 Mathematics
BISM 6051 Temple University DMBA / Mathematical Statistics Worksheet 10

BISM 6051

Temple University

BISM

### Question Description

I need help with a Statistics question. All explanations and answers will be used to help me learn.

Please find the following assignment attached below! You must have SAS and know how to use the software to proceed with the assignment. Thank you!

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Mathematical Statistics II (BISM 6051) Homework #10 1. Data in the table below represent the lifetimes in days of two groups of rats (distinguished by a pretreatment regimen) following exposure to the carcinogen DMBA (from Kalbfleisch and Prentice (1980)). Time in parenthesis indicates the right censored data. Group 1 (n=21) 142, 156, 163, 198, (204), 205, 232, 232, 233, 233, 233, 233, 239, 240, 261, 280, 280, 296, 296, 323, (344) Group 2 (n=19) 143, 164, 188, 188, 190, 192, 206, 209, 213, 216, (216), 220, 227, 230, 234, (244), 246, 265, 304 Using Kaplan-Meier method, a) Calculate the estimates of the survival function, 𝑆̂(𝑡𝑖 ) at 𝑡𝑖 . b) What are the median times for Group 1 and Group 2? c) Construct 95% confidence interval of 𝑆(𝑡𝑖 ) using the LogLog approach, where 𝑡𝑖 is the median time. Calculate it for each group. d) Use the logrank test to compare two survival curves. Use 𝛼 = 0.05. e) Interpret your results. f) Use the SAS procedure to compare the results. 1 ...
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Mathematical Statistics II
(BISM 6051)
Homework #10

1. Data in the table below represent the lifetimes in days of two groups of rats
(distinguished by a pretreatment regimen) following exposure to the carcinogen
DMBA (from Kalbfleisch and Prentice (1980)). Time in parenthesis indicates the
right censored data.
Group 1 (n=21)
142, 156, 163, 198, (204), 205, 232,
232, 233, 233, 233, 233, 239, 240,
261, 280, 280, 296, 296, 323, (344)

Group 2
(n=19)
143, 164, 188, 188, 190, 192, 206,
209, 213, 216, (216), 220, 227, 230,
234, (244), 246, 265, 304

Using Kaplan-Meier method,
a) Calculate the estimates of the survival function, 𝑆̂(𝑡𝑖) at 𝑡𝑖.
S(ti) =

𝑁𝑢𝑚𝑏𝑒𝑟 𝑙𝑖𝑣𝑖𝑛𝑔 𝑎𝑡 𝑠𝑡𝑎𝑟𝑡−𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑁𝑢𝑚𝑏𝑒𝑟 𝑑𝑒𝑎𝑑
𝑁𝑢𝑚𝑏𝑒𝑟 𝑙𝑖𝑣𝑖𝑛𝑔 𝑎𝑡 𝑠𝑡𝑎𝑟𝑡

When there is censoring at ti eg t1: at S(t2)
Number of start changes to the current number at risk whenever there is a censor,
and also cumulative dead starts again. For example, number of start becomes
number at risk at t2

S(t2)= (number at start- cumulative number dead)/ number at risk * S(t1)

Group 1
Number at start = 21

number
time (ti) n.risk
142
21
156
20
163
19
198
18
204
18
205
16
232
15
233
13
239
9
240
8
261
7
280
6
296
4
323
2
344
1

1
1
1
1
0
1
2
4
1
1
1
2
2
1
0

cumulative
censored
1
0
2
0
3
0
4
0
4
1
1
0
3
0
7
0
8
0
9
0
10
0
12
0
14
0
15
0
15
1

Formula: survival
survival
(S(ti))
(S(ti))
=(21-1)/21
0.9524
=(21-2)/21
0.9048
=(21-3)/21
0.8571
=(21-4)/21
0.8095
=(21-4)/21
0.8095
=(16-1)/16 * 0.8095
0.7589
=(16-3)/16 * 0.8095
0.6577
=(16-7)/16 * 0.8095
0.4554
=(16-8)/16 * 0.8095
0.4048
=(16-9)/16 * 0.8095
0.3542
=(16-10)/16 * 0.8095
0.3036
=(16-12)/16 * 0.8095
0.2024
=(16-14)/16 * 0.8095
0.1012
=(16-15)/16 * 0.8095
0.0506
=(16-15)/16 * 0.80... mickeygabz (5233)
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