ENGI 4421 Probability and Statistics Faculty of Engineering and Applied Science Problem Set 10 Solutions Chi-Square Tests; Simple Linear Regression

Similar documents
Use 10 m/s 2 for the acceleration due to gravity.

Chapter 2 Linear Mo on

PRINCE SULTAN UNIVERSITY Department of Mathematical Sciences Final Examination Second Semester (072) STAT 271.

PRINCE SULTAN UNIVERSITY Department of Mathematical Sciences Final Examination First Semester ( ) STAT 271.

Dishonest casino as an HMM

ENGI 4421 Probability & Statistics

In the complete model, these slopes are ANALYSIS OF VARIANCE FOR THE COMPLETE TWO-WAY MODEL. (! i+1 -! i ) + [(!") i+1,q - [(!

Comparing Several Means: ANOVA. Group Means and Grand Mean

Comparing Possibly Misspeci ed Forecasts

Applied Statistics Qualifier Examination

R th is the Thevenin equivalent at the capacitor terminals.

Ch 2: Simple Linear Regression

Estimating σ 2. We can do simple prediction of Y and estimation of the mean of Y at any value of X.

The Components of Vector B. The Components of Vector B. Vector Components. Component Method of Vector Addition. Vector Components

Exercises H /OOA> f Wo AJoTHS l^»-l S. m^ttrt /A/ ?C,0&L6M5 INFERENCE FOR DISTRIBUTIONS OF CATEGORICAL DATA. tts^e&n tai-ns 5 2%-cas-hews^, 27%

TSS = SST + SSE An orthogonal partition of the total SS

Statistics 423 Midterm Examination Winter 2009

Lecture 1 Linear Regression with One Predictor Variable.p2

4.8 Improper Integrals

Decompression diagram sampler_src (source files and makefiles) bin (binary files) --- sh (sample shells) --- input (sample input files)

Asynchronous Sequen<al Circuits

Lecture Notes for STATISTICAL METHODS FOR BUSINESS II BMGT 212. Chapters 14, 15 & 16. Professor Ahmadi, Ph.D. Department of Management

Lecture 4 ( ) Some points of vertical motion: Here we assumed t 0 =0 and the y axis to be vertical.

Natural Language Processing NLP Hidden Markov Models. Razvan C. Bunescu School of Electrical Engineering and Computer Science

The Buck Resonant Converter

Correlation Analysis

Experimental Design and the Analysis of Variance

The Perceptron. Nuno Vasconcelos ECE Department, UCSD

Topic 7: Analysis of Variance

Interval Estimation. Consider a random variable X with a mean of X. Let X be distributed as X X

e t dt e t dt = lim e t dt T (1 e T ) = 1

CHAPTER 24: INFERENCE IN REGRESSION. Chapter 24: Make inferences about the population from which the sample data came.

AP Statistics Practice Test Unit Three Exploring Relationships Between Variables. Name Period Date

2015 Sectional Physics Exam Solution Set

Confidence Interval for the mean response

Inference for Regression

Distribution of Mass and Energy in Five General Cosmic Models

17 - LINEAR REGRESSION II

PH2200 Practice Exam I Summer 2003

UniSuper s Approach to Risk Budgeting

Faculty of Engineering

10.7 Power and the Poynting Vector Electromagnetic Wave Propagation Power and the Poynting Vector

Lecture 3: Inference in SLR

Statistics for Managers Using Microsoft Excel/SPSS Chapter 14 Multiple Regression Models

DCDM BUSINESS SCHOOL NUMERICAL METHODS (COS 233-8) Solutions to Assignment 3. x f(x)

CHAPTER EIGHT Linear Regression

Inference for Regression Simple Linear Regression

Principle Component Analysis

Biostatistics 380 Multiple Regression 1. Multiple Regression

***SECTION 12.1*** Tests about a Population Mean

Group Theory Problems

Department of Quantitative Methods & Information Systems. Time Series and Their Components QMIS 320. Chapter 6

