Lecture 28 Chi-Square Analysis

Similar documents
POLI 443 Applied Political Research

Statistics for Managers Using Microsoft Excel

Chapter 10: Chi-Square and F Distributions

Lecture 9. Selected material from: Ch. 12 The analysis of categorical data and goodness of fit tests

Analysis of Categorical Data Three-Way Contingency Table

HYPOTHESIS TESTING: THE CHI-SQUARE STATISTIC

Parametric versus Nonparametric Statistics-when to use them and which is more powerful? Dr Mahmoud Alhussami

Introduction to Statistical Data Analysis Lecture 7: The Chi-Square Distribution

Inferences About Two Proportions

Topic 21 Goodness of Fit

Ron Heck, Fall Week 3: Notes Building a Two-Level Model

Example. χ 2 = Continued on the next page. All cells

Lecture 41 Sections Mon, Apr 7, 2008

The goodness-of-fit test Having discussed how to make comparisons between two proportions, we now consider comparisons of multiple proportions.

Chapter 11. Hypothesis Testing (II)

We know from STAT.1030 that the relevant test statistic for equality of proportions is:

Chapter 8 Student Lecture Notes 8-1. Department of Economics. Business Statistics. Chapter 12 Chi-square test of independence & Analysis of Variance

Statistics and Quantitative Analysis U4320

STAT Chapter 10: Analysis of Variance

STP 226 EXAMPLE EXAM #3 INSTRUCTOR:

4/6/16. Non-parametric Test. Overview. Stephen Opiyo. Distinguish Parametric and Nonparametric Test Procedures

MTS Degree Program: Ethnicity & Gender

STP 226 ELEMENTARY STATISTICS NOTES

POLI 443 Applied Political Research

Department of Economics. Business Statistics. Chapter 12 Chi-square test of independence & Analysis of Variance ECON 509. Dr.

Chapter 12: Estimation

10: Crosstabs & Independent Proportions

Section 9.4. Notation. Requirements. Definition. Inferences About Two Means (Matched Pairs) Examples

Basic Business Statistics, 10/e

Class 24. Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science. Marquette University MATH 1700

Inferential statistics

2 and F Distributions. Barrow, Statistics for Economics, Accounting and Business Studies, 4 th edition Pearson Education Limited 2006

Statistics 3858 : Contingency Tables

Introduction to Machine Learning CMU-10701

The One-Way Independent-Samples ANOVA. (For Between-Subjects Designs)

Psych 230. Psychological Measurement and Statistics

Statistical Analysis for QBIC Genetics Adapted by Ellen G. Dow 2017

STAT Chapter 13: Categorical Data. Recall we have studied binomial data, in which each trial falls into one of 2 categories (success/failure).

CHAPTER 17 CHI-SQUARE AND OTHER NONPARAMETRIC TESTS FROM: PAGANO, R. R. (2007)

13.1 Categorical Data and the Multinomial Experiment

10.2 Hypothesis Testing with Two-Way Tables

Rama Nada. -Ensherah Mokheemer. 1 P a g e

CHI SQUARE ANALYSIS 8/18/2011 HYPOTHESIS TESTS SO FAR PARAMETRIC VS. NON-PARAMETRIC

Statistics Handbook. All statistical tables were computed by the author.

Ch. 11 Inference for Distributions of Categorical Data

CIVL /8904 T R A F F I C F L O W T H E O R Y L E C T U R E - 8

CBA4 is live in practice mode this week exam mode from Saturday!

The One-Way Repeated-Measures ANOVA. (For Within-Subjects Designs)

Module 4 Practice problem and Homework answers

Lecture Slides. Elementary Statistics. by Mario F. Triola. and the Triola Statistics Series

REVIEW 8/2/2017 陈芳华东师大英语系

Chapter 11 - Lecture 1 Single Factor ANOVA

Lecture Slides. Section 13-1 Overview. Elementary Statistics Tenth Edition. Chapter 13 Nonparametric Statistics. by Mario F.

