816003137
Justin Singh
STAT 6160
Coursework #2
1) ℎ(𝑡) = 𝜃 + 𝛽𝑒 −𝛾𝑡
a) Assuming 𝜃 = 0 𝑎𝑛𝑑 𝛾 = 1
ℎ(𝑡) = 𝛽𝑒 −𝑡
𝑑(ln 𝑆(𝑡))
ℎ(𝑡) =
𝑑𝑡
(∫ ℎ(𝑡)𝑑𝑡)
⇒ 𝑆(𝑡) = 𝑒
𝑆(𝑡) = exp (∫ ℎ(𝑡)𝑑𝑡) = exp (∫ 𝛽𝑒 −𝑡 𝑑𝑡) = exp(−𝛽𝑒 −𝑡 ) = 𝑒 −𝛽𝑒
b) Median occurs at 𝑆(𝑡) = 0.5
0.5 = exp(−𝛽𝑒 −𝑡 )
⇒ −𝛽𝑒 −𝑡 = ln 0.5
ln 0.5
𝑒 −𝑡 =
−𝛽
∴ 𝑀𝑒𝑑𝑖𝑎𝑛, 𝑡 = − ln (
ln 0.5
)
−𝛽
2) 𝛼 = 1.5 𝑎𝑛𝑑 𝜆 = 0.01
1
a) 𝑆(𝑡) = 1+𝜆𝑡 𝛼
1
1 + 0.01𝑡1.5
1
1
𝑆(50) =
=
= 0.22
1.5
1 + 0.01(50)
4.54
1
1
𝑆(100) =
=
= 0.09
1.5
1 + 0.01(100)
11
1
1
𝑆(150) =
=
= 0.05
1.5
1 + 0.01(150)
19.37
b) Median occurs at 𝑆(𝑡) = 0.5
1
0.5 =
1 + 0.01𝑡1.5
0.5 + 0.005𝑡1.5 = 1
0.5
𝑡1.5 =
= 100
0.005
∴ 𝑡 = 21.54 𝑑𝑎𝑦𝑠
𝑆(𝑡) =
c) 𝐸(𝑥) =
𝜋
𝛼
𝜋 𝑐𝑜𝑠𝑒𝑐( )
1
𝛼𝜆𝛼
𝜋
𝜋 𝑐𝑜𝑠𝑒𝑐 ( ) 85.96
1.5
𝐸(𝑥) =
1 = 0.07 = 1235.06
1.5
1.5(0.01)
−𝑡
816003137
Justin Singh
STAT 6160
Coursework #2
3) 𝜇 = 1 𝑎𝑛𝑑 𝜎 = 1
𝜎2
a) 𝐸(𝑥) = 𝑒 𝜇+ 2
1
𝐸(𝑥) = 𝑒 1+2 = 4.48
2
2
𝑉(𝑥) = 𝑒 2𝜇+𝜎 (𝑒 𝜎 − 1)
𝑉(𝑥) = 𝑒 2+1 (𝑒1 − 1) = 34.51
b) Median occurs when 𝑆(𝑡) = 0.5
0.5 = 1 − Φ(ln 𝑡 − 1)
0.5 = Φ(ln 𝑡 − 1)
Φ−1 (0.5) = 0 = ln 𝑡 − 1
∴ 𝑡 = 𝑒 1 = 2.72
4)
a)
𝑑𝑖
Survival time
𝑛𝑖
3
1
7
9
1
4
Survival
time
4
6
9
𝑑𝑖
𝑛𝑖
̂ = ∏1 −
𝑆(𝑡)
𝑑𝑖
𝑛𝑖
̂
̂ 𝑆(𝑡)
𝑉𝑎𝑟
𝑑𝑖
𝑛𝑖 (𝑛𝑖 − 𝑑𝑖 )
1
0.8572 × (
) = 0.017
7×6
1
0.6432 × (
) = 0.034
4×3
̂ 2∑
= 𝑆(𝑡)
1
= 0.857
7
1
0.857 × (1 − ) = 0.643
4
1−
b)
c)
1
1
2
𝐻0 : 𝑆1 = 𝑆2
𝐻1 : 𝑆1 ≠ 𝑆2
𝑡𝑖
𝑑𝑖
3
4
6
9
1
1
1
3
𝑛𝑖1
7
6
6
4
8
7
5
̂
𝑆(𝑡)
0.875
0.75
0.45
𝑛𝑖2
8
8
7
6
2
𝜒𝑐𝑎𝑙𝑐
=
𝐸𝑖1
̂𝑁𝐴 (𝑡) = ∑
𝐻
𝑑𝑖
𝑛𝑖
0.125
0.268
0.668
𝑛𝑖1
=
∗ 𝑑𝑖
𝑛𝑖1 + 𝑛𝑖2
0.467
0.429
0.462
1.2
2.558
̂𝑁𝐴 (𝑡))
𝑆̂𝑁𝐴 (𝑡) = exp(−𝐻
0.882
0.765
0.513
𝐸𝑖2
𝑛𝑖2
∗ 𝑑𝑖
𝑛𝑖1 + 𝑛𝑖2
0.533
0.571
0.538
1.8
3.442
=
(2 − 2.558)2 (2 − 3.442)2
+
= 0.726
2.558
3.442
2
𝜒0.05,1
= 3.84
2
2
⇒ 𝜒0.05,1 > 𝜒𝑐𝑎𝑙𝑐
∴ 𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0