HP-41C PROGRAMS AND INSTRUCTIONS FOR PROBABILITY …HP-41C calculator.,, In particular, this means...

"AD-R1i48 573 HP-41C PROGRAMS AND INSTRUCTIONS FOR PROBABILITY AND 1/2STRTISTICS(U) NAVAL POSTGRADUATE SCHOOL MONTEREY CA

CLS P W ZEHNA FEB 84 NPS55-84-003

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AHP-41C PROGRAMS AND INSTRUCTIONS

FOR

PROBABILITY AND STATISTICS

by

Peter W. Zehna

C

LULFebruar 1984

Approved for public release; distribution unlimited.

--a'.. Prepared for:

,-3 Naval Postgraduate SchoolMonterey ,California 93943

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NAVAL POSTGRADUATE SCHOOLMonterey, California

Comwdore R. H. Shumaker David A. Schrady

Superintendent Provost

Reproduction of all or part of this report is authorized.

* This report was prepared by:

Peter W. Zehna, Prof esyorDepartment of Opera i Research

Reviewed b Released by:

A . Washburn, Chairman Kneale T. Marshallepartment of Operations Research Dean of Information and icy

Sciences

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REPORT DOCUMENTATION PAGE EAO CsOREucTIORs1. REPORT NUMBRN '2. GOVT ACCESSION NO . RECIPIENT'S CATALOG NUMeIr

NPS55-84-003

4. TITLE (and Subltile) S. TyPe O" REPORT A PERIOD COVERED

HP-41C PROGRAMS AND INSTRUCTIONS FOR PROBABILITYAND STATISTICS Technical

a. PERFORMING O01G. REPORT NMB~ER

W. uTNORM() S. CONTRACT OR GRANT NUMSER(I)

Peter W. Zehna

S. PERFORMING ORGANIZATION NAME AND ADDRESS 10. PROGRAM ELEMENT. PROJECT, TASK

Naval Postgraduate School AREA & WORN UNIT NUMS6 'Monterey, CA 93943

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Naval Postgradate ScIool February 1984Monterey, Ca 93943 3.1UM9R OF PAGES108

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Approved for public release; distribution unlimited.

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It. SUPPLEMENTARY NOTES

is. KEY WORDS (COMM . is tevoie o if nee)w, #Ad Idmllfr IW 6lA 1h01m)

HP ProgramsProbability and StatisticsCalculator Programs

I&. AUTRACI (Cosstloe an rove" olds It rnoowr aid fdsttl IV IMah 11006W)

S.. .A compendium of programs and instructions to solve problems typically encoun-

tered in probability and statistics. Programs are designed to operatedirectly on the HP-41C family of hand-held calculators.

DO , JA 143 EIINO O 55OSLT UNCLASSIFIED

S/N 0102- Ll- 014. 6601 S9CumTV CLASSIFICATION OF TWI PAGE f De 1 0

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RP-41C PROGRAMS AND INSTRUCTIONSFOR

PROBABILITY AND STATISTICS

by

Peter W. Zebna

I\&.~4~ ~INTRODUCTI ON/The pu'pose of this report is to make available a set of programs and the

correspondng user instructions so that the problem material found in thewriter's textbooks, Probability by Calculator and Statistics by Calculator-'(hereaf er referred to, respectively, as ZP and ZS) may be ri-elved--using theHP-41C calculator.,, In particular, this means that courses using those text-books, written entirely around the TI-59, need no longer be restricted to thatparticular machine as a prerequisite. It is almost essential, however, thatthe HP user have in his, or her possession either the HP-4ICV, or the HP-41Cwith the quad memory module installed, along with a card reader for recordingmagnetic cards. Also, as,ith the TI-59, it will be necessary to insert theHP applications module STAT PAC for use with the programs in ZS. Noadditional module is require4 for ZP.

The original intention wA to write the HP programs in such a way thatthe TI user instructions could be used with little or no modification. That

program was about 90% successful so that, in general, storage in various reg-isters are identical as are the main subroutines labeled with user definedkeys (with HP a,b,c, etc., replacing TI A', B', C', etc. in a natural way).There are, however, some special problems created by the differences in thetwo machines (RPN not being one of them, by the way) that made it impossibleto be 100% successful in that endeavor. For example, the TI random numbergenerator could not be duplicated in the HP because of the difference in accu-racy of the two machines. Since the TI carries more significant figures in-

.4 ternally than the HP, and that internal carriage is used to generate succes-sive seeds for repeated applications, the two machines soon differ in theiroutput. For true applications of random number generation that would be in-

significant, even a desirable difference perhaps, but for tutorial purposes,which is the main intent of the books, that makes it impossible to verify an-swers and that is a serious drawback for the learner. Otherwise, the differ-ence in accuracy created no special problems. The FIX 4 format is used in allof the HP output to follow and it will be found that the corresponding answersthen agree to within 4 decimal places (the maximum usually presented in ZP andZS) of the published answers given in the two books, almost without exception.

Writing the HP program to utilize essentially the same user instructionsas the TI eant not being able to take full advantage of the superior alpha-numerics and prompting facility of the HP41-C. The user may well want to ad-Just the programs presented here to take better advantage of that option butshould of course adjust the user instructions accordingly. That particularfeature in itself creates some special problems with regard to the use of HPapplications modules like STAT PAC. Almost all of the programs in that modulecontain pauses for prompts from the user. Unfortunately, when such programsare called as subroutines within a calculator program, there is no automaticreturn from the module program to the parent one. Much of the success of theTI program depended on precisely this feature utilizing the canned programsavailable in the master module for ZP and the statistics module for ZS. Thismade it necessary to replace several of the programs in the HP STAT PAC thatwould otherwise have been used, as well as to supply several key programs,such as the t and F distributions, that were missing. Fortunately, themassive memory capability furnished by the HP quad memory made it possible tofurnish these and still have enough room for the main programs of interest.For ZS then, a special program called ZSTAT has been supplied for which there

*Prentice-Hall, Inc., Englewood Cliffs, NJ, 1982

Is no direct TI analogue. The reader may view this as simply an addition tothe HP STAT PAC in order to bring it more in line with the TI statisticsmodule utilized throughout ZS.

In order to follow the textbooks as closely as possible with the leastamount of cross-referencing, the following format will be followed. Startingwith ZP, each chapter or section for which a separate program exists will bediscussed separately starting on a new page. After pointing out any generaldifferences that may exist for that chapter or section including the illustra-tive examples contained therein, the HP version of the User Instructions forthat program will be added, together with a set of examples for each subrou-tine such as presently found in the books for the TI programs. These modelexamples will show exactly what the user may expect to see in the display uponexecuting each step. In each case, the reader will find, in addition to theRegister Contents as currently published in the textbooks, a set of assign-ments used by the program along with a listing of labels used (which may alsobe seen in the complete listing of the programs in the appendix).

The reader should remember to assign, record (and subsequently read) themagnetic cards in USER mode so as to preserve those assignments. In thoseassignments, we often use lower case versions of capital letters even when theydo not, technically, exist. Thus, [I) is used for the alphanumeric [<j sincethe latter is located above [1] and is effected by pressing the gold shift key,then [M]. Similar remarks apply to [g] (really [%), [h] (really []) and [j](really [>]). Of course [a], [b], [c], etc. are actually listed in the alphakeyboard.

Since the [X<>YJ key is used in so many programs, and its execution is con-siderably slower in USER mode, it is advisable to assign the function X<>Y tothis key at the start of a session. Such an assignment cannot be made perma-nent in the programs, but will remain in effect unless the master clear isused.

2

PROBABILITY BY CALCULATOR

Section 1.3: The CalculatorIt is assumed here that the reader is reasonably familiar with the Owner's

Handbook and Programming Guide for the HP-41C. The general remarks found inthis section apply to the HP as well. It has already been remarked that a cardreader will be needed to follow the program outlined here. It is possible todo without the magnetic cards for some of the programs since they may be keyedin once and the continuous memory feature of the HP will preserve them. Buteven that generous memory allowance will soon be used up and programs will haveto be replaced to follow all of the subroutines presented in these textbooks.The magnetic cards removes the necessity of having to re-key so many separateprograms. Guidelines for recording magnetic cards will be found in the CardReader handbook and should be consulted.

Section 1.4: The ProgramsMany of the remarks in this section will not apply directly to the HP cal-

culator and, again, the Owner's Handbook should be consulted for specifics re-garding the related keys. The programs will appear in print-out (see Appendix)as numbered steps with the corresponding mmemonic code (no key code as with theTI). Most are self-explanatory and the Function Index given in the back of theHandbook will be found very helpful should the reader encounter any that arenot immediately recognized. Naturally, the programs should be identical withthe listings given in the Appendix before any recording takes place.

Section 2.4: Counting ProblemsThe internal function FACT in the HP will replace the use of label C in

Pgm 16 of the TI to display factorials as discussed on page 21. That functionhas exactly the same restriction, namely, that n must be any positive integerbetween 0 and 69 inclusive, displaying OUT OF RANGE for larger values. Thereare no internal programs to handle permutations and combinations directly sothey have been programmed in the first card program labeled ZP2. You will findthe instructions under Steps 7 and 8. Each scheme prompts you for an input offirst N and then R to compute the corresponding values. (The HP alphanu-merics do not permit lower case letters so the notation differs just slightlyfrom the book.) With these routines, the answers to the problem in thissection may be verified.

Section 2.5: Conditional ProbabilityThe rest of program ZP2 has to do with Bayes probabilities and the in-

structions match those for the TI exactly (with a,b,c, etc. replacing A',B',C')as previously remarked.

3

!.4

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W. -. -- -W'

ZP2 (Assigned [e]) USER INSTRUCTIONS (HP) SIZE > 090- -STEP PROCEDURE ENTER PRESS DISPLAY

1. Initialization xxx [e] 0.0000

2. Input probabilities(Repeat for j = 1,2,..., k) Pr(EIC4 ) (A] j

NOTE: If Pr(C) - /k, use Step 2' Pr(C )J [R/S] j

2'. a. Input partition size k [E] 1/k

b. Input given priors Pr(EIC ) [R/SI i

3. Compute Bayes posterior i [B] Pr(CiIE)probability Pr(Ci [ E)

4. a. Initialize for sensitivity analysis xxx [e] 0.0000

b. Recall given priors [D] Pr(EIC j)

c. Input new cause probabilities New Pr(C [R/SJ J(Repeat for j = 1,2,...,k)

NOTE: If new Pr(C ) 1/k, use Step 4'

4'. a. Initialize xxx [e] 0.0000

b. Input partition size k [d] k

5. Compute Pr(E) (Law of Total Probability) xxx [a] Pr(E)

6. Birthday Problem k [C] Pr(E )

(Ek is the event that two or more amongk people in a room have the samebirth date.)

7. Calculate N [b] N-?

N [R/Sl R - ?

R [R/S] P( N

8. Calculate [c]

N [R/S] R - ?

R [R/S] C( NR

4

q 4

Register Contents

00 Used 10 20 Pr(EIC 1)

01 11 Used 21 Pr(C I)

02 12 Used 22 Pr(EIC 203 k 13 I/N 23 Pr(C 204 EPr(EIC) 14 Used 24

05 15 25

06 16 26

07 17 27

08 18 28

09 19 29

AssieentsLabels Used

ZP2 le 02 A a

03 B b

04 C c

05 D d

08 E

09

10

11

12

5

EXAMPLES ZP2 (1) Suppose in medical diagnostics a particular symptom (E)

always occurs in conjunction with three diseases Ci,

C2 , C3 with respective probabilities 0.90, 0.09 and 0.009

or else occurs rarely (0.001) with no apparent reason

(C4 ) at all. National statistics show that most people

are free of the three diseases, Pr(C 4) - 0.99, and dis-

ease CI is fairly rare, Pr(CI ) - 0.0001 . Diseases C2

and C3 occur with respective probabilities 0.0045 and

0.0054.

Bayes Format:

Conditional Cause ConditionalEvents Priors Probabilities Posteriors

CI W Disease #1 0.90 0.0001 0.0587

C2 - Disease #2 0.09 0.0045 0.2641

. C3 - Disease #3 0.009 0.0054 0.0317

C4 - No Disease 0.001 0.9900 0.6455

E - Symptom Pr(E) - 0.0015

Calculator Solution:

ZP STEP ENTER PRESS DISPLAY COMMENTS

Step 1 [e) 0.0000 Initialization

Step 2 .9 [Al 1.0000 First conditional prior

..0001 [R/S] 1.0000 Shows one pair entered

..4 09 [A] 2.0000 Second conditional prior

.0045 [R/S] 2.0000 Shows two pairs entered

.009 [A] 3.0000 Third conditional prior

.0054 [R/S] 3.0000 Shows three pairs entered

.001 [Al 4.0000 Fourth conditional prior

.99 [R/S] 4.0000 Shows four pairs enteredI Step 3 1 [B) 0.0587 First conditional posterior

- 2 [B] 0.2641 Second conditional posterior

" 3 [B] 0.0317 Third conditional posterior

4 [B] 0.6455 Fourth conditional posterior

Step 4 [al 0.0015 Probability of E

6

A

EXAMPLES ZP2 (2) A manufacturer of hand-held calculators has three different

assembly plants F, M and T. These three plants historically

produce defective items with respective probabilities 0.01,

0.02 and 0.04. Plant F produces 50% of the calculators while

plants M and T produce, respectively, 30% and 20%.

*, Original Bayes Format:

Events Priors Causes Posteriors

C1 - Plant A 0.01 0.50 0.2632

C2 - Plant B 0.02 0.30 0.3158

C3 - Plant C 0.04 0.20 0.4211

E - Defective Pr(E) - 0.0190

Calculator Solution for Changing Priors to pi 1/3 (after original entry):


Step 4a. (e] 0.0000 Initialization

Step 4b. [D] 0.0100 First prior displayed

Step 4c. 1/3 [R/S] 1.0000 First cause prob. changed

Step 4b. (D] 0.0200 Second prior displayed

Step 4c. 1/3 [R/S] 2.0000 Second cause prob. changed

Step 4b. [D] 0.0400 Third prior displayed

Step 4c. 1/3 [R/S] 3.0000 Third cause prob. changed

Step 3 1 [BI 0.1429 New Pr(EIC 1)

2 [B] 0.2857 New Pr(E IC2 )

3 [B] 0.5714 New Pr(EIC 3 )

Alternate Solution:

Step 4'a. [e) 0.0000 Initialization

Step 4'b. 3 [d] 3.0000 Partition size entered

Step 3 1 [B] 0.1429 New Pr(EIC 1 )

2 [B] 0.2857 New Pr(EIC 2)

- 3 [B] 0.5714 New Pr(EIC 3)

7

,.

---.

EXAMPLES ZP2 (3) Calculate P( 1, 4! and C( 52

Solution:


Step 7. [b] N-? Prompts for entry of N

1 10 [R/S] R-? Asks for the value of R.

2 [R/S] 90.0000 Display P(2 90.

[b] N-? Initializes permutation routine.

4 [R/S] R-? Asks for the value of R.

4 [R/S] 24.0000 Displays 4! - P(4).

[c N-? Initializes combination routine.

52 [R/S] R-? Prompts for entry of R-5.

5 [R/S] 2,598,960 Displays C( 2), the total number5

of poker hands.

NOTE: 4! may also be computed by executing the function FACT.

.4

-3-l

N

4?" I,.

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* - -

Section 3.2: Moments of a Random VariableJust as with ZP2, the HP version of ZP3.2 (denoted ZP3-2 since a period

may not be used in an ALPHA label) is almost exactly the same as the TI ver-sion. In the discussion of the program on page 57, you may ignore the warn-

- ings concerning capacity limitations and repartitioning the calculator. Siz-ing the HP to allow for more memory registers will accomplish the same thing.

f: In any case, such problems will never arise in the applications presented-. here. You might observe the use of the alternate HP form, [X0YI, for the X

exchange Y key throughout this report. This is merely a concession to ease ofprinting. (HP Y-register is always used in place of TI T-register)

The one place where there is serious departure from the TI-59 is in re-peated application of LABS. To erase a previous application with the TI, oneneed only over-write the old algorithm with the new one, paying no attentionto what may or may not remain when the new algorithm is finished with a RETURNinstruction. But, because algorithms must be created as individual sub-routines with the HP, erasing is not so simple. At Step 4f. the beginning ofthe old algorithm is displayed at program step 88. The steps from this point

:. on need to be erased and this may be accomplished with the internal functionDEL. Then the new algorithm may be inserted where the old one resided and theprogram will function for the new case. As suggested in the footnote to theuser instructions that follow, you might assign DEL to a label like [g] if alot of erasing is to be done. Unfortunately, the DEL function cannot be re-corded as an instruction in program memory so this will only be helpful forgiven session.

On page 59, an HP version of the algorithm for g(x)-3x+19 would be

RCL 09,3,*,19,+,RTN

and g(x)n(x*x-4)/(6x+7) could be keyed in as

4,RCL 09, ENTER,*,-,CHS,RCL 09,6,*,7,+,/,RTN

Here we have taken the liberty of using the printed symbol / for the divisionoperator and the symbol * for multiplication.

Section 3.3: Hypergeometric and Binomial DistributionsSection 3.4: Other Discrete Distributions

For both of there sections, the HP programs are practically identical:- with the TI programs. The basic difference is that the HP initialization step

is to press [e] instead of RST. Having so used label e , label J is usedfor the number of trials, Y , to rth success at NB5 in program ZP3-4.

9'

• % o " • % . -• .. . . . . . . . . ... . .** L

ZP3-2 (Assigned [e]) USER INSTRUCTIONS (HP) SIZE 060

STEP PROCEDURE ENTER PRESS DISPLAY

1. Distribution Entry

a. Initialize [e] 0.0000

b. Enter (in order ) x., pi xi [A] xi

(Repeat for i - 1,2,...,N < 20; Pi [R/S] i.0000

x < x < ... < X)1 2 N

2. Calculate P(x) x-code [C] P(x)

(x-code - J where x < x < xj+ l )

3. Calculate E(X), V(X) [E] E(X)

(after Step 1) [X<>Y] V(X)

4. Calculate E[g(X)], V[g(X)](after Step 1)

a. Initialize [GTO] [B] x.xxxx

-' NOTE: It is understood that [ALPHA]must be used for label B.

b. Enter Program Mode [PRGM] 87 LBL B

c. Key in g(x) where x e R09

(Avoid labels already in use, endwith RTN)

d. Exit Program Mode [PRGM] x.xxxx

e. Calculate Moments. [D E[g(X)]

4 [X<>YI V[g(x)I

f. To ERASE Algorithm in [B], completeSteps a,b; then [SSTJ 88 yy

[g] DEL

let nnn be at least as large as the nnn 87 LBL Bnumber of Steps in [Bit [PRGM] x.xxxx

-For repeated uses of this step use ASN to assign DEL to g(%).

. .?

10

.-. .,C'.-V,* .**-**.f.. . . .-.. *-... S . . .. . . .• • ,,. . • .C--. .. . .. .- .. .. . . . . . . . *. . . S .., ,,. ?,. , -,- .-e.., .,..:., ........ •...... ... .... ......-........-.....

Register Contents

00 Used 10 P(x) 20 x 1

01 ,-address 11 21 p1

02 p-address 12 22 x2

03 N 13 23 p2

04 xip i 14 24 x305 x 2i 15 25 P3

06 Mean 16 26

07 2nd Moment 17 27

08 Variance 18 28

09 x-value 19 29

Assig;nments Labels Used

ZP3-2 I e 01 A

02 B

03 C

04 D

05 E

06

07

/ y.

_1, ° ,q- . i - , o - . - . - , • , o . o o . . . . .1 1

'fd EXAMPLE ZP3-2. X m # daily sales of a morning newspaper at a local drugstore.

x: 0 1 2 3 4 5

p(x): 0.01 0.01 0.04 0.03 0.67 0.24

Solution.:


Step la. [e] 0.0000 Initialization

Step lb. 0 [A] 0.0000 Enter first x-value

" .01 [R/SI 1.0000 Enter first p-value

1 [A] 1.0000 Enter second x-value

% .01 [R/S1 2.0000 Enter second p-value

2 [A) 2.0000 Enter third x-value

.04 [R/SI 3.0000 Enter third p-value

3 [A] 3.0000 Enter fourth x-value

.03 [R/S] 4.0000 Enter fifth p-value

4 [A] 4.0000 Enter fifth x-value

.67 [R/S1 5.0000 Enter fifth p-value

5 [Al 5.0000 Enter sixth x-value

.24 [R/S] 6.0000 Enter sixth p-value

SCalculate P(4.5) (x-code = 5 since x5 < 4.5 < x6 - 5)

Step 2 5 [C] 0.7600 Note that x1 -O so that x 5 i4.

Calculate v - E(X) and a2 V(X).

Step 3 (E] 4.0600 Display p m 4.06

[X<>Y1 0.6764 Display a2 - 0.6764

Calculate E[g(X)l and V[g(X)] where g(x) = 25x - 50 is net daily income.

