p we will use that fact in constructing CI n for population proportion p. The approximation gets better with increasing n.

Similar documents
To make comparisons for two populations, consider whether the samples are independent or dependent.

The Hong Kong University of Science & Technology ISOM551 Introductory Statistics for Business Assignment 3 Suggested Solution

STAT-UB.0103 NOTES for Wednesday 2012.APR.25. Here s a rehash on the p-value notion:

Confidence Intervals

Chapter 9, Part B Hypothesis Tests

Estimating Proportions

Overview. p 2. Chapter 9. Pooled Estimate of. q = 1 p. Notation for Two Proportions. Inferences about Two Proportions. Assumptions

18. Two-sample problems for population means (σ unknown)

A quick activity - Central Limit Theorem and Proportions. Lecture 21: Testing Proportions. Results from the GSS. Statistics and the General Population

Confidence intervals for proportions

Recall the study where we estimated the difference between mean systolic blood pressure levels of users of oral contraceptives and non-users, x - y.

Chapter 8: Estimating with Confidence

Stat 200 -Testing Summary Page 1

tests 17.1 Simple versus compound

1 Models for Matched Pairs

Econ 325 Notes on Point Estimator and Confidence Interval 1 By Hiro Kasahara

Distribution of Sample Proportions

- E < p. ˆ p q ˆ E = q ˆ = 1 - p ˆ = sample proportion of x failures in a sample size of n. where. x n sample proportion. population proportion

BIOS 4110: Introduction to Biostatistics. Breheny. Lab #9

Common Large/Small Sample Tests 1/55

STA Learning Objectives. Population Proportions. Module 10 Comparing Two Proportions. Upon completing this module, you should be able to:

Confidence Intervals for the Difference Between Two Proportions

Introduction to Econometrics (3 rd Updated Edition) Solutions to Odd- Numbered End- of- Chapter Exercises: Chapter 3

Comparing Two Populations. Topic 15 - Two Sample Inference I. Comparing Two Means. Comparing Two Pop Means. Background Reading

Frequentist Inference

Chapter 22: What is a Test of Significance?

1 Inferential Methods for Correlation and Regression Analysis

Agreement of CI and HT. Lecture 13 - Tests of Proportions. Example - Waiting Times

Chapter 22. Comparing Two Proportions. Copyright 2010, 2007, 2004 Pearson Education, Inc.

Statistics 20: Final Exam Solutions Summer Session 2007

Statistical Inference (Chapter 10) Statistical inference = learn about a population based on the information provided by a sample.

Chapter 18: Sampling Distribution Models

HYPOTHESIS TESTS FOR ONE POPULATION MEAN WORKSHEET MTH 1210, FALL 2018

This chapter focuses on two experimental designs that are crucial to comparative studies: (1) independent samples and (2) matched pair samples.

Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date: Confidence Interval Guesswork with Confidence

MATH/STAT 352: Lecture 15

2 1. The r.s., of size n2, from population 2 will be. 2 and 2. 2) The two populations are independent. This implies that all of the n1 n2

University of California, Los Angeles Department of Statistics. Hypothesis testing

MOST PEOPLE WOULD RATHER LIVE WITH A PROBLEM THEY CAN'T SOLVE, THAN ACCEPT A SOLUTION THEY CAN'T UNDERSTAND.

Examination Number: (a) (5 points) Compute the sample mean of these data. x = Practice Midterm 2_Spring2017.lwp Page 1 of KM

Interval Estimation (Confidence Interval = C.I.): An interval estimate of some population parameter is an interval of the form (, ),

Section 9.2. Tests About a Population Proportion 12/17/2014. Carrying Out a Significance Test H A N T. Parameters & Hypothesis

Chapter 22. Comparing Two Proportions. Copyright 2010 Pearson Education, Inc.

STAC51: Categorical data Analysis

Topic 9: Sampling Distributions of Estimators

STATISTICAL INFERENCE

S160 #12. Review of Large Sample Result for Sample Proportion

Math 140 Introductory Statistics

S160 #12. Sampling Distribution of the Proportion, Part 2. JC Wang. February 25, 2016

Topic 9: Sampling Distributions of Estimators

Topic 9: Sampling Distributions of Estimators

Class 23. Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science. Marquette University MATH 1700

Confidence Intervals for the Population Proportion p

Final Examination Solutions 17/6/2010

LESSON 20: HYPOTHESIS TESTING

Properties and Hypothesis Testing

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

TI-83/84 Calculator Instructions for Math Elementary Statistics

Estimation of a population proportion March 23,

Sample Size Determination (Two or More Samples)

5. A formulae page and two tables are provided at the end of Part A of the examination PART A

Successful HE applicants. Information sheet A Number of applicants. Gender Applicants Accepts Applicants Accepts. Age. Domicile

ST 305: Exam 3 ( ) = P(A)P(B A) ( ) = P(A) + P(B) ( ) = 1 P( A) ( ) = P(A) P(B) σ X 2 = σ a+bx. σ ˆp. σ X +Y. σ X Y. σ X. σ Y. σ n.