Minimum Squared Error

Design of Diaphragm Micro-Devices. (Due date: Nov 2, 2017)

Minimum Squared Error

OVERVIEW Using Similarity and Proving Triangle Theorems G.SRT.4

Modeling and Predicting Sequences: HMM and (may be) CRF. Amr Ahmed Feb 25

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

Lecture 10 Multiple Linear Regression

REVIEW OF ENGINEERING THERMODYNAMICS

Statistics for Managers using Microsoft Excel 6 th Edition

K The slowest step in a mechanism has this

11.2. Infinite Series

T h e C S E T I P r o j e c t

Physics 201 Lecture 2

Simplified Variance Estimation for Three-Stage Random Sampling

Determining the Accuracy of Modal Parameter Estimation Methods

Average & instantaneous velocity and acceleration Motion with constant acceleration

Linear models and their mathematical foundations: Simple linear regression

Solutions to Problems. Then, using the formula for the speed in a parabolic orbit (equation ), we have

SUMMER REV: Half-Life DUE DATE: JULY 2 nd

Overview Scatter Plot Example

Statistics MINITAB - Lab 2

AP Physics 1 MC Practice Kinematics 1D

Square law expression is non linear between I D and V GS. Need to operate in appropriate region for linear behaviour. W L

STA 302 H1F / 1001 HF Fall 2007 Test 1 October 24, 2007

where x and ȳ are the sample means of x 1,, x n

Basic Business Statistics, 10/e

Chapter 2. Continued. Proofs For ANOVA Proof of ANOVA Identity. the product term in the above equation can be simplified as n

Objective of curve fitting is to represent a set of discrete data by a function (curve). Consider a set of discrete data as given in table.

STA 108 Applied Linear Models: Regression Analysis Spring Solution for Homework #6

STAT 350 Final (new Material) Review Problems Key Spring 2016

Trigonometry. Trigonometry. Solutions. Curriculum Ready ACMMG: 223, 224, 245.

Chapter 14 Student Lecture Notes Department of Quantitative Methods & Information Systems. Business Statistics. Chapter 14 Multiple Regression

Chapter Summary. Mathematical Induction Strong Induction Recursive Definitions Structural Induction Recursive Algorithms

Multiple Regression. Inference for Multiple Regression and A Case Study. IPS Chapters 11.1 and W.H. Freeman and Company

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.

The Multiple Regression Model

Multiple Linear Regression

INTERNATIONAL JOURNAL OF CIVIL AND STRUCTURAL ENGINEERING Volume 1, No 3, 2010

STAT 3340 Assignment 1 solutions. 1. Find the equation of the line which passes through the points (1,1) and (4,5).

ELG3150 DGD 6 P9.5; P9.7; P9.13; P9.18; AP9.5; DP9.2

Probability and. Lecture 13: and Correlation

Statistics for Business and Economics

Supporting information How to concatenate the local attractors of subnetworks in the HPFP

Inference for the Regression Coefficient

ALGEBRA 2/TRIGONMETRY TOPIC REVIEW QUARTER 3 LOGS

ACE 564 Spring Lecture 7. Extensions of The Multiple Regression Model: Dummy Independent Variables. by Professor Scott H.

Hubble s Law PHYS 1301

CH.3. COMPATIBILITY EQUATIONS. Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Transcription:

ENGI 441 Prbbly nd Sscs Fculy f Engneerng nd Appled Scence Prblem Se 10 Sluns Ch-Squre Tess; Smple Lner Regressn 1. Is he fllwng se f bservns f bjecs n egh dfferen drecns cnssen wh unfrm dsrbun? Drecn Number f bjecs bserved N 4 NE 6 E 13 SE 11 S 1 SW 18 W 11 NW 5 There re egh cegres, cnnng l f O = 80 80 E = = 10 > 5 8 Tes : d re frm unfrm dsrbun vs. : n. Exendng he ble, Drecn O E O E ( O ( O E E E N 4 10 6 36 3.6 NE 6 10 4 16 1.6 E 13 10 3 9 0.9 SE 11 10 1 1 0.1 S 1 10 4 0.4 SW 18 10 8 64 6.4 W 11 10 1 1 0.1 NW 5 10 5 5.5 Tl: 80 80 0 15.6

