Chi-square problem solutions

Slides:

Advertisements

Similar presentations

What is Chi-Square? Used to examine differences in the distributions of nominal data A mathematical comparison between expected frequencies and observed.

Advertisements

Chi Square Example A researcher wants to determine if there is a relationship between gender and the type of training received. The gender question is.

Hypothesis Testing IV Chi Square.

Chapter 13: The Chi-Square Test

Chi-Square Test of Independence u Purpose: Test whether two nominal variables are related u Design: Individuals categorized in two ways.

Chi Square Test Dealing with categorical dependant variable.

CHI-SQUARE GOODNESS OF FIT TEST u A nonparametric statistic u Nonparametric: u does not test a hypothesis about a population value (parameter) u requires.

Crosstabs and Chi Squares Computer Applications in Psychology.

Chapter 11(1e), Ch. 10 (2/3e) Hypothesis Testing Using the Chi Square ( χ 2 ) Distribution.

Hypothesis Testing IV (Chi Square)

Cross Tabulation and Chi-Square Testing. Cross-Tabulation While a frequency distribution describes one variable at a time, a cross-tabulation describes.

Using SPSS for Chi Square UDP 520 Lab 5 Lin November 8 th, 2007.

1 Psych 5500/6500 Chi-Square (Part Two) Test for Association Fall, 2008.

HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Section 10.7.

Phi Coefficient Example A researcher wishes to determine if a significant relationship exists between the gender of the worker and if they experience pain.

Tests of Signiﬁcance June 11, 2008 Ivan Katchanovski, Ph.D. POL 242Y-Y.

Chapter 9: Non-parametric Tests n Parametric vs Non-parametric n Chi-Square –1 way –2 way.

Copyright © 2012 by Nelson Education Limited. Chapter 10 Hypothesis Testing IV: Chi Square 10-1.

Chapter 11 Hypothesis Testing IV (Chi Square). Chapter Outline  Introduction  Bivariate Tables  The Logic of Chi Square  The Computation of Chi Square.

CHI SQUARE TESTS.

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

Chapter 11, 12, 13, 14 and 16 Association at Nominal and Ordinal Level The Procedure in Steps.

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

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.

Chapter 11: Chi-Square  Chi-Square as a Statistical Test  Statistical Independence  Hypothesis Testing with Chi-Square The Assumptions Stating the Research.

Section 12.2: Tests for Homogeneity and Independence in a Two-Way Table.

Chi Square & Correlation

Chapter 13. The Chi Square Test ( ) : is a nonparametric test of significance - used with nominal data -it makes no assumptions about the shape of the.

Bullied as a child? Are you tall or short? 6’ 4” 5’ 10” 4’ 2’ 4”

Chapter 14 – 1 Chi-Square Chi-Square as a Statistical Test Statistical Independence Hypothesis Testing with Chi-Square The Assumptions Stating the Research.

COURSE: JUST 3900 INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE Chapter 17: Chi Square Peer Tutor Slides Instructor: Mr. Ethan W. Cooper, Lead Tutor © 2013.

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

Bivariate Association. Introduction This chapter is about measures of association This chapter is about measures of association These are designed to.

Seven Steps for Doing  2 1) State the hypothesis 2) Create data table 3) Find  2 critical 4) Calculate the expected frequencies 5) Calculate  2 6)

Statistical Analysis: Chi Square

Chi-Square (Association between categorical variables)

Chapter 9: Non-parametric Tests

Presentation 12 Chi-Square test.

10 Chapter Chi-Square Tests and the F-Distribution Chapter 10

Chapter 11 Chi-Square Tests.

INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE

Chapter Fifteen McGraw-Hill/Irwin

Hypothesis Testing Review

Community &family medicine

Qualitative data – tests of association

Hypothesis Testing Using the Chi Square (χ2) Distribution

Chapter 14 in 1e Ch. 12 in 2/3 Can. Ed.

Chapter 13 (1e), (Ch. 11 2/3e) Association Between Variables Measured at the Nominal Level: Phi, Cramer’s V, and Lambda.

Chapter 12: Inference about a Population Lecture 6b

PPA 501 – Analytical Methods in Administration

The Chi-Square Distribution and Test for Independence

Is a persons’ size related to if they were bullied

Different Scales, Different Measures of Association

Consider this table: The Χ2 Test of Independence

Is a persons’ size related to if they were bullied

Reasoning in Psychology Using Statistics

Contingency Tables (cross tabs)

Statistical Analysis Chi-Square.

