Download presentation

Presentation is loading. Please wait.

Published byGianni Kayton Modified over 3 years ago

1
Chi-Square Non-parametric test (distribution- free) Nominal level dependent measure

2
Categorical Variables Generally the count of objects falling in each of several categories.Generally the count of objects falling in each of several categories. Examples:Examples: Xnumber of fraternity, sorority, and nonaffiliated members of a class Xnumber of students choosing answers: 1, 2, 3, 4, or 5 Emphasis on frequency in each categoryEmphasis on frequency in each category

3
One-way Classification Observations sorted on only one dimensionObservations sorted on only one dimension Example:Example: XObserve children and count red, green, yellow, or orange Jello choices XAre these colors chosen equally often, or is there a preference for one over the other? Cont.

4
One-way--cont. Want to compare observed frequencies with frequencies predicted by null hypothesisWant to compare observed frequencies with frequencies predicted by null hypothesis Chi-square test used to compare expected and observedChi-square test used to compare expected and observed Called goodness-of-fit chi-square ( 2 ) Called goodness-of-fit chi-square ( 2 )

5
Goodness-of-Fit Chi-square Fombonne (1989) Season of birth and childhood psychosisFombonne (1989) Season of birth and childhood psychosis Are children born at particular times of year more likely to be diagnosed with childhood psychosisAre children born at particular times of year more likely to be diagnosed with childhood psychosis He knew the % normal children born in each monthHe knew the % normal children born in each month Xe.g..8.4% normal children born in January

6
Fombonnes Data

7
Chi-Square ( 2 ) Compare Observed (O) with Expected (E)Compare Observed (O) with Expected (E) Take size of E into accountTake size of E into account XWith large E, a large (O-E) is not unusual. XWith small E, a large (O-E) is unusual.

8
Calculation of 2 2.05 (11) = 19.68

10
Conclusions Obtained 2 = 14.58Obtained 2 = 14.58 df = c - 1, where c = # categoriesdf = c - 1, where c = # categories Critical value of 2 on 11 df = 19.68Critical value of 2 on 11 df = 19.68 Since 19.68 > 14.58, do not reject H 0Since 19.68 > 14.58, do not reject H 0 Conclude that birth month distribution of children with psychoses doesnt differ from normal.Conclude that birth month distribution of children with psychoses doesnt differ from normal.

11
Jello Choices Red GreenYellowOrangeRed GreenYellowOrange 35 20 25 20 35 20 25 20 Is there a significant preference for one color of jello over other colors?Is there a significant preference for one color of jello over other colors?

12
RedGreen YellowOrangeRedGreen YellowOrange O: 35 20 25 20 E: 25 25 25 25 X 2 = (35-25) 2 /25 + (20-25) 2 /25 + (25-25) 2 /25 + (20-25) 2 /25= 6 There was not one jello color chosen significantly more often than any other jello color, X 2 (3, N= 100) = 6, p >.05

13
Contingency Tables Two independent variablesTwo independent variables XAre men happier than women? Male vs. Female X Happy vs Not HappyMale vs. Female X Happy vs Not Happy XIntimacy (Yes/No) X Depression/Nondepression

14
Intimacy and Depression Everitt & Smith (1979)Everitt & Smith (1979) Asked depressed and non-depressed women about intimacy with boyfriend/husbandAsked depressed and non-depressed women about intimacy with boyfriend/husband Data on next slideData on next slide

15
Data

16
Chi-Square on Contingency Table Same formulaSame formula Expected frequenciesExpected frequencies XE = RT X CT GT RT = Row total, CT = Column total, GT = Grand totalRT = Row total, CT = Column total, GT = Grand total

17
Expected Frequencies E 11 = 37*138/419 = 12.19E 11 = 37*138/419 = 12.19 E 12 = 37*281/419 = 24.81E 12 = 37*281/419 = 24.81 E 21 = 382*138/419 = 125.81E 21 = 382*138/419 = 125.81 E 22 = 382*281/419 = 256.19E 22 = 382*281/419 = 256.19 Enter on following tableEnter on following table

18
Observed and Expected Freq.

19
Chi-Square Calculation

20
Degrees of Freedom For contingency table, df = (R - 1)(C - 1)For contingency table, df = (R - 1)(C - 1) For our example this is (2 - 1)(2 - 1) = 1For our example this is (2 - 1)(2 - 1) = 1 XNote that knowing any one cell and the marginal totals, you could reconstruct all other cells.

