1 2 Test for Independence 2 Test for Independence

2 Data Types

3 Hypothesis Tests Qualitative Data

4 2 Test of Independence 2 Test of Independence 1.Shows If a Relationship Exists Between 2 Qualitative Variables, but does Not Show Causality 2.Assumptions Multinomial Experiment All Expected Counts 5 3.Uses Two-Way Contingency Table

5 2 Test of Independence Contingency Table 2 Test of Independence Contingency Table 1.Shows # Observations From 1 Sample Jointly in 2 Qualitative Variables 1.Shows # Observations From 1 Sample Jointly in 2 Qualitative Variables

6 2 Test of Independence Contingency Table 2 Test of Independence Contingency Table 1.Shows # Observations From 1 Sample Jointly in 2 Qualitative Variables Levels of variable 2 Levels of variable 1

7 2 Test of Independence Hypotheses & Statistic 2 Test of Independence Hypotheses & Statistic 1.Hypotheses H 0 : Variables Are Independent H 0 : Variables Are Independent H a : Variables Are Related (Dependent) H a : Variables Are Related (Dependent)

8 2 Test of Independence Hypotheses & Statistic 2 Test of Independence Hypotheses & Statistic 1.Hypotheses H 0 : Variables Are Independent H a : Variables Are Related (Dependent) 2.Test Statistic Observed count Expected count

9 2 Test of Independence Hypotheses & Statistic 2 Test of Independence Hypotheses & Statistic 1.Hypotheses H 0 : Variables Are Independent H a : Variables Are Related (Dependent) 2.Test Statistic Degrees of Freedom: (r - 1)(c - 1) Rows Columns Observed count Expected count

10 Expected Count Example

11 Expected Count Example Marginal probability =

12 Expected Count Example Marginal probability =

13 Expected Count Example Marginal probability = Joint probability =

14 Expected Count Example Marginal probability = Joint probability = Expected count = 160· = 54.6

15 Expected Count Calculation

16 Expected Count Calculation

17 Expected Count Calculation 112x x x x78 160

18 You randomly sample 286 sexually active individuals and collect information on their HIV status and History of STDs. At the.05 level, is there evidence of a relationship? You randomly sample 286 sexually active individuals and collect information on their HIV status and History of STDs. At the.05 level, is there evidence of a relationship? 2 Test of Independence Example on HIV 2 Test of Independence Example on HIV

19 2 Test of Independence Solution 2 Test of Independence Solution

20 2 Test of Independence Solution 2 Test of Independence Solution H 0 : H a : = = df = Critical Value(s): Test Statistic: Decision:Conclusion:

21 2 Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship = = df = Critical Value(s): Test Statistic: Decision:Conclusion:

EPI809/Spring Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship =.05 =.05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision:Conclusion:

23 2 Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship =.05 =.05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision:Conclusion: =.05 =.05

24 E(n ij ) 5 in all cells 170x x x x Test of Independence Solution 2 Test of Independence Solution

25 2 Test of Independence Solution 2 Test of Independence Solution

26 2 Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship =.05 =.05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision:Conclusion: =.05 =.05 2 = = 54.29

27 2 Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship =.05 =.05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision:Conclusion: Reject at =.05 =.05 =.05 2 = = 54.29

28 2 Test of Independence Solution 2 Test of Independence Solution H 0 : No Relationship H a : Relationship =.05 =.05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision:Conclusion: Reject at =.05 There is evidence of a relationship =.05 =.05 2 = = 54.29

The procedure to test for the independence: 1. State a hypotheses based on the fit of the data 2. Make a table of the observed and expected values. You will most likely be given the observed values. 3. Calculate the chi-squared test statistic, this is 4. Look up the chi-squared critical value from your chi-squared tables in the information booklet. 5. Compare your test statistic with your critical value and make a conclusion. If the test statistic lies in the critical region then reject H 0 in favour of H 1. Otherwise do not reject H 0 in favour of H 1. At first glance this is similar to the goodness of fit test, but the test statistic is worked out differently.

Degrees of freedom, v. When undertaking a chi-squared test you will have a table of observed and expected values. The degrees of freedom will be defined as: v=(number of rows-1)(number of columns-1) The chi-squared distribution. The distribution will alter depending on the value of v. The general curve is shown opposite.

Example of Chi-squared independence test The headmaster of a large IB school is concerned that the maths results are dependent on the maths teacher. There are 3 SL teachers and the results for each class have been shown below. These are the observed values. Test at the 5% level of significance to see if the grades are independent of the teacher Total Mr. P Ms. Q Mrs. R Total Make your hypotheses: H 0 : the grade at maths SL is independent of the teacher. H 1 : the grade at maths SL is not independent of the teacher. Make a table of expected values. To do this take each row total x column total and divide by the grand total. This is shown opposite. This value is the expected value for this cell. Find the expected number of grade 2s that Mr. P gets. Complete a table of expected values.

Total Mr. P Ms. Q Mrs. R Total continued.... Observed Expected Total Mr. P Ms. Q Mrs. R Total Calculate the chi squared test statistic: Find the critical value from your tables. v=(7-1)(3-1)=12 Critical value = Make your conclusion: Do not reject the null hypothesis. At the 5% level of significance there is no evidence to suggest that the choice of teacher influences the grade achieved. the p value