Question

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Use the following information for questions 1-10: A researcher is creating a new treatment protocol for Myelodysplastic Syndrome (MDS), a form of preleukemia. Following the old treatment protocol, 32% of patients with MDS will develop leukemia within 5 years of MDS diagnosis. He believes his new treatment protocol will lead to fewer MDS patients developing leukemia. He takes a random sample of 100 individuals on his new treatment protocol. Of these 100 individuals, 27 develop leukemia within 5 years of MDS diagnosis. Create a 99.8% confidence interval for the true proportion of all MDS patients on this treatment who will develop leukemia within 5 years of MDS diagnosis. 1. What is the 99.8% confidence interval? PI2*/P(1-P) ß = 220 - 27 Z* = n=100

Answer 1

-solution:- 0 99.89 Condidence interval - BI Zur Þ Lib 2.)!30 0 3 (2) we are gg. 87. consident that toue probottion lies within 0.133 and 0.407. (3) np = 27 (expected number of success) nai-6 = 73 (Experted number of failion since, mi sn (1-6) 210 Honce condition are met. Ho: p=0.32 Ha: pco.32 45) C 0.05 Too = -1.07 (7) P-value = P(2ca107) - Jo.1423)