Step 4a. [GTOJ GTO Initialize

[ALPHA] GTO

[B] GTO B[ALPHA] x.xxxI

Step 4b. [PGRMI 87 LBL B

Step 4c. RCL) 88 RCL

09 88 RCL 09 Brings current x-value into

25 89 25 R- X

12

| •. I " " " '"'" "" " " - " " " ,t 9 ,b" " " " % " " " " " " .% " ".*-. "- °

* ""

ZP STEP ENTER PRESS DISPLAY COMMENTS. 90*

(xl 90 * Multiplies x by 25,

50 91 50 and subtracts

-1 92 - 50.

[RTNJ 93 RTN Ends algorithm.% Step 4d. (PRGM] x.xxxx Exit Program mode.

Step 4e. [D) 51.5000 Calculates "average" dailynet income as 51.5 cents.

[X<>Y] 422.7500 Exhibits variance in cents2

(USER]ix' 20.5609 Shows a as 20.56 cents.

Calculate E[g(X)] and Vtg(X)] where g(x) is daily profit.

Step 4f. [GTO] GTO Initialize

[ALPHA] GTO

[B] GTO B

[ALPHA] x.x

[PGRMJ 87 LBL B Enters ZP-3.2 program at B.

[SST] 88 yy Locates first step of lastalgorithm

[g] DEL __ Prepares to delete algorithmsteps.

. 010 87 LBL B Deletes to END statement.

4 Step 4b. 2 88 2 Enters 2 for comparison with x.

[RCL]09 89 RCL 09 Retrieves x.

[MY?] 90 x>y? Asks if x>y?

[GTO]20 91 GTO 20 Proceeds to subroutine to beconstructed for evaluating g(x).

0 92 0 Otherwise g(x)=O

[RTNJ 93 RTN Ends that part of algorithm

1[LBL]20 94 LBL 20 Prepares to define subroutine.

25 95 25 g(x) - 25x - 50.

[x] 96 * A return is not necessary

50 97 50 since it is controlled by END.

1-1 98 -[PRGM] x.xxxx Exits program mode.

Step 4c. (D] 52.2500 Calculates and exhibits "average"daily profit of 57.25 cents.

Step 4d. [X<>Y] 313.6875 Shows profit variance.

13

* ,', € ~.:.- y .. . % .. :

.... ... .. ..

ZP3-3 (Assigned [el) USER INSTRUCTIONS (HP) SIZE 030


Hypergeometric Distribution

11 Initialization [e] 0.0000

12 Enter Parameters N [STO114 N

(n < N and 0 < M < N) M [STO]15 M

n [STO]13 n

H3 Calculate P(k) - Pr(X < k) k [A] P(k)

p(k) - Pr(X - k) [X<>Y] p(k)

H4 Calculate Q(k) - Pr(X > k) k [a) Q(k)

p(k) - Pr(X - k) [X<>YJ p(k)

NOTE: Repeat H3 and/or H4 as often asdesired.

Binomial Distribution

B1 Initialization [e] 0.0000

B2 Enter Parameters (M < N) N [STO114 N

- H [STo115 M

n [STO113 n

B3 Calculate P(k) Pr(X < k) k [BJ P(k)

p(k) - Pr(X - k) [X<Y p(k)

B4 Calculate Q(k) - Pr(X > k) k [b] Q(k)

p(k) - Pr(X - k) [X<>YI p(k)

', NOTE: Repeat B3 and/or B4 as often asdesired.B

* E Display E(X) and V(X) (following any [E] E(X)application of H3(H4) or B3(B4 ) [X(>Y] V(X)

'.1

',-- ,€'-S,' ,,s e, ,, e'.'.'.'.". .;.2 ..... :. .'.' '.>... .'., . . . . ..., . . . .",",",".-., ,- - -,;,,

Register Contents

0 00 10 P(x) 20 Used

01 N( for 11 ji 21 1 - M/N

M02 r k MTON) 12 a2 22 M/N

03 13 n 23 Used

04 14 N 24 N-M

05 15 M 25

06 Used; p(k) 16 p(O) 26

07 Used; p(k) 17 27

08 Used 18 28

09 19 29

Assin ments Labels Used

ZP3-3 e 01 A a

PMTON ± 02 B b

CMBON h 03 E

04

Note: PMTON and 05CMBON requirestorage of u in 06JRol id k in R02,jfor execution 07

fusingxEo ' '08

11

12

18

19

15

77 . ,, ,.

EXAMPLES ZP3-3. An urn contains five black balls and seven white balls.

(1) A sample of size 3 is drawn without replacement. Calculate the

probability of obtaining exactly two black balls, at most two black

balls and at least two black balls. Answers are, respectively,

p(2)-0.32, P(2)-0.95 and O(1)-0.36.

(See display below.)

(2) Repeat (a) for a sample drawn with replacement. Answers are,

respectively, p(2)-0.30, P(2)-0.93, 0(1)-0.38.

(3) For each of (a) and (b) determine the mean and variance of X

the number of black balls in the sample.

.Aa (a) U - 1.25, a2 . 0.60; (b) u = 1.25, a2 0.73.

Solution (1), (3):

w ZP STEP ENTER PRESS DISPLAY COMMENTS

HI [e] 0.0000 Only necessary when starting-- a new problem.

12 12 !STO114 12.0000

5 [STO]15 5.0000

3 [STO113 3.0000

H3 2 [A] 0.9545 Displays CDF P(2) first

[X<>Y] 0.3182 Displays p(2).

H4 1 [a] 0.3636 Displays Q(1). No re-initial-ization necessary.

E [El 1.2500 Displays mean and variance

[X<>YJ 0.5966

Solution (2), (3):

BI lei 0.0000 Signals the start of a new pro-gram even though the sameparameters are involved (B2unnecessary)

B3 2 [B] 0.9277 Binomial CDF differs from H3

[X<>Y] 0.3038 Binomial p(2).

B4 1 [b] 0.3762 Q(1) Pr(X > 1) - pr(X ), 2)

E [E] 1.2500 Mean

[X<'>Y] 0.7292 Variance

16

-. ? ,*" . .. 'e. % ' %.*'%-*...'-'%*%.%.. %' * - .' - _' . .- *-

ZP3-4 (Assigned [e]) USER INSTRUCTIONS (HP) SIZE 030


Binomial Distributionbin 1 Initialization [e] 0.0000

* . bin 2 Enter Parameters n [STO113 n

p [ST0122 p

bin 3 Calculate P(k) - Pr(X < k) k (B] P(k)

,X<>Y] p(k)

bin 4 Calculate Q(k) w Pr(X > k) k [b] Q(k)

[X<>Y] p(k)

NOTE: Repeat 3 and 4 as often asdesired

Poisson Distribution

POI Initialization for Poisson [e] 0.0000

P02 Enter Parameters t [STO]13 t

S[STO122

P03 Calculate P(k) - Pr(X < k) k>O [C] P(k)

[X<>Y] p(k)

P04 Calculate Q(k) = Pr(X > k) k>O [c] 0(k)

[X<>0Y p(k)

NOTE: See Note in bin

Negative Binomial Distribution

NBI Initialization for Negative Binomial [e) 0.0000

NB2 Enter Parameters r [STO]13 r

p [ST0122 p

NB3 Calculate P(k) - Pr(X < k) k>O [A] P(k)

p(k) - Pr(X - k) [X<>Y p(k)

NB4 Calculate Q(k) - Pr(X > k) k>O [a] Q(k)

p(k) - Pr(X - k) [X<>Y] p(k)

- NB5 Calculate P(k) a Pr(Y < k) k>r [J] P(k)

p(k) - Pr(Y a k) [X<>Y] p(k)

NOTE: See Note in bin;

Y - X+r - # Trials

17

|'% - W .f*.'$-. ~

ZP3-4 USER INSTRUCTIONS (HP),#-


Geometric Distribution

GI Initialization for Geometric [e] 0.0000

G2 Enter Parameter p [STO122 p

G3 Calculate P(k) - Pr(Y < k) k>1 [D) P(k)

p(k) - Pr(Y = k) [X<>Y] p(k)

G4 Calculate Q(k) - Pr(Y > k) k>1 [d] O(k)

[X<>Y] p(k)

NOTE: See note under bin.

E Display E(X) and V(X) (after any of [E] E(X)the foregoing routines) [X<>YJ V(X)

Register Contents:

00 Used 10 z 20 Used01 11 4 21 q

02 12 a2 22 p(X)

03 13 n(t,r) 23

04 14 24

05 15 25

06 Used (p(k)) 16 p(O) 26

07 Used (p(k)) 17 27

08 18 28a.

09 19 29

Assignments Labels Used

ZP3-4 I e 05 A a

.- 07 B b

08 C c

11 D d

13 E

14 JI 151819

20

29

18

...-- ."

EXAMPLES ZP3-4

(1) (Binomial model) The probability of hitting a target in a single

trial is 0.3. Suppos4 10 independent firings are made. Calculate

the probability of 3 hits, no more than 4 hits, at least 6 hits and

the mean and variance of the number of hits.

Solution (1), (3):


bin 1 [e] 0.0000 Initialize program.

bin 2 10 [STO]13 10.0000 Enter parameters.

.3 [STO]22, 0.3000

bin 3 3 [B] 0.6496 Display the CDF at 3.

[X>YJ 0.2668 Required probability p(3).

bin 3 4 [B] 0.8497 Repeating to find P(4).

bin 4 5 [b] 0.0473 Required Q(5) - Pr(X > 6)

E [El 3.000 Mean value np - 3.[X<>Y] 2.1000 Variance of X - npq.

(2) Poisson model) Telephone calls arrive at a switchboard at

the rate of 10 per hour. What is the probability of at most

3 calls in the next 20 minutes? Exactly 3? The mean number

of calls?

I Solution:

POI [e] 0.0000 Initialize program.

P02 0.3333 [STO]13 0.3333 Enter total time period 20 min.

10 [ST0122 10.0000 Enter rate X - 10 per hour.

P03 3 [C] 0.5730 P(3) - Pr(X < 3).

[X<>Y] 0.2202 p(3) - Pr(X - 3).

E [El 3.3333 Mean number of calls in 20 mins.

44,.

19

44/

W 7_7 7 76 "- -

(3) (Negative Binomial model) A fly fisherman estimates that his prob-

ability of catching a fish on a given cast of his rod is 0.05. He

decides to keep trying until he catches three fish. What is the

probability that he will need to cast at least 10 times and what is

the expected number of failures? What is the probability of 9

trials? The mean number of trials?

Z __Solution_:


NBI [e] 0.0000 Initialize program.

NB2 3 [STO]13 3.0000 Enter r parameter of 3.

.05 [ST0122 0.0500 Enter probability of-obtaining 1.

NB4 6 [a] 0.9916 Probability that the number

of failures is at least 7,0(6).

[X<>Yl 0.0026 Probability of exactly 6

failures.

E [El 57.0000 Mean number of failures.

NB5 9 [J] 0.0084 Probability of no morethan 9 trials.

[X<>Y] 0.0026 Probability of exactly 9trials.

*(4) Geometric model) An item has failure probability 0.005 and is

cycled until it fails. What is the expected number and standard

deviation of the number of cycles? What is the probability that

number exceeds 10?

Solution:

GI [a] 0.0000 Initialize program.

G2 .0005 [STO122 0.0005 Enter single parameter.

G3 10 [d] 0.9950 Displays CCDF at 10,Pr(Y > 10).

SE [El 2000.0000 Mean cycles to failure.

[X<>Y] 3,998,000 Variance.

[USER] 7 1999.4999 Standard deviation.

[e] 0.0000 Clears program.

20

.. A

Section 4.3: Normal DistributionThe user instructions for the HP version of ZP4 are practically identical

to those for TI given in the book. Label [J] is used for initialization inplace of RST; otherwise, pressing the same labels produces the same; results.Using the parameter choices 0 and I at step Ni in ZP4 replaces ML-14 in theTI master module everywhere the discussion refers to the latter starting onpage 109.

Section 4.4: Uniform Family; SamplingAs previously indicated, the serious departure from the TI format occurs

in the random number generator and consequently, both the instructions and theresults will differ from those published in the book. The departure begins Okpage 121. The random number generator adopted for the HP programs is one de-veloped by Don Malm for the HP-65 User's Library and is referred to on page 24of the HP-41C Standard Applications manual. The algorithm used is the simpleone.

r n+ 1 W FRC (9821*r n + .211327)

It allegedly will generate one million random numbers when a seed between 0(inclusive) and I is used. This random number generator is initialized bypressing [I] whereupon you are prompted for a seed which is then entered with[R/SJ instead of TI [E'. For some degree of uniformity with the TI illustra-tions, you may use a decimal point in front of each of the seeds given in thebook, such as .419 in Example 4.10 on page 121. Subroutine RNDMU, assigned tolabel [i, replaces [SBR] [D.MS] and outputs a random number from the unit in-terval. For this illustration, the output of the HP program is .2104 insteadof 0.65816 as listed, and the corresponding value of x will accordingly be15,589.

In example 4.11, ML-15 is used to generate normal deviates. Here, StepN6, programmed as label [G) of the Normal Distribution program in ZP4, may beused in its place. For the example, using a seed of .793, the output shouldbe 56.2958. (Of course, the parameters must be suitably stored by Step Ni tobegin with.)

Continuing on page 122, the subroutine [P+R] replaces the TI key [x],while [R P] replaces [INV] [x]. In Example 4.12, the sample values will be47,30,56,48 with a mean of 47.3 and a standard deviation of 10.4. The nextsuccessive values are 49,45,57,50,61 with a mean of 49.8 and a standard devia-tion of 8.5. In Example 4.15, the ten successive values will be 727,708,417,3401,326,213,1770,686,825,2783 with running counts checked in Register 06rather than 03. The mean will be 1147.9 rather than the published 1311. InExample 4.16, using a seed of .66, the successive values will be 0,2,4,6,2.

If you have been able to check these examples, then, while your answerswill differ from the published ones whenever random number generation iscalled in the problems that follow, you may rely on the results nevertheless.

21

4!

y , . -. . . .. - . .. . " , : .... ..... .. .". " - .. .". ' • ". . . . " " ' ' i

ZP4 (Assigned [J) USER INSTRUCTIONS (HP) SIZE 030

R. REG 01


Exponential Distribution

El Initialization [J] 0.0000

E2 Enter Parameter X [STO]22 x

E3 Compute P(x) and Q(x) x > 0 [El P(x)

NOTE: Xx must not exceed 228 [X<>Y] O(x)

" E4 Calculate 100(l-a)th Percentile a [e] x

NOTE: Repeat E3 and E4 at will

E5 Generate sample of size n

a. Initialize Random Number Generator [I] SEED?

b. Enter Seed (0 < Seed < 1) Seed [R/S] Seed

c. Execute Step E2

d. Generate x (Repeat n times) [B] x

Normal Distribution

NI Enter Parameters [STO]I1*' 2 2

a 2 [STO112 a

N2 Compute P(x) and Q(x) x [C) P(x)

[X<>Y] O(x)

N3 Compute Pr(x < X < x 2

a. Enter xI xI [D] P(x1)

b. Enter x2 and compute. x2 [R/S] Pr(x 1<X<x 2)[X<>Y Pr(x )

+ Pr(X x 2 )

N4 Calculate Standard lO0(1-a)th a (c] zPercentile

NB5 Calculate General 100(l-a)th 01 [d] x

NOTE: Repeat N2-N5 as often asdesired

N6 Generate sample of size n

a. Initialize Random Number [I] SEED?Generator


c. Execute Step NI

d. Generate x (Repeat n times) [G] x

22

ZP4 USER INSTRUCTIONS (HP)i-.


Uniform Distribution

U1 Initialization [J] 0.0000

U2 Enter Parameters a [STO113

b [ST0114

U3 Compute P(x) and Q(x) x [Al P(x)

[X<>Y 0(x)

U4 Compute 100(100(1-a)th Percentile a (SF105 xxx

[A] xa

U5 Generate sample of size n from

RX {el,e 2 ... ,eN corresponding

to LABELS 00, 01,...,K.

a. Initialize Random Number [I] SEED?Generator

b. Enter Seed (0 4 Seed < 1) Seed [R/S] Seed

c. Execute Step U2 with a - 0,

b - K

d. Generate Random Label R [a] R

e. Enter e-value corresponding x i [R/S] i

to R. Repeat d and e for

NOTE I. Summary stats stored in

R01 - 06

NOTE 2. To generate from-AA+,...,B}

*' execute steps a-d with

a- A, b - B

M For each of the above distribu- [b]

tions w and a2 may be recovered [X<Y]a

after computing any P(x).

23

* .,..

Register Contents:

01 Used 11 0 21

02 by 12 a22 X~

031 ZREG 13 z (z) 230j 4 bK) 24

*05' 15 b-a 25 Used

06 p(x) 16 26 Used

07 P(x) 17 27

08 P(x )-P(x) 18 2809 Seed 19 29


ZP4 J

ZCDF HI 03 A a

GEN-INI 1 07 B b

RNDMU i 09 C c

XBAR P.R 11 D d

SD ReP 12 E e

15 G

16

17

s.N

El 24

..-- Lo%:.- ~%~ a.* i-~ ~.- ;.*K-W. ~ , *

.1

EXAMPLES ZP4

(1) Time to failure, X, is exponential with failure rate 0.0001.

a. Determine the reliability at x - 100 and at x 500.0 0

b. What would the failure rate have to be to achieve a reliability

of 0.99 at 500 hours?

c. Calculate mean and median time to failure and the variance of X.-J

Solution (1), (3):

ZP STEP ENTER PRESS DISPLAY COMNENTS

* El [J] 0.0000 Initialize exponential subroutine.

E2 .O001 [ST0122 0.0001 Single parameter stored in R22.E3 100 [E] 0.0100 Displays P(100)

[X(>Y 0.9900 Displays 0(100), the reliabilityat 100.

E3 500 [El 0.0488 P(500) displayed.

[X<>YJ 0.9512 Q(500) - reliability at 500

E2 500 [STO122 500.0000 Treating 500 as X temporarilyfor computation in b.

E5 .99 [el 2.0101-05 Value of X - An(0.99)/500E2 .0001 [ST0122 0.0001 Restores true X in R for

the model. 22

M [b] 10,000.0000 Displays mean time to failure22

[X<Y] 100,000,000 Displays a2 - Ui for this model

E4 0.5 [e] 6931.0000 The median time to failure

E3 [El 0.5000 Verifies that P(6931) - 0.50.

(2) A standardized test is administered to incoming freshmen at a univer-

sity. Scores, X, are assumed to be normally distributed and, based

2on thousands of past scores, it is assumed that u - 100 and a - 245.

For an incoming freshman chosen at random what is the probability

that the test score will be:

a) greater than 110? b) less than 90? c) between 75 and 125? If

only the top 80% of incoming freshmen are to be admitted on the basis

of this test, what would the minimum passing score be?

25

.................. ,,;,:'.,','''. .;.. ",. '.'."-'-.:.vs>.;.-'.-.....-.'-.:'

Solutions:

ZP STEP ENTER PRESS DISPLAY COMENTS

(J] 0.0000 Clears exponential problem.

NI 100 [STOJI1 100.0000 Enter the mean value.

245 [STOI12 245.0000 Enter the second parameter a2

* N2 110 [C] 0.7385 Displays P(110) - Pr(X < 110).A-

[X<>Y] 0.2615 Displays 0(110), the required

probability

[RCL]IO 0.6389 Shows the standardized valuefor x - 110, namely,z = (110-100)//24F

N2 90 [C] 0.2615 Displays P(90)

N3a 75 [PD 0.9449 Displays 0(75), of minor

interest

N3b 125 JR/SI 0.8898 Calculates and displays

Pr(75 < X < 125)

[X<>Y] 0.1102 Displays Pr(X<75) + Pr(X>125).

N5 .80 [d) 86.8291 Displays the 20th percentile

for X so that Pr(X>87) - 0.80.

(3) The time a passenger must wait for a commuter flight on arrival at

an airport is a uniform random variable over an inteval from 0 to

30 minutes.

a. What is the probability that the passenger will have to waitat least 10 minutes for a flight?

' b. What waiting time corresponds to a 90% chance of catching aflight?

c. What is the probability that the passenger will wait betweenS.. . 10 and 20 minutes?

d. What is the mean waiting time? a 2 and a?

Solutions:

UI [J] 0.0000 Initialize program (clears all

previous work).

U2 0 [STO]13 0.0000 Enters first parameter a - 0in R1

30 [STO]14 30.0000 Enters second parameter b = 30in R4

U3 10 (A] 0.3333 Displays P(0).

[X<>YJ 0.6667 Displays the required Q(10).

U4 .90 [SF]05 0.9000 Signals calculator thatpercentile is coming.

.4 U4 [A] 3.0000 Displays x.90

[A] 0.1000 Verifies that P(3) - .10 sothat Q(3) = .9

26

% . '*.* t*..* .. *E

r. " , ,'.: .' .. < .'.. .-. : ...- ... ,.-.. .- ,-..........- .. ,.... . ' ... *. .- ",- ..- .. . .. . . . . . , . . .

t ' . *,-,

-I. ,L


U4 10 [ENTER] 10.0000 Enters difference between 10and 20 min.

30 [*] 0.3333 Calculates and displays-, Pr(10 < X < 20) - (20-10)/30.

M [b] 15.0000 Recalls and displays u - 15from R 1

[X<>Y 75.0000 Displays the variance

a 2 . 302/12

[USER][ r] 8.6603 Displays the value of a.

[J] 0.0000 Clears the program.

