Question

A researcher is creating a new treatment protocol for Myelodysplastic Syndrome (MDS), a form of preleukemia. Following the old treatment protocol, 33% 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. 1.What is the 99.5% confidence level? 2.What is the correct interpretation of the confidence interval? 3.Are the assumptions met? why?

Answer 1

1)

Sample proportion = 27 / 100 = 0.27

99.5% confidence interval for p is

- Z/2 * Sqrt( ( 1 - ) / n) < p < + Z/2 * Sqrt( ( 1 - ) / n)

0.27 - 2.807 * sqrt( 0.27 * 0.73 / 100) < p < 0.27 + 2.807 * sqrt( 0.27 * 0.73 / 100)

0.145 < p < 0.395

99.5% CI is (0.145 , 0.395)

2)

We are 99.5% confident that population proportion of the the patients with MDS will develope Leukemia within 5 years

is between 0.145 and 0.395.

3)

Since claimed proportion 0.33 is lies in the confidence interval, asumption of the fewer MDS patients develope Leukemia

is not met.