Task 10: In the course of one year, 15 patients were admitted to a hospital ward because of a certain infection. Most of these patients have since been discharged, but some are still on the ward. The following are the lengths of stay in days: 3, 5, 5, 6, 8, 8+, 10, 16, 20, 22+, 24, 25+, 26+, 28, 33. A + indicates patients who could not be discharged by the specified length of stay. a) Calculate with the help of the formula Ś(t) = 11.1₂ ti 3 5 6 8 1₁.12.. M₁ M₂ the probability that a patient can be discharged within 28 days. Here is li the number of patients who have been treated up to the iten event time ti is had to stay in hospital and li the number of patients who must remain in hospital beyond the iten event time. "Event time" here means number of days in hospital regardless of the date of admission. The following applies $ (0)= 1 The following table shows the start of the calculation. *** n₁ 15 14 I.! n₁-1 n₁ li 14 12 12 11 li/ni 14/15=0.933 12/14=0.857 S(t₁)= S(t₁-1) ¹/₁ 1.0.933 0.933 0.933 0.857= 0.8 b) Determine the median length of stay in hospital from the calculation in part a).
Task 10: In the course of one year, 15 patients were admitted to a hospital ward because of a certain infection. Most of these patients have since been discharged, but some are still on the ward. The following are the lengths of stay in days: 3, 5, 5, 6, 8, 8+, 10, 16, 20, 22+, 24, 25+, 26+, 28, 33. A + indicates patients who could not be discharged by the specified length of stay. a) Calculate with the help of the formula Ś(t) = 11.1₂ ti 3 5 6 8 1₁.12.. M₁ M₂ the probability that a patient can be discharged within 28 days. Here is li the number of patients who have been treated up to the iten event time ti is had to stay in hospital and li the number of patients who must remain in hospital beyond the iten event time. "Event time" here means number of days in hospital regardless of the date of admission. The following applies $ (0)= 1 The following table shows the start of the calculation. *** n₁ 15 14 I.! n₁-1 n₁ li 14 12 12 11 li/ni 14/15=0.933 12/14=0.857 S(t₁)= S(t₁-1) ¹/₁ 1.0.933 0.933 0.933 0.857= 0.8 b) Determine the median length of stay in hospital from the calculation in part a).
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Task 10:
In the course of one year, 15 patients were admitted to a hospital ward because of a
certain infection. Most of these patients have since been discharged, but some are still
on the ward. The following are the lengths of stay in days:
3, 5, 5, 6, 8, 8+, 10, 16, 20, 22+, 24, 25+, 26+, 28, 33.
A+ indicates patients who could not be discharged by the specified length of stay.
a) Calculate with the help of the formula
Ŝ(t)=
4.12.
M₁ M₂
the probability that a patient can be discharged within 28 days. Here is li the number of
patients who have been treated up to the iten event time ti is had to stay in hospital and
li the number of patients who must remain in hospital beyond the iten event time. "Event
time" here means number of days in hospital regardless of the date of admission. The
following applies $ (0)= 1
The following table shows the start of the calculation.
ti
3
5
6
8
...
n₁
15
14
12
1-1. li
n₁-1 n₁
li
24 5
1
28 2
1
33 1 1
14
12
11
li/ni
14/15=0.933
12/14=0.857
b) Determine the median length of stay in hospital from the calculation in part a).
c) If the calculation is done with the R-function survfit, the last 3 rows of the table are output as:
time n.risk n.event survival std.err lower 95% CI upper 95% CI
0.356 0.1325
0.178 0.1421
0.000 NAN
S(t₁)= S(t₁-1).¹¹n₁
1.0.933 0.933
0.933 0.8570.8
0.1713
0.0371
ΝΑ
0.738
0.852
ΝΑ
Describe in words how to interpret the confidence limits.
Based on these results, could it be ruled out with a statistically high degree of certainty that the
median retention time could be longer than 28 days?
d) Are there any anomalies regarding the confidence limits in task part c)? Which of the figures
given there are questionable?
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