4

'p...U.V

27'0W

N67 ... 10..- ,,.-VIV- -7

Chapter 5. BIVARIATE DISTRIBUTIONS

The user instructions are practically identical to those given for theTI-59 so little has to be modified in this chapter. At Step 2 in the HP ver-sion a display of moments routine has been added which is effected by pressing[d] followed by successive presses of [R/S]. Of course, these characteristicsmay also be recalled manually from the respective registers just as instructedin the book.

As with ZP3-2, some modification of the routine for LABS is called forhere also. The HP instructions on the matter at Step 3 are reasonably clear.As a footnote, it is advised once more that if you will be involved in a lotof erasing of old algorithms, perhaps it would be advisable to assign the de-lete function DEL to an unused label, like [g] for a given session. Whenapplying LABS to various algorithms such as those found on page 142, naturallythey will have to be programmed in RPN here. It is assumed that the reader is

- already sufficiently familiar with the HP calculator that the translation forvarious examples can be made without additional instruction here. Consult the

a OWNERS HANDBOOK AND PROGRAMING GUIDE for any required assistance. As one ex-ample, the function g(x,y)-(x-I)(y-2) may be programmed at Step 3c as

RCL, 09, 1, -, RCL, 10, 2, -, *, RTN

Other cases can be handled in a similar fashion.

28

: ~... . ....... .. .............;','' ) A - " % : ' > ' ' ' 5 ' '.',2€,,., ',?i'',..'','':'" .- ,.:.-,.,'e'-".:..,28.'

ZP5 (Assigned [e]) USER INSTRUCTIONS (HP) SIZE 090


1. Distribution Entry

a. Initialize (e] 0.0000

b. Enter in order xi3 Yip P(xiY 1 ) xi [A](Repeat for each i through Y1 [B] iN < 19)

NOTE: p(xiy) should be positive. p(xi,yi) [C] i

2. a. Compile Distribution Characteristics [E] p

b. Display Characteristics (d] 11

[R/S 2x

[R/S] 14

[R/SI a 2y

V.. NOTE: To re-compile, enter N in [R/S] a%R3 after [e] R/SxY

3. Calculate E[g(X,Y)], V[g(X,Y)l

(after Step 1)

a. Initialize [GTOI[a] x. xxxx

NOTE: It is understood that[ALPHA] must be used for label a

b. Enter Program Mode [PRGMJ 147 LBL a

c. Key in g(x,y) with x c R09 ,y c R10

(Avoid labels already in use;

end with RTN)

d. Exit Program Mode [PRGM] x.xxxx

e. Calculate Moments [DI E[g(X,Y)

[X<>Y] V[g(X,Y)]

f. To ERASE Algorithm in [a], [SSTI 148 yy

complete Steps a, b; then ... [g] DEL(Let nnn be at least as large t

as the number of steps in (a].) nnn 147 LBL a(PRGM] x.Xxxx

4. tFor repeated applications, use ASN to assign DEL to g (%).

29

'':€ '-: "'% :, "-"- '':""'-: '.; : ': % , ".- '-; ".-.: ".: " '; --.; -M- ..... v ----;,: .: ,-.: ..-.- ,:

i ; , .1 1 .. ., , J . . . . . ., ,, .... ....... , , ,., .

Register Contents

00 Counter 10 last y 20 x

01 xp(x,y) 11 1A 21 y

02 yp(x,y) 12 a 2 22 p(x1 ,Y1 )x03 N 13 i 23 x2

04 x2p(xy) 14 2 24 y

05 y2p(xy) 15 a 25 p(x2,y2)06 xyp(x,y) 16 p 26

07 E -p j - l(g(x,y)p(x,y)) 17 Eg(XY)J 27

08 lastp(g2 (x,y)p(xy)) 18 V[g(X,Y)] 28

09 last x 19 Used 29


ZP5 e 01 A a

02 B d

03 C

a,.

D

E

*.3

.3

4

,

-P .T - .:oz ,t I), ' I. + .P •} + ,::, +- . ++ .- , - +. - .-- ;-.. .-.- ,. .:/. ,. ... _. .- ......... . ...... . , .- , .

t. 2 2EXAMPLES ZP5 (1) Calculate the moments ' a 2 y ay mad p for the Joint

distribution of Figure 5-2 duplicated below.

3 0 .2 0 0

2 0 0 .2 0

1 .1 .2 0 .3

y/x 1 2 3 4

Solutions:

ZP STEP ENTER PRESS DISPLAY COMENTS

Step la [e 0.0000 Initialize ZP-5.

Step lb I [A] 1.0000 First x-value for pair (1,1) entered.

" 1 [B] 1.0000 Corresponding y-value is entered.

.1 [C] 1.0000 Enter p(1,i) - .1; count of 1triplet displayed.

[A] 2.0000 Enter x-value of second pairselected, (2,1).

1 [B] 2.0000 Enter corresponding y-value.

.2 [Cl 2.0000 Enter p(2 ,1I); display shows 2triplets entered.

4 [A] 3.0000 Pass up cell (3,1) since p(3,1) - 0;enter next x - 4.

1 [B] 3.0000 Complete (4,1) entry.

.3 [C] 3.000 Enter p(4,1); record of 3 tripletsshown.

3 [A] 4.0000 Only positive entry in second row, x-3.

2 [B] 4.0000 Enter y-value for pair (3,2).

.2 [C] 4.0000 Enter p(3,2).

2 [A] 5.0000 Enter x - 2 for only positive entryin third row

3 [B] 5.0000 Enter y - 3

.2 [C] 5.0000 Complete entry with p(2,3) - 0.2

[RCL]07 1.0000 Check on data entry to see Zp(x,y) - 1

Step 2a [E] -0.2736 Displays the value of p after completecompilation and storage of moments.

Step 2b (d] 2.7000 Displays v2

[R/S] 1.0100 Displays a2OX

(R/S] 1.6000 Displays p

31

-[,S +,. i.. *~%

** .'. % *--t~ . - -

.. . . .y . .. . .. ..- .


- [R/S] 0.6400 Displays a¥2

[R/S] -0.2200 Displays Xy.

"R/S] -0.2736 Verifies again that p =-0.27.

(2) Calculate the mean and the variance for g(X,Y) XY.

Verify that a,, M -0.22

Solution:

Step 3a [GTO] GTO __ Preparing for entry to subroutine a

[ALPHA) GTO in order to program g(x,y) with

[a] GTO a- x e R0 9 , y e R10

[ALPHA] x.xxxx

Step 3b [PRGM] 147 LBL a Enters program mode

"RCL] 148 RCL

09 148 RCL 09

[RCL] 149 RCL

" 10 149 RCL 10

[x] 150 * Completes formula z - xy

N[RTN] 151 RTN Required return statement forsubroutine.

Step 3d [PRGM] x.xxxx Return to keyboard operation (ignoredisplay).

Step 3e [D] 4.1000 Calculates and displays E(XY).

[x<>yl 3.2900 Retrieves a2 from R0 7 to RX.

[x<>yJ 4.1000 Returns E(XY) to RX.

[RCLI RCL __- Prepares to subtract iX"Y to

11 2.7000 evaluate Eq. (5-4)

[RCLI13 1.6000 Recovers p and multiplies by px

" [x] 4.3200

- -0.2200 Calculation complete and s xy verified.

Step 3f [GTO] GTO Prepare to erase algorithm in a.

[ALPHA] GTO Sends pointer to subroutine a.

[a] GTO a

[ALPHA) x.xxxx

[PRMhJ 147 LBL a Enters program mode.

-elf [SSTJ 148 yyy Forward one step to beginning ofalgorithm.

[g] DEL Execute delete function.

010 147 LBL a Use 10 lines (more than enough)

[PRGM] x.xxxx Exit program mode. Return to calculatorcontrol.

[e] 0.0000 Erases program.

32

L ** % .

.-

STATISTICS BY CALCULATOR

Section 1.3: The CalculatorThis section is quite like that of ZP so that only remarks concerning the

statistics module need be added here. As mentioned in the introduction, theHP module STAT PAC will be needed for some of the ZS programs. In addition,the program ZSTAT, found in the appendix, will be needed for all of the ZSprograms starting in Chapter 4 since they contain the probability distribu-tions, among other things, that are missing in STAT PAC. Most of the applica-

*x tions of ZSTAT occur internally within ZS programs but, occasionally, some ofthe subroutines are called for individually. For that reason, suggestivealphanumeric labels have been included and the program has been assigned tolabel [SCI] to make it convenient to access from the keyboard.

Section 2.3: SimulationThe first departure from the TI format occurs on page 16 in the digression

for computing moments of discrete distributions. A subroutine called MU-SIGand assigned to label [J] has been inserted into program ZS-2 to replace theTI use of ST-03. As the reader can see from the User Instructions thatfollow, the pairs are entered in opposite (but more natural) order with xfirst, followed by p . Instead of a running count of the number of pairsbeing displayed at the end of each entry, the cumulated probabilities areshown; thus, the number 1 should be seen at the conclusion of all entries. Apress of [ii will then output the mean, and sigma will be found in theY-register. (It should be noted throughout that, as with ZP, the HPY-register replaces the TI T-register always).

Of course, the random number generator output will differ here, just aswas the case in ZP. The same 'HP user instructions apply here, however. Thus,the generator is initialized by pressing [I] as before and you are promptedfor a seed. The subroutine RNDMU, assigned to [H], will replace the TI [D.MS]routine to output a number between 0 and 1. If you will use a seed of .49instead of 49 in the example treated on page 18, the HP output will be .5014,with a second application yielding .2349. A second program, called RNDMAB(assigned to [hi) replaces Steps 4-6 of ST-02 to output a (continuous) randomnumber between A and B , provided A and B are stored in registers 13and 14, respectively. For the example, again on page 18, using A-10 and B-67,the respective values will be 16.0050, 59.2222, 16.6282 and 24.0426. Finally,the subroutine RNDMI, assigned to [g], will generate random labels. On page 19using a seed of .21, successive presses of (g] will produce labels 45, 53, 11and 20. That will take care of the problems for this section. The answerswill differ from those published of course. Be sure to press [J] when youwish to return to the main programs in ZS-2.

MSection 2.4: Simulating Continuous DistributionsIn Example 2.3, if a seed of .635 is used, the successive values of u

are: .5464, .1799, .9504, .6085, .7613, yielding x values of 791, 198,3004, 938 and 1435, respectively. The program instructions at Step E5 shouldbe modified according to the ones provided here.

Program ST-19 may be replaced entirely by using the N routine in ZS-2

with u - 0 and o - 1. (For that matter, P(z) may be found here by entering zand pressing (XEQ] 19, to mimic the TI program). Alternatively, programZNORMD in STAT PAC may be used to calculate Q(z). Try z = 2.695 as on page 24

33

- €;' ' "J""'"¢"¢" "-" " :"- '" -- " "..-" " "" -v- "°" "" -""' ""';' '-i.2 i..;-'''- ,-.'. ,'..<.< .. '.".-

to see that .9964 is the value of P(z). The value Q(z) -. 0036 will then befound in the Y-register. Generating random samples from both the exponentialand normal distributions has been automated in ZS-2 just as in the TI case andexamples follow the user instructions. *No further checks will be given here.

Section 25: Bernoulli Trials* As with ST-19, we have mimicked the TI binomial program ST-20 as subrou-

tine 20 here. The instructions are given under the code BIN in ZS-2 and thatprogram may be used to check all of the problems of this section. It might benoted that the standard deviation is found in the Y-register, pressing [X(>Y]after [a], rather than a separate label [B'], as with TI.

° .

4,.4

• o..%. ... Seto .5 enuliTil

" ." swt4T1,weh. iikdte Ibnma.rgamS-.a uru

4,,'-tn 0hr.Teintutosaegvnudrth oeBNi S2adta

prga-a eusdt4hc.4lo h rbem fti etin tmgtb

4.j oe httesadrddvaini on nte -eltr rsig[<Y

-. __. <34

ZS-2 (Assigned [J]) USER INSTRUCTIONS (HP) SIZE > 030

Z REG 01


E Exponential Distribution

I. Initialization [J] 0.0000

2. Enter Parameter X [ST0116

3. Compute P(x) and Q(x) x [E] P(x)

Note: Xx must not exceed 228 [X<>YJ Q(x)

4. Calculate 100(1-a)th Percentile a [e] xa

Note: Repeat E3 and E4 at will.

5. Generate sample of size n


. b. Enter Seed (0 < Seed < 1) Seed [R/SI Seed

c. Execute Step E2

d. Generate x (Repeat n times) [B] x

N Normal Distribution

1. Initialization [J] 0.0000

2. Enter Parameters [STO117 va [STO118

3. Compute P(x) and 0(x) x [C] P(x)[X<>Y] 0(x)

4. Compute Pr(x1 < X < x2) or 1-P(x 1 < X < x2 )

a. Enter x [D] O(x )

b. Enter x2 and compute x2 [R/S] Pr(xl<X<x2)

[X<>Y] Pr(X<x I ) +

Pr(X>x2 )

5. Calculate Standard 100(1-a)th Percentile a [c] z

6. Calculate General 100(1-i)th Percentile a [d] x

Note: Repeat N3-N6 as often as desired

7. Generate sample of size n



c. Execute Step NINP d. Generate x (Repeat n times) [b) x

8. Standard Normal (TI ST-19) z [XEO]19 P(z)

[X<>Y] 0(z)

35

ZS-2 USER INSTRUCTIONS 2.


BIN Binomial Distribution (TI ST-20)

1. Initialize [XEO]20 PMTERS?

2. Enter Parameters n [R/S] n

p [R/SI 0.0000

3. Calculate probabilities k [A] p(k)

[R+] P(k)

fR+J Q(k)

KU-SIG Discrete Moments

1. Initialize [ji ]BSTG

N 2. Enter discrete pairs x i [ENTER] xi(Repeat i-l,...,N) [A]

3. Calculate p and a . [i]

[x<>Yl a

MOM Recall Moments [a]

[X<>Y]

Register Contents

00 Used 10 K(label) 20 z (x-p)/a

01 11 21 n

02 Used 12 22 p

03 by 13 A 23 i-p

0 Z04 14 B 24 Used

05 15 25 Used

06 16 X 26

07 P(x) 17 U 27

08 P(x1 ) - P(x2 ) 18 a 28

09 Seed 19 z 29

36

'le ""eY-" ."- . ". - ."- .* . VA- . ...

Assignments Labels Used-4

ZS-2 J 03 A a

GEN-INI I 06 B b

RNDKU H 07 C c

RNDHAB h 08 D d

RNDMI g 09 E e

BSTG j 11

MU-SIG i 12

XBAR P*R 19

SD R+P 20

RD R+

'. .J.

"

D--

•

.. S.

Fi

-" 37

: .4.

EXAMPLES ZS-2.

1. Let X have an exponential distribution with parameter X - 0.001

and suppose X measures time to failure in hours.

(a) Calculate the probability that time to failure will exceed

1200 hours.

(b) Compare the mean time to failure with the median time to failure.

(c) How many hours may we reasonably depend upon for survival of 90%

of such items?

(d) Generate a random sample of five times to failure.

SOLUTIONS:


El [J] 0.0000 Initialize the exponential subroutine

E2 0.001 [STO]16 0.0010 Single parameter X stored in R

" E3 1200 [E] 0.6988 Displays P(1200) - Pr(X 4 1200).

[X<>Yj 0.3012 Displays Q(1200) - Pr(X > 1200)

which is the answer to (a).

E4 .50 [e] 693.1472 Calculates and displays the medianX.50 (in hours).

[RCL]17 1000.0000 Recall V, the mean time to failure.This answers (b).

E4 .10 [el 105.3605 Displays x.90

E5a [I] SEED? Initialize the random no. generator

E5b .635 [R/S] 0.6350 Enter Seed - 635 for illustrative

purposes.

E5c (0.001) ([STOI16) (0.001) Enter the parameter X if notalready entered.

E5d [B] 790 Displays the first generated sample

value, x1 (rounded).

[B] 198 The second simulated time to failure.

[B] 3003 Successive times to failure

[B] 938 (rounded to whole hours)

[B] 1432 ifor a random sample of size 5.

4 O

38

.fA

9 , ¢ , ' ° ", " - . * . o, . - . - - - -. - - - o - . . .. ..

ZS STEP ENTER PRESS DISPLAY COMMENTS

(2) A standardized test is administered to incoming freshmen at a

university. Scores, X, are assumed to be normally distributed

and, based on thousands of past scores, it is assumed that I - 100

and a - 16. For an incoming freshman chosen at random what is the

probability that the test score will be:

a) greater than 110? b) less than 90? c) between 75 and 125?

If only the top 80% of incoming freshmen are to be admitted on the

,... basis of this test, what would the minimum passing score be?

Solutions:

Ni [J] 0.0000 Initialize.

N2 100 [STOI17 100.0000 Enter the mean value.

16 [STO]I8 16.0000 Enter the second parameter a.

N3 110 [C) 0.7340 Displays P(110) - Pr(X 4 110).

[X<>YI 0.2660 Displays Q(110), the requiredprobability.

[RCLI20 0.6250 Shows the standardized value for

x - 110, namely, z - (110-100)/16.

N3 90 [C] 0.2660 Displays P(90).

N4a 75 [D] 0.9409 Displays Q(75), of minor interest.

N4b 125 [R/S] 0.8818 Calculates and displays Pr(75<X<125).

[X<>Y] 0.1182 Displays Pr(X<75) + Pr(X>125).

N6 .80 [d] 86.5367 Displays the 20th percentile for

X, so that Pr(X>87) - 0.80.

(3) Generate a random sample of size 5 from a normal distribution

N7a [I] SEED? Initialize random number generator.N7b .198 [R/S] 0.1980 Enter Seed - 198 for illustrative

purposes

N7c 50 [ST0117 50.0000 Enter the normal parameters

10 [ST0118 10.0000 and store in appropriate registers.

N7d [b] 42.63 Displays first generated sample

value x (rounded).

[b] 56.34 Successive sample values

[b] 63.21 are generated and

[b] 72.72 displayed (rounded).

(b] 46.84

39

ZS STEP ENTER PRESS DISPLAY COMMENTS3 3*

(4) Find the mean and standard deviation of the discrete probability

distribution.

x: 4.0 3.0 2.0 1.0 0.0

-- p(x): 0.13 0.21 0.43 0.14 0.09

SOLUTION:

MU-SIG 1. [i] ZBSTG Initialize module to start program.

MU-SIG 2. 4 [ENTERI 4.00 Enter distribution as data

.13 [A] 0.13 pairs(xi.Pi) i - 1,2,...,n.

3 [ENTER] 3.00 See Epi accumulated in RX with

.21 [A] 0.34 1 indicating final data entry.

2 [ENTER] 2.00.43 [A] 0.77

1 [ENTER] 1.00

.14 [A] 0.91

0 [ENTER] 0.00

.09 [A] 1.00

MU-SIG 3. [i] 2.1500 Displays p.

[X<>Y] 1.0989 Displays a from Ry.

(5) Example 2-7.

SOLUTION:

BIN 1. [XEQ]20 PMTERS? Initialize binomial program.

Prompt is for n and p.

BIN 2. 4 [R/SI 4.0000

.51 [R/S] 0.0000 Parameter entry complete.

BIN 3. 0 [A] 0.0576 Display P - p(O) - Pr(Y=0).

2 [A] 0.3747 Display is p(2) so P(2) is found

[R+] 0.6724 in R.

[R+] 0.3267 Q(2) is found in Rz

I [A] 0.2400 p(l) is displayed.

MOM [a] 2.0400 Displays p -np

[X<>YI 0.9998 Displays a - npq

[USER][x 2 0.9996 Calculates a2

40

pb ' m *" ** .P* .. .

, . ( -- . ..-. ' w ' *- , . - . . - .. . .

Chapter 3 Data Processing

This chapter is rather independent of the others and, as the name suggests,deals with the processing of numerical data to produce traditional statisticalsummaries as well as grouping data into different patterns. Three programshave been created for this purpose, ZS-3 and two separate ones that are revi-sions of corresponding TI programs ST-03, ST-07 and ST-09. The latter werecreated and so named in order to follow the textbook material with the leastamount of revision of instructions. The three programs should be loaded simul-taneously for solving the problems here. Since some partitioning (using theSIZE function) may be called for, it is advisable that all other programs becleared from calculator memory. The labels to which the programs have beenassigned make it very convenient to move from one to the other when necessary.

Section 3.1: Sample CharacteristicsPicking up the discussion on page 41, the HP, like the TI, is hard wired to

compute means and standard deviations when data are entered on the keyboardwith the Z + key. Consult the Owner's Handbook for details. The basic dif-ferences are that (be sure you are not in USER mode) the registers are clearedwith the CLE key rather than using Pgm 01 and you execute the functions MEANand SDEV instead of (1 and (INVI[i], respectively. Even so, the TI programST-03, here assigned to (I1, will allow for data storage as it does in the TImodule. You see from the User Instructions that follow, you must initialize bypressing [e) and then enter the data one-by-one using label [A]. At the con-clusion you will find the data stored beginning in register 31. In addition,you may press [P RJ in place of TI (2J and use [R+P] instead of TI [INV][].To find the range, press [J] to enter program ZS-03 and then press [C] as perthe instructions for that program. (Do not forget to press [I again if youwish to return to ST-03 for any reason.) The remarks regarding repartioningmay be easily transferred to appropriate remarks using the SIZE function forthe HP. When data have been entered using program ST-03, you may find MSD bypressing (ENG] (the key that the subroutine MSD has been assigned to). MAD iscomputed by pressing [J] to enter ZS-03 and then press (B]. In this way, theseinstructions practically follow those of the TI to the letter. Verify the so-lution on page 47 for Example 3.1 following these instructions.

