Question

Thirty nine women with a diagnosis of inoperable or metastatic breast cancer have been followed-up for a number of years, while under continuous treatment with a medication called trastzuzumab. The time in months) each patient remained in remission was recorded. Remission is is a decrease or disappearance of signs and symptoms of cancer. Below is the data. 50 74 35 39 21 37 27 35 30 35 26 38 34 34 26 41 61 33 33 26 25 41 35 34 44 33 60 61 42 30 80 31 24 49 26 31 28 41 37. 1. Are there any outliers in the sample? If so, indicate their values. 2. Compute the 45-th percentile of the remission time. 3. Find a 97% confidence interval for the average remission time.

Answer 1

1) We can make a boxplot using R to detect outliers.

The boxplot is given below.

Time in montas 20 30 40 50 90 70 80 1 BOXPLOT

From the boxplot, we can see that 4 values are outside the range. Those four values are the outliers here.

So the 4 outlier points are:

60, 61, 74, 80.

2)

2) Let $0:45 be the 45th percentile of the data X, X2 .- .X 39. P[#xis < 80.45] >, 0.45 & P[#xis > $0.45] > 0.55 Here, we calculate the asth percentile using the quantile ()" function in R. R code : quantile (data, 0.45)

Using R, the value of 45 th percentile of the data is 34.

3)

Let X., ... Xn id NCM, 02), Mo both unknown. = 39.,M= Average remission time. So, 10(-)% ci for Juu: (x-s. tyyn-ykt s tymt) Here, I=sta 2** = 38.12821, 32, 3-3 = 28-5358 5=13.36173. X=0:03 , 7 +0:03; 39-) = 2:2546 to 01s; 38 x-s. to-015; 38 = 33•3043 , Xt si = 42.9521. √39 √39 2
So a 97% confidence interval for average remission time is (33.3043, 42.9521).