VIlat are the 95% confidence limits for the differenc e between the two means for the company A resistors and company B resistors (i. e., A- B)?

(A) -1.1 W;-1.0W

(B) -1.1 W; -0.93 W

(e) -1.1 W; -0.90 W

(D) -1.0W;-0.99W

Solution From Table 12.3, the value of the standard normal variable for a two-tail test with 95% confidence is 1.9600

From Eq. 12.52, the confidence limits for the differenc e between the two means are

12.28

The tmSWeT is (B).

- - !f0;' .r, = X 1 - X 1 - Z,,/2 - + -

n, "' = 2W-3W

- 1.9600 (0.25 W)2 (0.30 W)2

100 + 150

= - U )686 W (-1.1 W)

- - !f0;' .r, = X 1 - X 2 + Z,,/2 - + -n, "' = 2W-3W

+ 1.9600 (0.25 W)2 (0.30 W)2

100 + 150

= -0,9314 W (-0,93 W)

EtpUJtion 12.54: Co,gukncr LimiJstmillnU"ulford~ V~ oj"4 Nonnnl Disuilnliion

(n _ 1)82

r,'/2~'-1 12. 54

Equation 12.54 gives the limits of a confidenc e interval (confidenc e C - 1 - (X) for an estimate of the population varianc e calculated as the sample varianc e from Eq. 12.19 with a sample size ofn drawn from a normal distribution. Sinc e the varianc e is a squared variable, it will be distributed as a chi-squared distribution with n - 1 degr""s of freedom Therefore, the denominators are the i' values taken from Table 12.5 at the end of this chapter. (The values in Table 12. 5 are already squared and should not be squared again.) Sinc e the chi-squared distribution is not symmetrical, the table values for r:JJ2 and for 1 - (r:JJ2) will be different for the two confidenc e limits

15. HYPO THESIS TESTING

A hypot""'i' t05t is a procedure that answers the question, "Did these data come from [a particular typ e of] distributionT There are many typ es of tests, depending on the distribution and parameter being evaluated. The most simple hypothesis test determines whether an average value obtained from n repetitions of an experiment could have come from a population with known mean If- and standard deviation (5. A practical application of this question is whether a manufacturing process has changed from what it used to be or should be Of course, the answer (i. e., yes or no) cannot be given with absolute certainty-there will be a confidence level associated with the answer

The following procedure is used to determine whether the average of n measurements can be assumed (with a given confidence level) to have come from a known normal population, or to determine the sample size required to make the decision with the desired confidence level

12.29

Etpmtion 12.55 l1um'6" Eq. 12.60: TMt on ~han of Nonnnl DisUill11Jion, POJIIUntion ~ tmiI V/Jrianct!o Kn_n

,tep J Assume random sampling from a normal population

The n~!! hypothe'i' is

12. 55

The altemativo hypothe'i' is

12. 56

A typo lorror is rejecting Ho when it is true The probability ofa typ e I error is the Iovd oj'ignijiccmco

a = probability(type I error) 12.57

A typo llorror is accepting Ho when it is fals e

fJ = probability(type II error) 12. 58

,tep 2 Choose the desired confidence level, C.

,tep 3 Decide on a one-tail or two-tail test If the hypothesis being tested is that the average has or has not incrw,od or has not

12.30

Etpllliion 12.61 Tlum'6" Eq. 12.tiB: Stunpk SiuforNormal Distri1l11Jion, tmiI p Known

(Z,,/2 + Z,9 )2

12.31

(A) There is at least a 5% probability that the plant is operating properly

(B) There is at least a 95% probability that the plant is operating properly

(C) There is at least a 5% probability that the plant is not operating properly

(D) There is at least a 95% probability that the plant is not operating properly

Solution Since a specific direction in the variation is not given (i. e , the example do es not ask if the average has decreased), us e a two-tail hypothesis test

From Table 12.3. x - 1.9600

Us e Eq. 12.59 to calculate the actual standard normal variable

= 871 - 880 = - 3.03

"

'"

Since -3.03 < 1.9600, the distributions are not the same. There is at least a 95% probability that the plant is not operating correctly

The tmSWeT is (D).