Section 3.2: Grouping DataData are grouped and recovered in cells suitable for histogram construction

by means of program ST-07/9 (assigned to label [i]), a program resembling thecorresponding TI programs ST-07 and ST-09 discussed in the book. The same re-marks regarding conventions and parameter limitations discussed on pages 52 and53 apply here as well.

After pressing [il to enter the program, you initialize with [e] just aswith the TI program, only here you will be prompted for the number of cells.When you enter that number with a press of [R/SI you will then be prompted forlowest class limit XMIN and, after entering that, for the width, w , of eachcell. These instructions conform to the TI instructions. At this point youhave two options. If data have been entered previously, either with programST-03 or with ST-07 itself, you have merely to press [d] whereupon you areprompted for the sample size n . Entering this number and pressing [R/S]causes the program to automatically group the data into cells as per the entry

41

It

. - .- .. .. - -. - - - . -..- '

in steps P1,2,3. Otherwise, you enter the data one-by-one using [A] just aswith ST-03. Once the data have been entered, the histogram is constructed bythe steps under code H. After initializing with [E], the successive cell fre-quencies and boundaries are displayed with a STOP at the end tosignal comple-tion of the display. This replaces the discussion on pages 53 and 54 of thetext.

As for computing grouped moments, the version of ST-03 presented here isinitialized the same way ([e]), and pairs are entered as discussed under code G(same as the TI entry). Moments are then displayed in the X-register when XBAR([P R]), SD(ReP]) and MSD ([ENG]) are used. You may then proceed to ZS-3 to findMAD and the range as discussed on page 55. The last two paragraphs on thatpage may be safely ignored.

Section 3.3: TransformationsStep 5 of ZS-3 presented here allows for data transformations just as with

the TI version. As with ZP programs, it may be advisable now and then to erasesome of the algorithms used in [a] to create transformations if many applica-tions happen to be used. Again, the DEL function will have to be used and thisshould be assigned to [g) if many such erasures will be taking place. You mayalso have to repartition your calculator with the SIZE function if there is noroom for the data. For the small data sets illustrated here, that situation is

* not likely to arise. The answers to the problems given at the end of the sec-tion may all be verified with the program instructions on the following page.

Section 3.4: The Central Limit TheoremThe program ZNORMD in STAT PAC will have to be used in this section in

place of ST-19, or, as remarked on page 63, you may use ZS-2 with the cautionmentioned there. Since there is no binomial program in STAT PAC, the lattermight be the advisable thing to do for resolving some of the problems in thissection.

44

'.

', . 42

'C.

'/ , •• m w="° % ° o

" ,"J"• " • a " ...

7.

ZS-3 (Assigned [J]) USER INSTRUCTIONS (HP) SIZE 090 1.

Z REG 01

* STEP PROCEDURE ENTER PRESS DISPLAY

1. Calculate the range of a sample whenraw data have been entered using ST-03 [C] R

2. Calculate the range of a sample whendata are grouped and have been enteredusing ST-03 (w - cell width). w [c] R

3. Compute MAD when ungrouped data have" been entered using ST-03. [B] MAD

4. Compute MAD when grouped data havebeen entered using ST-03 [b] MAD

5. Transform data by the transformationx' - f(x):a. Initialization. [e] 0.0000

b. Enter program mode [GTO] [a] x.xxxx

[PRGM] 160 LBL a

c. Enter f(x) using parentheses wherenecessary and always end with[INV][SBRJ. Exit program mode. [PRGM] x.xxxx

d. (1) Keyboard Entry (repeat foreach i). xi (A] i.0000

(2) Original Data Stored by ST-03(n - sample size). n [El n.0000

NOTE 1: Steps I and 3 apply followingStep 5d.

NOTE 2: To ERASE Algorithm in [a], [SST] 161 yycomplete Step b; then... [XEO][DEL]t [g] DEL(Let nnn be at least as large nun 161 LBL aas the number of steps in [a]) [PRGM] x.xxxx

6. Recall transformed data. [d] 0.0000[DI x'

NOTE: May be repeated at any time. [D] x' 2

[D] X'3

7. Clear Step 6 [CFOI x.xxxx

For repeated applications of Step 5, use ASN to assign the function DEL to label g(%)before executing this step. (DEL is not programmable and cannot be preserved in usermode by the card reader.)

43

' ' %,; %, . .. "-.-."4,," ." .- --.. . . . . . .- - -. . . . .. " * , - ,q . .' %. % "% - ,% .. "*-o - -.-. ' • ' '. ',. . .. - --. ,. -

ST-03 (Assigned (I]) USER INSTRUCTIONS (HP) SIZE 060-089

Z REG 01


I Initialization [el 0.0000'S _ __ ___- _ _

U Ungrouped Data Entry xi (Al i.0000Repeat i - 1,2,...,n.

G Grouped Data Entry f [B] f

'I Repeat i - 1,2,...,n. xA] i.0000IiMOM 1. Calculate sample mean [P+R] x

and sample Standard Deviation [R+P]

2. Calculate MSD [ENG] MSD

REGISTER CONTENTS (Grouped data in parentheses)

00 Used 10 1(f ) 20 30 Pointer

01 x(zfx) 11 w 21 31 x (x )

02 Ex ( fx 2 ) 12 Xmin 22 32 x2(f1)

03 13 Xmax 23 33 x 3 (x 2 )

04 14 Used 24 34 x 4 (f 2 )

05 15 25 35

06 n(Zf1 ) 16 26 36 .

07 Zjxi-XI 17 27 37

08 Used 18 Used 28 38

09 Lastx 19 xcount 29 39


ZS-3 J ZS-3 ST-03 ST-07/9

ST-03 I 01 A a 01 A e 01 A c

ST-07/9 i 02 B b 02 B 02 E d

XBAR P.R 03 C c 03 03 e

SD R+P 04 D d 04

MSD ENG 05 E e

12

13

44

av*

ST-07/9 (Assigned [i]) USER INSTRUCTIONS SIZE 060-089Z REG 01


I Initialization [i] 0.0000

P Enter Parameters [el CELLS?

1. Enter number of cells (< 15) Cells [R/S] XMIN?

2. Enter Lowest Class Limit xmin [R/SI W = ?

3. Enter Interval Width w [R/S] 0.0000

DE Data Entry and Compilation

1. Original Data (Repeat i 1,2,...,n) xi [A] i.0000OR:

2. If Data Are Previously Stored [d] N - ?

n JR/S] n.0000

H Histogram Construction (after DE)

1. Initialization [E] 0.0000

2. Display Cell Frequency [c] f

3. Display Upper Limit, Bi, of Interval [R/S] Bi

(Repeat i - 1,2,...,Cells)

STOP

Note: Ungrouped moments after DE may be computed by XBAR, SD and MSD in ZS-3.Corresponding grouped moments are then found in R% if Histogram has

been constructed. In either case, you must press [i] again to returnto ST-07/9.

REGISTER CONTENTS

00 Used 10 Used 20 f7 30 Pointer

01 Zx 11 w 21 f 31 xi2

02 Zx 12 xmin 22 f9 32 x

03 Zfixi 13 Xmax 23 f10 33 x

04 Zfixi 14 f1 24 f 11 34

05 Used 15 f2 25 f12 35 :

06 n 16 f3 26 f 36

07 Used 17 f4 27 f14 37

08 Used 18 f5 28 f15 38

09 CELLS 19 f6 29 xcount 39

45

* * *.. . .- .- ... .5... - .* . * * * - * ' ' * *, . . ..,.......,:,. .. ..-. > --..... -.... , . p *,, - . , ... ..-- ... -. -a.. -. - h... . -. S , --

EXAMPLES ZS-3

I. For the ungrouped data below, calculate 5, s, MSD, MAD and R. Then trans-

form the data by x' - I/x and calculate the same statistics for thetransformed data.

" 5, 10, 6, 4, 3, 8, 12

Recall the actual values of the first three data points.

.... ZS STEP ENTER PRESS DISPLAY COMMENTS

ST-03 [I] 0.0000 Select ST-03

- . I [e] 0.0000 Initialize for data entry.

U 5 [A] 1.0000 Enter data.

10 [A] 2.0000

6 [A] 3.0000

4 [A] 4.0000

3 [A] 5.0000

8 [A] 6.0000,..,12 [A] 7.0000 Data Entry complete.

[P*R] 6.8571 The value of the sample mean.

[R.P] 3.2878 The value of the sample standard

" deviation.

[ENG] 9.2653 Value of MSD.

ZS-3 [J] 9.2653 Enter program ZS-03.

".".3. [B] 2.6939 MAD calculated and displayed.

1. [C] 9.0000 The range R - 9

- 5a. [e] 0.0000 Initialize ZS-3 for data trans-formation.

5b. [GTOI[a][PRGM] 160 LBLa Preparation for transformation.

- 5c. [I/xl 161 1/x Simple algorithm.

[RTN] 162 RTN Necessary return instruction.

[PRGMJ x.xxxx Exit program mode for ZS-3

operation.

5d.(2) 7 [El 7.0000 Data automatically transformed andstored in R3 1, R3 2,

(P*Rj 0.1798 Value of K' rounded.

[R P] 0.0892 Rounded value of s'.

[ENG] 0.0068 Rounded value of MSD'.

3. [B] 0.0697 Rounded value of MAD'.

1 1. [C] 0.2500 Value of R', the new range.

.9 6. [d] 0.0000 Initialize to recall transformed-N data.

46

!;,-

-" . .' " '- - -" ' ' .. - - .'. " '-. "--*. "J . ". . " " " , . . . . .*..**. , * ,- . - . ,t - , -;


[D] 0.2000 Recall value of x' 1 = 1/x

[D] 0.10000 Recall value of x'2 = I/x2.

[D] 0.1667 Recall value of x'3 = 1/x3.

[CF]01 0.1667 Clear display program

2. For the grouped data below, calculate R, s, MSD, MAD and the range.

Frequency: 3 4 9 4 5

ClassInterval: 0-10 10-20 20-30 30-40 40-50

SOLUTION:

ST-03 [I] 0.0000 Select program ST-03.

I Eel 0.0000 Initialize ST-03 for data entry.

G 3 [B] 3.0000 Enter first frequency.

5 [A] 1.0000 Enter first midpoint; running count

4 (B] 4.0000 displayed.

15 [A] 2.0000 Repeat for each pair.

9 [BI 9.0000

25 [A] 3.0000

*... 4 [B] 4.0000

35 [A] 4.0000

5 JBJ 5.0000

45 [A] 5.0000 Data entry concluded.[P R] 26.6000 Grouped mean value X.

[R PI 12.8062 Rounded value of s.

[ENG] 157.4400 Value of MSD.

ZS-3 [J] 157.4400 Enter Program ZS-3

4. [b] 10.0480 Value of MAD.

2. 10 [ci 50.0000 Value of grouped range R-based on

a class width of 10.

3. Group the following data into a histogram consisting of 6 cells of widthw - 10 starting at x -mn 70.

120 86 87 75 100 120 100 80110 105 95 90 100 85 95 85

Calculate: R, s, MSD for both grouped and ungrouped data.

SOLUTION:

ST-07/9 [i] 0.0000 Select program ST-07-9

I [e] CELLS? Initialize for parameter entry.

PI 6 [R/S] XMIN? Enter total number of cells.

P2 70 [R/S] W = ? Enter x , lowest data limit.

47

7..,


P3 10 [R/S] 0.0000 Enter cell width.

DE 120 [A] 1.0000 Enter first data value

86 [A] 2.0000 Enter second data value

85 [A) 16.0000 Enter last data value

Hi [El 0.0000 Initialize histogram display.

H2 [c] I First cell frequency% H3 [R/SI 80.0000 B so Cell 1 runs from 70 to 80.

12 [c] 5 Second cell frequency

H3 [R/SJ 90.0000 B2 establishing interval 80 to 90.

H2 [c] 3 Third cell frequency

H3 [R/S] 100.0000 Third cell upper limit.

H2 [c] 4 Fourth cell frequency

113 [R/S] 110.0000 Fourth cell boundary.

H2 [c] I Fifth frequency for cell

H3 [R/S] 120.0000 running from 110 to 120

H2 [c] 2 Sixth frequency for last cell.

H3 [R/SI 130.0000 Upper bound on all data (notincluded as a possible value)

[c] STOP Indicates conclusion of program.

[P+R] 95.8125 R for ungrouped data

[X<>Y] 98.1250 R for grouped data

[ReP] 13.2525 s for ungrouped data

[X<>Y] 14.9304 s for grouped data

[ENG] 164.6523 MSD for ungrouped data

[X<>Y] 208.9844 MSD for grouped data

[i 0.0000 Ensures return to ST-07/9

.48

48

Ao,.

Chapter 4 Estimation

Chapter 5 Hypothesis TestingThe problems in both of these chapters are covered by a single program,

called ZS-4/5. This was one of the more successful translations from TI to HPso that very little needs to be added in the way of remarks. As the readerwill see from the User Instructions that follow, the directions are practi-cally identical to those published in the text. One small difference is thatraw data will not be entered by ST-03, but rather by a self-contained dataentry scheme (DE) which is much simpler and covers all of the cases treated.Naturally, any TI reference to the T-register should be translated to the HPY-register, and the display register, RD referred to so often, becomes the HP

X-register. Another important point that is universally true of thedifference between the two calculators is that R06 is used by the HP routines

V- for storing sample sizes while TI used R0 3* That change should be noted

throughout the instructions that follow.

As previously remarked, the program ZSTAT should be loaded into programmemory for all of the ZS programs from this point on in the text. It will beconvenient to assign ZSTAT to a label, say [SCI], for easy access to the pro-grams that are referred to occasionally in these chapters.

On page 85 reference is made to the formula for the t-density in ASM. Itis really not particularly instructive for the applications presented here toactually see the formula but it may be found in most standard textbooks, anda picture of the typical density is shown on page 103. In any case the valueof the CDF P(t) may be found by storing degrees of freedom, v , in R 15 enter-

ing t and then [XE0][TFI in ZSTAT. On page 86 it should be noted that thesubroutine ZA in ZSTAT replaces the subroutine [sin] in TI. (See also the Notein the User Instructions that follow.)

One of the few distributions provided by STAT PAC is the Chi-square, re-fferred to on page 92. This distribution is labeled ZCHISOD and is discussedon page 70 of the STAT PAC handbook. It may also be found as the subroutine[CHISD] in ZSTAT (requiring, again, only that degrees of freedom be store in

-I R15 ). Either replaces references to (C] in TI ST-21. A typical Chi-square

density is depicted in the legend to Table C on page 104, where percentilesare located. It should be observed that the footnote regarding large degreesof freedom applies verbatim to the HP program ZS-4/5.

That takes care of all of the differences in these two chapters. Follow-ing the User Instructions on the next three pages will be found the typicalmodel problems for verifying program output.

w49

!7.*_444**. . . . . .49

4, . * *S*%" . . . 4

ZS-4/5 Assigned [JI) USER INSTRUCTIONS (HP) SIZE 050

E REG 01


DE ORIGINAL DATA ENTRY

1. Enter Data

a. Initialize [J] DATA?

b. xi Repeat i - 1,2,...,n [R/S] i.0000

2. Process Data for Storage [d] 0.0000

N(w) NORMAL MEAN - a UNKNOWN

1. Enter Data using DE OR:

a. Enter Sample Size n [STO]06 n

b. Enter Sample Mean 5 [STO137

c. Enter Sample Standard Deviation s [STO]38 s

2. Test H0 i 0

a. Enter H -code* H -code [a] H -code

b. Enter P and Compute P-value 10 [R/S] P

3. CI for U

a. Calculate Degrees of Freedom [A] V

b. Enter ta/2 with d.f. - v tca/2 [R/S] 9.

and calculate limits [X<>Y] u

NOTE: For One-sided intervals, enter t at Step 3b and ignore L or uas the case may be.

N(pja) NORMAL MEAN - a KNOWN

1. Enter Data Using DE OR:

a. Enter Sample Size n [STO106 n

b. Enter Sample Mean 3 [STO]37 R

3. Test H0 0

a. Enter Hi-Code H1 -code [bi H1 -code

b. Enter p0 and Compute P-value U0 [R/S] P

4. Calculate 100(1-a)% CI for u a/2 [B] 9

[X<>YI u

Note: Enter a for one-sided intervals and ignore X or u as the case may be.

@11•1 fH:e oe In ZSTAT (assigned [SCI]),

H -code - 0 if H : 8 * 8 [XEOI[ZA] displays z if P is in R

-1 if H1 : < 9P

77 50

:' ', 2"~~~f- j" '"" "" "" P. """" .

,,, - . - - 7 1 7 •

ZS-4/5 USER INSTRUCTIONS


2N( a) NORMAL VARIANCE

.. Enter Data Using DE OR:


b. Enter Sample Standard Deviation s [STO]38 s

2 2

a. Enter H -code Hi-code c] H1 -code

b. Enter aI and Compute P-value a2 [R/S] P

3. CI for 2

a. Calculate Degrees of Freedom [C] vb. Enter Chi-square Percentiles x2 [STO4i X

2 2

(v-n-1) and Calculate Limits 2 [STO131 X2

(R/S]

[X<>Y u

NOTE: For Upper One-sided Itra, enter XA_ in R4 1 and in R'31 and see u

2

displayed. For Lower One-sided interval, enter X in R3 1 and R see

see I displayed.

Exp(p) Exponential Mean

1. Enter data using DE or:


b. Enter Sample Mean R [STO137

2. Test H 0 : 0

a. Enter H -code H -code [e] Hi-code

b. Enter p 0 and Compute P-value 10 [R/S] P

3. CI for

a. Calculate Degrees of Freedom (El V2 2

b. Enter Chi-square Percentile v-2n 2 STO41 Xl_c/2

2 2and Calculate Limits x2 [STOI31 2

NOTE: SeePreviousx<>Y] NtfrnSd IevNOTE-: See Previous Note for One-Sided Intervals.

..

*. .. . . _ ... .

REGISTER CONTENTS

00 10 20 30 ts 40 801 E ii 21 31 ta/ 2 '(/ 2 41 2

Xl-a/2

02 Exi2 12 22 32 SE 42

03 13 23 33 Used 43

04 14 Used 24 1 34 60 44

05 Ue 15 v 25 35 45

06 n 16 26 Used 36 46

07 17 27 37 x 47

08 18 28 H-COde 38 s 48

09 19 29 P(ts) 39 e(O±e) 49 Used

SAssinments Labels Used

ZS-4/5 IJ 01 A a

02 B b

03 C c

04 D d

05 E e

06

* 07

52

4. , - , o . . . O . o O . . . .4,, ... . - : . . , . .. . - _ , . , , . , , . . .. . . . . . , .. .Z

7 7d TO - --- -7

EXAMPLES ZS-4/5

(1) To study the effects of a drug, nine athletes were timed in a series of

physical tests and yielded an average of R - 10.13 minutes. It was assumed

in the study that a - 1 and that reaction times are normally distributed.

.2 a. Find a 90% CI for the mean reaction time p.

b. Deterriine a 99% lower one-sided interval for W.

c. Find a one-sided upper bound on p having risk 15%.

(2) Four specimens of an expensive cloth were subjected to strength

tests and the breaking strengths in lbs./sq. in. were recorded

as 181, 173, 176, 175. The standard deviation based on past

experience is 5 lbs./sq in. Assume normality.

a. Find a 95% CI for u, the mean breaking strength.

b. What is a lower one-sided bound for U with contidence 90%?

Solution (1):

ZP STEP ENTER PRESS DISPLAY CONMENTS

N(Il) Ia. 9 [STO106 9.0000 Places the sample size in R06'

lb. 10.13 [ST037 10.1300 Stores the sample average in R 3 7.

2. 1 (STO]48 1.000 Stores known a-value in R48'

4. .05 [B] 9.5816 Enter a/2 - .10/2; display £.

[X0Y 10.6784 Exchange and display u.90% Ct for P is (9.58, 10.68).

4. .01 [B] 9.3544 Enter a - .01 and find 1 9.35

so confidence is 99% that > 9.35solving (b) (Ry is not examined)

4. .15 [B 9.78 Using a - .15, 2 is calculated butignored.

[X<>Y] 10.4755 The Y-register yields required

upper limit on u, solving (c).

Solution (2):

DE 1. [J] DATA?

[R/S] 0.0000 Initialize ZS-4 for data entry.

181 [R/S] 1.0000 First breaking strength entered.

173 [R/S] 2.0000 Second breaking strength entered.

176 [R/SI 3.0000 Third breaking strength entered.

175 [R/S] 4.0000 Fourth breaking strength entered.

DE 2. [d] 0.0000 Data processed.

N(pla) 2. 5 [ST048 5.0000 Stores known a-value in R48.

53

. ......... ... , ....- . .,-~ .. . ., ,..', L . * ,- . .,_. . ,

tI

-, -ww .~ - -. . . ...... . . . . .. •.


