Chapter 9: Inferences based on Two samples: Confidence intervals and tests of hypotheses

Similar documents
Comparison Procedures

Tests for the Ratio of Two Poisson Rates

For the percentage of full time students at RCC the symbols would be:

14.3 comparing two populations: based on independent samples

Student Activity 3: Single Factor ANOVA

Section 11.5 Estimation of difference of two proportions

Lecture 21: Order statistics

The steps of the hypothesis test

Normal Distribution. Lecture 6: More Binomial Distribution. Properties of the Unit Normal Distribution. Unit Normal Distribution

Continuous Random Variable X:

MIXED MODELS (Sections ) I) In the unrestricted model, interactions are treated as in the random effects model:

Lecture 3 Gaussian Probability Distribution

Continuous Random Variables

Acceptance Sampling by Attributes

Continuous Random Variables

Chapter 2 Organizing and Summarizing Data. Chapter 3 Numerically Summarizing Data. Chapter 4 Describing the Relation between Two Variables

4.1. Probability Density Functions

Chapter 5 : Continuous Random Variables

Design and Analysis of Single-Factor Experiments: The Analysis of Variance

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Solution for Assignment 1 : Intro to Probability and Statistics, PAC learning

( ) 1. Algebra 2: Final Exam Review. y e + e e ) 4 x 10 = 10,000 = 9) Name

Lecture INF4350 October 12008

Non-Linear & Logistic Regression

Expectation and Variance

8 Laplace s Method and Local Limit Theorems

CS667 Lecture 6: Monte Carlo Integration 02/10/05

Hybrid Group Acceptance Sampling Plan Based on Size Biased Lomax Model

1 Probability Density Functions

Experiments with a Single Factor: The Analysis of Variance (ANOVA) Dr. Mohammad Abuhaiba 1

INTRODUCTION TO INTEGRATION

3.4 Numerical integration

Chapter 6 Continuous Random Variables and Distributions

Vyacheslav Telnin. Search for New Numbers.

Math 135, Spring 2012: HW 7

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

38.2. The Uniform Distribution. Introduction. Prerequisites. Learning Outcomes

Credibility Hypothesis Testing of Fuzzy Triangular Distributions

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Predict Global Earth Temperature using Linier Regression

than 1. It means in particular that the function is decreasing and approaching the x-

The graphs of Rational Functions

Coimisiún na Scrúduithe Stáit State Examinations Commission

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Lesson 1.6 Exercises, pages 68 73

7 - Continuous random variables

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

The Regulated and Riemann Integrals

APPROXIMATE INTEGRATION

p-adic Egyptian Fractions

1B40 Practical Skills

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:

Math 116 Final Exam April 26, 2013

Physics 202H - Introductory Quantum Physics I Homework #08 - Solutions Fall 2004 Due 5:01 PM, Monday 2004/11/15

: : 8.2. Test About a Population Mean. STT 351 Hypotheses Testing Case I: A Normal Population with Known. - null hypothesis states 0

The Shortest Confidence Interval for the Mean of a Normal Distribution

( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that

MATH 144: Business Calculus Final Review

Monte Carlo method in solving numerical integration and differential equation

X Z Y Table 1: Possibles values for Y = XZ. 1, p

Chapter 3. Extensions

Method: Step 1: Step 2: Find f. Step 3: = Y dy. Solution: 0, ( ) 0, y. Assume

Penalized least squares smoothing of two-dimensional mortality tables with imposed smoothness

5 Probability densities

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

Genetic Programming. Outline. Evolutionary Strategies. Evolutionary strategies Genetic programming Summary

Testing categorized bivariate normality with two-stage. polychoric correlation estimates

Section 6.1 Definite Integral

Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2.

A Brief Review on Akkar, Sandikkaya and Bommer (ASB13) GMPE

Arithmetic & Algebra. NCTM National Conference, 2017

Bayesian Networks: Approximate Inference

4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve

A New Statistic Feature of the Short-Time Amplitude Spectrum Values for Human s Unvoiced Pronunciation

Math 360: A primitive integral and elementary functions

Quantum Physics II (8.05) Fall 2013 Assignment 2

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph.