ENGI 441 Prblem Se 10 Sluns Pge f 14 1 (cnnued The ch-squre es ssc s χ bs = 15.6. The number f degrees f freedm s ν = n 1= 7 The crcl vlue s χ α,7. A α =.05, c = χ = 14.067, bu α =.01, c = χ = 18.475.05, 7.01, 7 χ bs = 15.6 The decsn depends n he level f sgnfcnce α chsen. A α =.05, χ > c rejec, bu bs bs A α =.01, χ < c d n rejec. Ths resul s mrgnlly sgnfcn. Frm he ssced Excel fle, he p-vlue s.09, s h.01 < p <.05. The nswer s NO YES f α =.05 f α =.01. A mdel f chemcl prcess predcs he numbers f runs f prcess h shuld fll n ech f fve me nervls. Are he vlues shwn n he ble belw cnssen wh he vlues expeced frm he mdel 5% level f sgnfcnce? Tme nervl bserved expeced 0-5 7 10 5-10 19 5 10-15 7 30 15-0 31 5 > 0 16 10 O = E = 100 nd every expeced vlue s > 5. Tes : d re frm he mdel dsrbun vs. Exendng he ble, : n, α =.05.

ENGI 441 Prblem Se 10 Sluns Pge 3 f 14 (cnnued Tme nervl O E O E O ( E ( O E E 0-5 7 10 3 9 0.90 5-10 19 5 6 36 1.44 10-15 7 30 3 9 0.30 15-0 31 5 +6 36 1.44 > 0 16 10 +6 36 3.60 Tl: 100 100 0 7.68 The ch-squre es ssc s χ bs = 7.68. The number f degrees f freedm s ν = n 1= 4 The crcl vlue s c = χα,4 = χ.05,4 = 9.48. A α =.05, χ < c d n rejec. bs Frm he ssced Excel fle, he p-vlue s.104, s h p >.05. The nswer s YES 3. Rndm smples f cbles re ken frm shpmens frm fur mnufcurers. Ech cble n ech smple s grded ccrdng he ld ppled when h cble breks. The resuls re summrzed n hs cnngency ble. Mnufcurer Quly f cble A B C D X 8 14 6 Y 14 61 19 6 Z 10 1 10 9 Are hese d cnssen wh ndependence beween mnufcurer nd quly f cble? Tes : he fcrs mnufcurer nd quly re ndependen vs. ν = I 1 J 1 = 3 = 6 ( ( : n.

ENGI 441 Prblem Se 10 Sluns Pge 4 f 14 3 (cnnued Exendng he ble, Mnufcurer Quly f cble A B C D Tl X 8 14 6 50 Y 14 61 19 6 100 Z 10 1 10 9 50 Tl 3 104 43 1 00 The ble f vlues expeced when s rue s clculed by ej = 3 100 43 Fr exmple, e3 = = = 1.5. 00 Expeced vlues: Quly f cble A B C D Tl X 8 6 10.75 5.5 50 Mnufcurer Y 16 5 1.50 10.50 100 Z 8 6 10.75 5.5 50 Tl 3 104 43 1 00 ( j ej All expeced vlues re > 5. The vlues f ej re: Quly f cble A B C D X 0.000 0.615... 0.98... 0.107... Mnufcurer Y 0.50 1.557... 0.90... 1.98... Z 0.500 0.961... 0.05....678... j. 3 4 = 1 j= 1 ( j ej χbs = = e j 9.94 c= χ = χ = 1.59 r χ = 16.81 χ < c,6.05,6.01,6 bs α Frm he ssced Excel fle, he p-vlue s.17, s h p >.05. D n rejec. Therefre YES These d re cnssen wh ndependence beween mnufcurer nd quly f cble.