Notes on Hypothesis Testing, Type I and Type II Errors

Y i n. i=1. = 1 [number of successes] number of successes = n

Expectation and Variance of a random variable

Announcements. Unit 5: Inference for Categorical Data Lecture 1: Inference for a single proportion

STAT 155 Introductory Statistics Chapter 6: Introduction to Inference. Lecture 18: Estimation with Confidence

General Instructions:

Data Analysis and Statistical Methods Statistics 651

Inferential Statistics. Inference Process. Inferential Statistics and Probability a Holistic Approach. Inference Process.

This is an introductory course in Analysis of Variance and Design of Experiments.

If, for instance, we were required to test whether the population mean μ could be equal to a certain value μ

Basics of Inference. Lecture 21: Bayesian Inference. Review - Example - Defective Parts, cont. Review - Example - Defective Parts

Hypothesis Testing. Evaluation of Performance of Learned h. Issues. Trade-off Between Bias and Variance

Confidence Intervals รศ.ดร. อน นต ผลเพ ม Assoc.Prof. Anan Phonphoem, Ph.D. Intelligent Wireless Network Group (IWING Lab)

Last Lecture. Wald Test

Big Picture. 5. Data, Estimates, and Models: quantifying the accuracy of estimates.

Lecture 6 Simple alternatives and the Neyman-Pearson lemma

Mathematical Statistics - MS

Because it tests for differences between multiple pairs of means in one test, it is called an omnibus test.

1036: Probability & Statistics

Stat 319 Theory of Statistics (2) Exercises

7-1. Chapter 4. Part I. Sampling Distributions and Confidence Intervals

DS 100: Principles and Techniques of Data Science Date: April 13, Discussion #10

Math 152. Rumbos Fall Solutions to Review Problems for Exam #2. Number of Heads Frequency

Chapter 13: Tests of Hypothesis Section 13.1 Introduction

Understanding Dissimilarity Among Samples

Estimating the Population Mean - when a sample average is calculated we can create an interval centered on this average

April 18, 2017 CONFIDENCE INTERVALS AND HYPOTHESIS TESTING, UNDERGRADUATE MATH 526 STYLE

Resampling Methods. X (1/2), i.e., Pr (X i m) = 1/2. We order the data: X (1) X (2) X (n). Define the sample median: ( n.

A Confidence Interval for μ

Worksheet 23 ( ) Introduction to Simple Linear Regression (continued)

1 Constructing and Interpreting a Confidence Interval

Introduction There are two really interesting things to do in statistics.

Chapter 20. Comparing Two Proportions. BPS - 5th Ed. Chapter 20 1

Data Analysis and Statistical Methods Statistics 651

Samples from Normal Populations with Known Variances

x z Increasing the size of the sample increases the power (reduces the probability of a Type II error) when the significance level remains fixed.

Continuous Data that can take on any real number (time/length) based on sample data. Categorical data can only be named or categorised

Transcription:

Estimatig oulatio roortio: We will cosider a dichotomous categorical variable(s) ( classes: A, ot A) i a large oulatio(s). Poulatio(s) should be at least 0 times larger tha the samle(s). We will discuss a samlig distributio of a estimate of a oulatio roortio =P(A) i our oulatio(s). Suose we take a SRS of size, ad deote X=# of subjects with characteristics A i our samle, the we ca estimate by usig two differet samle statistics: ordiary samle roortio (-hat): = X (Note: sometimes researchers use also Wilso-Adjusted Proortio (-tilde): = X + which gives +4 CI-s more reliable tha those based o -hat, but wee will oly use -hat i our comutatios) Ifereces for a sigle roortio: is a ubiased estimate of. For large samlig distributio of is aroximately ormal, = with mea μ = ad stadard deviatio σ (1 ) we will use that fact i costructig CI for oulatio roortio. The aroximatio gets better with icreasig. ) Oe samle CI ( z-iterval) for : ±z α/ (1 ) Stadard Error of : SE = (1, assume X ad -X both 15 or greater EX1 Durig a 1998 race for state seator a ewsaer coducted a oll ad foud that 607 of 100 registered voters samled would vote for the Reublica cadidate. Let be the oulatio roortio of registered voters who would vote for the Reublica today. a. Give a 90 % level cofidece iterval for. ^= 607 100 =0.5058 0.5058±1.645 0.5058(1 0.5058) gives: (0.48, 0.53) 100 b. Based o your CI from art a, ca you coclude that a Reublica is ow likely to wi if more tha 50% is eeded for a wi. Exlai. Aswer: Not coclusive, CI cotais umbers below ad above 50%. c. What samle size is eeded to cut margi of error i your iterval to 1%?

