2G1503 Simulation and Modeling Exercise #5
Ali Ghodsi [email protected]
1
Point and Interval Estimates
• Assume Xi denotes the results from a real-world
system. Yi denotes the result from a simulation of
the same system. D is the difference between Yi
and Xi. 6 experiments are shown and the values of
Xi and Yi are IID. Calculate the point estimate and
a confidence interval of 90%.
i
1
2
3
4
5
6
Xi
3.06
2.79
2.21
2.54
9.27
0.31
Yi
3.81
3.37
2.61
3.59
9.02
0.71
Di
0.75
0.58
0.40
1.05
-0.25
0.40
2
Point Estimate: mean
• The point estimate for the mean is:
1 n
D = ∑ Di
n i =1
6
1
Di
∑
6 i =1
1
D = (0.75 + 0.58 + 0.40 + 1.05 − 0.25 + 0.40 ) = 0.488
6
D=
i
1
Di
0.75
2
0.58
3
0.40
4
1.05
5
-0.25
6
0.40
3
Interval Estimate: confidence interval 1/4
• The variance for the sample is:
1
S [ D] =
∑ (D − D )
n
2
2
n − 1 i =1
1
S 2 [ D] =
(0.75 − 0.488) 2 + (0.58 − 0.488) 2 + ... + (0.40 − 0.488) 2 ≈ 0.1907
6 −1
0.069 + 0.008 + 0.008 + 0.316 + 0.545 + 0.008
S 2 [ D] =
≈ 0.1907
5
i
(
)
i
1
2
3
4
5
6
Di
0.75
0.58
0.40
1.05
-0.25
0.40
4
Interval Estimate: confidence interval 2/4
• The variance of the sample distribution is:
σ2
2
σ (D ) =
n
• But we do not know 2, for the underlying
population. It is estimated by s2[D]:
1 n
2
(
)
D
−
D
i
n
S 2 [ D] n − 1 ∑
1
2
2
i =1
(
)
σ (D ) =
=
D
−
D
=
∑
i
n
n
n(n − 1) i =1
0.1907
σ 2 (D ) =
≈ 0.0318
6
5
Interval Estimate: confidence interval 3/4
• Since 2 was replaced by s2[D] the variance of the
sampling distribution is no longer normally
distributed, but instead t-distributed with n-1
degrees of freedom.
• For 90% confidence interval we lookup the tdistribution:
100 − 90
= 0.10
α=
100
v = 6 −1 = 5
tα / 2,v = t0.10 / 2,5 = 2.02
6
Interval Estimate: confidence interval 4/4
• The confidence interval is calculated by:
D ± t0.05,5σ ( D )
0.488 ± 2.02 * 0.0318
0.128 ≤ µ ≤ 0.848
7
Summary
• Use the given formula:
D ± tα / 2,n −1
S 2 [ D]
n
• where:
1
(0.75 + 0.58 + 0.40 + 1.05 − 0.25 + 0.40) = 0.488
6
0.069 + 0.008 + 0.008 + 0.316 + 0.545 + 0.008
≈ 0.1907
S 2 [ D] =
5
tα / 2,v = t0.10 / 2,5 = 2.02
D=
D ± t0.05,5σ ( D )
0.1907
6
0.128 ≤ µ ≤ 0.848
0.488 ± 2.02 *
8