4. .025 [B] 171.3490 Entering a/2 for a = .05, X isdisplayed.

[X<>Y] 181.1510 Y-register yields u. CI:(171.3,181.2)

is reported and (a) is resolved.

4. .10 [B] 173.0457 The 90% lower limit for (b) of 173.0

is found using a = .10.

(3) Five specimens of coke tested for porosity showed weight gains of

2.16, 2.19, 2.31, 2.30 and 2.21, all in pounds. The variance of the

process is unknown. Find a 90% C.I. for the mean weight gain.

Find estimates of V and a and SE.

Solution:

DE 1. [J] DATA?

[R/S] 0.0000 Initialize ZS-4 Ungrouped for dataentry.

2.16 [R/S] 1.0000 First weight entered.

2.19 [R/S] 2.0000 Successive weights entered.

2.31 [R/S] 3.0000

2.30 [R/S] 4.0000

2.21 [R/S] 5.0000

DE 2. [d] 0.0000 Process data.

N(W) 3a. [A] 4.0000 Display v = 4.

3b. 2.132 [R/S] 2.1698 Lower confidence limit displayed

[X<>YI 2.2982 Upper Limit retrieved from Ry.

[RCL140 2.2340 Retrieve -2 R, the estimate of v.

[RCL]38 0.0673 Retrieve o, the estimate of a.

[RCL132 0.0301 Retrieve s/n-, the estimate of SE.

Report 2.17 < u < 2.30 or 90% C.I. for V is (2.17, 2.30).

(4) Summary data for a problem are K - 2.268 and s - 0.225. Determine a

90% lower one-sided C.I. for v and an upper 99% C.I. for V.

Solution:

N(O) la. 5 [STO]06 5.0000 Enter the sample size in R0 3 .

lb. 2.268 [STO]37 2.2680 Enter the sample average in R40.

. Ic. .225 [STO]38 0.2250 Enter the sample s.d. in R 38.

3a. [A] 4.0000 Display v - 4.

3b. 1.533 [R/S] 2.1137 Lower limit is displayed; Ry ignored.

3a. [A] 4.3b. 3.747 [R/S] (1.890) t .01 entered and lower limit ignored;

[X<>YI 2.6450 R yields the upper one-sided limit.

54


(7) Times to failure for six expensive pieces of electronic equipment were

recorded in hours as 233.6, 1402.7, 3119.0, 612.9, 258.3 and 2211.2.

(a) Find a 95% C.I. for mean time to failure.

(b) Determine a point estimate and a lower one-sided 95% confidence limit

on the reliability at 500 hrs.

Solution:

DE 1. [JI DATA?

[R/S] 0.0000 Initialize ZS-4 for raw data entry.

233.6 [R/SI 1.0000 First time to failure entered.

1402.7 [R/SI 2.0000 Succeeding times to failure

-... 3119 IR/S] 3.0000 entered and processed.

612.9 [R/SI 4.0000

,. 258.3 [R/SI 5.0000

2211.2 fR/S] 6.0000

DE 2. [d) 0.0000 Data processed.

Exp(p) 3a. [E] 12.0000 Display v - 2n

3b. 4.4 [STO]41 4.4 Storing lower percentile in R

23.4 [STO]31 23.3 Storing upper percentile in R31

[R/S] 669.8889 Display I

[X<>Y] 3562.5909 Find u so CI is (673,3563)

$.-" [RCLj40 1306.2833 R , - 1306

[ul[/xI[uI 0.0008 ) r -/ 0.0008

9.-, 500 [x] 0.3828 Multiplying by 500 to find 500

% [CHS] -0.3828 Change sign for exponentiation

[U][e x ] [U] 0.6820 to yield estimate of R(500).

[E] 12.0000 Display v to start new problem.

21.0 [STO131 21.0000 Store required X2 in 1[,,..,x. 0 5 R3

[STO]41 1.0000 and 1 in R4 1 for one-sided limit.

[R/S] 746.4476 Display required lower limit on 4

[U][I/x][U] 0.0013 Upper limit on X

500 [x] 0.6698 Multiplying by 500 to find upper

limit on -500X

[CHS] -0.6698 Lower limit on -500X

[U][eX][U] 0.5118 Lower bound on R(500).

NOTE: [U] stands for the [USER] key.

• f -,-, %,," % . " W* %5 . . , ., * , . . . . . . .. . .- : Z ~ - & x :- ~ * ~ .

C--

Examples ZS-4 (Testing, Chapter 5)

(1) Seven observations of measured radiation intensity at a nuclear plant were

3.6, 4.2, 4.0, 4.1, 3.8, 3.9, 4.0. Conduct a significance test of

H 0 3.8 against H1 u > 3.8..

Solution:


DE 1 [J] DATA? Select and initialize ZS-4

3.6 [R/S] 1.0000

4.2 [R/S] 2.0000

* -C-En t er Data

4:0 [R/S] 7.0000J

DE 2 [d] 0.0000 Process data.

N(u) 2a. I [a] 1.0000 Enter H -code (+I).

2b. 3.8 [R/S] 0.0530 Enter boundary value and computeP - .053 from t-density.

(2) A water meter has variance 14 (cu. ft)2 . Twenty monthly readings indicate

a sample mean of 1284 cu. ft. per month.

(a) Test the hypothesis H0 : i = 1286 against H1 : i * 1286, using

a - .05.

(b) Calculate the significance level for the one sided alternative

HI' : < 1286.

Solution:

4 N(ula)la 20 [STO]06 20.0000 Enter sample size.

V N(UIO)lb 1284 [STO37 1284.0000 Enter sample average.

N(uaIclc 14 [x][ST0148 3.7417 Enter known a.

N(uIa)4 .025 [BI 1284.3598 x

[X<>Y] 1285.6402 u (Since (Z,u) does not contain01 H 0 is rejected.)

N(ula)3a -1 [b] -1.0000 H1-code for part (b).

N(sila)3b 1286 [R/S] 0.0084 P-value (data are inconsistentwith H0 ).

56

-. 5 C, 4 - . 5 •% • :• . . . .. .P -C ..


(3) The standard deviation in GRE scores nationwide has been 40 points.

The GRE scores for 86 Smith High School students this year has (sample)

standard deviation 35.2. What is the significance of this result?

Solution:

N(O 2) Ia. 86 [STO]06 86.0000 Enter sample size.

N(a 2) lb. 35.2 [STO]38 35.2000 Enter sample standard deviation.

N(a 2) 2a. 0 [c] 0.0000 Enter H -code for H : a * 40.

N(a 2 ) 2b. 1600 [R/S] 0.1220 P-value. (Data are somewhatconsistent with H : a = 40.)

0

(4) Times to failure of a sample of 12 unused D-cells were (in weeks):

27, 41, 29, 33, 30, 33, 26, 37, 29, 11, 20, 29. The shelf life is

claimed to be at least 35 weeks. Conduct a significance test of

H0 : u < 35 vs. H 1 : i > 35.

Solution:

DE 1 [J] DATA? Select and initialize ZS-4.

[R/S] 0.0000

27 IR/S] 1.0000

41 (R/SI 2.0000

* Enter data

29 [R/S] 12.0000

DE 2 [d] 0.0000 Process data.

Exp(m) 2a. 1 [e] 1.0000 Enter H -code

Exp(IA) 35 JR/S] 0.7129 P-value (data are consistent withH0.)

57

°. . . . . . . . . . ., .p

Chapter 6 Bivariate Populations

Program ZS-6 is another very successful transfer from the TI version andis assigned to [J] which also serves to initialize data entry and will ulti-

mately replace references to ST-04. For matters discussed in Section 6.2,however, it is more convenient to use the program EBSTAT in STAT PAC. Theprocedure for inputting paired data is discussed on page 11 of the STAT PAC

handbook. Output is then displayed by successive [R/S]'s, some of which areof no interest here. It should be noted that the output labeled GXY is simplythe correlation coefficient referred to on page 130 of ZS. Also, in the nora-tion of ZS, the HP output labeled SX. is RMSD for X, while SY. is RMSD for Y.

The STAT PAC program EBSTAT does not appear to be suitable for enteringindependent data of the type discussed on page 131 of ZS. Nor is any provi-sion made for entering univariate data in any of the programs published inSTAT PAC. The simplest solution is to start with the x-data and enter the

data twice at Step 2 (that is, let yi - xi) in BSTAT, in which case all of the

moments are X-moments and the correlation is 1; alternatively, the [ENTER]portion of Step 2 may be ignored, each x entered with [A] in which case you

should ignore all X-outputs in the list and copy only those for Y and ignoreGXY altogether. Then the whole process needs to be repeated for the y-data.

Section 6.3; Paired DataFor implementation of the programs in ZS-6, raw data will be entered via a

* self-contained subroutine, called DE in the User Instructions that follow, andreplaces references to ST-04 in the rest of the chapter. That subroutine isdivided into two parts depending on whether the data are paired or independ-ent. For this section, the data are paired so that option P will be usedand the user instructions make it clear how the data are to be entered. Besure to process the data after entry by pressing [d]. Otherwise, the instruc-tions are identical to those provided in the book for TI.

Section 6.4: Independent DataIn this section the I option of data entry DE is to be used and, at the

conclusion of data entry once more [d] must be used to process the data.Please keep in mind also that R is to be used in place of TI R throughout.

06 03The rest of the instructions are identical.

Section 6.5: Equality of VariancesNo F-distribution is provided by STAT PAC so that distribution has been

programmed into ZSTAT. Again, no formula is provided in ZS, nor is one reallyneeded in this context. But the subroutine FCDF in ZSTAT will output P(F),while FCCDF will output Q(F) provided v1 is in R15 and v2 is in R16. For ex-

ample, if V1 - 2 and v2 24, you may verify by executing FCCDF in ZSTAT that

Q(2.63) - .0927; if v - 20 and v - 7, then P(.4) - .0510. Again, the rest

of the remarks in the book apply to the HP programs verbatim.

58

,p * ***, - ' IN -¢ q-. ,'% o"":" --', -.-'.:,-. '/'.:-'. ., . . -.. \ ' -- ...P A. .

-a77 --

ZS-6 (Assigned [JI) USER INSTRUCTIONS (HP) SIZE 050

- REG 01

a STEP PROCEDURE ENTER PRESS DISPLAY

DE ORIGINAL DATA ENTRY

0. Initialize [J] DATA?

1. Enter Data

a. Paired Data IP 0.0000

(Repeat i - 1,2,...,n) x i [ENTER] xI

yi [R/SJ i.0000

b. Independent Data [II 0.0000

(1) Repeat i - 1,2 ,...,n x R/SI] i.0000

[j ] 0.000

(2) Repeat j = 1,2,...,ny yj [R/S]

2. Process Data [d] 0.0000

PN Paired Data: V-

x y1. Enter Data Using DE OR:

i a. Enter Sample Size n [STO006 n

b. Enter Sample Means

(1) Original Means R [STO]47

OR: [STO137

(2) Mean Difference d [STO147

0 [ST0137 0

c. Enter Sample Standard Deviation sd [STO127 a

2. Test H0 : ix - 1 8 0

a. Enter H -code H -code [bi H -code

b. Enter e0 and Computer P-value 00 [R/S] P

3. CI for u 11

a. Calculate degrees of freedom [B]

b. Enter t / 2 with d.f. - v and t /2 [R/S] I

Calculate Limits X<>Y] u

Note: For one-sided intervals, enter t at 3b and ignore X or u as the

case ,may be.

I 1if H1 :6H -code 0 if H : 6 * e0

-lifH : 8< 81 0

59%""-

.' d '•*~ ' -- . ,.. J,. ,.. te'. .- ,, _ ,"



INA Independent Data: x -~(axay)y)

I. Enter data Using DE OR:Clear Memory and [J] DATA?

a. Enter Sample Sizes n [STO]13 n

n [STO]06 n

b. Enter Sample Averages 5 [STO]47

[ STO]37

c. Enter Standard Deviation

(I) Pooled Estimate Available s [STO133 sOR: P p

* (2) Original S.D.'s Available s [STO]48 ss y STO138 sY Yy

2. Test H0 : - 0

a. Enter H1 -Code H -code [c] H - co de

b. Enter 60 and Compute P-value 60 [R/S] P

3. CI for x - y

a. Calculate degrees of freedom [C] vb. Enter t a 2 with d.f. - v ta/2 [R/S]

and Calculate Limits [X<>Y] u

See Previous Note for One-sided Limits.

INB Independent Data: px- y(a X oy) (Welch Approximate t)

1. Enter Data using DE OR:a. Enter Sample Sizes n [STO]13 nx x

n (STO106 n

b. Enter Sample Means [STO47 5

[STOJ37

c. Enter Sample Standard Deviations s [STO148 sx

a [STO138 sy

60



2. Test H0 : x - 1 0

a. Enter H -code H -code [a] Hi-code

b. Enter 80 and Compute P-value 00 [R/S] P

3. CI for -

I a. Calculate Degrees of Freedom [A] V

b. Enter t /2 with d.f. - v and t /2 [R/S]

Calculate Limits [X<>Y] u

I See Previous Note for One-Sided Limits

LSN LARGE SAMPLE NORMAL Vx -_y

* OR: a , a knowno y

1. Enter Summary Data Only:

a. Enter Sample Sizes n [STO113 nx x

n [STO]06 nY Yb. Enter Sample Means [STO]47

[STO]37 o

c. Enter Standard Deviations ax or sx [STO48 a or s

a y or s [STO]38 a or s

2. Test H0 : P - y - 10

a. Enter H1-code Hi-code (a] HI-code

b. Set Flag 5 [SF]05 H1-code

c. Enter Focal and Calculate 0 ER/S] PP-value

3. CI for W -1,x y

a. Initialize (Ignore output) [Al xx

b. Enter za/2 z /2 [R/S] £

and Calculate Limits [X<>YI u

See Previous Note for One-Sided Limits

61

*b~~~~W 777" --. .7 -

Or I



NV Independent Data a/x y

1. Enter Data using DE OR:

a. Enter Sample Sizes n x[STO]13n

n [STO106 ny y

b. Enter Sample Standard Deviations s [ST0148 sXs [ST0138 s

y y

2. Test H: a2 =a 20 X y

a. Enter H 1-code; Set Flag 4. H 1-code [SF104 H -code

b. Calculate P-value [D] P

3. CI fora 2 a2

a. Compute Degrees of Freedom [DI 1

From Accompanying Table: [R/SJ V 2

b. Enter F-value with d.f. - v~ )F a2 [STO131 F a 2

c. Enter F-value with d.f. - v2= F a/ [T0141 F a 2

d. Calculate Limits [R/S] k.

[X(>Y]

Note: a. For Lower One-Sided Interval, enter F aat Step 3b,

1 at Step 3c and ignore u.

b. For Upper One-Sided Interval, enter I at Step 3b,

F at Step 3c and ignore Z.

Exp Independent Exponential u /U

1. Enter Data Using DE OR:

.4.a. Enter Sample Sizes n x[ST0113 nXn [ST0106 n

b. Enter Sample Means R[STO]47R

[5TO137

2. Test H0 :.La. Enter H 1-code; Set Flag 4. H -code [SF]04 H cdb. Calculate P-value [E] P

(See Note under W,2)j

62

S' ,



3. CI for P/

a. Compute Degrees of Freedom [E] V

From Accompanying Table: [R/S] V2

b. Enter F-value with d.f. = (v,,v 2 ) F [STO]31 F/ 2

c. Enter F-value with d.f. - (v2V F /2 [STO]41 Fa/2

d. Calculate Limits [R/S I

[X<>Y] u

See Previous Note (NV) for One-Sided Limits

Register Contents

00 xxx 10 20 30 40

01 Ey 11 1 21 31 t / 2,F a/2 41 F/2

02 Ey 12 22 32 SE 42

03 Ex Used 13 n 23 Used 33 s 432 x p

04 Ex2 14 CCDF 24 34 60 44

05 Exy 15 v(v ) 25 35 45

S06 n 16 v 2 26 36 46y2

07 Used 17 Used 27 sd 37 47 R,

08 Used 18 28 H1 -code 38 s 48 s

09 Used 19 29 P(ts) 39 e(O±e) 49


ZS-6 J 00 A aDEP P 01 B b

DEI I 02 C cOPlI ENG 04 D d

X TO Y j 05 E060809

12

13

63I... .. \*

.-..-. -. 4,'-.,,... . .- ." .'. -. .. " ..4, , .. .,,% .-.- ,. .

- T T

Examples ZS-6

(1) Before (X) and After (Y) weights were recorded in lbs. after two weeks of

dieting. Find a 95% CI for the mean difference v - v and conduct a sig-x y

nificance test of equality test of equality of means. Test for a weight

.'*. loss of at least 2 lbs.

x: 119 122 136 130 129 136 134 133 119 115

. , y: 114 119 134 126 119 137 124 127 119 107

Solution:


DE 0. IJ] DATA?

la. [P] 0.0000 Initialize ZS-6 for paired data

entry.

119 [ENTER] 119.0000 Enter first x-value.

114 [R/S] 1.0000 Follow with first y-value.

122 [ENTER] 122.0000 Enter second x-value.

119 [R/S] 2.0000 Follow with second y-value.

136 [ENTER] 136.0000 Enter succeeding pairs.

134 [R/S] 3.0000

130 [ENTER] 130.0000

126 [R/S] 4.0000

129 [ENTER] 129.0000

119 [R/S] 5.0000136 [ENTER] 136.0000

137 [R/S] 6.0000

134 [ENTER] 134.0000

124 [R/S] 7.0000

133 [ENTER] 133.0000

127 [R/S] 8.0000

119 [ENTER] 119.0000

119 [R/S] 9.0000

115 [ENTER] 115.0000 Enter last x-value

107 [R/S] 10. Follow with last y-value (n 10)

DE 2. [d] 0.0000 ZS program processes data.

PN 3a. [B] 9.0000 Calculates and displays v - 9 d.f.

3b. 2.262 [R/S] 1.9389 Enter t from t-table anddisplay' = 1.94

[X<>Y] 7.4611 Display u so CI is (1.94, 7.46).

PN 2a. 0 [b] 0.0000 Enter H -code 0 for two-sided

test.

64

% '. .. %, -" - . .. ".. * -. °.J L.. . .. . .. . ... ... . • •- . % •%% %" -%%% %" - . . . .".. .. .r

_. ._W 7.7 .- 7! 0 7 7 77 71' 7 7 7 , . .,

ZS STEP ENTER PRESS DISPLAY COMENTS

0 [R/S] 0.0039 Use 8_ = 0 for this case and findP = 0.0039; reject at usual levels.

[RCL]30 3.8504 Display value of ts.

PN 2a. 1 [b] 1.0000 Use H -code of I making

H - 1y _ 2 the disclaimer.0 x y

2 [R/SJ 0.0271 With 60 = 2, P-value is enough to

reject at a = 5%.

(2) A test of color perception was administered to a control group (X)

and an experimental group (Y) with results:

x: 16.3 14.7 12.3 13.5 16.0 17.1 17.3

4. y: 14.0 16.5 17.7 15.9 18.0 16.3

Analyze the two groups for dirferences. Also test for equality of

variances.

Solution assuming a - ax y

DE 0. [J] DATA? Initialize ZS-6 for independent

data entry.

lb. 16.3 [R/S] 1.000 Enter first x-value.

14.7 [R/S] 2.0000 Continue x-values assuming data

12.3 [R/SI 3.0000 are independent.

13.5 [RIS] 4.0000

16.0 [R/S] 5.0000

17.1 [R/S] 6.0000

17.3 [R/S] 7.0000 Last x-value entered; n = 7.

[j] 0.0000 Prepare for y-values.

14 [R/S] 1.0000 Begin entering y-value

16.5 [R/S] 2.0000 (as with label B in ST-04)

17.7 [R/SI 3.000015.9 [R/S] 4.0000

18 [R/S] 5.0000

16.3 [R/S] 6.0000 Conclude y-entries; n 6

DE 2. [d] 0.0000 Process data.

INA 2. 0 [c] 0.0000 Enter H -code for two-sided test.

0 [R/S] 0.2740 Display P-value (for 60 0) of 0.27;

[RCLJ30 -1.1512 Accept H0 (ts - -1.15).

INA 3. [C] 11.0000 Reveal d.f. - n +n -2 - 11 for thisx ycase.

2.201 [R/S] -3.1615 Entering t - 2.201, CI runs from

.025

[X<>YI 0.9900 Z. -3.16 to u - 0.99 which doesinclude 0.

65

...... . .. , _ , , ., ,_


Solution assuming a * a- Y

INB 1. Same as INA so data are already entered and processed.

INB 2. [A] 10.0000 Calculates degrees of freedom for

2.228 [R/S] -3.1404 approximate CI based on Welch t.

, [X<>Y] .9690 Comes close to preceding solution.

INB 3. 0 [a] 0.0000 Begins Welch t-test with H -code

0 [R/S] 0.2663 followed by e0 = 0 to yield about

the same P-value.

.' 2 2To test for a = a

x yNV 1. Same as INA

NV 2. 0 [SF]04 0.0000 HI-COde for two-sided test. Flag

4 signals NV a test is being calledfor.

o2 2[D] .5645 Large P-value; accept aY = a 2with

x y(RCL]30 1.7290 ts = 1.73.