DOING PHYSICS WITH MATLAB MATHEMATICAL ROUTINES

Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date: , the DeMoivre-Laplace Theorem states: ( ) n x.

Inference on One Population Mean Hypothesis Testing

Thermal Diffusivity. Paul Hughes. Department of Physics and Astronomy The University of Manchester Manchester M13 9PL. Second Year Laboratory Report

Probabilistic Investigation of Sensitivities of Advanced Test- Analysis Model Correlation Methods

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

pivot F 2 F 3 F 1 AP Physics 1 Practice Exam #3 (2/11/16)

!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise.

Multi-Armed Bandits: Non-adaptive and Adaptive Sampling

Chapter = 6.9, and for red and brown is ( ) ( ) ( ) ( ) ( ) ( )

Problem. Statement. variable Y. Method: Step 1: Step 2: y d dy. Find F ( Step 3: Find f = Y. Solution: Assume

Best Approximation. Chapter The General Case

Riemann is the Mann! (But Lebesgue may besgue to differ.)

#6A&B Magnetic Field Mapping

New Expansion and Infinite Series

Discrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17

Data Structures and Algorithms CMPSC 465

Recitation 3: More Applications of the Derivative

Strategy: Use the Gibbs phase rule (Equation 5.3). How many components are present?

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

Read section 3.3, 3.4 Announcements:

Transcription:

Chpter 9: Inferences bsed on Two smples: Confidence intervls nd tests of hypotheses 9.1 The trget prmeter : difference between two popultion mens : difference between two popultion proportions : rtio of two popultion vrinces 9. Compring two popultion mens: independent smpling Cse 1, Lrge smples Conditions required for vlid lrge smple: 1. two smple nd selected from two independent popultion,. nd n 30 x x Smpling distribution of ( 1 ) is pproximtely norml with: Men: x1 x µ = ( ) 1 Stndrd error: σ ( x1 x) = + n n σ σ 1 (1- α ) 100% Confidence intervl for ( µ 1 µ ) (the difference of two popultion mens): Interpret: We re (1- α ) 100% confident tht the true between these two popultions will flls in this intervl. Exmple1, IETSTUY, To investigte the effect of new low-ft diet on weight loss, two rndom smples of 100 people ech re selected. One group of 100 is plced on the low-ft diet, while the other group with regulr diet. For ech person, the mount of weight lost (or gined) in 3-week period is recorded. iet Low-ft diet (1) Regulr diet () Weight loss 8, 1, 13,.., 10 (100 observtions) 6, 14, 4,, 8 (100 observtions) 1

Q: Form 95% confidence intervl for ( µ 1 µ ), the difference between the popultion men weight losses for the two diets. Interpret the result. Group Sttistics IET N Men Std. evition Std. Error Men WTLOSS LOWFAT 100 9.31 4.668.467 REGULAR 100 7.40 4.035.404 How cn we mke inference bsed on (1- α ) 100% Confidence intervl for ( µ )? If the confidence intervl, it implies tht there is difference between these two popultion mens; If the confidence intervl, it implies tht there is difference between these two popultion mens. Exmple: A confidence intervl for ( µ ) is (-10, 4), wht inference cn we mke? A confidence intervl for ( µ ) is (-18, -9), wht inference cn we mke? A confidence intervl for ( µ ) is (3.6, 1.4), wht inference cn we mke?

