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Shkiper50 [21]
3 years ago
8

A large number of disease-free individuals were enrolled in a study

Biology
1 answer:
Nuetrik [128]3 years ago
3 0

Answer:

a,Answer: Type I right censoring, censoring time = 60.

b.This is called an Interval censoring,

c.this is called Random censoring because the year in consideration was chosen at random, censoring time = 61.

d.The observations are left truncated respectively at age 30, 40, 50 and 42.

e.L\alpha \frac{S(60)}{S(30)} *\frac{S(52)-S(55)}{S(40)} *\frac{S(61)}{S(42)} *\frac{S(55)}{S(50)}

Explanation:

a) A healthy individual, enrolled in the study at age 30, never developed

breast cancer during the study.

Answer: Type I censoring, censoring time = 60.

(b) A healthy individual, enrolled in the study at age 40, was diagnosed

with breast cancer at the fifth exam after enrollment (i.e., the disease

started sometime between 12 and 15 years after enrollment).

THis is called an Interval censoring,

if the clinical exams is conducted every three years , then 3*4=12, and 3*5=15

add this to 40 years respectively, we have

censoring time = (52, 55]

c) A healthy individual, enrolled in the study at age 50, died from a

cause unrelated to the disease (i.e., not diagnosed with breast cancer

at any time during the study) at age 61.

this is called Random censoring because the year in consideration was chosen at random, censoring time = 61.

(d) An individual, enrolled in the study at age 42, moved away from

the community at age 55 and was never diagnosed with breast cancer

during the period of observation.

Random censoring, censoring time = 55.

The observations are left truncated respectively at age 30, 40, 50 and 42.

e.  Confining your attention to the four individuals described above,

write down the likelihood for this portion of the study.

L\alpha \frac{S(60)}{S(30)} *\frac{S(52)-S(55)}{S(40)} *\frac{S(61)}{S(42)} *\frac{S(55)}{S(50)}  

the above represent the likelihood

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