NV 3a. To take a CI point of view

[D] 6.0000 Displays vI = n -1.

[R/S] 5.0000 Displays v 2 = n -1

NV 3b. 6.98 [STO]31 6.9800 Enter F with d.f. = (6,5)..025

NV 3c. 5.99 [STO]41 5.9900 Enter F with reversed d.f. = (5,6)..025

[R/S] 0.2477 Shows a 95% CI that includes the

[X<>Y 10.3565 value 1. Accept a2 a ax y

(3) A sample of 60 exponential times to failure averaged R = 1306 hrs.

Six independent times averaged 1247 hours. Test H :x< I0 x- y

Solution:

Exp 1. 60 (ST0113 60.0000 Enter Summary Data

6 [STOJ06 6.0000

1306 [STO]47 1,306.0000

1247 [STO137 1,247.0000 Data entry concluded.

Exp 2. 1 [SF]04 1.0000 Enter H -code for H 1 > U and

set flag 4 to signal H-test.

[E] 0.5043 P-value of 0.50 obtained; do notreject.

Exp 3a. [E] 120.0000 V1 " 2nx displayed.

[R/S] 12.0000 2 2n displayed.2 y

66

.- .. . . . . .

'-.+''-,-' '.. '.L". i _"-,-" -,," - L" " " -""" " - ,,'"", "- -""'" ""- . v .. .., . ... .- , ,,.......,-,,. . ,..,.,,


3b. 2.35 [STO]31 2.3500 F.0 5 for v 1 = 100, v 2 =10 entered

1 [STQI4I 1.0000 1 store in R 41to compute lower CI

(R/SI 0.4457 Lower bound on pi /p displayed.x y

04.

V %

* .- , %°.4.. - . °- °- . . 7

Chapter 7 Proportions

This chapter represents the most successful transfer of programs of all.Indeed, the only remarks that need to be added to the existing programs is to

remind you once more that all references to register R03 in TI are to be re-

- placed with R06 in HP, that [X<>Y] is the HP version of [x t] (so that any

reference tp TI R should be replaced by Ri). Finally, since ZS-7 has been

assigned to [J], you should press the latter key whenever you need to access the.4. programs here and is the only initialization necessary.

* 68

, %

' 5 .W '. * . . . . .r _ °• - .•r C .o °-. _ r r ° % - . . ° . , . . . .

ZS-7 (Assigned [J]) USER INSTRUCTIONS (HP) SIZE 050

Z REG 01

* STEP PROCEDURE ENTER PRESS DISPLAY

B(p) Bernoulli Parameter

0. Initialization (if not already in ZS-7) [J] 0.0000

1. Data Entry


b. Enter Proportion Estimate p [STO]40 p

2. Test H 0 : p - p0

a. Enter H -code H -code [b] Hi-codeb. Enter pO and Compute P-value OR: P0 [R/S] P

b'. For n Large (Normal Test) PO [ci P

3. CI for p for n large

Enter Risk and Calculate Limits a/2 [C] k

Note: For One-sided Limits, Enter a [X<>YJ u

-S "* and Ignore £ or u.

4. CI for p for n small

a. Find first d.f. for F. [B] v1

[R/S] v5 b. Enter F with d.f. - F [STOJ 31 2

c../ /2 " V l jV 2) Fa/2 [T31 Fa/2

c. Find second d.f. for F [R/S] v1• S. [R/S 1 2

d. Enter Fa/2 with new d.f. Fa/2 [STO]41 Fa/2

e. Calculate limits [R/S] I

[X<>Y] u

Note: For Lower One-sided Limit, enter F at Step 3b, 1 at Step 3dand Ignore u.

For Upper One-sided Limit, enter I at Step 3b, F at Step 3d

and Ignore X.

S(p -p Two Bernoulli ParametersV. x y1. Data Entry

a. Enter Sample Sizes n x [STO113 nx.:-n [STO]06 n

b. Enter Proportion Estimates px [STO]47 P

p [STO]37 p

1 if HI : > 6H -code 0 if H : 0 * 0

1 0-if HI : < a 6

~69

'"Jv *' 'L * '- " -• %-' "" "," " '"" '-:' " • """, "'" " " " " . . .. """- " """, ,"" " ,""" .- "•" V.'



2. Test H0 p P

Enter H1-code and Calculate P-value H 1-code [a] P

3. CI for p - p

Enter Risk and Calculate Limits a/2 [Al .

[X>Yl u

Note: For One-sided Limits Enter a and Ignore I or u.

Register Contents

00 10 20 30 40 0

01 11 Used 21 31 z /2,F /2 41 F /2

02 12 Used 22 32 SE 42

03 13 n 23 Used 33 Used 43

04 u 14 24 34 0 44

05 1 ) 15 25 35 45

06 n(n ) 16 26 36 46

07 17 Used 27 _ 37 p 47 py Px

08 BIN p(O) 18 Used 28 H -code 38 48

09* 19 Used 29 p(ts) 39 e(0+e) 49


ZS-7 J 01 A a

02 B b

03 C

04

05

06

i%.-4

70

F,7£

ZS-7 EXAMPLES

(1) In nine independent Bernoulli trials, there were exactly four successes.

Find a 95% CI for p and test H0 : p 0.5.

Solution:

ZP STEP ENTER PRESS DISPLAY COIMENTS

B(p) 1. 9. [ST0106 9.0000 Enter sample size.

.44... [STO]40 0.44... Enter p - 4/9, the estimate of p.

4a. [B] 12.0000 Since n is small find first pair

[R/S] 8.0000 of d.f. = (10,10).

4b. 4.20 (ST0131 4.2000 Enter first F 0 2 5 percentile..025

4c. fR/S] 10.0000 Discover revised d.f. - (12,8).

IR/S] 10.0000

4d. 3.72 [STO]41 3.72 Store second F 0 2 5 percentile.

[R/S] 0.1370 Lower confidence limit displayed

IXOY] 0.7881 and u found in %.

3. 0. [b] 0.0000 Enter Hl-code for H1 : p * 0.5.

0.5 [R/S] 1.0000 Significance level 1; accept H

(2) A device was tested 25 times and passed 23 times. Find a lower one-sided

CI on p, the probability of passing. Test H0 p > 0.95.

Solution:

B(p) 1. 25 [ST006 25.0000 Enter data as above.23+25 [ST0140 0.9200 p - 23/25 = 0.92

2a. -1 [b] -1.0000 Enter H1-code for H: p < 0.95.

2b. 0.95 (ci 0.2456 Comparing with large sample test.

4a. [B] 6.0000 Initial d.f. for small n CI (to

[R/S] 46.0000 be ignored along with v2 .

4b. 2.29 [STO]31 2.2900 Following instructions store 1 in

4c. [R/SI 48.0000 R3 1 . Calculate new d.f. v i W 6

[R/S] 4.0000 and v 2 = 46.

4d. I [STO]41 1.0000 Enter F.05 for d.f. - (6,50) and[R/S] 0.7700 calculate lower confidence bound.

(4) A sample of size 100 was taken from a lot with replacement and 2

defective items were found. Test the manufactures claim that

p < 0.05 at a - .01.

71

V V


Solution:

B(p) 1. 100. (STO06 100.0000 Enter data as usual.

.02 [STO]40 0,0200

2a. -1. [b] -1.0000 H1 -code for H. : p < .05, the nullhypothesis being H0 p > .05, the

disclaimer.

2b. .05 JR/S] 0.1183 Enter p0 and find P - 0.12

supporting H0 not HI

2b'. .05 [c] 0.0843 Compare normal test.

(3) In a random sample of 500 men (X) 350 were found to favor a certain

political issue. In a similar sam'ple of 300 women (Y) 200 were

so inclined. Is there any real difference between sexes on this

issue?

Solution: To test H0 : Px y

B(p X-p y)Ia. 500 [STO]13 500.0000 Enter first sample size.

300 [STO106 300.0000 Enter second sample size.

lb. 0.7 [STO]47 0.7000 Enter first proportion estimate Px"

2/3 [ST0137 0.6667 Enter second proportion estimate py.

2. 0. [a] 0.3248 Enter H1-code for HI : Px F Py

[RCL]30 .9847 and find P - 0.32 with ts = .98.

Data supports H0 .

3. .025 [A] -0.0335 A 95% CI for the difference u -P

[X<>Y] 0.1001 extends from -.03 to + .10; includes0.

72

Chapter 8 Analysis of Variance

The big change here is the data entry which is via STAT PAC through theAnalysis of Variance routines provided there. Unfortunately, those routinesare not complete enough to accomplish all of the goals set out in the text sothat they too had to be supplemented with program ZS-8, whose user instructionsfollow.

Section 8.2: One-Way ClassificationsOn page 95, you may replace the reference to ST-22 with execution of FCCDF

in ZSTAT. If you will consult the user instructions, you will see that theprogram utilizes subroutine ZAOVONE, assigned to CHI for convenience, in placeof the TI program ST-06, referred to on page 197. After pressing [HI and see-ing the display EAOVONE, you follow Steps 3-5 for inputting data (a model ex-ample is provided following the user instructions). A press of [El while stillin ZAOVONE will then output most of the AOV table. The only, but important,missing item is the prob-value and that is calculated at Step 3 in ZS-8 byexiting ZAOVONE with a press of [JI followed by [A). The Scheffe' confidenceintervals discussed in the very next section follow precisely the same user in-structions as the TI and are duplicated in the HP User Instructions thatfollow.

Section 8.4: Two-Way ClassificationsIn this section, the program EAOVTWO in STAT PAC is used for data entry in

place of ST-06. This subroutine is assigned to [I] in ZS-8 and, once pressed,the instructions for data entry and output discussed on page 23 of the STAT PAChandbook should be followed. (Again, a model problem is provided at the end ofthe user instructions for ZS-8). This will provide for only part of the Two-Way table as displayed in this section of the text (and most other textbooks onthe subject). To complete the table, you need to exit ZAOVTWO by pressing [JIand then [C] will output the remaining items needed for the table including theall-important prob-values. Once again, the instructions for implementing theSheffe' confidence interval formulas discussed in the next section are identi-cal to those for the TI and are duplicated in the user instructions thatfollow.

.7

'.

.

a.73

...'p..-," ' -I"-,I. - ' - " "" - . 'i ." " " •" " " " " " " " " '" " "" '* " "" """ '"' ," " .

ZS-8 (Assigned [JI) USER INSTRUCTIONS (HP) SIZE 060


AOV-1 One-Way Analysis of Variance' 0. INITIALIZATION (if not already in ZS-8) [JI xxxx

1. Enter Data Using EAOVONE [H] ZAOVONE

NOTE: Record Each Row Mean

2. Calculate AOV Table Entries

NOTE: These Steps may NOT be repeated [E] SSonce Step 3 is executed. [R/S] RSS

[R/SI ESS[R/S] K-i"R/S] N-K[R/S) N-1[R/SI ] MRSS[R/S] MESSfR/S] F

3. Exit rOVONE and compute P-value F [J] F

[Al P

4. Confidence Intervals for Contrasts

(After Step 2)

a. Initialize [e] 0.0000

b. Enter Contrast Data (Repeat for

each i; ignore any ci - 0) ci [R/S] ci

Rii R ]

n i [R/S] i

c. Enter F-percentile d.f. - (K-1,N-K),

and calculate CI F [a] £

,*,,4 NOTE 1: Steps 3abc may be repeated.

NOTE 2: These Steps are also valid if R0 3 and R48 are manually stored.

- "REGISTER CONTENTS

00 SS 10 Used 20 30 F 40 50 Zcix ii i01 RSS 11 N-K 21 FCDF 31 P 41 51 Last

02 ESS 12 Used 22 32 42 52 LastR2

03 K-i 13 23 33 43 53 Zc /n i

4 04 Used 14 24 34 44 54 Used

4 % 05 Used 15 R-1 25 35 45 55

06 M 16 N-K 26 36 46 56

07 Used 17 Used 27 37 47 57

08 Used 18 28 38 48 MESS 58

09 K 19 for 29 39 49 e 59

74

~ ***.*- *,.*. p...--. .7. - p.. * .~ . p ~ .. ~ - *~. N


STEP PROCEDURE ENTER PPESS DISPLAY

AOV-2 Two-Way Analysis of Variance

0. Initialize (if not in ZS-8) [J] x.xxxx

1. Enter Data Using ZAOVTWO [I] ZAOVTWO

2. Calculate Row and Column Means

Calculate Row Means After each Row IRIS] sum

entry (Record). Repeat i - 1,...,R. C [f] Ri.

Calculate Coluim Means After each [R/S]

Colum entry (Record). Repeat R [W .

j - 1,2....C.

3. Calculate AOV Table Entries [El RSS

-- NOTE: These Steps may NOT be repeated [R/S] CSS

once Step 4 is executed [R/S] SS

'R/S] ESS

-RIS] R-I

[R/S] C-1[R/S] (R-1)(C-1)

-,'? [IRIS!]F

IRIS] FC

4. Exit EAOVTWO [J] Fc

5. Complete the AOV output FC [C] MRSS

[R/S] MCSS

[RIS] MESS

[R/S] , R

IR/S] PC

4_,.

-•75

4 -,. . .. ,

. A- . .



6. I Confidence Intervals for Posterior II Contrasts I

a. Initialize [E] 0.0000

b. Enter Contrast Data c [R/SJ c2i

(Repeat for each i (or j)) 3i (or 5 E [R/S] i (or j)

c. CI for Row Contrast c iPi. F [D] L

d.f. - (R-I,(R-1)(C-1)) [X>Y] u

d. C1 for Column Contrast Eci u F [d] Z

d.f. = (C-1,(R-1)(T-1)) [X<>Y] u

NOTE 1: Steps 4abc or 4 abd may be

repeated.

NOTE 2: These steps are also validif the contents of registersR 48 R below are

\I~. stored manua2ly.

REGISTER CONTENTS:

00 Used 10 20 30 ts 40 50 ZcR

01 R 11 (R-1)(C-1) 21 31 41 51 last c

02 C 12 RSS 22 32 42 52 lastR

03 RC 13 CSS 23 33 43 53 Zc

04 x.. 14 R-1 24 34 44 54 Used2

05 x.. 15 C-I 25 35 45 55

06 x.. 16 (R-1)(C-1) 26 36 46 56 FR

07 MESS 17 Used 27 37 47 57 FC

08 18 28 38 48 MESS 58 R-1

09 19 29 39 49 e 59 C-i

Assi ments Labels Used

ZS-8 J 01 A aEAOVONE IH 02 C d-AOVTWO 1 03 D e

E

76

... EXAMPLES ZS-8

(1) Three types of solvents are tested on grease-soaked material and the amount

of grease removed in milligrams is noted for several specimens with the

following results:

Solvent A 11 12 12

Solvent B 13 15

Solvent C 12 10 11 11

Test the hypothesis of no differences in solvents.

Solution:


[J x.xxxx

AOV-1, 1. (HI EAOVONE Call for ZAOVONE in the module.

ZAOV 2. 11 [A] 1.00 Enter data for the first row.

12 [A] 2.00 Enter data for the first row.

12 [A] 3.00 Row i data entry concluded.

ZAOV 5. [R/S] 11.67 First row mean x calculated.

Record!1

"R/Sl 0.58 s for row I and row sum; ignore and

[R/S] 35.00 go on to enter data for second row.

ZAOV 2. 13 [A) 1.00 Running count begins anew.

15 [A] 2.00 Row 2 data entry concluded.

ZAOV 5. [R/S] 14.00 Row mean x2 calculated. Record!

[R/S] 1.41 Row 2 s and sum; ignore and proceed

|R/S] 28.00 to enter data for third row.

-AOV 2. 12 (A] 1.00

10 [A] 2.00

11 [A] 3.00

11 [A] 4.00 Row 3 data entry concluded.

EAOV 5. [R/S] 11.00 Value of x3. (Record)

[R/S] 0.82 Value of s and sum; ignore. Data

'R/S] 44.00 entry concluded.

ZAOV 6. (E] 16.89 Value of SS displayed for total.

[R/SI 12.22 Value of RSS

"'.- [R/S] 4.67 ESS displayed

-R/S] 2.00 d.f. for RSS

"R/S] 6.00 d.f. for ESS displayed

.R/S] 8.00 d.f. for SS displayed

-R/S] 6.11 Value of MRSS

[R/S] 0.78 Value of MESS

-77

;- %"€'-

. -.. .. .. % ',.%. 7 - . .-. . * *-. *- -. ._-.2 i . ".K . " . K . x . . *::*-:. -.".. . " .' ". " " . -" - .. . ." .


-R/S] 7.86 F-ratio = MRSS/HESS.

[J] 7.8571 Exit ZAOVONE.

[Al 0.0211 P-value of the F-ratio.

(2) Find Scheffe 95% confidence intervals for the contrast pi-0.5p -0.5111 2 3

where MESS 105.97 and K = 3, N-k = 18; Ml = 103.71, R = 92,

x 3 = 89, n -7.

Solution:

AOV-l. 2. 2 [STO]14 2.0000 V1

= K-I 2 stored in R14

105.97 [STO]48 105.9700 MESS stored in R48 as required.o48

AOV-I. 3. [e] 0.0000 Initialize CI routine.

1 [R/S] 1.0000 First of triple triple Cix,n,".- entered.

103.71 [R/S] 103.7100 Second member of triple

7 ER/SI 1.0000 Sample size n ;count of I (triple)displayed.

-0.5 [K/S] -0.5000 Beginning entry of c2,.R2,n

92 [R/S] 92.0000

7 [R/SI 2.0000 Running count of 2 lis~laved.-- -0.5 [R/S] -0.5000 Entering final triple

89 [R/S] 89.0000

7 [R/S] 3.0000 Entry completed.

3.49 [a] 0.6203 F for d.f. - (2,20) entered.05

[X<>Y] 25.7997 and confidence limits displayed.

Conclude contrast significantly

different from 0.

(3) Five teachers were matched with three schools to produce the following

average scores on a standardized examination after a unit of

instructions

School Teachers A B C D E R ii Schools

1 j53 47 461 50 4 46155II52 58 54 56

III 51 51 49 54 50 51

x 55 51 49 54 51

.78

O ZS STEP ENTER PRESS DISPLAY COMMENTS

.- Construct a two-way AOV table and find Cl's for I.1- 2. and .i- .3.

-,.-. Solution :

[J] x.xxxx

AOV-2 1. 11 ZAOVTWO Call for ZAOVTWO from the modL

ZAOV 3. 53 [A] 1.00 Enter first data value from rc

47 [A] 2.00 Continue entering data from rc

46 [A] 3.00

50 [A] 4.00

49 [A] 5.00 until all the data from row Ientered

[R/S] 245.00

ZAOV 5. 5 [.1 49.00 Calculate and record row mean

ZAOV 3. 61 [A] 1.00 Go on with first value from rc

55 [Al 2.00 and continue

52 [A] 3.00 until all of

58 [A] 4.00 the data from row 2

54 [A] 5.00 have been entered.

ZAOV 5. [R/S] 280.00 Calculate and

5 [+1 56.00 record R2 .

EAOV 3. 51 [A] 1.00 Continue non-stop with data

51 [A] 2.00 entry from the third and

49 [A] 3.00 last row

54 [A] 4.00

50 [A] 5.00

EAOV 5. [R/S] 255.00 Calculate and

5 [ ] 51.00 record x 3 "

EAOV 6. [R/S] COLUMN-WISE Prepare for column computatioi

EAOV 8. 53 [A] 1.00 Enter first value from column

61 [Al 2.00 Enter second value from colum

51 [Al 3.00 Enter last value from column

ZAOV 10. [R/S] 165.00 Calculate? R

[ 1 55.00 and record.

.AOV 8. 47 [A] 1.00 Repeat for colum 2

55 [A] 2.00

51 [A] 3.00

10. [R/S] 153.00 Calculate and

3 [+] 51.00 record R.2

D 79

- .,, ' - - -.. - . , . . -oi..

. .. .. . -2-.. . -


"AOV 8. 46 [A] 46.00 Repeat for colum 3

52 [A] 52.00

49 [A] 49.00


3 [ ] 49.00 record

EAOV 8. 50 [A] 50.00 Repeat for column 4

58 [A] 58.00

54 [A] 54.00


3 [ ] 54.00 record x.4

ZAOV 8. 49 [A] 49.00 Repeat for colum 5.

54 [A] 54.00

50 (A] 50.00


3 [ 1 51.00 record

EAOV 11. [El 130.00 Data compiled and RSS displayed

[R/S1 72.00 CSS displayed.

[R/S] 224.00 SS total sum of squares.

fR/SI 22.00 ESS displayed

[R/S] 2.00 Row d.f. = R-I = 2 displayed.IR/S] 4.00 Colum d.f. = C-I = 4 displayed

[R/S] 8.00 Error d.f. = (R-1)(C-1) displayed

[R/S] 23.64 FR displayed

IR/S] 6.55 F displayedC

[J] 6.5455 Exit ZAOVTWO and enter ZS-8

ZS-7 5. [C] 65.0000 MRSS displayed

(R/S] 18.0000 MCSS displayed

[R/Sj 2.7500 MESS displayed

[R/S] 0.0004 P-value for FR computed and

displayed

[R/S] 0.0122 P-value for FC

- 4. [El 0.0000 Initi-alize for Scheffe Cl's

1 [R/S] 1.0000 Enter c = 1 to find CI for' I" - '

49 [R/S] 1.0000 Enter row mean R

-1 [R/S] 1.0000 Enter c2 = -1 for contrast

U1.- U2.

