Contingency Tables

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Chapter 11: Goodness-of-Fit and Contingency Tables
11.2: Contingency Tables

1. Elementary Statistics Chapter 11: Goodness-of-Fit and Contingency Tables 11.2 Contingency Tables 1

2. Chapter 11: Goodness-of-Fit and Contingency Tables 11.1 Goodness-of-Fit 11.2 Contingency Tables 2 Objectives: 1 Test a distribution for goodness of fit, using chi-square. 2 Test two variables for independence, using chi-square. 3 Test proportions for homogeneity, using chi-square.

3. 3 11.2 Contingency Tables Contingency Table: A contingency table (or two-way frequency table ) is a table consisting of frequency counts of categorical data corresponding to two different variables. (One variable is used to categorize rows, and a second variable is used to categorize columns.) When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested by using the chi-square test. Test of Independence (of variables): In a test of independence, we test the null hypothesis that in a contingency table, the row and column variables are independent. (That is, there is no dependency between the row variable and the column variable.) The Test of Independence of Variables is used to determine whether two variables are independent of or related to each other when a single sample is selected. Chi-Square Test of Homogeneity: ( Test of Homogeneity of Proportions ) A chi-square test of homogeneity is a test of the claim that different populations have the same proportions of some characteristics. The Test of Homogeneity of Proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations.

4. 4 11.2 Contingency Tables Objective, Notation & Requirements Conduct a hypothesis test of independence between the row variable and column variable in a contingency table. O represents the observed frequency in a cell of a contingency table. E represents the expected frequency in a cell, found by assuming that the row and column variables are independent. r represents the number of rows in a contingency table (not including labels or row totals). c represents the number of columns in a contingency table (not including labels or column totals). Requirements The sample data are randomly selected. The sample data are represented as frequency counts in a two-way table. For every cell in the contingency table, the expected frequency E is at least 5 . (There is no requirement that every observed frequency must be at least 5.) H : The row and column variables are independent. (There is no relationship between two variables.) H 1 : The row and column variables are dependent. (There is a relationship between two variables.) The critical values are found in Chi-Square Table ( ), Df = ( r − 1) ( c − 1) where r is the number of rows and c is the number of columns. Tests of independence with a contingency table are always right-tailed .       TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate Note: Matrix B gets populated as Expected Matrix  

5. 5 11.2 Contingency Tables d.f. = (rows – 1) (columns – 1) = ( R – 1)( C – 1) Test Statistic: d.f. = ( R – 1)( C – 1) O = observed frequency E = expected frequency :  

6. 6 The cells of this table contain frequency counts. The frequency counts are the observed values, and the expected values are shown in parentheses. The row variable identifies the treatment used for a stress fracture in a foot bone, and the column variable identifies the outcome as a success or failure. Refer to the table and find the expected frequency for the cell in the first row and first column, where the observed frequency is 54 . Example 1: 11.2 Contingency Tables: Finding Expected Frequency blank Success Failure Surgery 54 ( E = 47.478) 12( E = 18.522) Weight-Bearing Cast 41( E = 66.182) 51( E = 25.818) Non-Weight-Bearing Cast for 6 Weeks 70( E = 52.514) 3( E = 20.486) Non-Weight-Bearing Cast for Less Than 6 Weeks 17( E = 15.826) 5( E = 6.174)   Total: Sum =182 Sum = 71 66 92 73 22 253   Interpretation: Assuming that success is independent of the treatment, then we expect to find that 47.478 of the subjects would be treated with surgery and that treatment would be successful. There is a discrepancy between O = 54 and E = 47.478, and such discrepancies are key components of the test statistic that is a collective measure of the overall disagreement between the observed frequencies and the frequencies expected with independence between the row and column variables.   Note: Example: Row 1:  

