Chapter 14: Chi-Square Procedures

14.1 – Test for Goodness of Fit

Chi-square test for goodness of fit: Used to determine if what outcome you expect to happen actually does happen Observed Counts: Count of actual results Expected Counts:Count of expected results To find the expected counts multiply the proportion you expect times the sample size (2)(2)

Note: Sometimes the probabilities will be expected to be the same and sometimes they will be expected to be different

Ho: All of the proportions are as expected Chi-square test for goodness of fit: H A : One or more of the proportions are different from expected

Chi-square test for goodness of fit: df = k – 1 categories

Properties of the Chi-distribution: Always positive (being squared) Skewed to the right Distribution changes as degrees of freedom change

Area is shaded to the right Properties of the Chi-distribution: Always positive (being squared) Skewed to the right Distribution changes as degrees of freedom change

Chi-square test for goodness of fit: Conditions: all expected counts are ≥ 5

Calculator Tip!Goodness of Fit L1: Observed L2: Expected L3: (L1 – L2) 2 / L2 List – Math – Sum – L3

Calculator Tip! P(  2 > #) 2nd dist -  2 cdf(  2, big #, df)

Example #1 A study yields a chi-square statistic value of 20 (  2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories?

Example #1 A study yields a chi-square statistic value of 20 (  2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories? < P < Or by calc:  2 cdf(20, , 11) = 2nd dist -  2 cdf(  2, big #, df)

Example #2 Find the expected values and calculate the  2 of the 96 rolls of the die. Then find the probability. Face Value Observed Expected 16 (19 – 16) 2 16 = (15 – 16) 2 16 =

Example #2 Find the expected values and calculate the  2 of the 96 rolls of the die. Then find the probability. Face Value Observed Expected = 4.75

P(  2 > 4.75) = df =6 – 1 = 5

P(  2 > 4.75) = df =6 – 1 = 5 P(  2 > 4.75) > 0.25 Or by calc:  2 cdf(4.75, , 5) =

Example #3 The number of defects from a manufacturing process by day of the week are as follows: The manufacturer is concerned that the number of defects is greater on Monday and Friday. Test, at the 0.05 level of significance, the claim that the proportion of defects is the same each day of the week. MondayTuesdayWednesdayThursdayFriday # P: The true proportion of defects from a manufacturing process per day

H: The proportion of defects from a manufacturing process is the same Mon-Fri Ho: The proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more) HA:HA:

A: MondayTuesdayWednesdayThursdayFriday # Expected 150 errors total = (All expected counts > 5) N:Chi-Square Goodness of Fit

T:  2 = MondayTuesdayWednesdayThursdayFriday # Expected  (O – E) 2 E = = (36 – 30) (23 – 30) …

O: P(  2 > 7.533) = df =5 – 1 = 4

P(  2 > 7.533) = df =5 – 1 = ) > 0.15 Or by calc:  2 cdf(7.533, , 4) = O:

M: P ___________  > Accept the Null

S: There is not enough evidence to claim the proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more)

14.2 – Inference for Two-Way Tables

To compare two proportions, we use a 2-Proportion Z Test. If we want to compare three or more proportions, we need a new procedure.

Two – Way Table:  Organize the data for several proportions  R rows and C columns  Dimensions are r x c

To calculate the expected counts, multiply the row total by the column total, and divide by the table total: Expected count = Row total x Column total table total Degrees of Freedom: (r – 1)(c – 1)

Chi-Square test for Homogeneity: Ho: The proportions are the same among 2 or more populations Ha: The proportions are different among 2 or more populations Expected Counts are ≥ 5 SRS Conditions: Compare two or more populations on one categorical variable

Chi-Square test for Association/Independence: Ho: There is no association between two categorical variables Ha: There is an association Expected Counts are ≥ 5 SRS Conditions: Two categorical variables collected from a single population

Calculator Tip!Test for Homogeneity/Independence 2 nd – matrx – edit – [A] – rxc – Table info Stat – tests –  2 –test Observed: [A] Expected: [B] Calculate Then: Note: Expected: [B] is done automatically!