Chapter 14 Chi Square - 2 1/10/2019 Chi square.

Chi-Square Test For nominal/qualitative data

Chapter 11 Chi-Square Tests.

The 2 (chi-squared) test for independence

Chapter 14 Chi Square - 2 2/27/2019 Chi square.

Chi-square = 2.85 Chi-square crit = 5.99 Achievement is unrelated to whether or not a child attended preschool.

Reasoning in Psychology Using Statistics

Parametric versus Nonparametric (Chi-square)

Reasoning in Psychology Using Statistics

Chapter Outline Goodness of Fit test Test of Independence.

Chapter 11 Chi-Square Tests.

Contingency Tables (cross tabs)

Presentation transcript:

Chi-square problem solutions #10.4 and 10.6 (2/3e) or 9.4, 9.6 (1e) Chi-square problem solutions

#10.4 2/3e (9.4 1e) Is there a gender gap in support for the Liberal Party? Party Preference Male Female Totals Liberal 10 40% 15 60% 25 Other 100% N = 50

#10.4 2/3e (9.4 1e) cont. Step 1: Step 2: H0: fo = fe H1: fo ≠ fe Independent random samples Level of measurement is nominal Step 2: H0: fo = fe H1: fo ≠ fe Step 3: Sampling Distribution = χ2 Alpha = .05 df = (r-1)(c-1) = 1 χ2 (critical) = 3.841

#10.4 2/3e (9.4 1e) cont. Step 4: Find expected frequencies for each cell: fe = cell 1: fe=25x25/50=12.5 cell 2: fe=25x25/50=12.5 cell 3: fe=25x25/50=12.5 cell 4: fe=25x25/50=12.5 Computational table to find χ2 (obtained): Step 5: χ2obt.< χ2cr. Fail to reject Ho. The % pattern shows that generally women prefer Liberals (60% vs 40%), but the association is not significant. fo fe fo - fe (fo - fe)2 (fo - fe)2 /fe 10 12.5 -2.5 6.25 .5 15 2.5 Total 50 - Χ2 = 2.0

#10.6 2/3e (9.6 1e) Are pet therapy program participants more alert? Alertness Participants Non-participants Totals High 23 67.6% 15 45.5% 38 Low 11 32.4% 18 54.5% 29 34 100% 33 N = 67

#10.6 2/3e (9.6 1e) cont. Step 1: Step 2: H0: fo = fe H1: fo ≠ fe Independent random samples Level of measurement is nominal Step 2: H0: fo = fe H1: fo ≠ fe Step 3: Sampling Distribution = χ2 Alpha = .05 df = (r-1)(c-1) = 1 χ2 (critical) = 3.841

#10.6 2/3e (9.6 1e) cont. Step 4: Find expected frequencies for each cell: fe = cell 1: fe=34x38/67=19.28 cell 2: fe=33x38/67=18.72 cell 3: fe=34x29/67= 14.72 cell 4: fe=33x29/67=14.28 Computational table to find χ2 (obtained): Step 5: χ2obt.< χ2cr. Fail to reject Ho. The % pattern shows that generally participants are more alert (67.6% vs 45.5%), but the association is not significant. fo fe fo - fe (fo - fe)2 (fo - fe)2 /fe 23 19.28 3.72 13.838 .718 15 18.72 -3.72 .739 11 14.72 .940 18 14.28 .969 Total 67 - Χ2 = 3.366

Going Further by Answering the 3 Questions Posed in Ch Going Further by Answering the 3 Questions Posed in Ch. 11 in 2/3e (12 in 1e) #9.4 (10.4) Liberal Party support by gender Is there an association? Not a significant association. How strong is the association? Calculate Phi = .20. A moderate association (table 11.12/12.12) What is the pattern of the association? Generally the pattern of percentages shows that females are more likely to support the Liberal Party (60% vs 40%).

Going Further cont. #9.6 (10.6) Alertness by program participation Is there an association? Not a significant association. How strong is the association? Calculate Phi = .22. A moderate association (table 11.12/12.12) What is the pattern of the association? Generally the pattern of percentages shows that program participants are more alert (67.6% vs 45.5%).