Question

48. A study is designed to investigate whether there is a difference in response to various treatments in patients with Rheumatoid arthritis. The outcome is the patient's self-reported effect of treatment. The data are shown below. If you plan to run a test at a 5% level of significance to test whether there is a significant difference in effect of treatment, what is your critical value for this test? (2 decimal places) Symptoms Total] Worsened No Effect Symptoms Improved 14 50 Treatment 1 Treatment 2 Treatment 3 22 14 9 14 15 12 21 50 50 49. Based on the data and hypothesis testing you performed in Q48, do you reject the null hypothesis which states that the patients' self-reported effect of treatment is not independent of the treatment type? (enter 1 for "yes", and 0 for "no")

Answer 1

Answer Date: 08/10/2019 48) To test the hypothesis is that the patients' self-reported effect of treatment is not independent of the treatment type at 5% significance level. The null and alternative hypothesis is, The null hypothesis is that the patients' self-reported effect of treatment is independent of the treatment type. The alternative hypothesis is that the patients' self-reported effect of treatment is not independent of the treatment type. The chi-square critical value is, First, compute the degree of freedom then find chi-square critical value. The information in given matrix is three rows and three columns for compute degrees of freedom.

d.f. =(r - 1)*(c-1) = (3 – 1)*(3-1) = 4 From the critical value table, the degrees of freedom which is 4 and the significance level 0.05, so chi-square critical values is 9.49. The chi-square critical value is 9.49). 49) The chi-square statistics is, By using Excel software, find chi-square test statistics with the help of following steps: 1) Import the data. 2) Select Contingency table from Chi-Square in MegaStat option from Add-Ins menu. 3) Select Chi-Square and Expected frequencies in Output options. Click Ok.

Chi-square Contingency Table Test for Independence Worsened No effect 22 14 Improved 14 21.333 13.667 15.000 14 15.000 Total 50 50.000 50 50.000 15 21 13.667 21.333 Treatment 1 Observed Expected Treatment 2 Observed Expected Treatment 3 Observed Expected Total Observed Expected 12 29 50 50.000 15.000 45 45.000 13.667 41 41.000 21.333 64 64.000 150 150.000 11.356 chi-square 4 df .0228 p-value From the Excel output, the chi-square test statistic is 11.36. Decision

The conclusion is that chi-square test statistic 11.36 greater than chi-square critical value 9.49, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the patients' self-reported effect of treatment is not independent of the treatment type. The result is statistically significant.