Basic Business Statistics, 11e 2009 Prentice-Hall, Inc.Chap 12-*

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Learning ObjectivesIn this chapter, you learn:

How and when to use the chi-square test for contingency tablesList the characteristics of the Chi-Square Distribution.Conduct a chi-square test for association / independence

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Contingency TablesContingency TablesUseful in situations involving multiple population proportionsUsed to classify sample observations according to two or more characteristicsAlso called a cross-classification table.

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Contingency Table ExampleLeft-Handed vs. Gender Dominant Hand: Left vs. Right Gender: Male vs. Female

2 categories for each variable, so called a 2 x 2 table

Suppose we examine a sample of 300 children

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Contingency Table ExampleSample results organized in a contingency table:

(continued)120 Females, 12 were left handed180 Males, 24 were left handedsample size = n =300:

GenderHand PreferenceLeftRightFemale12108120Male2415618036264300

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Characteristics of the Chi-Square DistributionThe major characteristics of the chi-square distribution are:It is positively skewedIt is non-negativeIt is based on degrees of freedomWhen the degrees of freedom change a new distribution is created14-2

CHI-SQUARE DISTRIBUTION df = 3df = 5df = 10c22-2

It is used to test whether two traits or variables are related.Each observation is classified according to two variables.The usual hypothesis testing procedure is used.The degrees of freedom is equal to: (number of rows-1)(number of columns-1).The expected frequency is computed as: Expected Frequency = (row total)(column total)/grand total 14-152 Test of Independence

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*2 Test of IndependenceHYPOTHESESH0: The two categorical variables are independent(i.e., there is no relationship between them)H1: The two categorical variables are dependent(i.e., there is a relationship between them)

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*2 Test of Independencewhere:fo = observed frequency in a particular cell of the r x c tablefe = expected frequency in a particular cell if H0 is true

(Assumed: each cell in the contingency table has expectedfrequency of at least 1)The Chi-square test statistic is:(continued)

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Expected Cell FrequenciesExpected cell frequencies:Where:row total = sum of all frequencies in the rowcolumn total = sum of all frequencies in the columnn = overall sample size

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..Chap 12-*Decision RuleThe decision rule isWhere is from the chi-squared distribution with (r 1)(c 1) degrees of freedomIf , reject H0,otherwise, do not reject H0

Basic Business Statistics, 11e 2009 Prentice-Hall, Inc..

EXAMPLE 1Is there a relationship between the location of an accident and the sex of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the .05 level of significance, can we conclude that gender and the location of the accident are related?14-16

EXAMPLE 11 continuedNote: The expected frequency for the work-male intersection is computed as (90)(80)/150=48.Similarly, you can compute the expected frequencies for the other cells.14-17

Sex

Work

Home

Other

Total

Male

60

20

10

90

Female

20

30

10

60

Total

80

50

20

150

EXAMPLE 1 continuedStep 1: : Gender and location are not related. : Gender and location are related.

Step 2: is rejected ifStep 3:Step 4: is rejected. Gender and location are related.14-18

ExampleThe meal plan selected by 200 students is shown below:

ClassStandingNumber of meals per weekTotal20/week10/weeknoneFresh.24321470Soph.22261260Junior1014630Senior14161040Total 708842200

ExampleThe hypothesis to be tested is:(continued)H0: Meal plan and class standing are independent(i.e., there is no relationship between them)H1: Meal plan and class standing are dependent(i.e., there is a relationship between them)

Observed:Expected cell frequencies if H0 is true:Example for one cell:Example: Expected Cell Frequencies(continued)

ClassStandingNumber of meals per weekTotal20/wk10/wknoneFresh.24321470Soph.22261260Junior1014630Senior14161040Total 708842200

ClassStandingNumber of meals per weekTotal20/wk10/wknoneFresh.24.530.814.770Soph.21.026.412.660Junior10.513.26.330Senior14.017.68.440Total 708842200

Example: The Test StatisticThe test statistic value is:(continued) = 12.592 from the chi-squared distribution with (4 1)(3 1) = 6 degrees of freedom

Example: Decision and Interpretation(continued)Decision Rule:If > 12.592, reject H0, otherwise, do not reject H0Here, = 0.709 < = 12.592, so do not reject H0 Conclusion: there is no sufficient evidence that meal plan and class standing are related at = 0.05220.05=12.5920 0.05Reject H0Do not reject H0

Problem 1The makers of the movie Titanic imply that lower-class passengers were treated unfairly when the lifeboats were being filled. We want to determine whether that portrayal is accurate. The following table contains the survival data by passenger class for the 1316 passengers.

Use a chi-square test to determine whether there is a relationship between survival and passenger class. Use 1% level of significance.

ClassSurvivedLostFirst203122Second118167Third178528