Question

Solve the problem.

A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill 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 a liquid placebo (no dosage at all). 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:

C Diagnosis Treatment Better Worse Total Control 4 33 37 DM 12 21 33 Honey 24 11 35 Total 40 105

Diagnosis

   

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 alternative 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

Given table data is as below MATRIX col1 col2 TOTALS row 1 4 33 37 row 2 12 21 33 row 3 24 11 35 TOTALS 40 65 N = 105 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N row 3 row3*col1/N row3*col2/N ------------------------------------------------------------------ expected frequencies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 14.095 22.905 row 2 12.571 20.429 row 3 13.333 21.667 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 4 14.095 -10.095 101.909 7.23 33 22.905 10.095 101.909 4.449 12 12.571 -0.571 0.326 0.026 21 20.429 0.571 0.326 0.016 24 13.333 10.667 113.785 8.534 11 21.667 -10.667 113.785 5.252 ᴪ^2 o = 25.507 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, α = 0.05 from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =5.991 since our test is right tailed,reject Ho when ᴪ^2 o > 5.991 we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei from the table , ᴪ^2 o = 25.507 critical value the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 2 - 1 ) = 2 * 1 = 2 is 5.991 we got | ᴪ^2| =25.507 & | ᴪ^2 α | =5.991 make decision hence value of | ᴪ^2 o | > | ᴪ^2 α| and here we reject Ho ᴪ^2 p_value =0 ANSWERS --------------- 1. null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent 2. test statistic: 25.507 3. critical value: 5.991 4. p-value:0 5.decision: reject Ho

we have enough evidence to support the claim that   whether the treatment group is independent of the coughing diagnosis,