ECI 114
Fall 2011
University of California, Davis
Department of Civil and Environmental Engineering
Homework Assignment 6 (67 points)
Turn in to Wei Zou by 10:45 a.m. Wednesday, Nov. 30, 2011 – NO EXTENSIONS
Show all work and explain your answers; no credit will be given for solutions only (especially true for problems in the
back of the book and on the accompanying CDROM). Write your name on the front of the first page and back of the
last page of your assignment; do not fold it. A twopoint penalty will be assessed for an inability to follow these
instructions. A penalty of up to three points may also be assessed for lack of neatness.
Preliminary note:
With any hypothesis test asked for by these problems, you need to provide the
null and alternative hypotheses, the test statistic, the rejection region, the outcome of the test, and an
interpretation of the outcome (i.e. not just “reject” or “FTR”, but what that means in the context of
the problem).
1.
(3+2+3 = 8 pts) Problem 996, p. 332.
In addition:
(c)
Explain what you would do if you were not given the information “with a mean of 1.2”.
Solution:
(a)
Given that E[X] = λ= 1.2
P
0
= Pr (X=0) =
= 0.3012
P
1
= Pr (X=1) =
= 0.3614
P
2
= Pr (X=2) =
= 0.2169
P
3
= Pr (X=3) =
= 0.0867
P
4
= P (X4) = 1 P
0
P
1
P
2
= 1  0.3012  0.3614  0.2169  0.0867 = 0.0338
E
i
= nP
i
where n = 100
Value of X
Number of
Observations
Taking this
Value
Probability of
Taking this
Value
Expected Frequency
of Taking this Value
out of a Sample of
100 Observations
0
24
0.3012
30.12
1
30
0.3614
36.14
2
31
0.2169
21.69
3
11
0.0867
8.67
≥
4
4
0.0338
3.38
d.f. = k – p – 1 = 5 – 0 – 1 = 4
a.1.
Question of interest: The question of interest is the form of the distribution of the
observations.
a.2.
Null Hypothesis: H
0
 The observation follows the form of the Poisson distribution
with a mean of 1.2.
1
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a.3.
Alternative Hypothesis: H
1
– The observation doesn’t follow the form of the Poisson
distribution with a mean of 1.2.
a.4.
Test Statistic:
a.5.
Reject H
0
if the Pvalue is less than 0.05
a.6.
Computations
=
a.7.
Conclusion
The critical value , so we FTR H
0
, and conclude that distribution of the data is
consistent with that of a Poisson distribution with mean 1.2.
(b)
Therefore 0.05 < pvalue < 0.1 (the exact value, obtained from Excel, is 0.094).
(c)
If “a mean of 1.2” is not given:
c.1.
We need to estimate the mean by using the sample mean:
(24*0 + 30*1 + 31*2 + 11*3 + 4*4) / 100 = 1.41.
c.2.
DOF = kp1 = 511 = 3.
c.3.
The last two columns of the table above would have to be recalculated, using the
estimate for λ.
2.
(3+3+2 = 8 pts) Using the data of 9108, p. 336:
(a)Under the assumption of independence, calculate the estimated expected number of students
in each of the cells of the table.
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 Fall '08
 Mocktarian
 Normal Distribution, Null hypothesis, Statistical hypothesis testing

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