Hypothesis test for ( µ ): Criticl-vlue pproch: 1.. significnce level α 3. test sttistic: 4. rejection region : when H 1 µ 0 when H < 0 when H > 0 5. conclusion: if the vlue of test sttistic flls in R.R, reject H 0,nd conclude tht t α level, there is sufficient evidence to conclude H is true. if the vlue of test sttistic does not fll in R.R, do not reject H 0, nd conclude tht t α level, there is insufficient evidence to conclude P-vlue pproch: 1. H µ = 0 1 0 H µ ( or µ µ < or µ µ > ) 1 0 1 0 1 0. significnce level α ( x1 x) 0 3. test sttistic z0 = S1 S + n n 1 H is true. 4. p-vlue = when H 1 µ 0; when H < 0; when H > 0. 5. Conclusion: if p-vlue is α, reject H 0, nd conclude tht t α level, there is sufficient evidence to conclude H is true. If p-vlue is no less thnα, H 0, nd conclude tht t α level, there is insufficient evidence to conclude H is true. 3

Exmples for compring two popultion mens: independent, lrge-smples: Exmple1, IETSTUY, To investigte the effect of new low-ft diet on weight loss, two rndom smples of 100 people ech re selected. One group of 100 is plced on the low-ft diet, while the other group with regulr diet. For ech person, the mount of weight lost (or gined) in 3-week period is recorded. iet Low-ft diet (1) Regulr diet () Weight loss 8, 1, 13,.., 10 (100 observtions) 6, 14, 4,, 8 (100 observtions). At =0.05, conduct test of hypothesis to determine whether the men weight loss for low-ft diet is different from tht of regulr diet. b. At =0.05, conduct test of hypothesis to determine whether the men weight loss for low-ft diet is greter thn tht of regulr diet. 4

SPSS output for IETSTUY, Group Sttistics IET N Men Std. evition Std. Error Men WTLOSS LOWFAT 100 9.31 4.668.467 REGULAR 100 7.40 4.035.404 Independent Smples Test Levene's Test for Equlity of Vrinces t-test for Equlity of Mens 95% Confidence F Sig. t df Sig. (-tiled) Men ifference Std. Error ifference Intervl of the ifference Lower Upper WTLOSS Equl vrinces ssumed Equl vrinces not ssumed 1.367.44 3.095 198.00 1.910.617.693 3.17 3.095 193.940.00 1.910.617.693 3.17 Cse, Smll smples with equl vrinces Conditions required for vlid smll smple: 1. The two smples re nd selected from the two trget popultion,. Both smpled popultions hve distributions tht re pprox., 3. The popultion vrinces re. ( σ 4. Smple size is smll ( ). = σ ) 1 Since these two popultions hve equl vrince, ( σ = σ ), it is resonble to construct 1 for use in confidence intervls nd test sttistics. 5

(1- )100% confidence intervl for ( µ 1 µ ): t α with df = n1+ n. Hypothesis test for ( µ 1 µ ): 1. H0 = 0 H µ ( or µ µ < or µ µ > ) 1 0 1 0 1 0. level of significnce α ; 3. test sttistic: t = with df = 4. rejection region : when H 1 µ 0 when H < 0 when H > 0; 5. conclusion: if the vlue of test sttistic flls in R.R, H 0,nd conclude tht t α level, there is sufficient evidence to conclude H is true. if the vlue of test sttistic does not fll in R.R, H 0, nd conclude tht t α level, there is insufficient evidence to conclude H is true. 6

Exmples for compring two popultion mens: independent, smll-smples: Exmple1: REAING, Suppose we wish to compre new method of teching reding to slow lerners to the current stndrd method. The response vrible is the reding test score fter 6 months. slow lerners re rndomly selected, 10 re tught by the new method, 1 by the stndrd method. The test score is listed below. New method (1) 80, 80, 79, 81, 76, 66, 71, 76, 70, 85 Stndrd method () 79, 6, 70, 68, 73, 76, 86, 73, 7, 68, 75, 66. Use 95% confidence intervl to estimte the true men difference between the test score for the new method nd the stndrd method. Interpret the intervl. b. Conduct test of hypothesis to determine whether the stndrd method leds to lower test score thn new method. Useα = 0.05. 7

