Contingency tables. X = (Y, Z)T is 2-dimensional random vector.

Slides:

Advertisements

Similar presentations

Introduction to Probability

Advertisements

McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Chi-Square Tests Chapter 12.

CHI-SQUARE(X2) DISTRIBUTION

 2 Test of Independence. Hypothesis Tests Categorical Data.

Inference about the Difference Between the

Introduction to Categorical Data Analysis

Analysis of frequency counts with Chi square

Two-Way Tables Two-way tables come about when we are interested in the relationship between two categorical variables. –One of the variables is the row.

Statistics 303 Chapter 9 Two-Way Tables. Relationships Between Two Categorical Variables Relationships between two categorical variables –Depending on.

Irwin/McGraw-Hill © The McGraw-Hill Companies, Inc., 2000 LIND MASON MARCHAL 1-1 Chapter Thirteen Nonparametric Methods: Chi-Square Applications GOALS.

Quiz 10  Goodness-of-fit test  Test of independence.

Presentation 12 Chi-Square test.

The table shows a random sample of 100 hikers and the area of hiking preferred. Are hiking area preference and gender independent? Hiking Preference Area.

Analysis of Categorical Data

Chapter 11 Chi-Square Procedures 11.3 Chi-Square Test for Independence; Homogeneity of Proportions.

Dr.Shaikh Shaffi Ahamed Ph.D., Dept. of Family & Community Medicine

Maximum Likelihood Estimator of Proportion Let {s 1,s 2,…,s n } be a set of independent outcomes from a Bernoulli experiment with unknown probability.

Chapter 12 The Analysis of Categorical Data and Goodness-of-Fit Tests.

Chi-Square Procedures Chi-Square Test for Goodness of Fit, Independence of Variables, and Homogeneity of Proportions.

Chapter 11 The Chi-Square Test of Association/Independence Target Goal: I can perform a chi-square test for association/independence to determine whether.

FPP 28 Chi-square test. More types of inference for nominal variables Nominal data is categorical with more than two categories Compare observed frequencies.

Analysis of Qualitative Data Dr Azmi Mohd Tamil Dept of Community Health Universiti Kebangsaan Malaysia FK6163.

6.1 - One Sample One Sample  Mean μ, Variance σ 2, Proportion π Two Samples Two Samples  Means, Variances, Proportions μ 1 vs. μ 2.

HYPOTHESIS TESTING BETWEEN TWO OR MORE CATEGORICAL VARIABLES The Chi-Square Distribution and Test for Independence.

Section 10.2 Independence. Section 10.2 Objectives Use a chi-square distribution to test whether two variables are independent Use a contingency table.

CHAPTER INTRODUCTORY CHI-SQUARE TEST Objectives:- Concerning with the methods of analyzing the categorical data In chi-square test, there are 2 methods.

The table shows a random sample of 100 hikers and the area of hiking preferred. Are hiking area preference and gender independent? Hiking Preference Area.

Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 11 Analyzing the Association Between Categorical Variables Section 11.2 Testing Categorical.

ContentFurther guidance  Hypothesis testing involves making a conjecture (assumption) about some facet of our world, collecting data from a sample,

Chi Square Tests PhD Özgür Tosun. IMPORTANCE OF EVIDENCE BASED MEDICINE.

1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial.

Contingency table analysis

Chapter 12 Chi-Square Tests and Nonparametric Tests.

The Birthday Problem. The Problem In a group of 50 students, what is the probability that at least two students share the same birthday?

Exercise 18 D.2 Null Hypothesis, H 0 : There is a relationship between smoking marijuana as a teenager and suffering schizophrenia within the next 15 years.

Chapter 11 Chi Square Distribution and Its applications.

Introduction to probability (6)

Section 10.2 Objectives Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent.

Copyright © Cengage Learning. All rights reserved. 14 Goodness-of-Fit Tests and Categorical Data Analysis.

Chi Square Test Dr. Asif Rehman.