ENGI 441 Prblem Se 10 Sluns Pge 5 f 14 4. A prculr ype f mr s knwn hve n upu rque whse rnge n nrml pern fllws nrml dsrbun. Seven mrs re chsen rndm nd re esed wh he ld nd new mehds f cnrllng he rnge f rque vlues. The resuls f he ess re s fllws: Mr: 1 3 4 5 6 7 New mehd: 5.5 3.16 4.43 6.1 5.75.1 6.01 Old mehd: 7.83 6. 7.46 8.83 8.19 5.64 8.88 ( Jusfy yur chce f mehd n (b belw. The d re prs f mesuremens n sngle se f ndvduls (he seven mrs. Therefre he pprpre mehd s pred w-smple -es (b Cnduc n pprpre hyphess es deermne wheher here s suffcen evdence cnclude h he rnge f rques wh he new mehd s les uns less hn wh he ld mehd. Le X = rque (new mehd Y = rque (ld mehd nd D = X Y Tes : µ D = vs. : µ D < Free chce f α. I chse α =.01. Mr: 1 3 4 5 6 7 Sum SSq x = New Mehd: 5.5 3.16 4.43 6.1 5.75.1 6.01 y = Old Mehd: 7.83 6. 7.46 8.83 8.19 5.64 8.88 d = Dfference:.58 3.06 3.03.71.44 3.43.87 0.1 58.5004 D: n= 7, d = 0.1, d = 58.5004 4.688 4 d =.874 85, sd = = 0.1116, sd = 0.334108 4

ENGI 441 Prblem Se 10 Sluns Pge 6 f 14 4 (b (cnnued Mehd 1 s c = µ D.01, 6 n 0.334 = 3.14 7 = 3.14 0.16 = 0.396 =.396 OR OR d =.87 < c Rejec n fvur f. Mehd d µ D.87 ( bs = = = 6.9 s 0.16 n = = c bs <.01, 6 3.14 c Rejec n fvur f. Mehd 3 bs = = 6.9 Clerly P[T < 6.9] <<.01 [Frm sfwre, P[T < 6.9] < 0.0003] Rejec n fvur f ny resnble vlue f α. An Excel spredshee fle s ls vlble llusre hs slun. (c Use he smple lner regressn mdel n hese d fnd he equn f he lne f bes f hese d. Ne h he lne f bes f depends n whch vrble s he predcr nd whch s he respnse. The rles f X nd Y re nerchnged f he new mehd s he respnse, (whch s he mre nurl ssgnmen.

ENGI 441 Prblem Se 10 Sluns Pge 7 f 14 4 (c (cnnued Summry sscs (respnse = Y = ld mehd: n= 7 x= 3.93 x = 168.6941 nsxx = n x ( x = 96.4738 n S xy = n xy ( x( y = 78.4941 nsyy = n y ( y = 65.08 xy= y= y = 60.7758 53.05 411.3579 S ˆ xy 78.4941 β1 = = = 0.813631656 S 96.4738 xx nd β 1 0 ( y β1 x ˆ = ˆ = n 3.751017 490 Therefre, (crrec 3 s.f. n ech ceffcen, OR y = 0.814 x + 3.75 Summry sscs (respnse = Y = new mehd: n= 7 x= 53.05 x = 411.3579 xy= y= y = 60.7758 3.93 168.6941 xx ( 65.08 ( ( 78.4941 ( 96.4738 ns = n x x = n S = n xy x y = xy ns = n y y = yy S ˆ xy 78.4941 β1 = = = 1.03845540 S 65.08 xx nd β 1 0 ( y β1 x ˆ = ˆ = n 4.41914370 Therefre, (crrec 3 s.f. n ech ceffcen, y = 1.0 x 4.4 An Excel spredshee fle s vlble llusre he secnd versn f hs slun.

