1. Textbook Section 11.2

2. * We already know how to compare two proportions for two populations/groups. * What if we want to compare the distributions of a single categorical variable over several populations or treatments? * This information is usually organized into a two-way table. * There are two tests we can use with these tables. * Chi Square Test for Homogeneity tests to see if the distribution of the categorical variable is the same for several populations * Chi Square Test for Independence tests to see if there is a relationship between the row and column variables of the table.

3. * Market researchers think that the background music in a restaurant affects what people order. * What is the conditional distribution of entrees ordered for each treatment? * No music: French: 0.357; Italian: 0.131; Other: 0.512 * French music: French: 0.520; Italian: 0.013; Other: 0.467 * Italian music: French: 0.357; Italian: 0.226; Other: 0.417 * Did music affect what people ordered? Type of Music EntréeNoneFrenchItalianTotal French30393099 Italian1111931 Other4335 113 Total847584243

4. * In this case the null hypothesis is * H 0 : There is no difference in the true distribution of entrees ordered at this restaurant when no music, French music or Italian music is played. * If the null hypothesis is true, the observed differences are simply due to chance. * The alternative hypothesis, on the other hand, * H a : There is a difference in the true distribution of entrees ordered at this restaurant when no music, French music, or Italian music is played. * Notice ANY difference would favor the alternative.

5. * We could conduct several two-sample tests of proportion for all of the different pairs possible… * That’s 36 tests! And it still doesn’t give us an overall test of music played vs. entrée ordered * The overall test needed uses the Chi-Square statistic…which means we need to calculate expected counts.

6. * For each cell in the table, the expected count can be calculated by (row total)×(column total)÷(table total) * All expected counts are > 5 Type of Music EntréeNoneFrenchItalianTotal French34.2230.5634.2299 Italian10.729.5710.7231 Other39.0634.8839.06113 Total847584243

7. * State hypotheses * Check conditions for a Χ 2 test for homogeneity * Random – stated in problem * Expected counts > 5 – ALWAYS draw expected counts table! * Calculations  Calculator directions p. ??? * Report df = (rows – 1)(columns – 1), χ 2, and P-value * Conclusion: Because the P-value of 0.001 is less than α = 0.05, we reject H 0. We have convincing evidence that there is a difference in entrees ordered if there is no music, French music or Italian music playing.

8. * Let’s compare the actual counts vs. expected counts.. * The largest differences appear to be in Italian entrees. * If we calculate those individual χ 2 contributions: * Italian entrée + French music = 7.672 * Italian entrée + Italian music = 6.404 * Together they make up approximately 14 of the total χ 2 of 18! * Therefore: Orders of Italian entrees are strongly affected by French and Italian music. Type of Music EntréeNoneFrenchItalian French34.2230.5634.22 Italian10.729.5710.72 Other39.0634.8839.06 Type of Music EntréeNoneFrenchItalian French303930 Italian11119 Other4335

9. 1. What is the expected count for students who use Facebook at least once a week and live on the main campus? 2. Calculate the Χ 2 statistic. 3. Calculate the p-value Use FacebookMain CampusCommonwealth Several times a month or less5576 At least once a week215157 At least once a day640394 Total Facebook Users910627

10. * Another common situation is when a single population is classified based on two categorical variables. In this case, we are examining the potential relationship between the variables. * Test = Chi Square Test for Independence * Hypotheses: * H 0 : There is no association between the two variables * H a : There is an association between the two variables.

11. * A random sample of healthy adults were given the Spielberger Anger Scale Test which measures how prone a person is to sudden anger. Researchers then followed the subjects to determine if they developed Coronary Heart Disease and recorded the data: * H 0 : There is no association between anger levels and Coronary Heart Disease. * H a : There is an association between anger levels and Coronary Heart Disease. Low Anger Moderate Anger High Anger Total CHD5311027190 No CHD305746216068284 Total311047316338474

12. * If conditions are met, we will conduct a Chi-Square test of Independence. * Random – Stated random sample * Expected Counts Greater than 5? * Calculate: df = 2, P = 0.0003, χ 2 = 16.0767 * With a P-value of 0.0003 less than α = 0.05, we reject H 0. We have convincing evidence that there is an association between a person’s propensity to sudden anger and coronary health. With a χ 2 component of 11.564 which is 72% of the total χ 2 statistic, it appears that there is a relationship between high levels of anger and coronary heart disease. LAMAHA CHD69.73106.0714.19 No CHD3040.274624.92618.81

13. * Because tests with two-way tables are carried out with the same methods, it is sometimes difficult to tell them apart. * It’s important to look at how data was gathered. * Two or more populations with an explanatory and response variable  Homogeneity * One population or sample with two different categorical variables where explanatory/response is not clear – just testing a relationship  Independence