STAT 135 Lab 11 Tests for Categorical Data (Fisher s Exact test, χ 2 tests for Homogeneity and Independence) and Linear Regression

Chapter 7 Comparison of two independent samples

Chapter 11 - Lecture 1 Single Factor ANOVA

Nominal Data. Parametric Statistics. Nonparametric Statistics. Parametric vs Nonparametric Tests. Greg C Elvers

Conditional Probability Solutions STAT-UB.0103 Statistics for Business Control and Regression Models

GROUPED DATA E.G. FOR SAMPLE OF RAW DATA (E.G. 4, 12, 7, 5, MEAN G x / n STANDARD DEVIATION MEDIAN AND QUARTILES STANDARD DEVIATION

Psych Jan. 5, 2005

7.2 One-Sample Correlation ( = a) Introduction. Correlation analysis measures the strength and direction of association between

Contingency Tables. Safety equipment in use Fatal Non-fatal Total. None 1, , ,128 Seat belt , ,878

Chapter 10. Chapter 10. Multinomial Experiments and. Multinomial Experiments and Contingency Tables. Contingency Tables.

Testing Independence

16.3 One-Way ANOVA: The Procedure

AMS7: WEEK 7. CLASS 1. More on Hypothesis Testing Monday May 11th, 2015

Lecture 25. STAT 225 Introduction to Probability Models April 16, Whitney Huang Purdue University. Agenda. Notes. Notes.

Averages How difficult is QM1? What is the average mark? Week 1b, Lecture 2

An Analysis of College Algebra Exam Scores December 14, James D Jones Math Section 01

Analysis of variance (ANOVA) Comparing the means of more than two groups

Is there a connection between gender, maths grade, hair colour and eye colour? Contents

Lecture 41 Sections Wed, Nov 12, 2008

Hypothesis testing (cont d)

Hypothesis Testing One Sample Tests

Sociology 6Z03 Review II

Lecture 22. December 19, Department of Biostatistics Johns Hopkins Bloomberg School of Public Health Johns Hopkins University.

Unit 22 deals with statistical inference, which uses the concepts of probability to explain

10/4/2013. Hypothesis Testing & z-test. Hypothesis Testing. Hypothesis Testing

One-Way ANOVA. Some examples of when ANOVA would be appropriate include:

Lecture 6: Linear Regression

8.1-4 Test of Hypotheses Based on a Single Sample

Lecture 14: ANOVA and the F-test

Goodness of Fit Tests

Review for Final. Chapter 1 Type of studies: anecdotal, observational, experimental Random sampling

Question. Hypothesis testing. Example. Answer: hypothesis. Test: true or not? Question. Average is not the mean! μ average. Random deviation or not?

Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date: χ = npq

Case-Control Association Testing. Case-Control Association Testing

Difference between means - t-test /25

their contents. If the sample mean is 15.2 oz. and the sample standard deviation is 0.50 oz., find the 95% confidence interval of the true mean.

Calculate the volume of the sphere. Give your answer correct to two decimal places. (3)

Contingency Tables. Contingency tables are used when we want to looking at two (or more) factors. Each factor might have two more or levels.

STAT Chapter 8: Hypothesis Tests

15: CHI SQUARED TESTS

An Introduction to Path Analysis

Lecture 7: Hypothesis Testing and ANOVA

Hypothesis testing. Data to decisions

Sampling Distributions: Central Limit Theorem

There are statistical tests that compare prediction of a model with reality and measures how significant the difference.

Lecture 5: ANOVA and Correlation

Mathematical Notation Math Introduction to Applied Statistics

Transcription:

Lecture 28 STAT 225 Introduction to Probability Models April 23, 2014 Whitney Huang Purdue University 28.1

χ 2 test for For a given contingency table, we want to test if two have a relationship or not To answer this question, we need to perform a statistical hypothesis test A statistical hypothesis test is a method of making decisions using data from a scientific study A χ 2 test is suitable for answering whether two qualitative have a relationship or not 28.2

χ 2 test for For a given contingency table, we want to test if two have a relationship or not To answer this question, we need to perform a statistical hypothesis test A statistical hypothesis test is a method of making decisions using data from a scientific study A χ 2 test is suitable for answering whether two qualitative have a relationship or not 28.2

χ 2 test for For a given contingency table, we want to test if two have a relationship or not To answer this question, we need to perform a statistical hypothesis test A statistical hypothesis test is a method of making decisions using data from a scientific study A χ 2 test is suitable for answering whether two qualitative have a relationship or not 28.2

χ 2 test for For a given contingency table, we want to test if two have a relationship or not To answer this question, we need to perform a statistical hypothesis test A statistical hypothesis test is a method of making decisions using data from a scientific study A χ 2 test is suitable for answering whether two qualitative have a relationship or not 28.2