16. LINEAR REGRESSION

EqrUJtiOIt 1Z.U Tlutn16" Eq. 1Z. 75: ~kd.od of LnJSt 8tpUUWi

Ifit is necessary to draw a ,traight line (11 = Ii + b;l) throughn two-dimensioml data points (Xj,n),(X2,n). .('",y"), the followilg method based on the _thad oj Iw" "'f'Kl'"' can be used ,top J Cc1culate the following seven quanti~s

, ., ~"

' ~ (I/n) (i>.) .-,

12.69

' ~ (I/n)(ty.) .-,

12.70

12.32

,top 2 Calculate the slop e, &, of the line

12.71

12.72

12.73

,top 3 Calculate they-intercept, a

12.74

The equation of the straight line is

12.75

The least squares method is used to plot a straight line through the data points (1 6), (2,7), (3, 11), and (5, 13). The ,lope of the line is most nearly

(A) 0.87

(B) 1.7

(C) 1.9

(D) 2.0

Solution First, calculate the following values

I>; =1 + 2 + 3 + 5 = 11 L: y; = 6 + 7 + 11 + 13=37 L::r~ = (1)2 + (2)' + (3)2 + (5)2 = 39

L: :r,y; = (1)(6) + (2)(7) + (3)( 11)+ (5)(1 3) = 118

Find the value of S~~ using Eq. 12.72

12.33

= liS - O}ll)(37) = 16.25

Find the value of Sa from Eq. 12.73

= S.75 From Eq 12.71 . the slop e is

16.25 j, = Sz. / Szz = - -8.75 = 1.857 (1.9)

The tmSWeT is (q.

Etpmtion 12. 76 tmiI Eq. 12. 77: SttuulnnJ Error of Estinttlu

12.76

12.77

Equation 12.76 gives the ""''''' ;quar.d .rror. S; or MSE. wl:ich estimates th, likelihood ofa V3lue being clos e to an observed value by averaging fr.e square of the mors (i. e .. the differenc e between the estimated value and observed value) Small MSE values are fa.orable, as they indicate a smaller likelihood of error

Etpmtion 12. 7B tmiI Eq. 12.7J: Cotifitkncr lllknulsfor S/~ tmillllkrcqll

12.73

- + - MSE ( 1 ") n Su 12.79

12.34

Etpllliion 12.BO tmiI Eq. 12.B1: Stunpk COnlntion CH.JIicwlII

12. 80

12. 81

Onc e the slop e of the line is calculated using the l""t squares meth,d, the goodn

12.35

- 32 - (t) (9)(3) ~ ~=;=;=~===;====;=

.j (39 - (J) (9)' )(87 - (1)(3)') = - 0.972 (-0.91)

The tmSWeT is (A).

II l: 0.1

2

3

,

6

o

o

o

2

o

2

J

o

2

)