ENGI 441 Prblem Se 10 Sluns Pge 8 f 14 4 (d Fnd he ceffcen f deermnn R nd use cmmen n yur nswer pr ( bve. Irrespecve f whch f ld r new mehds s he regressr, ( nsxy ( nsxx ( nsyy ( 78.4941 r = = =.979 49 96.4738 65.08 r = 97.9% (crrec 3 s.f. Very hgh crreln cnn use unpred -es Chce n ( s crrec. A Mnb prjec fle nd Mnb Repr Pd RTF fle re ls vlble llusre hs slun. 5. A sudy ws cnduced nlyze he relnshp beween dversng expendure nd sles. The fllwng d were recrded: X Y Adversng ($ Sles ($ 0 310 4 340 30 400 3 40 35 490 Assume smple lner regressn beween sles Y nd dversng X. Clcule he ceffcens β 0 nd β 1 f he lne f bes f hese d nd esme he sles when $8 re spen n dversng. Is here sgnfcn lner sscn beween Y nd X? Exendng he ble: x y x x y y 0 310 400 600 96100 4 340 576 8160 115600 30 400 900 1000 160000 3 40 104 13440 176400 35 490 15 17150 40100 Sum: 141 1960 415 56950 78800

ENGI 441 Prblem Se 10 Sluns Pge 9 f 14 5 (cnnued n S xx = n Σ x (Σ x = 5 415 (141 = 065 19881 = 744 n S xy = n Σ xy (Σ x Σ y = 5 56950 141 1960 = 84750 76360 = 8390 n S yy = n Σ y (Σ y = 5 78800 (1960 = 3941000 3841600 = 99400 S ˆ xy 8390 β1 = = = 11.7688170 S 744 xx 1 nd β0 ( y β1 x ( ˆ ˆ 1960 11. 141 = = = 73.991935 483 n 5 The regressn lne ( 4 s.f. s y = 11.8 x + 73.99 x = 8 y = 11.77... 8 + 73.99... = 389.744 63 7... When $8 s spen n dversng, we predc $390 f sles (crrec he neres dllr. There re mny chces f mehd fr n hyphess es fr lner sscn. Tes : ρ = 0 vs. : ρ 0 r, equvlenly, es : β 1 = 0 vs. : β1 0 ( nsxy ( nsxx ( nsyy 8390 r = = = 744 99 400.95184 jus ver 95% f ll vrn n y s explned by he lner regressn. The crreln (.975... s very srng, (whch suggess h here s lner sscn, bu he smple sze s very smll, s we shuld prceed wh frml hyphess es. r n.95 3 = = 1 r 1.95 7.700 r

ENGI 441 Prblem Se 10 Sluns Pge 10 f 14 5 (cnnued ( nsxy ( n ( nsxx ( nsyy ( nsxy 8390 3 = = 744 99 440 8390 7.700 r clcule he enres n he ANOVA ble s fllws: ( nsxy n( nsxx 8390 SSR = ( yˆ y = = = 189.607 56 5 744 nsyy SST = y y = S yy = = = n 5 ( 99 440 19880 SSE = SST SSR = 957.39 473... MSR = SSR / ν R = SSR = 18 9.607 56... MSE = SSE / ν E = SSE / 3 = 319.130 84 37... f = MSR / MSE = 59.94... d.f. SS MS f R 1 18 9.607 5... 18 9.607 5... 59.94... E 3 957. 39 48... 319.130 86 6... T 4 19 880 = + f = 7.700 r MSE n MSE 5 319.13 sb = = = = S 744 xx ( nsxx ˆ β 0 11.7 s.14 1 = = = b 7.700.144969 Cmpre he bserved = 7.700....005, 3 = 5.841... > α/, 3 fr ny resnble chce f α. [The p-vlue s less hn.0046.] Rejec n fvur f YES, here s sgnfcn lner sscn beween Y nd X. [See ls he ssced Excel, Mnb prjec nd Mnb RTF fles.]