7. Example 2: Test of Independence 7 Step 1: H : Success is independent of the treatment. ( Claim ) H 1 : Success and the treatment are dependent. RTT Step 3: df = (4 − 1)(2 − 1) = 3 & α = 0.05   Step 4: Decision: Reject H The claim is False There is NOT sufficient evidence to support the claim that success of the treatment is independent of the type of treatment.   Use the same sample data from the previous example with a 0.05 significance level to test the claim that success of the treatment is independent of the type of treatment. What does the result indicate about the increasing trend to use surgery? REQUIREMENT CHECK: 1. On the basis of the study description, let’s assume random selection & assignment of subjects to the different treatment groups. 2. Results are f requency counts. 3. E > 5 for all. (lowest = 6.174.) OR: P -Value from Table: TS: = 58.393 > highest value (12.838) in Table P -value < 0.005.   Interpretation: Success is dependent on the treatment and the success rates of 81.8% (54/54+12 or 66), 44.6% (41/92), 95.9% (70/73), and 77.3% (17/22) suggest that the best treatment is to use a non–weight-bearing cast for 6 weeks. These results suggest that the increasing use of surgery is a treatment strategy that is not supported by the evidence. ( from the table of example 1 ) Step 1: H , H 1 , claim & Tails Step 2: TS Calculate ( TS ) Step 3: CV using α Step 4: Make the decision to a. Reject or not H b. The claim is true or false c. Restate this decision: There is / is not sufficient evidence to support the claim that… CV: = 7.815   TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate     Step 2:  

8. 8 Example 3: Test of Independence of Variables A random sample of 3 hospitals was selected, and the number of infections for a specific year has been reported. Test the claim that there is a relationship between the hospital and the number of patient infections. ( Number of patient infections depends on the hospital ). TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate Note: Matrix B gets populated as Expected Matrix    

9. 9 Example 3 Continued: Step 1: H : The number of infections is independent of the hospital. H 1 : The number of infections is dependent on the hospital ( claim), RTT Step 3: df = (3 – 1)(3 – 1) = 4 & α = 0.05 Step 4: Decision: Reject H The claim is True There is sufficient evidence to support the claim that the number of infections is related to the hospital where they occurred. CV: = 9.488     Step 2:   CV TS

10. 10 A researcher wishes to determine whether there is a relationship between the gender of an individual and the amount of alcohol consumed. A sample of 68 people is selected, and the following data are obtained. At α = 0.10, can the researcher conclude that alcohol consumption is related to gender? Example 4: Test of Independence of Variables   Gender Alcohol Consumption Total Low Moderate High Male 10 9 8 27 Female 13 16 12 41 Total 23 25 20 68 (9.13) (9.93) (7.94) (13.87) (15.07) (12.06)     Step 1: H : The amount of alcohol that a person consumes is independent of the individual’s gender. H 1 : The amount of alcohol that a person consumes is dependent on the individual’s gender (claim), RTT Step 3: df = (2 – 1 )(3 – 1) = 2 & α = 0.10 CV: = 4.605   Step 4: Decision: Do not Reject H The claim is False There is not sufficient evidence to support the claim that the amount of alcohol a person consumes is dependent on the individual’s gender.     Step 2:   TS: χ ² = CV: χ ² =

11. 11 11.2 Contingency Tables: Test for Homogeneity of Proportions Contingency Tables: Chi-Square Test of Homogeneity: A chi-square test of homogeneity is a test of the claim that different populations have the same proportions of some characteristics. Assumptions for Homogeneity of Proportions: In conducting a test of homogeneity, we can use the same notation, requirements, test statistic, critical value, and procedures given previously, with this exception: Instead of testing the null hypothesis of independence between the row and column variables, we test the null hypothesis that the different populations have the same proportion of some characteristic. The data are obtained from a random sample. The expected frequency for each category must be 5 or more. In a typical test of independence , sample subjects are randomly selected from one population and values of two different variables are observed. In a typical chi-square test of Homogeneity , subjects are randomly selected from different populations separately. W e test the null hypothesis that the different populations have the same proportion of some characteristic.

12. REQUIREMENT CHECK Based on the description of the study, we will treat the subjects as being randomly selected and randomly assigned to the different cities. The results are expressed as frequency counts. The expected frequencies are all at least 5 . (All expected values are either 5.625 or 6.375.) The requirements are satisfied. 12 This table lists results from an experiment in which 12 wallets were intentionally lost in each of 16 different cities , including New York City, London, Amsterdam, and so on. Use a 0.05 significance level with the data to test the null hypothesis that the cities have the same proportion of returned wallets . (Note: This Lost Wallet Test” implies that whether a wallet is returned is dependent on the city in which it was lost.) Test the claim that the proportion of returned wallets is not the same in the 16 different cities. City (16) A B C D E F G H I J K L M N O P Wallet Returned 8 5 7 11 5 8 6 7 3 1 4 2 4 6 4 9 Wallet Not Returned 4 7 5 1 7 4 6 5 9 11 8 10 8 6 8 3 Example 5: Test of Homogeneity