χ 2 test The procedure of χ 2 test for : 1 Define the Null (H 0 ) and Alternative (H A ) hypotheses H 0 : there is no relationship between the 2 H A : there is a relationship between the 2 2 (If necessary) Calculate the marginal totals, and the grand total 3 Calculate the expected cell counts expected cell count = Row Total Column Total Grand Total 4 Calculate the partial χ 2 values (a χ 2 value for each cell of the table) partial χ 2 value = (observed - expected)2 expected 28.3

χ 2 test The procedure of χ 2 test for : 1 Define the Null (H 0 ) and Alternative (H A ) hypotheses H 0 : there is no relationship between the 2 H A : there is a relationship between the 2 2 (If necessary) Calculate the marginal totals, and the grand total 3 Calculate the expected cell counts expected cell count = Row Total Column Total Grand Total 4 Calculate the partial χ 2 values (a χ 2 value for each cell of the table) partial χ 2 value = (observed - expected)2 expected 28.3

χ 2 test The procedure of χ 2 test for : 1 Define the Null (H 0 ) and Alternative (H A ) hypotheses H 0 : there is no relationship between the 2 H A : there is a relationship between the 2 2 (If necessary) Calculate the marginal totals, and the grand total 3 Calculate the expected cell counts expected cell count = Row Total Column Total Grand Total 4 Calculate the partial χ 2 values (a χ 2 value for each cell of the table) partial χ 2 value = (observed - expected)2 expected 28.3

χ 2 test The procedure of χ 2 test for : 1 Define the Null (H 0 ) and Alternative (H A ) hypotheses H 0 : there is no relationship between the 2 H A : there is a relationship between the 2 2 (If necessary) Calculate the marginal totals, and the grand total 3 Calculate the expected cell counts expected cell count = Row Total Column Total Grand Total 4 Calculate the partial χ 2 values (a χ 2 value for each cell of the table) partial χ 2 value = (observed - expected)2 expected 28.3

χ 2 test cont d 5 Calculate the χ 2 statistic χ 2 = partial χ 2 value 6 Calculate the degrees of freedom (df ) df = (#of rows 1) (#of columns 1) 7 Find the χ 2 critical value with respect to α from the χ 2 table 8 Draw your conclusion: Reject H 0 if your χ 2 statistic is bigger than the χ 2 critical value There is an statistical evidence that there is a relationship between the 2 at α level 28.4

χ 2 test cont d 5 Calculate the χ 2 statistic χ 2 = partial χ 2 value 6 Calculate the degrees of freedom (df ) df = (#of rows 1) (#of columns 1) 7 Find the χ 2 critical value with respect to α from the χ 2 table 8 Draw your conclusion: Reject H 0 if your χ 2 statistic is bigger than the χ 2 critical value There is an statistical evidence that there is a relationship between the 2 at α level 28.4

χ 2 test cont d 5 Calculate the χ 2 statistic χ 2 = partial χ 2 value 6 Calculate the degrees of freedom (df ) df = (#of rows 1) (#of columns 1) 7 Find the χ 2 critical value with respect to α from the χ 2 table 8 Draw your conclusion: Reject H 0 if your χ 2 statistic is bigger than the χ 2 critical value There is an statistical evidence that there is a relationship between the 2 at α level 28.4

χ 2 test cont d 5 Calculate the χ 2 statistic χ 2 = partial χ 2 value 6 Calculate the degrees of freedom (df ) df = (#of rows 1) (#of columns 1) 7 Find the χ 2 critical value with respect to α from the χ 2 table 8 Draw your conclusion: Reject H 0 if your χ 2 statistic is bigger than the χ 2 critical value There is an statistical evidence that there is a relationship between the 2 at α level 28.4

Example 67 A 2011 study was conducted in Kalamazoo, Michigan. The objective was to determine if parents marital status affects children s marital status later in their life. In total, 2,000 children were interviewed. The columns refer to the parents marital status. Use the contingency table below to conduct a χ 2 test from beginning to end. Use α =.10 (Observed) Married Divorced Total Married 581 487 Divorced 455 477 Total 28.5

Example 67 cont d 1 Define the Null and Alternative hypotheses: H 0 : there is no relationship between parents marital status and childrens marital status H A : there is a relationship between parents marital status and childrens marital status 2 Calculate the marginal totals, and the grand total (Observed) Married Divorced Total Married 581 487 1068 Divorced 455 477 932 Total 1036 964 2000 28.6

Example 67 cont d 1 Define the Null and Alternative hypotheses: H 0 : there is no relationship between parents marital status and childrens marital status H A : there is a relationship between parents marital status and childrens marital status 2 Calculate the marginal totals, and the grand total (Observed) Married Divorced Total Married 581 487 1068 Divorced 455 477 932 Total 1036 964 2000 28.6