80

4 . - o . , . . . .

V.. * . R :,'I


56 [R/S] 2.0000 All non-zero c's now entered.

4.46 [D] -10.1324 F.05 = 4.46 for d.f. = (2,8)

.. [X<>Y] -3.8676 Z is displayed followed by u.

[RCL]49 3.1324 The value of e retrieved from

5,* R49 for further comparisons.

[E] 0.0000 Re-initialize C.I. program.

1 [R/S] 1.0000 Enter cI for contrast w.I -1.3"

55 [RIS] 1.0000 R., the first column mean is

entered.

-1 [R/S] 1.0000 Enter c2 = -1

49 [R/S] 2.0000 Enter third column mean x'3"

3.84 [d] 0.6934 Enter F.05 for d.f. = (4,8) and

'"" calculate Lower limit.

"X<>YI 11.3066 Upper limit retrieved from Ry.

[RCL]49 5.3066 Value of e found in R49 for. ~ further comparisons.

5

81

• . --" .- f . . - - - . - - - - - , - . " . - . - - . - . - . . - . , .. . . . .

81%

.Chapter 9 Simple Linear Regression

It is rather surprising that the HP is not hard-wired for at least simplelinear regression as is the TI and many lesser hand-held calculators. There isa routine in STAT PAC, but, just as with the TI statistics module, no provi-sion is made for confidence intervals, tests of hypotheses, etc.. In order to

*make the HP output match the discussions given in the text, we have created asimple data entry scheme in a program called ZS-9 (assigned the label [II forconvenient access). Once that program is accessed, you have only to press (D](for data) and enter the successive pairs of numbers as per Step 2 of the in-structions. At the conclusion of entry, press [e] to compile the data where-upon the degrees of freedom will be displayed for you. At this point, you mayenter a t-percentile if you like. In any event, the effect of entering datathis way will force the register contents to almost agree with those of the TIentry, with a couple of notable exceptions. Referring to page 237, HP R06 asusual must replace TI R and then the TI R R R become HP R R

03 R0 4 ' R05 ' 06 03' 4R0 5 , respectively. As usual, the HP functions MEAN and SDEV replace TI [21

and [INV][] and will output the same quantities. There is a subroutine with-in ZS-9 called [OpIll whose execution will exactly match that of TI [Op]11 asreferred to in this chapter. Try this on the data in Example 9.1 to verify theresults published on page 238. Similarly, there are subroutines in ZS-9 called

V. [Op]14 and [Op]15 that will function in exactly the same way as their TI coun-terparts referred to in the text. In Note 2 on page 238, HP will display themessage DATA ERROR if the data all have the same carrier value. Similarly, inNote I on page 250, HP will display the message ALL REALS to signify that theCI does not exist. Otherwise, all of the instructions for the various regres-sion routines through Section 9.4 are identical to those given in the book forthe TI. For that reason only Step 1 needs to be modified and that has beentaken care of in the User Instructions that follow on the next page.

Section 9.5: Curve FittingThe procedures in this section utilize the TI statistics module and, for-

tunately, most of them are duplicated in the HP STAT PAC under the same title,Curve Fitting, beginning on page 32 of the STAT PAC handbook. The only problem

is that the HP notation differs slightly from that of TI. Thus, TI b is HP aand TI m is HP b . You will have to make that adjustment in order to use yourHP for solving problems in this section. The output of label [El in that pro-gram, however, will produce the right estimated equations and can be used toverify the numbers given in Example 9.9 as well as most of the exercises. Theone big departure is that HP makes no allowance for creating your own user de-fined transformation so that examples like 9.10 on page 265 cannot be checked.Those are not too common, however, so that for the main type of transformationsyou are likely to run into in practice, what is provided by STAT PAC will suf-fice. All of the answers to the problems, with the exception of 40e, can beverified with those routines.

82

. .- . . . .. . - . .. . . , , J .. _I . , .e ~ ~ , . . .; - . ..

ZS-9 (Assigned [I]) USER INSTRUCTIONS (HP) SIZE 050

Z REG 01


I 0. Initialization (if not already in ZS-9) [I] 0.0000

1. Clear registers [D] 0.0000

2. Enter data (repeat i = 1,2,...,n) x [ENTER] x

Yi [R/S] i

3. Compile data [e] n-2

4. Enter percentile for Cl's (d.f. - n-2) t [R/SJ t(may also store manually in R31 atany time)

SLOPE

I CI for m [A] Z

[X<>Y] u

2 Test Ho:m -Mi.

a. Enter H -code.* H -code [a] H -code

b. Enter hypothesized value. m0 [R/S] P

INT

1 C1 for b . [B] 2

[X<>Y] u

2 Test H0 :b - b0 .

a. Enter H -code. H -code [b] H -code

b. Enter hypothesized value. b [R/S] P0

V at x CIlfor mx0 +b x[C] L.

[X<>Yj u

Y at x0 PI for Y 0 .mx 0 + b + e x0 [c] I

_ _ ___,_[x<>Y] u

DISC C1 for x*, when y* is

observed y [d] 2

[ X< >Y ] u

CORR Test H0 :P - 0.

Enter H1-code. H -code [E] P

Note: Valid whenever n-2 eR15

I for H :6 > 1 0

83

6' "€ -""" """ """""" '" "e "" " -"" -". - - " . . - . . .- -'-" " ," - -" "v -. . .-- - . . . .

REGISTER CONTENTS:

00 Used 10 20 30 ts 40

01 y 11 21 31 t /2 41 Used

02 Eyi 12 22 Used 32 42 d

03 Ex 13 23 by 33 s 43 y*-y

04 Ex 14 n-I 24 TCDF 34 0 44 ri 0

05 -x iy 15 v - n-2 25 35 s- 45 m

06 n 16 26 36 si 46

07 -(Y 1)2 17 27 37 s.i 47

08 Z(xi-R) 18 28 Hl-code 38 six0 48

09 19 29 P(ts) 39 e(i±e) 494" "

For curve fitting, consult STAT PAC p. 32.


ZS-9 i 01 A aoP12 H 02 B bOP13 IP 03 C cOP14 Ih 04 D dOP15 Ii 05 E e

iI 84

"V

2,, v" ,"5 ,-,,, , .. - , -, .,. , - ... ,. . .. . ..4.

EXAMPLE ZS-9

The resistance of a length of wire is thought to be a linear function of the

.r. temperature of the wire. For a given temperature, errors in readings of resistance2

are normally distributed with mean 0 and variance a . The following readigs were made

at the temperatures indicated.

Temperature 0 10 20 30 40 50

Resistance 22.6 25.1 29.0 29.9 33.4 34.8

(a) Estimate the regression of resistance on temperature.

(b) Estimate the resistance if temperature is 25.

(c) Estimate the temperature if resistance is 30.

(d) Find a 95% confidence interval for the slope, m.

(e) Find a 95% confidence interval for the intercept b.

(f) Find a 95% confidence interval for the expected resistance whentemperature is 25.

C (g) Find a 95% prediction interval for the measured response whentemperature is 25.

(h) Find a 95% discrimination interval for the temperature at which aresistance of 30 is observed.

(i) Test the hypotheses H 0: m - 0 vs. H m * 0.

(j) Test the hypotheses H0 : b < 20 vs. HI: b > 20.

(k) Calculate the coefficient of determination.

(1) Test H0 : p -0 vs. H1: p * 0.

SOLUTIONS:


I 0. (I] x.xxxx

1. [D] 0.0000 Clear data registers

I 2. 0 [ENTER] 0.0000 Enter x1 .

22.6 [R/S] 1.0000 Enter y update data base; (x,y)count displayed.

10 [ENTER] 10.0000 Enter x2.

25.1 [R/S] 2.0000 Enter y2 9 update data base

20 [ENTER] 20.0000 Enter x3e.' 29 IR/S] 3.0000 (x,y) - count displayed.

30 [ENTER] 30.0000

29.9 [R/S] 4.0000 (x,y) - count displayed.

40 [ENTER] 40.0000

33.4 [R/S] 5.0000 (x,y) - count displayed.

50 [ENTER] 50.0000

" 34.8 [R/S] 6.0000 Value of n - 6 concludes data entry.

85

,. . . . . . . . . . . • - .- . - . . . .. . . - . .- .- . . - . - . . . .,, : ,. .- . .. .. . .

*ZS STEP ENTER PRESS DISPLAY COIfrINTS

I 3. [e) 4.0000 Compile data; display d.f. = 4.

I 4. 2.776 [R/SJ 2.7760 Enter t for d.f. = 4.(H] 22.9333 b

[X<>YI 0.2480 m; y - .248x + 22.93 answers (a).25 [h] 29.1333 y for x - 25 answers (b).

30 [i] 28.4946 x for y - 30 answers (c).

SLOPE 1. [A] 0.1983 X

[X(>Y] .2977 u, so CI is .198 < m < .298,answering (d).

INT 1. [B] 21.4294 £

[X(>Y] 24.4373 u, so CI is 21.4 < b < 24.4,answering (e).

p at x0 25 [C) 28.2850 £

IX<>Y] 29.9817 u, so C1 is 28.28 < 25m + b < 29.98,answering (f).

Y at x0 25 [c] 26.8889 £

[X(>Y] 31.3778 u, so PI is 26.89 < Y0 < 31.38,

answering (g)

DISC 30 [d] 19.3744 9

[X<>Y] 37.9069 u, so CI is 19.37 < x* < 37.91,answering (h).

SLOPE 2a. 0 [a] 0.0000 Enter Hl-code for H m m # 0

2b. 0 [R/S] 0.0002 Significance of test; reject H0 ;answers (i).

INT 2a. 1 [b] 1.000 Enter H -code for HI: b > 20

2b. 20 [R/S] 0.0028 Significance of test; reject H0;

answers (j).

[P 0.9897 Calculates and displays r[x2] 2[x 2 0.9796 r 2 0.98 is the answer to (k).

CORR 0 [U] [El[U 0.0002 Significance of test of H P 0(must agree with (i)).

(Ul (USERI

864%

'4,% ... .-. "-"v -- '',' .. .. ,), ,'-"."-".'-,_. - . . - . .. ... ;.-. ,-. . - .. , ,-, . , ,-,-.. . .. . ,. .

Chapter 10 Multiple Regression

Only the data entry scheme differs from the TI version of this program.The regression program EMLRXY in STAT PAC is utilized for entering the data inthe HP version" and partial processing takes place in that program. Furtherprocessing takes place in program ZS-10 (assigned to label [J] for easy entryfrom STAT PAC) so that even the register contents (with the slight modificationgiven below) and the remaining instructions will match those given in the book.

Once ZS-10 is entered, a press of [d] will force the pointer to STAT PACprogram EMLRXY. Data are then entered as follows: first, x and y are suc-cessively entered with the [ENTER] key and then the value of z with [A]. At

. the conclusion of data entry, pressing [El will cause partial processing, end-ing with a display of the coefficient of determination. It is at this pointthat STAT PAC must be exited and ZS-1O entered with a press of [J]. Then,pressing [el will cause the rest of the processing to take place. Thereafter,the user instructions for ZS-10 may be followed to the letter. For that rea-son, only the DE instructions need to be modified and are summarized below.The usual sample problem is presented starting on the following page.

SIZE 050


DE Data Entry

0 Initialization [JI 0.0000[d] ZMLRXY

1 ENTER DATA xi [ENTER] xi

(repeat i1l,...,n) yi [ENTER] yizi [A] i

2 Compile Data (partial) [E] R2

Complete compile [J] 0.0000-e] n-3

3 Enter t-percentile ta/2 [R/S] ta/2(d.f. = n-3)

Assignments Labels Used Register Contents

ZS-o {J 01 A a 40ZM..RXY e 02 B b 41

C C 42D d 43E e 44

4546 a0

% 47 a48 a 2

*. .".". 49

87

'- .. .. . ... .... ....-k 5 -1 . . .. ; ~~...,. . . ,. ,., ,. . . .' .. o . . . .. .'..'..'.

S - . .. . . . . . .. . . . • _- . . . . - . . o . . . . • . . •a777

ZS-l0 EXAMPLE.

The data below represent characteristics of a sample of automobiles.

Weight 3810 4220 2900 3290 3400 3920 4350

Horsepower 255 180 16 120 100 140 150

Cost 7999 9221 8222 9010 10099 11019 11219

(a) Regress cost on weight and horsepower.

(b) Predict the cost of an automobile weighting 5,000 lbs. and having160 horsepower.

(c) Determine the significance of horsepower for predicting cost.

(d) Find a 95% confidence interval for the coefficient of weight.

(e) Estimate a.

(f) Find the coeffficient of determination.

Solution:


DE 0. [J] 0.0000 Enter program ZS-10

1. (d] LMLRXY Initialize MLRXY for data entry.

3810 [ENTER] 3810.00 Enter x1.

255 [ENTER] 255.00 Enter y1.

7999 [A] 1.00 Enter z completing one triple.

4220 [ENTER] 4220.00 Enter x2 .

180 [ENTER] 180.00 Enter y2.

9221 [A] 2.00 Enter z2 completing two triples.

2900 [ENTER] 2900.00 Enter x3.

96 [ENTER] 96.00 Enter y3 "

8222 [A] 3.00 Enter z3 completing three tri,'es.

3290 [ENTER] 3290.00 Enter x4.

120 [ENTER] 120.00 Enter y4 "

9010 [A] 4.00 Enter z4, completing four triples.44

3400 [ENTER] 3400.00 Enter x5.100 [ENTER] 100.00 Enter y5 "

10099 [A] 5.00 Enter zs, completing five triplets.

3920 [ENTER] 3920.00 Enter x6 .

140 [ENTER] 140.00 Enter y6 '

11019 [A] 6.00 Enter z69 completing six triplets.

4350 [ENTER] 4350.00 Enter x7.

150 [ENTER] 150.00 Enter y7 "

11219 [A] 7.00 Enter z7P completing last triplet.

88

• -~ A".** X

W. 7:


[E] 0.80 Process data and display R.

(J] 0.8050 Exit ZMLRXY and enter ZS-10

1 2. (e] 4.0000 Process trivariate data furtherand display v = 4.

1 3. 2.776 [R/S] 2.7760 Enter t. for CI's..025

D [D] 3361.7167 Display a0.

[RIS] 2.4657 Display a1

[R/Si -19.7686 Recall and display a 2Regression equation: z -3361.72 + 2.466x -19.76

9 y

E 5000 [E] 0.0000 Prepare to predict Z.

160 [R/Sl 12,527.0724 Predicted cost.

C 0 tc] 0.0000 Testing H0 :2 0 vs.H0 201 2

0 [R/SI 0.0318 Significance level for a2 (ts =-3.2).

B [B] 0.6786 Lower 95% limit for a1

[X(>YI 4.2527 Upper 95% limit for a1.

[RCL133 691.7057 ;=s.(RCL127 0.8050 Recall and display R .80.

89

APPENDIX

In the appendix that follows, you will find a complete listing of the

programs discussed in the previous sets of User Instructions. These programsare named according to their ZP or ZS application by chapter and, occasionally,by section. To implement the programs, the first step is to key in each stepinto your calculator exactly as it appears in the listing. (Consult theOwner's Handbook for any instructions that may be unfamiliar.) Next, youshould assign various subroutines using the function ASN according to the

S:- assignments listed just after the register contents in the User Instructionsfor each program. Then place your HP41-C in USER mode and record the programon a magnetic card for future reference.

S.

---S

",'.5 ,- ' . i' % ,' e ., ,_-. , - , .,- . . . , _ . , , . . . . . .. . . . , . . . .. . , . . .

,,',

ZP 2

" I*L L "P2"99 101481. 51L81. d 188 STO 14''"12 EREG 01 52 STO 08 191 RCL 1263 CLZ 53 IX 182 8

04 54 STO 13 183 X=Y?

85 STO 07 184 GTO 18

8 6 21 5541. 83 195 RI".'OO 56 RCL IND 9788 57 RCL 13 186LE1 b89 RTI 58 XEQ 82 197 XEQ 88

59 DSE 801It 181 I 68 CTO 83 2881L 8912 RCL 13 61 RTN 189 RCL 11

118 ST* 1413 1 62#LOL E 111 114 + 63 1/X 112 ST- 11Is STOP 64 STO 13 113 DSE 12

16 STO IND 88 114 CTO 917.RDN 65#LI 84 14~O818 X)Y 66 STOP 115#LEL 1819 Rt 67 STO IND 87 116 RCL 14

8 8 RC. 13 117 STOP28LBL e2 (9 XEQ V2

* 21 STO IND C3 78 CTO 84 .18.18L c22 2+ 119 XEQ e823 2 7!#LBL C 128 X)Y?24 ST+8? 721 121CTO1225 S+ 83 121 I?26 RC! . 73 STO 83 122 RD1

I 2IRTH 74 - 123 CL 11' .27 RIM 75 STO 88 124 WY

28. 1 9 76. 365 125 X)Y?29 , 77 STO 8 126 CT 1229 2 78 STO 62

i'' 39 * 127.L81 1131 18 79.LL e5 128 RCL I!32 881 129 ST* 14

"33 S7 130 134 1 82 RCL S1 131 ST- It3T5 + 83RM 02 132 RcL 1236 SID 10 84 / 133ST/ 14

37 RCL IHND 9 85 ST* 83 134 USE 12,38 ICL IND le 96 )SE 9e 135 GTO 11

. 3*O 05 136 GTG 1846RC195 8? 3 141 89 RCL 83 137*LBL 1242RT 98- 138 84L 91 RIM 139 STOP* *i 43,181 , 148 END

* 44 RCL 05 2 3.. 45 RTN 93 *H:?"

41!194 PROMPT46#LM 1 95 STO It

47 RCL IND 7 96 *:?"48 STOP 97 PRGPT

49 0 INDo 98 slo 1258 G10 V2

91

*: 4 ................... .. ,-.,,........., -,,...,I, ,- -*:*U 4 . - . " . . : .-: .. - -/ :, . . ..,..*- .. ,.