SPSS output for REAING, Group Sttistics METHO N Men Std. evition Std. Error Men SCORE NEW 10 76.40 5.835 1.845 ST 1 7.33 6.344 1.831 Independent Smples Test Levene's Test for Equlity of Vrinces t-test for Equlity of Mens 95% Confidence F Sig. t df Sig. (-tiled) Men ifference Std. Error ifference Intervl of the ifference Lower Upper SCORE Equl vrinces ssumed Equl vrinces not ssumed.00.967 1.55 0.136 4.067.60-1.399 9.533 1.564 19.769.134 4.067.600-1.360 9.493 Cse 3, Smll smples with unequl vrince: Conditions: 1. two smples re rndomly nd independently selected from the two trget popultion,. both smpled popultions re pprox. norml, 3. the popultions vrince re not equl ( σ Procedure is on textbook P4-43. σ ). 1 8

9.3 Compring two popultion mens: pired difference experiments Two smpling compring: Exmple1: To investigte the effect of new teching method on reding. 1. Rndomly select slow lerner students, 10 re ssigned to new method, while the other 1 re ssigned to the stndrd method, the response vrible is the reding test score fter 6 months. (independent smpling). 8 pirs slow lerner re selected, not rndomly, two lerners in ech pir with the similr reding IQs; in ech pir, one use new method, the other one use stndrd method, then the pired test score difference could be used to mke inference bout ( µ ). Exmple: To investigte the effect of new protein diet on weight loss. 1. Two rndom smples of 100 people ech re selected. One group of 100 is plced on the low-ft diet, while the other group with regulr diet. For ech person, the mount of weight lost (or gined) in 3-week period is recorded. (independent smpling). FA rndomly choose five individuls with regulr diet nd record their weight (in pounds), then instruct them to follow the protein diet for three weeks. At the end of this period, their weights re recorded gin. The pired weight differences between these two diets could be used to mke inference bout ( µ ). (two subjects in ech pir with similr level, then ssign tretments, to see the effect) Pired difference experiment: Ech pir hs two experimentl units, re pired nd the re nlyzed. : mking comprisons within groups of similr experimentl units. Pired difference experiment is simple exmple of rndomized block design. (Red textbook P43, 433) Exmple dt (NEW PROTEIN IET): Person Weight before (1) Weight fter () 1 148 141 193 188 3 186 183 4 195 189 5 0 198 Note: The vrible we re interested is. 9

Inference bsed on pired difference (lrge smple): Conditions required for vlid lrge-smple inference bout µ : 1. A rndom smple of is selected from the trget popultion of differences;. The smple size is n. Pired difference (1- )100% confidence intervl for µ µ 1 µ = ( ): Pired difference Test of hypothesis for µ = ( µ ) : 1.. significnce level α ; x 0 3. test sttistic: z = σ n 4. rejection region: Z > when H : Z α Z < Z α when H : Z > when H : Z α 5. conclusion. Exmples of mking inference bsed on pired difference (lrge smple): Exmple, To investigte which supermrket (A or B) hs the lower prices in town, gency rndomly selected 100 items common to ech of the two supermrkets nd recorded the prices chrged by ech supermrket. The summry results re provided below. x =.09 x = 1.99 x = 0.10 A B S = 0.4 S = 0.19 S = 0.03 A B. Form 95% confidence intervl for µ = µ A µ B. Interpret the result. 10

b. Conduct test of hypothesis to determine whether the men price for supermrket B is cheper thn tht for supermrket B? Useα = 0.05. 11

Inference bsed on pired difference (smll smple): Conditions required for vlid smll-smple inference bout µ : 1. A of difference is selected from the trget popultion of differences;. The popultion of differences is pproximtely distributed; 3. The smple size n < 30. 1. (1- )100% confidence intervl for Pired difference µ µ 1 µ = ( ):, t α with df =. Test of hypothesis for Pired difference µ µ 1 µ 1. H = 0 0 = ( ): H ( or H < or H > ) 0 0 0. significnce level α ; 3. test sttistic: t = 4. rejection region: when H 0 5. conclusion. when H < 0 when H > 0 1