Example 67 cont d 3 Calculate the expected cell counts (Expected) Married Divorced 1068 1036 1068 964 Married 2000 = 553.224 2000 = 514.776 932 1036 932 964 Divorced 2000 = 482.776 2000 = 449.224 4 Calculate the partial χ 2 values partial χ 2 Married Divorced Married Divorced (581 553.224) 2 553.224 = 1.39 (455 482.776) 2 482.776 = 1.60 (487 514.776) 2 514.776 = 1.50 (477 449.224) 2 449.224 = 1.72 28.7

Example 67 cont d 3 Calculate the expected cell counts (Expected) Married Divorced 1068 1036 1068 964 Married 2000 = 553.224 2000 = 514.776 932 1036 932 964 Divorced 2000 = 482.776 2000 = 449.224 4 Calculate the partial χ 2 values partial χ 2 Married Divorced Married Divorced (581 553.224) 2 553.224 = 1.39 (455 482.776) 2 482.776 = 1.60 (487 514.776) 2 514.776 = 1.50 (477 449.224) 2 449.224 = 1.72 28.7

Example 67 cont d 5 Calculate the χ 2 statistic χ 2 = 1.39 + 1.50 + 1.60 + 1.72 = 6.21 6 Calculate the degrees of freedom (df ) The df is (2 1) (2 1) = 1 7 Find the χ 2 critical value with respect to α from the χ 2 table The χ 2 α=0.1,df =1 = 2.71 8 Draw your conclusion: We reject H 0 and conclude that there is a relationship between parents marital status and childrens marital status. 28.8

Example 67 cont d 5 Calculate the χ 2 statistic χ 2 = 1.39 + 1.50 + 1.60 + 1.72 = 6.21 6 Calculate the degrees of freedom (df ) The df is (2 1) (2 1) = 1 7 Find the χ 2 critical value with respect to α from the χ 2 table The χ 2 α=0.1,df =1 = 2.71 8 Draw your conclusion: We reject H 0 and conclude that there is a relationship between parents marital status and childrens marital status. 28.8

Example 67 cont d 5 Calculate the χ 2 statistic χ 2 = 1.39 + 1.50 + 1.60 + 1.72 = 6.21 6 Calculate the degrees of freedom (df ) The df is (2 1) (2 1) = 1 7 Find the χ 2 critical value with respect to α from the χ 2 table The χ 2 α=0.1,df =1 = 2.71 8 Draw your conclusion: We reject H 0 and conclude that there is a relationship between parents marital status and childrens marital status. 28.8

Example 67 cont d 5 Calculate the χ 2 statistic χ 2 = 1.39 + 1.50 + 1.60 + 1.72 = 6.21 6 Calculate the degrees of freedom (df ) The df is (2 1) (2 1) = 1 7 Find the χ 2 critical value with respect to α from the χ 2 table The χ 2 α=0.1,df =1 = 2.71 8 Draw your conclusion: We reject H 0 and conclude that there is a relationship between parents marital status and childrens marital status. 28.8

Example 68 The following contingency table contains enrollment data for a random sample of students from several colleges at Purdue University during the 2006-2007 academic year. The table lists the number of male and female students enrolled in each college. Use the two-way table to conduct a χ 2 test from beginning to end. Use α =.01 (Observed) Female Male Total Liberal Arts 378 262 640 Science 99 175 274 Engineering 104 510 614 Total 581 947 1528 28.9

Example 68 cont d (Expected) Female Male 640 581 640 947 Liberal Arts 1528 = 243.35 Science 274 581 1528 = 104.18 274 947 Engineering 614 581 1528 = 233.46 614 947 1528 = 396.65 1528 = 169.82 1528 = 380.54 partial χ 2 Female Male Lib Arts Sci Eng (378 243.35) 2 243.35 = 74.50 (99 104.18) 2 104.18 = 0.26 (104 233.46) 2 233.46 = 71.79 (262 396.65) 2 396.65 = 45.71 (175 169.82) 2 169.82 = 0.16 (510 380.54) 2 380.54 = 44.05 χ 2 = 74.50 + 45.71 + 0.26 + 0.16 + 71.79 + 44.05 = 236.47 The df = (3 1) (2 1) = 2 The χ 2 α=.01,df =2 = 9.21 We reject H 0 and conclude that there is a relationship between gender and major. 28.10