13. 13 Example 5 Continued: Step 1: H : Whether a lost wallet is returned is independent of the city in which it was lost. ( p 1 = p 2 = p 3 = … = p n ), Claim H 1 : A lost wallet being returned depends on the city in which it was lost. (At least one proportion is different from the others), RTT Step 3: df = (16 – 1 )(2 – 1) = 15 & α = 0.05 Step 4: Decision: Reject H 0, (Reject independence) The claim is False There is not sufficient evidence to support the claim that the proportion of returned wallets is independent of the city in which it was lost. (The proportion of returned wallets depends on the city in which they were lost.) (There is sufficient evidence to conclude that the proportion of returned wallets is not the same in the 16 different cities.) P -value = 0.002 χ ² = 35.388 CV : χ ² = 24.996 TS: χ ² = 35.388 CV: = 24.996     Step 2:   TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate Note: Matrix B gets populated as Expected Matrix  

14. 14 Example 6: Test of Homogeneity 100 people were selected from 4 income groups. They were asked if they were “very happy.” The percent for each group who responded yes and the number from the survey are shown in the table. At α = 0.05 test the claim that there is no difference in the proportions .     TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate Note: Matrix B gets populated as Expected Matrix  

15. 15 Example 6: Step 1: H : p 1 = p 2 = p 3 = p 4 , Claim H 1 : At least one of the proportions differs from the other. RTT Step 3: df = (2 – 1)(4 – 1) = 3 & α = 0.05 Step 4: Decision: Reject H The claim is False There is not enough evidence to support the claim that there is no difference in the proportions. Hence the incomes seem to make a difference in the proportions. χ ² = 14.149 CV: = 7.815   TI Calculator: Contingency Table Access Matrix (2 nd & Press ) Edit, Enter Dimensions & cell entries Stat Tests Observed Matrix must be A Calculate Note: Matrix B gets populated as Expected Matrix     Step 2:   CV: TS:

16. 16 Every cell must have an expected frequency of at least 5. Fisher’s Exact Test is often used for a 2 × 2 contingency table with one or more expected frequencies that are below 5. Fisher’s exact test provides an exact P -value. Because the calculations are quite complex, it’s a good idea to use technology. 11.2 Contingency Tables, Fisher’s Exact Test ( Skip ) Example 7: The MythBusters show on the Discovery Channel tested the theory that when someone yawns, others are more likely to yawn. The results are summarized below. Blank Subject Exposed to Yawning? Yes Subject Exposed to Yawning? No Did Subject Yawn? Yes 10 4 Did Subject Yawn? No 24 12 Using Fisher’s exact test results in a P -value of 0.513, so there is not sufficient evidence to support the myth that people exposed to yawning actually yawn more than those not exposed to yawning.

17. 17 For 2 × 2 tables consisting of frequency counts that result from matched pairs, the frequency counts within each matched pair are not independent and, for such cases, we can use McNemar’s Test of the null hypothesis that the frequencies from the discordant (different) categories occur in the same proportion. 11.2 Contingency Tables, McNemar’s Test for Matched Pairs ( Skip ) Blank Treatment X : Cured Treatment X : Not Cured Treatment Y : Cured a b Treatment Y : Not Cured c d McNemar’s test requires that for a table as shown, the frequencies are such that b + c ≥ 10. The test is a right-tailed chi-square test with the following test statistic:  

18. 18 Example 8: Are Hip Protector’s Effective? A randomized controlled trial was designed to test the effectiveness of hip protectors in preventing hip fractures in the elderly. Nursing home residents each wore protection on one hip, but not the other. Results are as follows. Blank No Hip Protector Worn: No Hip Fracture No Hip Protector Worn: Hip Fracture Hip Protector Worn: No Hip Fracture a = 309 b = 10 Hip Protector Worn: Hip Fracture c = 15 d = 2 McNemar’s Test can be used to test the null hypothesis that the following two proportions are the same: The proportion of subjects with no hip fracture on the protected hip and a hip fracture on the unprotected hip. The proportion of subjects with a hip fracture on the protected hip and no hip fracture on the unprotected hip. Solution: b = 10 and c = 15     α = 0.05 & df = (2 – 1)(2 – 1) = 1 CV: = 3.841 for this right-tailed test. does not exceed the critical value of χ ² = 3.841, so we fail to reject the null hypothesis. The proportion of hip fractures with the protectors worn is not significantly different from the proportion of hip fractures without the protectors worn.