Assignment 4
Due date: 16 November 2009
Total number of points: 27
Q
1.
(6.39) In a natural population of mice near Ann Arbor, Michigan, the coats of some individuals are whitespotted
on the belly. In a sample of 580 mice from the population, 28 individuals were found to have whitespotted
nellies. Construct 95% confidence interval for the population proportion of this trait.
Solution to Q1:
ˆ
p
= 28
/
580 = 0
.
048, estimated standard error=
0
.
048
*
(1

0
.
048)
/
580 = 0
.
0088, so that the 95% confi
dence interval is
0
.
048
±
1
.
96
*
0
.
0088
.
Marking scheme for Q1:
2 points (1 point for the appropriate formula, 1 for the correct final answer).
Q
2.
Nine measurements of ozone concentration are obtained during a year:
3.5 5.1 6.6 6.0 4.2 4.4 5.3 5.6 4.4
Assuming the measurements come from a normal population, provide two 95% confidence intervals for the
population mean
μ
:
(a)
one assuming that the population variance
σ
2
= 1
.
21;
(b)
another
relaxing this assumption
.
(c)
For the situation in part (b), test
H
0
:
μ
= 5
.
74 against
H
A
:
μ
= 5
.
74 taking
α
= 0
.
05. (You have to
use confidence intervals to answer this here).
Solution to Q2:
First,
α
= 1

0
.
95. Second, the sample mean is ¯
y
= 5
.
01 and the sample standard deviation is
s
= 0
.
9765.
(a)
Assuming that the population variance is 1.21:
5
.
01
±
(1
.
96
×
√
1
.
21
/
√
9)
⇒
5
.
01
±
0
.
72 or
(4
.
29
,
5
.
73)
.
(b)
If we relax the assumption, we have to compute
¯
y
±
t
α/
2
,n

1
s
√
n
= 5
.
01
±
2
.
306
0
.
9765
√
9
.
Indeed, from tables we have that
P
(
t
8
>
2
.
306) = 0
.
025
⇒
P
(
t
8
<

2
.
306) = 0
.
025
Thus,
t
0
.
025
,
8
= 2
.
306. The confidence interval is
5
.
01
±
2
.
306
0
.
97
√
9
or
(4
.
26
,
5
.
76)
.
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 Spring '08
 Normal Distribution, Standard Deviation, correct final answer

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