37. Calculate the observed value of the test statistic for the test of H0 : 1
2 = 0 versus Ha : 1 2 < 0 on the basis of the following information.
Test the hypotheses at the 5% level of significance.
Sample statistics for group 1:
sample size
50
sample v

130 THE HQC FILTER
where Q I diag(q,0), and q is the variance of wk. Multiplying out this
equation and equating the elements on both sides gives
(P11 + P12)“ — 1312/3)
1 + Pll/R
*(P11 + P12)P12/R
l-l- Pl‘l/R
P12 — Fizz/R
1+ Pll/R
mph/R
1+ Jpn/R
P11 +P12+P

134 THE Hac FlLTER
This is true for 6 S 1.25 and 6 2 3.25. Another condition for the existence of
the H00 ﬁlter is that Equation (11.90) must be satisﬁed.
P’l — as. + Hfsglnk > 0
P‘1 i 6 + 1 > 0
This is always true if 6 < 1. If 6 > 1 then this inequality

132 THE Hcc FILTER
Setting these partial derivatives equal to zero and solving gives
1'1 = 4
2:2 = 0
3:3 : 4
A = —8
11.6 Maximize (142: —— $2 + 6y i y2 + 7) subject to the constraints + y 5 2)
and (a: + 2y S 3) [Lue84].
Solution:
One of four situations ho

124 ADDlTlONAL TOPICS IN KALMAN FiLTERiNG
where f 2 0.8, and wk is zero-mean white noise with a variance of 0.01. We measure
the tread height every 7 weeks with zero-mean white measurement noise that has
a variance of 0.01. The initial tread height is kno

128 THE Hog FILTER
So we see that (I + ATIA 2 AU + Ari.
QED
11.2 Consider a scalar system with F = H = 1 and with process noise and
measurement noise variances Q and R. Suppose a state estimator of the form
iE+l=iE+K®riD
is used to estimate the state, whe

136 THE am FILTER
Solution:
The steady-state a pastem'om‘ Kalman ﬁlter Riccati equation solution is obtained
from Equation (5.19) as
P+ : [(P‘)_1 + HTH]_1
where P“ is the steadyrstat—e a priori Kalman ﬁlter Riccati equation solution, and
we have substitut

122
ADDITIONAL TOPICS IN KALMAN FILTERING
Pmy(k, k A 1) is found as
[P(k) A PmUc, k —1)]FT(k — 1, MHT
= 11/21
0
130‘ch — 1) is found as
Pmy(k,k — 1)
PUc, k — 1) = PUc) — 33903,}: e1)3*1(k —1)Pg;(k,k ~ 1)
_ 1/3 0
_ 0 1
c). In part (b) we saw that P‘Uc) = I

12D ADDITIONAL TOPICS IN KALMAN FILTERING
Where all quantities are deﬁned at time k. The second expression can be derived as
follows.
P9 = Elly - ﬁlly - @lTl
= E[(H:c+ v w ch')(Ha:+v — HijT]
: HEW: — we: — :E‘)T]HT + HEKJ: — sniff] + Ewe i 52-)T1HT + E(va

126 ADD‘TIONAL TOPICS IN KALMAN FILTERING
r. L, _ _‘ _ w 7 _ .
3 ‘ V
0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7
K
Figure 10.2 Solution to Probiem 10.16 CHAPTER 11
\
The H00 filter
Problems
Written exercises
11.1 Show that. (I + A)‘1.4 : AU + Ari.
Sol

(b) H: X c = X p A: X c < X p
(c) H: pc = pp A: pc < pp A: c < p
(d) H: X c = X p A: X c 6= X p
Solution: a
Past performance 1998 Dec - 95%
45. Refer to the JMP output above. Which is correct?
(a) We are about 95% confident that the rats in the poisoned g

EATS 1011 3.0 Introduction to Atmospheric Science
e. 10 m
21) What would the earth's surface temperature be if there was only N2 and O2 and no
radiatively active gases, such as H2O and CO2 in the atmosphere and the albedo
was the same as now? (Select from

(a) H: 1 2 =0 A: 1 2 6= 0
(b) H: X 1 X 2 =0 A: X 1 X 2 < 0
(c) H: 1 2 =0 A: 1 2 < 0
(d) H: X 1 X 2 =0 A: X 1 X 2 > 0
(e) H: 1 2 =0 A: 1 2 < 0.
Solution: c
27. The test statistic, rejection region (=.05), and the p-value are:
(a) T =1.2173; reject if T 1.7

(e) H: 1 2 = 0 A: 1 2 > 0
Solution: e
Past performance 1990 Feb - 97%
31. The value of the proper test statistic and rejection region (= 0.05) are:
(a) T = 1.38; reject H if T > 1.71
(b) T = 1.26; reject H if T > 1.75
(c) T = 1.38; reject H if T > 2.14
(d

Solution: a
Past performance 2006 Dec - 87%
48. A Type I (false positive) error would occur if:
(a) We conclude that larger nests have the same size eggs (on average)
when in fact they are larger.
(b) We conclude that larger nests have larger eggs (on ave

(a) 1.896, 0.033
(b) 1.896, 0.131
(c) 1.896, 0.065
(d) 1.887, 0.059
(e) 1.887, 0.118
Solution: c
Past performance 1993 Feb - 38% (a-53%)
41. A Type II error would occur if:
(a) We conclude malathion is ineffective when in fact it was effective.
(b) We con

138 THE HDO FlLTER
Computer exercises
11.14 Generate the time-varying solution to Pk for Problem 11.? with P0 = 1.
\Nhat is the largest value of 6 for which Equation (11.90) will be satisﬁed for all k
up to and including it = 20? Answer to the nearest 0.0