ENGI 441 Prblem Se 10 Sluns Pge 11 f 14 6. [Ths s n exensn f Exmple 1.06 frm he lecure nes.] A smple f 10 desel rucks were run bh h nd cld esme he dfference n fuel ecnmy. The resuls, n mles per glln, re presened n he fllwng ble. (frm In-use Emssns frm evy-duy Desel Vehcles, J. Ynwz, Ph.D. hess, Clrd Schl f Mnes, 001. Truck Cld 1 4.56 4.6 4.46 4.08 3 6.49 5.83 4 5.37 4.96 5 6.5 5.87 6 5.90 5.3 7 4.1 3.9 8 3.85 3.69 9 4.15 3.74 10 4.69 4.19 ( Use he smple lner regressn mdel n hese d fnd he equn f he lne f bes f (fr s he respnse nd Cld s he regressr hese d mnully. Summry sscs: n= 10, x= 45.86, y = 49.84 ns ( xx = n x x = 63.5604 nsxy = n xy ( x( y = 71.356 ns ( yy = n y y = 81.764 x = 16.6700, xy= 35.6988, y = 56.530 S ˆ xy 71.356 β1 = = = 1.1170 Sxx 63.5604 1 ˆ β0 = ( y ˆ β1 x = n 0.16 73 Therefre, crrec hree sgnfcn fgures n ech ceffcen, he regressn lne s y = 1.1 x 0.16

ENGI 441 Prblem Se 10 Sluns Pge 1 f 14 6 (b Clcule he 95% predcn nervl fr sngle fuure bservn f fuel effcency when run h, fr ruck whse fuel effcency when run cld s 4.00 mles per glln. The frmul fr he predcn nervl s ( ˆ ˆ 1 n x β0 + β1x ± α /, ( n s 1 + + n ( nsxx ( nsyy ( nsxy ( ( ns ( x ( ns xx ( 63.5604 81.764 71.356 s = MSE = = n n 10 8 63.5604 = 0.015 461 s = 0.14344 α =.05 = =.306 01 α /, n x 45.86 x = = = 4.586 n 10 xx.05, 8 ( ( x = 4.00 x x = 0.586 ˆ = ˆ + ˆ = 0.16 + 1.1 4 = 4.36 408 Therefre he 95% PI x = 4 s nd y β0 β1x 1 10 0.343396 4.36 408 ±.306 01 0.015 461 1 + + 10 63.5604 = 4.36 408 ± 0.30806 = ( 4.018, 4.634] ( 3 d.p.

ENGI 441 Prblem Se 10 Sluns Pge 13 f 14 6 (c Use Mnb (r nher sfwre pckge check h he dsrbun f he resduls s cnssen wh nrml dsrbun. Ths nrml prbbly pl clerly demnsres cnssency wh nrml dsrbun. (d Use Mnb (r nher sfwre pckge shw he ANOVA ble, he equn f he lne f bes f nd dsply he lne f bes f, he 95% cnfdence nervls nd he 95% predcn nervls n sngle dgrm. Regressn Anlyss: versus Cld The regressn equn s = - 0.163 + 1.1 Cld S = 0.14344 R-Sq = 98.5% R-Sq(dj = 98.3% Anlyss f Vrnce Surce DF SS MS F P Regressn 1 8.00395 8.00395 517.67 0.000 Errr 8 0.1369 0.01546 Tl 9 8.1764

ENGI 441 Prblem Se 10 Sluns Pge 14 f 14 6 (d (cnnued 7.0 6.5 6.0 ENGI 441 Prblem Se 10 Quesn 6 = - 0.163 + 1.1 Cld Regressn 95% CI 95% PI S 0.14344 R-Sq 98.5% R-Sq(dj 98.3% 5.5 5.0 4.5 4.0 3.5 4.0 4.5 Cld 5.0 5.5 6.0 (e Fnd he smple crreln ceffcen r beween h nd cld fuel effcences, crrec hree sgnfcn fgures. SSR 8.00395 r = = =.984 =.99 SST 8.17 64 Crrec hree sgnfcn fgures, r =.99 (whch verfes he clm f srng crreln mde n Exmple 1.06. An Excel fle, Mnb prjec fle nd Mnb RTF fle re vlble fr hs quesn. Bck he ndex f sluns