Samle size cosideratios: =.5 z /,if we have o guess for -hat or E = g g z /,where E g = best guess for -hat Usig secod formula: Aswer:.5058(1.5058)( 1.645.01 ) =6764., =6765 EX A oll was take of 1010 U.S. emloyees, they were asked whether they lay hooky from work at least oce er year, 0 resoded yes. a) Fid 95% CI for =roortio of all U.S. emloyees that lay hooky from work at least oce er year. ^= 0 1010 0.±1.96 =0. 0.(0.8) Gives CI: (0.1753, 0.47) 1010 b) What samle size will esure the margi of error i 95% CI for of o more tha 1%? Assume 0% to be a best guess for hat. =0.(0.8)(1.96/0.01)^=6146.56, roud it u to 6147. Hyothesis Tests for oe Poulatio Proortio := 0 vs H a : 0 or H a : 0 or H a : 0 Samle roortio, ( hat) : = x attribute: Test statistics: z= 0 1 0 0 assume 0, 1 0 both 5 or greater, where x=umber of members i a samle with secified has N(0,1) distributio if is true, so it is a Z-test. Suose i ( EX) we ask followig questio: Is there evidece at 5 % sigificace level that less tha 5 % of U.S emloyees lay hooky from work at least oce er year? Test aroriate hyothesis. : =0.5 H a :<0.5 ^=0., z=-3.67. P=0.0001<0.05

Reject, there is evidece at 5% sigificace level that less tha 5% of U.S. emloyees will lay hooky from work at least oce er year. EX3 A Harris Poll asked 150 U.S. adults their views o baig hadgu sales. Of those samled, 650 favored a ba. At 5% sigificace level, do the data rovide sufficiet evidece that a majority of U.S. adults favor baig hadgu sales? : =0.5 H a :>0.5 ^=0.5 Z=1.41, -value=0.079>0.05 Do ot reject, there is o evidece at 5% sigificace level that majority of U.S. adults favor baig hadgus. The Relatio Betwee Hyothesis Tests ad Cofidece Itervals If Null hyothesis : = 0 agaist two tailed alterative is rejected at sigificace level, the (1- )*100% CI for will ot cotai 0, otherwise (if ot rejected) 0 will be iside of CI. I EX 95% CI for is (0.1753, 0.47), our left tailed test : =0.5 H a :<0.5 had a -value =0.0001, so two tailed test : =0.5 H a : 0.5 would have - value=*0.0001=0.0004<0.05. We ca see that our CI ad test of two sided hyothesis are coected. Our CI does ot cotai 0.5 ad we reject ull hyotheses at 5% sigificace level. I EX3 two tailed test : =0.5 H a : 0.5 would have -value = *0.079=0.158>5%, so ull hyotheses would ot be rejected at α=0.05. At the same time 95% CI for is (0.49, 0.55) ad clearly cotais 0 =0.50 OPTIONAL: Ifereces about Two Poulatios Proortios. Both oulatios are two-category oulatios, ideedet samles give couts with desired attribute i each oulatio. All couts: x 1, 1 x 1, x, x must be 5 or greater 1 ad estimate roortios with desired attribute i each oulatio Two samles z-iterval for 1 : 1 ±z / 1 1 1

EX1. Suose that we wat to kow to what extet is the frequecy of arole violatio related to the tye of crime? Out of 4 ersos who had served time for imulsive murder, 9 violated their arole. Out of 40 ersos who had served time for remeditated murder, 18 violated their arole. Let 1 = roortio of all imulsive murderers who violate arole = roortio of all remeditated murderers who violate arole Obtai 95% CI for 1-. Based o that iterval do you thig there is a differece betwee the roortios? Aswer: (0.0335, 0.4486). Yes, there is a differece, CI has o zero iside. Hyothesis Tests for Two Poulatios Proortios Two samles z test: = vs H a or H a or H a test statistics: z= 1 1 1 1 1 where = x x 1 1 =ooled samle roortio. Test statistics has N(01) distributio if ull hyothesis is true. I the examle about arole violatios we ca ask the followig questio: Are imulsive murderers more likely to violate their arole tha remeditated murderers? Test aroriate hyothesis at 5% sigificace level. =, H a, 1 =.69, =.45 =.573 z =.0 =.014 < 0.05. Reject, evidece for alterative hyothesis. Imulsive murderers will violate arole more ofte tha remeditated oes. 1. The Relatio Betwee Hyothesis Tests ad Cofidece Itervals If = is ot rejected agaist two tailed alterative, at give α level tha (1- α)*100% CI for the differece betwee two meas will cotai 0, otherwise it will ot cotai 0. If CI cotais 0, o evidece that roortios are differet.

Usig Calculator (TI 83, 84) 1 Proortio Z iterval use STAT meu the TESTS otio A is 1-ProZIterval It will use -hat method, just iut x ad If we wat 95% CI usig -tilde, we ca iut x=x+ ad =+4, for other cofidece levels it will ot work Proortios Z iterval use STAT meu the TESTS otio B is -ProZIterval It will use -hat method, just iut x ad for each oulatio Testig other Hyothesis All tests are available i STATS TESTS otio: I each test rocedure Data otio ca be used if samle statistics are ot comuted. It works the same as cofidece itervals rocedures. Test for 1 oulatio roortio: 1-ProZTest Test for oulatios roortios: -ProZTest