21
Conclusions Since 25.61 > 3.84, reject H 0Since 25.61 > 3.84, reject H 0 Conclude that depression and intimacy are not independent.Conclude that depression and intimacy are not independent. XHow one responds to satisfaction with intimacy depends on whether they are depressed. XCould be depression-->dissatisfaction, lack of intimacy --> depression, depressed people see world as not meeting needs, etc.

22
Larger Contingency Tables Jankowski & Leitenberg(pers. comm.)Jankowski & Leitenberg(pers. comm.) XDoes abuse continue? Do adults who are, and are not, being abused differ in childhood history of abuse?Do adults who are, and are not, being abused differ in childhood history of abuse? One variable = adult abuse (yes or no)One variable = adult abuse (yes or no) Other variable = number of abuse categories (out of 4) suffered as childrenOther variable = number of abuse categories (out of 4) suffered as children XSexual, Physical, Alcohol, or Personal violence

24
Chi-Square Calculation

25
Conclusions 29.62 > 7.8229.62 > 7.82 XReject H 0 XConclude that adult abuse is related to childhood abuse XIncreasing levels of childhood abuse are associated with greater levels of adult abuse. e.g. Approximately 10% of nonabused children are later abused as adults.e.g. Approximately 10% of nonabused children are later abused as adults. Cont.

26
Nonindependent Observations We require that observations be independent.We require that observations be independent. XOnly one score from each respondent XSum of frequencies must equal number of respondents If we dont have independence of observations, test is not valid.If we dont have independence of observations, test is not valid.

27
Small Expected Frequencies Rule of thumb: E > 5 in each cellRule of thumb: E > 5 in each cell XNot firm rule XViolated in earlier example, but probably not a problem More of a problem in tables with few cells.More of a problem in tables with few cells. Never have expected frequency of 0.Never have expected frequency of 0. Collapse adjacent cells if necessary.Collapse adjacent cells if necessary. Cont.

28
Expected Frequencies--cont. More of a problem in tables with few cells.More of a problem in tables with few cells. Never have expected frequency of 0.Never have expected frequency of 0. Collapse adjacent cells if necessary.Collapse adjacent cells if necessary.

29
Effect Size Phi and Cramers PhiPhi and Cramers Phi XDefine N and k XNot limited to 2X2 tables Cont.

30
Effect Sizecont. Everitt & cc dataEveritt & cc data Cont.

31
Effect SizeOdss Ratio. Odds Dep|Lack IntimacyOdds Dep|Lack Intimacy X26/112 =.232 Odds Dep | No LackOdds Dep | No Lack X11/270 =.041 Odds Ratio =.232/.041 = 5.69Odds Ratio =.232/.041 = 5.69 Odds Depressed = 5.69 times great if experiencing lack of intimacy.Odds Depressed = 5.69 times great if experiencing lack of intimacy.

32
Effect SizeRisk Ratio. Risk Depression/Lack IntimacyRisk Depression/Lack Intimacy X26/138 =.188 Risk Depression | No LackRisk Depression | No Lack X11/281 =.039 Odds Ratio =.188/.039 = 4.83Odds Ratio =.188/.039 = 4.83 Risk of Depressed = 4.83 times greater if experiencing lack of intimacy.Risk of Depressed = 4.83 times greater if experiencing lack of intimacy.

Similar presentations

OK

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.

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.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google