Exmples of mking inference bsed on pired difference (smll smple): Exmple 1, NEW PROTEIN IET: To investigte new protein diet on weight-loss, FA rndomly choose five individuls nd record their weight (in pounds), then instruct them to follow the protein diet for three weeks. At the end of this period, their weights re recorded gin. person Weight before (1) Weight fter () 1 148 141 193 188 3 186 183 4 195 189 5 0 198. Clculte 95% confidence intervl for the difference between the men weights before nd fter the diet is used. Interpret the intervl. b. o the dt provide sufficient evidence tht the protein diet hs effect on the weight loss? Useα = 0.05. (p-vlue =? ) 13

SPSS output for Exmple1, NEW PROTEIN IET, Pired Smples Sttistics Men N Std. evition Std. Error Men Pir 1 W1 184.80 5 1.347 9.547 W 179.80 5.354 9.997 Pired Smples Test Pired ifferences t df Sig. (-tiled) Std. Error 95% Confidence Intervl Men Std. evition Men of the ifference Pir 1 W1 - W 5.000 1.581.707 3.037 6.963 7.071 4.00 Lower Upper SPSS output for Exmple, PAIRESCORES, Pired Smples Sttistics Men N Std. evition Std. Error Men Pir 1 NEW 76.00 8 6.98.449 ST 71.63 8 7.009.478 Pired Smples Test Pired ifferences t df Sig. (-tiled) Std. Error 95% Confidence Intervl Men Std. evition Men of the ifference Pir 1 NEW - ST 4.375 1.685.596.966 5.784 7.344 7.000 Lower Upper 14

Exmple, PAIRESCORES, To investigte the effect of new teching method on improving reding test score, 8 pirs slow lerner re selected, not rndomly, two lerners in ech pir with the similr reding IQs; in ech pir, one use new method, the other one use stndrd method. Then fter 6 months, the test scores re recorded. Pir New method (1) Stndrd method () 1 77 7 74 68 3 8 76 4 73 68 5 87 84 6 69 68 7 66 61 8 80 76. Construct 95% confidence intervl to estimte the difference of men test scores between new method nd stndrd method. Interpret the result b. o the dt provide sufficient evidence tht the new method leds to higher test scores thn the stndrd method? Useα = 0.05. (p-vlue =? ) 15

9.4 Compring two popultion proportions: independent smpling Conditions required for vlid lrge-smple inferences bout ( p1 p) : 1. The two smples re rndomly nd independently selected from the two trget popultions.. The smple size n 1 nd n re both lrge. (This condition will be stisfied if both.) Under lrge smple size, by the Centrl Limit Theorem, the smpling distribution of ( pˆˆ1 p) is pproximtely norml with: men: stndrd devition: ( p p ): Lrge smple 100(1- )% confidence intervl for 1 ( p p ): Lrge smple test of hypothesis bout 1 1.. Level of significnce α ; 3. Test sttistic: 4. rejection region : Z < Z or Z > Z when H : p1 p 0 α α Z < Z α when H : p1 p < 0 H : p p > 0; Z > Z α when 1 5. conclusion. 16

Exmples for compring two popultion proportions (lrge-smple): Exmple: Smoking Survey, Suppose the Americn cncer Society rndomly smples 1500 dults in 1995 nd then smpled 1750 dults in 005 to do smoking survey to determine whether there ws evidence tht the percentge of smokers hd decresed. 1995 (1) 005 () n 1 = 1500 n = 1750 x 1 = 555 x = 578 efine: : the true proportion of dult smokers in 1995; : the true proportion of dult smokers in 005.. Give point estimte of the percentge difference of dult smokers between 1995 nd 005? b. o the dt indicte tht the proportion of dult smokers decresed over this 10-yer period? Useα = 0.05. c. Form 95% confidence intervl for ( p1 p) to estimte the extent of the decrese. Interpret it. 17