MA22S6
Assignment 6 - modelling data
Mike Peardon (mjp@maths.tcd.ie)
May 5, 2010
Answer all four parts. Each part is worth five marks. Hand the homework in by the end of next
Thursday’s last lecture (1st April).
Question 1
Two professors, Prof. Tootoo and Prof. Essex have competing models that explain the growth rate g of
daffodils as different doses h of a hormone are administered. Prof Tootoo’s model suggests
g 22 (h) = e h
while Prof. Essex believes
g S6 (h) = 1 + 2h 2 .
Experiments are performed that measure growth rate for three different doses of the hormone that
yield the following data:
h
0.0
1.0
3.0
G
1.01
2.82
20.3
σG
0.02
0.16
0.15
1. Evaluate the χ2 -statistic for the two models
2. Based on this evaluation, what does the data indicate about the validity of both models?
The two values of χ2 are 2.70 and 76.6. This suggests Prof. Tootoo’s model is not ruled out by the
data, while Prof. Essex’s model is very unlikely to be correct.
25
Tootoo’s model
Essex’s model
20
15
10
5
0
0
1
2
3
Question 2
Prof Essex modifies his model to predict
(2)
g S6
(h) = 1 + αh 2 .
What value of the unknown parameter α gives the model that most closely agrees with the data given
in Question 1? What does the data indicate about the validity of the model?
To find the model that best fits the data, find α∗ , the value of α that minimises χ2 (α). Differentiating χ2 w.r.t α gives
3 2(1 + αh 2 −G ) · h 2
d χ2 X
i
i
i
=
2
d α i =1
σi
setting this to zero gives
α∗
Computing A =
P3
h i4
i =1 σ2
i
and b =
P3
i =1
2
3 h4
X
i
2
i =1 σi
(G i −1)·h i2
σ2i
=
3 (G − 1) · h 2
X
i
i
i =1
σ2i
gives A = 3639.1 and b = 7791.1 and so α∗ = b/A = 2.14.
For this value of α, recomputing χ gives χ2 (α∗ ) = 4.3 and now the model is not strongly ruled out by
the data, but looks shaky. More experiments needed.
25
Essex’s new model, with alpha=2.14
20
15
10
5
0
0
1
2
3
Question 3
Find the value of α that gives the best-fit straight line y = αx that models:
x
0.2
1.5
13.6
Y
0.68
4.21
42.6
σY
0.09
0.32
3.12
Give an estimate of the 1σ confidence interval of the parameter α computed by this means.
As before, α∗ the value of the free parameter that minimises χ2 can be found by differentiating and
setting d χ2 /d α = 0.
3 2(Y − αx ) · −x
d χ2 X
i
i
i
=
d α i =1
σ2i
2
so
3 x2
X
i
∗
α
Computing A =
P3
x i2
i =1 σ2
i
and b =
Yi x i
i =1 σ2
i
P3
2
i =1 σi
=
3 Y x
X
i i
σ2i
i =1
gives α∗ = 138.0/45.9 = 3.00. Here, chi 2 = 1.94, so this
model is not ruled out by the data. The uncertainty in this determination of α is given (see the notes)
by 1/A = 0.02, so the 1σ confidence interval for α is [2.98, 3.02]. The range of “good” models is shown
in the green band in the figure below.
50
40
30
20
10
0
0
1
2
3
4
5
7
6
8
9
10
11
12
13
14
15
Question 4
In Homework 3, T , the time between subsequent arrivals of the 22S6 bus was described by a random
number with probability density function given by
f T (t ) =


t
300
2
(30 − t )
300
0

0 ≤ t ≤ 20
20 < t ≤ 30
otherwise
A large sample of inter-arrival times is taken. What is the expected value of M T , the median of this
data-set?
With a large sampling, the expected value of the median, M T is the solution to the equation
F T (M T ) =
1
2
Start finding F T :
Z
F T (t ) =
t
0
s
d s when 0 ≤ t ≤ 20
300
s2 t
]
600 0
t2
600
= [
=
note that F T (20) > 1/2, so 0 ≤ M T ≤ 20. In fact,
M T2
600
= 12 implies M T =
3
p
300 = 17.32