CHI-SQUARE(X2) DISTRIBUTION

Test of independence: Contingency Table

Topic 6: Proportions and Probabilities

Chapter 9: Non-parametric Tests

Presentation 12 Chi-Square test.

Lecture8 Test forcomparison of proportion

5.1 INTRODUCTORY CHI-SQUARE TEST

Chapter 12 Chi-Square Tests.

Contingency (frequency) tables

Chapter Fifteen McGraw-Hill/Irwin

Association between two categorical variables

Chapter 12 Tests with Qualitative Data

Qualitative data – tests of association

Sun. Mon. Tues. Wed. Thurs. Fri. Sat.

Inferential Statistics and Probability a Holistic Approach

Inferential Statistics and Probability a Holistic Approach

Hypothesis testing. Chi-square test

Comparing Populations

Chi Square (2) Dr. Richard Jackson

STAT 312 Introduction Z-Tests and Confidence Intervals for a

Addition of Independent Normal Random Variables

Inference on Categorical Data

CHI SQUARE TEST OF INDEPENDENCE

Analyzing the Association Between Categorical Variables

Goodness of Fit.

7.3 Conditional Probability, Intersection and Independence

Looks at differences in frequencies between groups

Summary Table of Influence Procedures for a Single Sample (I)

The Binomial Probability Distribution and Related Topics

Karl L. Wuensch Department of Psychology East Carolina University

Presentation transcript:

Contingency tables. X = (Y, Z)T is 2-dimensional random vector. Y contains r objects, Z contains c objects. Probabilities pij = P(Y= i, Z = j). Let nij are number of cases, where 𝑖∈𝑌, 𝑗∈𝑍. Example. There were randomly selected 200 Czechs, 300 Norwegians and 150 Turks. 50 Czechs, 70 Norwegians and 80 Turks are smokers. Does smoking depend on the state? Randomly selected people are divided according to 2 criteria: Y … we have 3 groups according to the state. Z … we have 2 possibilities for smoking r = 3, c = 2 Criteria Y and Z are independent if pij = pi. p.j, where pi. means the raw i in the table, p.j denotes the column j in the table. pij means the probability that random selected person is from the state i and his smoking status is j.

We are testing H0: Y and Z are independent and H1: Y and Z are dependent (are not independent) Assuming the validity of H0, the frequency table is (expected frequencies): P(Czech,yes) = P(Czech)P(yes) = 200/650 * 200/650 = 0.0961 (expected frequency: 0.0961*650) P(Czech,no) = P(Czech)P(no) = 200/650 * 450/650 = 0.2139 (expected frequency: 0.2139*650) P(Norway,yes) = P(Norway)P(yes) = 300/650 * 200/650 = 0.1426 (expected frequency: 0.1426*650) P(Norway,no) = P(Norway)P(no) =300/650 * 450/650 = 0.3174 (expected frequency: 0.3174*650) P(Turkey,yes) = P(Turkey)P(yes) = 150/650 *200/650 = 0.0713 (expected frequency: 0.0713*650) P(Turkey,no) = P(Turkey)P(no) = 150/650 *450/650 = 0.1587 (expected frequency: 0.1587*650) Expected frequencies are:

c2 test with df = (number of rows – 1)*(number of columns -1): In our example: , P < 0.0001 Conclusion: We will reject H0, therefore smoking is depending on the state. (It was found that smoking in Norway is less than expected and in Turkey more than expected.) Fisher’s factorial test is performed for frequencies lesser or equal to 5.

Examples. In 27 randomly selected patients with a disease, it was determined whether they had been vaccinated against the disease and the course of the disease. Vaccination + heavy course 2, vaccination + light course 10, non-vaccination + severe course 11, non-vaccination + moderate course 4 There were selected 200 inhabitants of Ostrava, 150 inhabitants of České Budějovice and 500 inhabitants of Prague. It was found that 20 inhabitants of Ostrava, 20 inhabitants of Budějovice and 100 inhabitants of Prague suffer from kidney disease. Does kidney disease depend on the place?