o

2

3

,

09000

08100

0.9900

01190

09120

0.9990

06561

09417

09963

0.9999

05905

09185

0.9914

09995

I 0000

05314

0.8857

09842

09987

09999

I 0000

o . ~ ' .3 , .. 08000 0.7000 06000

0.6400

0.9600

05120

0.8960

0.9920

0 4096

0.4900 0.3600

0.9100 0.8400

0,3430 02160

0,7840 06480

0.9730 0.9360

02401 o 1296

08]910,6517 04752 0.9128

0. 9984

,m

0.91 63 0.8208

0. 991 9 0.9144

0, 1681 00778

07373 0,5282 03370

0.9421 0.8369 0.68 26

09933 09692 091 30

09991 0,9976 09898

0.2621

0. 6554

0. 1176 0.0467

0.4202 0.2333

090110,744305443

09830 0,9295 08208

0.9984 0.9891 0.9590

0.9999 0.9993 09959

,s

05000

02500

8.7500

0]250 05000

0.8150

00625

03115

06815

8.9315

00313

01815

0.5000

08125

06988

0.0156

8.1094

03438

06563

0.9806

09844

p

' .6 0.4000 0 3000

0.1 600 0.0900

0.6400 0.51 00

0,0640 00270

0,3520 02 160

0.7840 0.6570

00256 00081

0,1792 00837

0. 5248 0.3483

0.8704 0.7599

0,0]02 00024 0,0810 00308

0.3 ]]4 0.1631 06630 04718

0,9222 083 19

0. 0041 0.0007

0.04 10 0.01 09

0,[792 00705 0.4557 0 2557

0.7667 0.5798

09533 0.3824

01000

00400

8.3600

00080

o 1040

0.4880

00016

00271

o 1808

0.5904

00003

00067

0.0519

02627

067lJ 00001

0.0016

00170

00989

03 ...

0.7319

, ..

0]000 0.0100

01900

000]0 00180

0.2110

00001

00031

0.0523

0.3439

00000

00005

0.0086

00815

04095

00000

0 0001

00013

00159

01143

04686

0 .9~

0,0500

0.00 25

0. 0975

0,00 0] 0,0073

o 1426

00000

0,0005

0.0140

0.1 855

0,00 00

0,0000

0.00 1l

00226

0,2262

00000

0 0000

0,00 0] 0,00 22

00328

0.2649

' .99 001 00

0.00 01

0.0 199

00000

00003

0.0297

00000

00000

0.0006

0.0394

00000

00000

00000

000 10

00490

00000

00000

00000

00000

0.00 15

00585

12.36

,

,

,

1 1 ' 8 < ;

l K I '

[ S I T

' U

8 9 1

0 < ; 1

S L 6 0

< ; 9 L O

,

< ; [ 6 6 < ; 9 6 9

0 1 ' O e n 9 8 8 1 9 8 1 1 9 0 1 9 1 8 0

,

L < ; 9 9 1 [ 8 1

9 0 L l l

1 ' 1 [ 9

S L O T

[ 9 6 1 9 L [ 1

0 0 0 1

~OOO

1 0 ' 0

~;:OO

~O'O

o r o

~T'O _ D

0 ; : ' 0 -

0

~Z'O _ D

. .

,

_ 0 _ 0 _ 0 _ 0 _ 0

I l L ' ! ~ad

12.37

0.683

0.683

0.674

0.854

0.854

0.842

1055

1055

1036

1.311

1.310

1282

1.699

1.697

1645

H145

H142

1.960

* -The numb er of indep endent degrees of freedom, v, is always one less than the sample ,ize, n

Tahle 12.5 Critical Val"'" ofChi-Squar.d n,triDution

( (XL)

,

. - I

""~;'"-~ ~!': .. .., ... "

:Ii:;;~i;1! .. !'::!::!: .., ...... .. ~ .. ~ .... ~

",:;;u~ ~~ :!~~ ~~~~~ ~~!':~~ :>~ .. "' ... "

~ ... ~ .. .. .. .... .. .. .. .... .. ..

",~~i:~ ~~~5~ ~ ~~~~ ~~~~~ ~!': .. "' ... ~~ ... ~.. .. .... .. .. .. .... .. ..

"',.;'!;;~ ~!': .. "' ... ,,~~ii;:: g!!!wi";"; .., ill.; 8:::: 51:!!""'''' "

e~ f: !3: ;; ... ~.,~ ..

~~~~~ ... .,.,~ ..

~U"-~ .. .... .. ..

i~::S ;~~~~ ~~~~: ~ ~ a ~ ~~~~~ -;;~~~ a~ .. ~ ... .,;~i~~ ~~ .. ~~ ;~!;5 55~~~ ~~5~~ " ~~:~! "

.. _~Ji;lq::;; :1 ~~".., "

~1!~!I;e "' ... .........

-.. ~ ..... -----

>!:~u~ ..........

"Ii\~!!~ "' ... ~.., ...

ea5U .. ~ .. --

e~8ac; .. .... .. ..

!3:SUc; .. ........ .. ........

~ r1!~;::I!:, ..........

U: $Ii;;; .. ........

2.462

2.457

2.326

.;:t:p~ Ii

........

2.756

2.750

2.576

Mof