Example #1 The table shows the number of people in each grade of high school who preferred a different color of socks. a. What is the expected value for the number of 12th graders who prefer red socks? Expected count = Row total x Column total table total 20 x = =5

Example #1 The table shows the number of people in each grade of high school who preferred a different color of socks. b. Find the degrees of freedom. (r – 1)(c – 1) (3 – 1)(4 – 1) (2)(3) 6

Example #2 An SRS of a group of teens enrolled in alternative schooling programs was asked if they smoked or not. The information is classified by gender in the table. Find the expected counts for each cell, and then find the chi-square statistic, degrees of freedom, and its corresponding probability. Expected Counts:Row total x Column total table total 70 x x x x = = = =

 2 =  (O – E) 2 E = Expected Counts: SmokerNon-Smoker Male Female (23 – ) (47 – ) (56 – ) (91 – )

 2 =  (O – E) 2 E = Expected Counts: SmokerNon-Smoker Male Female (r – 1)(c – 1) =(2 – 1)(2 – 1) =(1)(1) =1Degrees of Freedom: P(  2 > ) =

 2 =  (O – E) 2 E = Expected Counts: SmokerNon-Smoker Male Female (r – 1)(c – 1) =(2 – 1)(2 – 1) =(1)(1) =1Degrees of Freedom: P(  2 > ) = More than 0.25OR:0.4535

Example #3 At a school a random sample of 20 male and 16 females were asked to classify which political party they identified with. DemocratRepublicanIndependent Male1172 Female781 Are the proportions of Democrats, Republicans, and Independents the same within both populations? Conduct a test of significance at the α = 0.05 level. P: The proportion of Democrats, Republicans, and Independents that are Male and Female.

H: Ho: The proportions are the same among males and females and their political party H A : The proportions are different among males and females and their political party

A: SRS(says) Expected Counts  5 DemocratRepublicanIndependent Male1172 Female781 Row total x Column total table total 20 x x x x = = = = x x 3 36 = = Not all are expected counts are  5, proceed with caution!

N:Chi-Square test for Homogeneity

T:  2 =  (O – E) 2 E = (11 – 10) (7 – 8) 2 8 DemocratRepublicanIndependent Male1172 Female781 Male Female Expected + (7 – 8.33) (8 – 6.67) (2 – 1.67) (1 – 1.33)

O: (r – 1)(c – 1) =(2 – 1)(3 – 1) =(1)(2) =2Degrees of Freedom: P(  2 > ) =

More than 0.25 OR: O: (r – 1)(c – 1) =(2 – 1)(3 – 1) =(1)(2) =2Degrees of Freedom: P(  2 > ) =

M: P ___________  > Accept the Null

S: There is not enough evidence to claim the proportions are different among males and females and their political party

Example #4 The following chart represents the score distribution on the AP Exams for different subjects at a certain high school. Is there evidence that the score distribution is dependent from the subject? P: Determine if AP scores are independent from the subject areas.

H: Ho: AP scores and AP test are independent. H A : AP scores and AP test are not independent.

A: SRS(says) Expected Counts  5 Row total x Column total table total

34 x x x x = = = = x x = = x x == x = 8.8

18 x x x x = = = = x x = = Not all are expected counts are  5, proceed with caution!

N:Chi-Square test for Independence

T: Expected  2 =  (O – E) 2 E =

O: (r – 1)(c – 1) =(5 – 1)(3 – 1) =(4)(2) =8Degrees of Freedom: P(  2 > ) =

More than 0.25 OR: O: P(  2 > ) = (r – 1)(c – 1) =(5 – 1)(3 – 1) =(4)(2) =8Degrees of Freedom:

M: P ___________  > Accept the Null

S: There is not enough evidence to claim the AP scores and AP test are dependent