Tugas Kelompok 3

• From: Tugas Kelompok 3.pdf

Question

Penelitian psikologi mencatat kemampuan 25 mahasiswa untuk berkonsentrasi pada pembelajaran online melalui Zoom meeting.

Variabel yang diamati adalah waktu (dalam jam) dari awal belajar sampai kehilangan konsentrasi.

Tanda + menunjukkan data tersensor (mahasiswa belum kehilangan konsentrasi saat pengamatan berakhir).

Data (disortir berdasarkan waktu):

No	Waktu (jam)	Tersensor
1	0.1
2	0.2
3	0.2
4	0.3
5	0.3
6	0.3
7	0.4
8	0.7
9	0.8
10	0.9	+
11	1.0
12	1.0
13	1.0	+
14	1.1
15	1.3
16	1.6
17	1.6
18	1.8
19	2.0	+
20	2.0	+
21	2.0	+
22	2.0	+
23	2.0	+
24	2.0	+
25	2.0	+

Total: 25 mahasiswa, 9 tersensor, 16 event (kehilangan konsentrasi)

• a. Tentukan taksiran untuk fungsi survival berdasarkan metode Kaplan-Meier, kemudian plot grafiknya.
• b. Hitunglah juga taksiran variansinya dan standar errornya.

Answer

a.

Survival function at time $t$ can be estimated with Kaplan-Meier Estimator using:

$$\hat{S}(t)=\prod \_t\_i\le t\left(1-\frac{d\_i}{Y\_i}\right)$$

where:

• $\hat{S}(t)$ : Estimated survival time at time $t$
• $d\_i$ : Event occuring at time $t\_i$
• $Y\_i$ : Event remaining at time $t\_i$

Let

• $c\_i$ : Censored data at time $t\_i$
• $Y\_1=25$
• $Y\_i=Y\_i-1-d\_i-1-c\_i-1$

Applying this estimator to the data:

$t\_i$	$d\_i$	$c\_i$	$Y\_i$	$\left(1-\frac{d\_i}{Y\_i}\right)$	$\hat{S}(t\_i)$
0.1	1	0	25	24/25	0.96
0.2	2	0	24	22/24	0.88
0.3	3	0	22	19/22	0.76
0.4	1	0	19	18/19	0.72
0.7	1	0	18	17/18	0.68
0.8	1	0	17	16/17	0.64
0.9	0	1	16	1	0.64
1.0	2	1	15	13/15	0.55
1.1	1	0	12	11/12	0.51
1.3	1	0	11	10/11	0.46
1.6	2	0	10	8/10	0.37
1.8	1	0	8	7/8	0.32
2.0	0	7	7	1	0.32

We get:

Interval	$\hat{S}(t)$
$t<$ 0.1	1.0
0.1 $\le t<$ 0.2	0.96
0.2 $\le t<$ 0.3	0.88
0.3 $\le t<$ 0.4	0.76
0.4 $\le t<$ 0.7	0.72
0.7 $\le t<$ 0.8	0.68
0.8 $\le t<$ 0.9	0.64
0.9 $\le t<$ 1.0	0.64
1.0 $\le t<$ 1.1	0.55
1.1 $\le t<$ 1.3	0.51
1.3 $\le t<$ 1.6	0.46
1.6 $\le t<$ 1.8	0.37
1.8 $\le t<$ 2.0	0.32
$t\ge$ 2.0	0.32

Plotting using Python:

import seaborn
 
seaborn.lineplot(
    ests, x="Interval", y="Estimated Survival Time", 
    drawstyle="steps-post", marker='o'
).figure.savefig("/home/fazuh/Notes/assets/1774435528.png")

b.

The variance and standard error of the estimated survival function can also be estimated using Greenwood’s Formula:

$$\begin{array}{rl}\widehat{\mathrm{Var}}\text{[}\hat{S}(t)]& =\text{[}\hat{S}(t){]}^2\sum \_t\_i\le t\frac{d\_i}{Y\_i(Y\_i-d\_i)}\\ \widehat{\mathrm{SE}}\text{[}\hat{S}(t)]& =\sqrt{\widehat{\mathrm{Var}}\text{[}\hat{S}(t)]}\end{array}$$

