1. CHI-SQUARE AND ANALYSIS OF VARIANCE

5. - the B-school Chi-Square as a test of independence Two attributes A and B are independent if the value of one variable has no influence on the value of another variable. Test H 0 : ‘A and B are independent’ against H A : ‘A and B are not independent’. Assume H 0 to be true and calculate the expected frequencies Chi-square statistic,

7. - the B-school Contingency Table - Example A brand manager is concerned that her brand’s share may be unevenly distributed throughout the country. In a survey in which the country was divided into four geographical regions, a random sampling of 100 consumers in each region was surveyed, with the following results. (a) Construct the contingency table and calculate the chi-square statistic. (b) State the null and alternative hypothesis. (c) At α = 0.05 , test whether brand share depends on the region OR the brand share is the same across the 4 regions. 400 100 100 100 100 Total 210 50 55 45 60 Do not purchase 190 50 45 55 40 Purchase the brand TOTAL SW SE NW NE

8. - the B-school Contingency Table H 0 : Purchasing is independent of region. H A : Purchasing depends on the region. OR H 0 : P NE =P NW =P SE =P SW H A : All proportions are not equal (at least two are unequal) 400 100 100 100 100 Total 210 O 24 = 50 E 24 =52.5 O 23 = 55 E 23 =52.5 O 22 = 45 E 22 =52.5 O 21 = 60 E 21 =52.5 Do not purchase 190 O 14 = 50 E 14 =47.5 O 13 = 45 E 13 =47.5 O 12 = 55 E 12 =47.5 O 11 = 40 E 11 =47.5 Purchase the brand TOTAL SW SE NW NE

10. - the B-school Exercise A researcher, studying the relationship between having a particular disease and addiction of the individuals, interviewed 32 male subjects. For each individual, the researcher recorded his disease status (Y = yes, N = No), and addiction type (Type – I, Type –2, Type – 3) as shown. From the above dataset can the researcher conclude that the males with addiction types 2 or 3 are more likely to have the disease than those with addiction type 1? 2 1 2 3 1 3 1 1 1 1 1 3 2 1 3 1 Addiction type N N Y N Y N N N N N Y N Y Y N Y Disease status 1 2 1 1 2 1 1 1 1 3 1 1 1 2 1 1 Addiction type N N N N Y N N Y N N N N Y N N N Disease status

11. - the B-school Exercise H 0 : Disease is independent of addiction type H A : Disease is prevalent among those who have addiction types 2 and 3 32 5 6 21 Column totals 9 O 23 = 0 E 23 = 0 O 22 = 3 E 22 = 1.69 O 21 = 6 E 21 = 5.91 Y 23 O 13 = 5 E 13 = 3.59 O 12 = 3 E 12 = 4.31 O 11 = 15 E 11 = 15.09 N Row totals TYPE 3 TYPE 2 TYPE 1

13. - the B-school Chi-Square as a Test of Goodness of Fit: Testing the Appropriateness of a Distribution The salesman of a Paper Company has five accounts to visit per day. It is suggested that the variable, sales by him be described by a binomial distribution, with the probability of selling each account being 0.4. Given the following frequency distribution of the number of sales per day, can we conclude that the data does not follow the suggested distribution? Use 0.05 significance level. No. of sales per day 0 1 2 3 4 5 Frequency 10 41 60 20 6 3

14. - the B-school Chi-Square as a Test of Goodness of Fit: Testing the Appropriateness of a Distribution Solution: H 0 : A binomial dist. With p = 0.4 is a good description of the sales. H A : A binomial dist. With p = 0.4 is NOT a good description of the sales. k (degrees of freedom) = 5, α = 0.05 Chi-Square Statistic, We reject H 0 (that the distribution is well described by a binomial distribution with p = 0.4 and n = 5) since

16. - the B-school ANOVA – An Example Three training methods were compared to see if they led to greater productivity after training. The following are the productivity measures for individuals trained by each method. Method 1 45 40 50 39 53 44 Method 2 59 43 47 51 39 49 Method 3 41 37 43 40 52 37 Use ANOVA at 0.05 level of significance, to determine whether these training methods lead to different levels of productivity?

17. - the B-school ANOVA –AN EXAMPLE Solution Method: Statement of Hypothesis : H 0 : µ 1 = µ 2 = µ 3 H A : µ 1 , µ 2 and µ 3 are not all equal Step 1: Calculate the grand mean , Step 2: Calculate the three sample means , Step 3: Calculate the three sample variances Step 4: Estimate the between-column-variance , i.e variance among sample means Step 5: Estimate the within-column-variance , i.e variance within the sample means n i = size of i th sample n T = total sample size Step 6: Calculate F-ratio/F-statistic ,

18. - the B-school ANOVA –AN EXAMPLE Solution Method: Statement of Hypothesis: H 0 : µ 1 = µ 2 = µ 3 H A : µ 1 , µ 2 and µ 3 are not all equal Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: F-ratio/F-statistic,

21. - the B-school ANOVA – EXERCISE The following data show the number of claims processed per day for a group of four insurance company employees observed for a number of days. Using ANOVA test the hypothesis that the employee’s mean claims per day are all the same. Use the 0.05 level of significance. Employee 1 15 17 14 12 Employee 2 12 10 13 17 Employee 3 11 14 13 15 12 Employee 4 13 12 12 14 10 9

22. - the B-school ANOVA – EXERCISE Solution Method: Statement of Hypothesis: H 0 : µ 1 = µ 2 = µ 3 = µ 4 H A : µ 1 , µ 2 , µ 3 , µ 4 are not all equal Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: F-ratio/F-statistic, Step 7: Critical F value (with df (3,15), α = 0.05) = 3.29