.4/J ZP 3-2

6-,191 "ZP23-2 58'.LBL E"-~ 92 CLR 51 XEQ 6l03 20

e4 $TO e' 52#LL 9385 21 53 RCL IND 81•6 STO 82 54 XEQ 9567 55 DSE 898 RIM 56 CTO 83

91LL A 57LBL 861I STO IND 91 58 RCL e7It STOP 59 RCL 86

12 STO IND 92 68 M13 1 61 -14 ST+ 93 62 5"0 8915 2 63 RCL 8616 ST+ 91 64RT17 ST# 8218 RCL 83 65LBL D19 P.TN 66 XEQ 81

28*L.L 01 67.LBL e421 RCL e3 68 XEO 8222 STO ri 69 DSE 8823 9 78 GTO 8424 STO 86 .71 CIO 8625 STO 0726 20 72*LBL C27 SO 81 73 TO 88

.28 21 74 829 STO 02 75 STO 1638 RTH 76 21

431.121 92 7? SIG V232 RCL IND 81 78*LBL 67

33 STO 89 79 RCL IND @234 XEQ 9 88 ST+ 18

81 235"-1. 85 82 ST+ 8236 sTO e4 83 DSE 8937 XtK 84 GIO 8738 S70 95 85 RCL I

39 RCL IND 82 66 RTH46 ST* 8441 ST" 85 87.LBL B42 RCL 94 88 END43 S7+ 0644 RCL 85

4" 45 ST+ 0746 247 S1 9148 ST+ 0249 RIH

92

• . ," .- ." ," " . . - . . • - *". . - -. " .- , - . " •".- "% - % . - *. . *% -. .44

D-A148 573 HP-41C PROGRAMS AND INSTRUCTIONS FOR PROBABILITY AND 2/2STRTISTICS(U) NAVAL POSTGRADUATE SCHOOL MONTEREY CAP N ZENA FEB 84 NPS55-84-003

UNCLASSIFIED F/G 9

* fkm

1.0 It I EJ,6 111.02.

MICROCOPY RESOLUTION TEST CHARTNATIO*A1 BUREAU-OF STANADARDS-1963-A

..-

"

,* *,,,*,*,, ,,**J ,,.z -,1 ,L.r~~ ,e,,,, .,r*,, ...- , , .,,. - - ' ': "JF ,",""

ZP 3-3

#I.L1L "Z73-3" 5 STO 16 185 RCL 29 ISLBL 18 293 mL 13z CF: 02 53 WCL 14 106 0 1531 204 STO 11 250.81. b

NO p 54 S0 II9 197 la 159 sto 04 205 STO 12 256 XE B04 SIOP 55 RCL 13 lo ENTERt 169 570 92 206 RCL 15 256 -

56 STO 02 199 RCIL 13 161 8 207 RCL 14 25? -

SULBL A 57 XEQ *PTCN" l1 ENTERt 16228 258 CH

06 STO 0 58 ST/ 16 111 RCL 20 163 CTO 18 299 STO 22 259 RTH

I? FS? 92 59 SF 02 112- 164 IN 229 Sit it0 CTO I 2 113 I 211 Sit 12 2.1LB E19 RL 14 60LBL 02 114 # 165#LOL PHTOH ° 212 1 261 RCL 12is STO 24 61 Its * 166 XEG 18 213 - 263 CL 11ItIJX 62 STO 20 116 ENTERt 214 CHS 263 RTM

12 SO 22 63 RM. 16 11? Ra 15 167,118 19 215 STO 21 264 END

13 RO. 15 64 STO 86 118 EKTERt 168 RCL o1 216 ST. 1214 ST- 24 653 07 II, 9 2 169 St 04 217 RCL 1315 St 22 6 0 120 - 179 1 213 YtX16 RCL 22 67 ENTERt 121 1 171 ST- 01 219 STO 1H1? STO 11 68 RCL 69 122 172 DSE 92 229 STO 0618 S10 12 69 X=Y? 123* 173 GTO 19 221 STO e719 1 78 G70 0 222SF e229- 124.1. 84 174*LL 1921 CHS 71*1L. 83 "25 ST 66. 175 RCL 04 2234LCL 6722 STO 21 72 1 12 CL 06 176 RTI 24 023 STS 12 73 ST+ 29 127 ST+ to 225 STO 2124 RCI. 13 74 RCL 23 123 DSE 69 177LBL -C1o05 26 RCL 1625 S10 83 75 ENTER? 129 GIG 03 178 XEQ 18 227 STO e&26 Sit 11 76 RE 15 179 X)Y? 223 STO 0727 St 12 77 - 130oLBL 05 186 GTO 12 229 RL 0823 ENTER? 79 1 131 RM e 181 RI 238 9291 R 14 79 - 132 RCL 87 231 X-Y?39 - tIe CZ 133 RTN 182 I . 2

31 CHS 81 ETERt 184 X)Y?.

V RML 14 82 0 134oLL 86 185 GIG 12 233#'BL 0333 1 83 XY 1351 234 134 - 04 X('Y? 136 STO 6 18641.81. 11 235 ST+ 2935/ 85 CTO e4 137 RCL 15 187 RC O 236 RL 2036 STS 12 86 CS 138 STO011885 04 237 RM 13

8? 72 139IRC20 238 -L 1 88of 148 STO 2b 198 ST- i 239 Cs

38 RC 24 89 RL 14 141 C£Ec tcBOH 191 RL 02 249 139RL 13 93 4 142 S70 23 192St'04 241+4X(Y? 91 R 13 143 RM 14 192DSt024242 RL 2041 CIO 92- 144 STO 1 194 GIG 1 24342 11- 93 EI'Rt 145 RCL 13 195 TO 1 244 RM2243 StO 16 94 0 146 S10 02 245 *44 SF 92 95 XOY 147 XEQ CNSH" 1969LBL 12 246 Rd. 2145 GIO 2 96 X(=y? 148 STf 23 197 9 247 1

97. GG 04 149 RCL 23 248 STO 0646*1.1. 99 98 1 150 CIT 04 249 RCL 6647 RE. 24 99 - 199*481. 251 ST* 6748 STO01 100 ENTERt 1514L91. a 209 SO 00 251 ISE ,a49 RCL 13 lot 0 152 XEQ A 291 FS? 92 252 GIG 9350 STO 92 192 X:Y? 153 1 202 CTO 97 253 CTO 85

51 XE£ "PNTOH" 163 G70 86 154 22IG,104 XOY 155 CIHS

156 RTH

93.oo

* " ' t,,'' , " ,\ '.'',.''. .""'",;..- " '''-,' ', ',,4",'- '4 '.'.,, ,. ".,..,, ). ",.,', "' - ,"

ZP 3-4

52 CHS 103 - 153#LBL. 285 RCL 20

01OLK0 CF041 53 RCL 13 104 CHS 154 XED C 206 /

0 CF 01 54 + 185 RIN 155 ENTERt 207 RM 21

03 CF 02 55 1 1561 209 9

04 CF 63 56 186#LR. E 157 - 209 ST* 86

.0 OF 04 57 RCL 20 107 RCL 12 158 CHS 210 RCL

666- ST58/ IfS RCL 11 159 RTH 211 ST. 67

59 RCL 22 109 RTN 212 DSE 80

i OeL. 8 *0 * 16*LBL A 213 GTO 19042 8O 61 RCL 21 11O#L1 C 161 STO 08 214 GTO 05O Fs 92 62 / fit 5706 9 162 FS? 04

11 GS 02 63 ST* 66 112 FS. 83 163 CTO 15 215LB. 1512 FS? 64 64 RCL 66 113 GTO 13 164 FS? 01 216 RCL 2113 CTO 14 65 ST, B? 1 FS17 64 165 CTO 18 217 ST. 1114 RCL 22 66 USE 06 115 GTO 1! 166 RCL 22 218 CF 0415 T+ 22 67 G1, 68 116 RCL 22 167 !14 219 RCL 80£6 511 117 RCL 13 168 STO 11 220 GTO A1 6 sT+ 12I? I 68#LEL 05 118 * 169 Xt2

1 - 69 RCL 96 119 STO 11 170 ST( 12 2212tL a10 - 78 RL 87 128 STO 12 171 J/X

26 STO 21 71 RTH 121 22 172 SORT 223 ENTiRt21 ST. 1 122 X(Y?. 173 1 224 121 ST, 12 72*LL 11 123 GTO II 174 - 225 -

72 L 73 edL 06 124 XOY 1'. S S 226 CHS23 RT 13 74 .5 125 CHS 176 STO 21 227 RTH24i S1 75 + 126 EtX 177 ST* I!25 S* 12 76 R. It 127 STO 16 173 ST* 12 228.19' J

27 77 - 123 SID 96 179 RCL 22 229 RCL :328228 78 R. 12 129 1097 18 RCL 13 230 -29 X=Y? 79 SORT 136 SF 83 181 ST I I 231 XE0 A39 GOI So 1 82 ST* 12 232 RCL 2131 Rt. 21 81 STO to 131#LBL 13 183 YtX 233 ST/ 1132 RE 21 82 XEO *ZCDF- 132 184 STO 16 234 SF 0433 RtX 83 so 23 133 STO 26 185 ST106 235 eRC 8634 T0 16 04 Rc. N0 134 RCL 16 186 STO i( 236 RMC 6?35 4s O6 85 .5 135 STO 06 187 SF 81 237 RIN35STO 7 86 - 136 ST 08737 ST02 87 C 11 137 RL S8 188LBL 18 232#LBL I

V8- 138 Xz6? 1898 239 STO 6

8937 39 RMt 12 139 GTO 85 190 STO 29 240 FS? 02390 W V SRT 191 RCL 16 241 G1,O20

.40 T0 20 91 1401LBL 14 192 STO 06 242 RCL 2241 RT 16 92 XEO *ZCDF" 141 1 193 ST0 07 24i 1

4251006 93 RCL 23 142 ST, 29 194 RCL 244-43 SO 06 94 - 143 RCL 20 195 Xse? 245 CHS44 S7 I? 5 CHS 144 l/X 196 GTO 05 246 STO 2145 RO 0 96 LRSTX 145 RCL 11 24? 510 1245 ? 97 SF 04 146 * 1974891 19 248 RCL 2246 Xz1, 9 RTN 147 ST, 66 196 1 249 l/X47CT005 148 RCL 06 199 ST+ 29 250 STO 11

49LBL 08 990.1. b 149 ST+ 07 290 RCL 20 251 Xt2

49 1 10 XED 8 158 ISE ea 261 Rd 13 252 STO 12

so S11 20 101 ENTERt 151 GIO 14 202 + 253 SF 82

51 RCL 20 102 1 152 GTO e5 293 1<, . 5 Rt, ;)2@4 -

94

ZF 3-4 (cont'd)

254LL.. 20 301 .361255 RCL 21 362256 EIITEP? 303 +257 RCL 00 304 1/X258 1 305 ENTEPt259 - I306 ENTEPt268 YtX 30? ENTER?261 RCL 22 30 I;338?442262 39263 EN TERt 310 -1.821255978264 ENTER? 311 +265 RCL 21 312 *266 RO. N 313 1.781477937267 Ytx 314 +268 ENTER? 315 *269 1 316 -.356563782270 - 317 +2 71 CMS 318*e2)2 RTh 314 .31939153

273#LBi. d 321274 MEO 1 322 PCL P4275 ENTER. 323 42"76 1 324 FS7 P277 - 325 CTO 29278 CR8 326 RTM279 RTr

3'!7#LBL 2?2ofLBL *ZCIF" 32-8 RCL 03

281 STO 93 329 CM

28? ENTER? 330 STO 03263 4 331 XEO 23284 2 V.2 1285 - 333 XOY286 CM 334-267 EtX281 Pi 335#LBL 29289?2 336 CF i2" 0 337 ENTER?

292' 3391 EMm /S . . M90-

*29430.6 4 341CM295 UP 342 EN)

29? SF 6

298OLIL 232" 10.6

95

ZP 4

33 8.8 216' 258LSL I0' CF 1 $4 RM 23 "S$ 159- 211ST061? 259 XEQ RNNU034 INcu 1a16N 212 1 2616LK046 54510T25 111 - 161 CLI1? 213 - 261 RCL 2205 STOP 57 XEQ 83 112 Cos 162 RTN 214CNS 262 /

e .1~~u 5,xo1 113 Rim 6EO 215 RCL 6? 263 CN$06O1 SICB 59 My#.31 163.1St. 3 216 RIM 264 STO 686ENTER? 115 CF 61 165 ST 821?#St 1526Rd6

85 *41.81. 116 ST~sO 19 166 MY) 213 RCL is 266 RCL 012 62 CF6 10it? Ri 16?SIOP 219 2? T

/I 63 ENTER? 168 XEG C 228 RCL 1412 CPIS 64 ENTER? * I#B *RNNU 169 SI- 08 221 - 8#OC13 EtX 65 1 178IRM ff 222 CNS 29XD*H.U14 PI 1295321 171 1 2C5279 XEG d

12192521CF0 271 S10 60is267015 121 * 172* 224 R1 221i 48311N4 122 .21137 173 RM 8 273 It 017 SORT 14CS25LLb23RL9

I/ 9B c123. 174 dM5 226 1St1b 274 RTN!9SO26 71. 24 FRC 55RM26RL1

78 RC 55? M 25 STO @9-LB 227 3d! 11 275*LCL OX9AR2 10 eO 126 RIM17.3! 228 RIM 276 MEAN22 GTO r, 72 X)l? 12.8 77 XEQ c 22*SE27? RIM23 SF 08 73XG1 2#B 78 - 229tO

74 ENEg in ~ Q &ER -OM0 19SRT 236 510 CO

2 0" 3 75 129 1 ISO * 231 FS? 01 2?0*s.t so*13301 131 /dle !. 11 232 CTO 1? 279 SDEV2.La1 77 11 13R14 182* +31 2 69END

2? K .25 7 132 4 283 RIM 2 221

28. ?srO 0 133 CL 13 2Z3551011.9S .616323 134 184491k Q 236 Xt2

30 1/X 1 s 35Il 13551068H 2ri751012*31 ENTEP? 12 .882853 136FIX1 166 FS7 85 238 SF 81

32 EibIERt 3 137n STOP 13? GIG 1S33 ENTER? 64 RCL N 133 FIX 4 1I8 FS? 61 239#LBL 17

34 1.33274429 85. 39I 119. ItCIO 6 248 RCL0935 s 86 2.515517 140 RIM 1" RU. 13 241 3d! 22

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22 RCL 33 76 RCL 48 136 R£H23 STO 82 7? Rd. 12 176,L. b24 RCL 31 78 * 131.81. A 177 XEO 1r'P*25 STO 93 79 ST- 33 132 RCL 46 178 RCL 47

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33 RCL 36 87 • 139 RCL 48 1O5 XEQ *HYP"34 STO 86 8 RL 06 146 8 186 RCL 4835 RCL 39 89 Xt2 141 EXTERt 187 -36 STO 13 99 - 142 EXTERt Ise CH37 RML 3 91 Id 49 143 R. 39 189 9L3838 RL 85 92 144 2 198039* 93 SRT 145 191 T0 6241 RCL 82 94 RL 33 146-41 95. 147 RTH 19291LL E42 S70 49 96 STO 36 193 RCL 4?43 2 97 RL 03 1484L 9 194.44 RL 04 98 RCL 02 149 RCL 47 19SSTO 0a45 * 99 * 150 STO 49 196 946 Rd. 81 too Rd. 81 151 RCL 31 197 STOP47 *l81 Xt2 152 RCL 3. 198 RcL 4848 RI 86 102 - 153 . 199449 * 193 RCL 49 154 STO 39 200 ST. o59 ST+ 49 164 1 155 CTO 61 201 RCL 4651 RCL 83 195 SORT 262 ST+ 00

52 RCL 96 186 RCL 33 203 RCL 0853 Xf2 197 204 RIM

106

ZSTAT

52*LE 02 153 24-O

01LLIS4'53 RCL 25 165 GT 0 154 YtA ii7 i02'SA* 4CS155 RCL 38 20k.- '3 PROWP 55 STO 25 IA#B 94 Mi 25 20'

i4 RTH Si XEQ 6! 17 S5P #1 211 &

e~tLBL *ZCBF ss -%) 15. EfX 1 ~4 STO 25 584 Cos 16 *I4 3'67 ENTERt III 11 TN 1 213 G',CF@8* 60*LBL 56 12 RCL 21 211 fi~#9 2 61 CF Be lI2*LBL 05 220 ytx

5i/ 13 CF O1 1.4 / 216#1St 'FCCDF11 US 62 ENTERf 114 RT" 165 RCL 24 1 I'1ICS63 1 16 218 STO 1512 Etk 64 - *~ cr~ 6 3 5218 2C 11. P1 65 CHS 115 Th 30IP17ST !7114i PH16.T 166 RCL 30;.5 6 T 11? 40 109 RCL 21 Z FR

11* KIIL :5 N1~S ~67#LBL *ZA* 419 * $. 2 0H TO266 MY 120 GTO 0 1g 12 2 24 5F 1

!~ S T 26 6 ~ 0 Y1 2 1 73 k C L 2 1 2 5 ~ L 1iRCL 25 70 XWV ^;25 MT 13

.zi cr0 04 1 OY 175 STO 26 26212SF 0E 73 44 2 U'E F;c2S6 135' 1 7 7 ST024 Z9 A1/hi502 20IY4.LL l42 12iLE 09 21

is 7 LII 12? INT MOLL HC3 230 X5 Oe1RL 25 77. SQT 128 .3. e~5 C L 5 7 S T 2 51 2 9 X S Y 7 1 6 0 R C L .2 6 2 3 G O 1~ .l4~7$ A0328 J36 121 2 M4 FS 1

79. * 24 s111o.228. 235 GTO 11L i * 0 .8 2 0 5 3 3 2 -1 8 3 S T O 2 6 6 C L 12^. IIX 81 +236 FCL 16

7; FhTE V ~ 04L 2T 30 1 ?w,- RCL 24 238 Ac(:?7£hTE4ERt 8 2.515517 139 GT 6i8 TO 17

27 IW74M85 +187STO 2444I3a2749, *. i36*LB. 66 188 8i RLB 2 11

35 -1.8Z125"7?8 817 .6C 25 13? .5 1og XtY? 241 CF 0237.8 618 139 T 9Y7 390 RC 2 242 RCL 13

3889* 13 G7821969L 243 STO 18

41 + 1RL2 141 1 193 PIN 245 ST0 19

41-366n 31.432M 143L9 6s 262 R4L 11 2547 /E1

44.1313%*1 I#B 9 5 ST. 21

45~9. +%*~* q?1 46PI19

.- - . . . ti .7-.,.

ZSTAT (cont'd)

.28 ST022 3 139T01 361 2 412 - 462 RCL 06 513'258 STO 23 399 STO 2S 362 ' 413 STO 25 463 RCL 80 514 STO 3925.S023., M 546 515 RCL 40

264 STO 24 316 XEQ 15 363RCL25 464

311 STO 21 364 ' 414Lt. 24 465 ST/ 35 516

26t41BL 13 312 CHS 365 SI. 23 415 2 4662 517 ENTERt-v1 B t346 5 18 £ TERt

262 RCL 22 313 ST0 2 366 RCL 23 416 ST+ 5 467 519 RT3

263 RCL 00 314 1 367 ST 22 417 ST* 26 468 STO 16 51, RCL 3?26-- 31 ST 29 368 418 RCL 26 469 RCt 35 529 2

265 CO 14 316 ST- 21 369 ST* 24 419 RCL 15 47914 52266 XOY 317 XEO 12 376 CTO 1 420 X(SY? 471 RC. 34 522 -

26714 318 CHS 421 C70 25 472 s3RT268 RCL 21 319 1 371.LBL -9 422 RC 26 473 1269 * 320 4 372 RCL 29 423 IIX 474 RCL 34 5?44LBL-YP

.j 278 CM 19 321 RTN 373 ST* 22 424 R 25 475- 5225 STO 28

476/ 26 STOPii271 374 RML 22 425 •476 527 ST0OP

272 2 322#LBL 18 375 ST+ 1? 426 Id 26 477 XED "FccF 527 310 34273STI 323 CF 12 .427 X1?2 478 RT 5:3 SF 04

274 / 324 RCL 17 37*LBL 21 428 s 525 RTHRM Is,, l 1 2 ST* ,V4'#BL*

275 ST* 23 , RCL,5 3771 4PS?47;L3i"F"276 1 326 * 373 STO 22 430 RCL 27 480 STO 30 53*L91 "1Y "

277 ST+ 22 327 RC. 16 379 1 431 ST# 22 481 40 531STD2}273RCL23 328 / 380 STO 24 432 GTO 24 482 RM15 5 CF8 4273 ST 24 329 SRT 381 IC1. 15 483 XY? 533 RCL. 21289 ST+ 13 339 RDW 382 X-Y? 4331 . 25 484 CTO 28 534 X,87

331 ATN 323 GTO 26 434 10 22 45 STO 16 5 5T036

2818LL 14 332 STO 17 364 RN 435 RCL 24 4861 $36 X*6?

282M .2 333 SI 35 CL 16 436 * 487 STO 15 537 GTO -99

25 SOT 334 STO 26 386 XY? 437 ST- 17 488 RC.3 538 RIM

284 EIERt 3353 1. 17 387 GTO 23 4" Xt2 539 2

2m R 25 336 COS 388 STO 23 43803BL 26 4" XEG FCC1I 548 6

286 YX 33 STO 21 439 CL 17 491 2 541 1

287 RC 24 333 STO 22 389*LBL 22 440 2 492 542 X)Y?

33 0 STO 23 30 1 441' 493S1066 543 P3mI UTh 340 KG 391 ST- 23 442 PI 494 RCL 16

341 1 392 R 23 443 / 495 STO 1 5 545 2

29LBL 15 342 STO 24 393 S1 24 444 1 496 36 546 -

291 CL 15 343 ST0 25 394 1 445 - 497 6 547 KS

22 CL 16 344 t. 16 3" ST- 23 446 CIS 4" X) 548 RN

3 / 345 XY? 396 RL 23 447 RTN 4"9 GTO 27294 RCL 17 346 GTO 21 397 ST/ 24 50 RelO0 a S lM * 347 2 398 XY?. 448LBL "FCW* 561 1

2% 1 341 - 399 CT0 22 449 XEO FCCVF 52 - 551 1

29 #7 349 STO 14 458 1 583 CPS 552 -2 14 48PeLBL 23 451 - 564 RTN 53 CNS

299 KIN M.IL3 19 401 RCL 21 452 CHS 554 ITN

351 RCL14 462 EHTEPt 33 R 6 565I.L 27

364Ut. 16 352 C 25 493 RCL 16 596 a 55*L'L 36

31 CF o 353 XKY? 404 YtX 454*L181 "NF 56iTN 556 RIo

3W 6 02 354 C70 2m 45 R0 26 455S 9Og 557 RN

3 CTO I 5 52 496 * 456 1 5884L 216356 ST* 2S 407 ST* 24 457 * 569 9c' 3 56.LSH -iS

364OLi 17 357 . 21 48 RCL 16 458 STO 35 511 CTO "ZCIF 55.0 . 4?

301S R 6 358 Xt2 4"91 459 2 56810 '47

3K6 STO It 359 RCL 24 410 S70 26 466 ' . 511,LB1 "1* 561 -

3W0 R 15 360' 411 STO 27 461 STO is 512 RC1 32 562 STO 4%3 L}iI

108

-1u I-r--9

I- 4 - -- - -

DISTRIBUTION LIST

NO. OF COPIES

Library, Code 0142 4

Naval Postgraduate SchoolMonterey, CA 93943

Office of Research AdministrationICode 012ANaval Postgraduate School

$1., Monterey, CA 93943

Library, Code 551Naval Postgraduate SchoolMonterey, CA 93943

Professor P. w. Zehna 10

V Code M5eNaval Postgraduate SchoolMonterey, CA 93943

Professor D. Rt. BarrICode 55BnNaval Postgraduate SchoolMonterey, CA 93943

Professor Rt. Shudde1Code 55SuNaval Postgraduate School

* Monterey, CA 93943

Professor P. A. W. Lewis1Code 55LvNaval Postgraduate SchoolMonterey, CA 93943

109

-. ~V .0~

-. 44. k, .

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