Continuing from point a.,

Let

$$v\_i=\{\begin{array}{ll}\frac{d\_i}{Y\_i(Y\_i-d\_i)},& Y\_i\ne 0\\ 0,& Y\_i=0\end{array}$$

Then the estimated variance and standard error for the estimated survival function are:

$t\_i$	$d\_i$	$Y\_i$	$\hat{S}(t\_i)$	$v\_i$	$\sum \_t\_i\le tv\_i$	$\text{[}\hat{S}(t\_i){]}^2$	$\widehat{\text{Var}}\text{[}\hat{S}(t\_i)]$	$\widehat{\text{SE}}\text{[}\hat{S}(t\_i)]$
0.1	1	25	0.9600	0.0017	0.0017	0.9216	0.0015	0.0392
0.2	2	24	0.8800	0.0038	0.0055	0.7744	0.0042	0.0650
0.3	3	22	0.7600	0.0072	0.0126	0.5776	0.0073	0.0854
0.4	1	19	0.7200	0.0029	0.0156	0.5184	0.0081	0.0898
0.7	1	18	0.6800	0.0033	0.0188	0.4624	0.0087	0.0933
0.8	1	17	0.6400	0.0037	0.0225	0.4096	0.0092	0.0960
0.9	0	16	0.6400	0.0000	0.0225	0.4096	0.0092	0.0960
1.0	2	15	0.5547	0.0103	0.0328	0.3077	0.0101	0.1004
1.1	1	12	0.5084	0.0076	0.0403	0.2585	0.0104	0.1021
1.3	1	11	0.4622	0.0091	0.0494	0.2136	0.0106	0.1028
1.6	2	10	0.3698	0.0250	0.0744	0.1367	0.0102	0.1009
1.8	1	8	0.3236	0.0179	0.0923	0.1047	0.0097	0.0983
2.0	0	7	0.3236	0.0000	0.0923	0.1047	0.0097	0.0983

Appendix

Python version

from itertools import groupby
 
data: list[tuple[float, bool]] = [ (1.8, False), (2.0, True), (1.0, True), (2.0, True), (1.6, False), (0.7, False), (0.9, True), (1.6, False), (0.2, False), (0.4, False), (2.0, True), (2.0, True), (2.0, True), (1.0, False), (1.1, False), (0.3, False), (0.1, False), (0.3, False), (0.3, False), (2.0, True), (1.0, False), (1.3, False), (0.8, False), (0.2, False), (2.0, True) ]
 
def kaplan_meier(subjects: list[tuple[float, bool]]) -> dict[str, list[float]]:
    subjects_sorted = sorted(subjects, key=lambda x: x[0])
    remaining = len(subjects_sorted)
    prev = 1.0
    ests: dict[str, list[float]] = {
        "Interval": [],
        "Estimated Survival Time": []
    }
 
    for t, group in groupby(subjects_sorted, key=lambda x: x[0]):
        group = list(group)
        d = sum(1 for (_, censored) in group if not censored)
        
        s_t = prev * (1 - d / remaining)
        ests["Interval"].append(t)
        ests["Estimated Survival Time"].append(s_t)
        prev = s_t
        remaining -= len(group)
 
    return ests
 
ests = kaplan_meier(data)
 
print("| Interval | Estimated | \n| --- | --- |")
for i, (t, est) in enumerate(zip(ests["Interval"], ests["Estimated Survival Time"])):
    prev_est = 1.0 if i == 0 else ests["Estimated Survival Time"][i-1]
    prev_est = round(prev_est, 2)
    if i == 0:
        print(f"| $t <$ {t} | {prev_est:.2f} |")
    else:
        print(f"| {ests['Interval'][i-1]} $\\leq t <$ {t} | {prev_est:.2f} |")
# last row: t >= last event time
print(f"| $t \\geq$ {ests['Interval'][-1]} | {round(ests["Estimated Survival Time"][-1], 2):.2f} |")

Data: Original ordering

No	Waktu (jam)	Tersensor
1	1.8
2	2.0	+
3	1.0	+
4	2.0	+
5	1.6
6	0.7
7	0.9	+
8	1.6
9	0.2
10	0.4
11	2.0	+
12	2.0	+
13	2.0	+
14	1.0
15	1.1
16	0.3
17	0.1
18	0.3
19	0.3
20	2.0	+
21	1.0
22	1.3
23	0.8
24	0.2
25	2.0	+