Question

A drug company developed a honey based liquid medicine designed to calm a chids cough at night. To test the drug. 105 children who were it with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups one group was given a dosage of the honey drug the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third control group) received aliquid placebo (no dosage a) After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below Diagnosis Treatment Better Worse Total Control 33 37 DM 21 33 Honey 24 11 35 Total 40 105 In order to determine whether the treatment group is independent of the coughing diagnosis a two way chi-square test was conducted 1. State the alterative and the null hypothesis 2. Calculate test statistic 3. Determine the chi square critical value 4 Write the decision rule 5. Decision and Conclusion

Answer 1

Solution There are 2 attributes : leuels (1) Treatment group haning kz 29 ( Control /om/ Honey) Coughing Diagnosis having lz2 lends ( Better / worse) cell Define, m the table. fi; frequency of the Cisjgth ginen i=1,2,.pk with corresponding j 21, 2, .gl population proportion bij l e fio E bij I fij, Pio ak ; i=1,2,. je j=1 foj { fij, Boj Boj - i=1 s bij ; j 21,2,. od ia k l l K and =n=sample size & El bij = 1 I E fij ia j=1 i=1jzi Part 1 test the following: he are

the Ho: muu hypothesis treatment group is independent of coughing diagnosis alternative hypothesis the against Not Ho H: ie, Ho: Pij bio toj (ره) + against Pij & Piotoj for at least one pair of (iss), Part 2 The test Ho against ti statistie for testing LÊ ( fis - fio toj) T = Ho X (kise-) i=1=1 a fiofoj ful Here, = 4 fiz 2 33 f22 fai 2 = 12 f32 2 11 31 = 24 for = 40, for=65, fio=37, f20 = 33, +30=35 =los 4.449 +0.026 +0.016 Tobs - 7.230 + + 8.533 +5.251 25.506

not independent Part 3 chi-square The critical value at significance level ginen by : = o.o5 is 2 X; (kainen) = upper & point of Xck-3-1) Olety distribution Here, k= 3 el =2 L=0.05 2 2 = 7.814728 Xo.oss 0.05;$343(2-1) = X 0.05;2 (using qchisq (0.05, 2, lower. tail e & , command in R) Part 4 Decision Rule : we reject Ho against Hi at level of significance L if Jobs > X kick-ocet). Part 5 -7.814 728 Decision Since Tobs = 25.500 > X 0.0572 me reject to against hi L = 0.05 level of significance. at Conclusion the concinde on the basis of the ginen dath that is sufficient enidence to beliene that treatment group and coughing diagnosis ane there