STA4502 Section 2906
EXAM 1. Solutions
1. Below are recorded the white blood cell counts for a sample of 6 adult males with
leukemia (recorded in thousands of cells).
19.3
32.0
1.8
43.0
14.1
1.4
Making no assumptions about the distribution of white cell counts in this population,
find a point estimate of the median white blood cell count for the population of adult
male with leukemia. Also, find roughly 80% confidence intervals for the population
median cell count and interpret it.
(10 pts)
ˆ
θ
0
.
5
=
14
.
1 + 19
.
3
2
= 16
.
7
From
R
, a roughly 80% confidence interval for the population median is given by
[
X
(2)
, X
(5)
) = [1
.
8
,
32
.
0)
and the achieve confidence level is
4
summationdisplay
k
=2
parenleftbigg
6
k
parenrightbigg parenleftbigg
1
2
parenrightbigg
6
= 0
.
78125
.
We are 78
.
13% confident that the median white blood cell count for the population
of adult male with leukemia falls in the interval [1
.
8
,
32
.
0).
2. Suppose
X
1
, . . . , X
10
∼
iid
F
. Let
θ
0
.
5
= median of
F
. Find
Pr
bracketleftbig
X
(2)
≤
θ
0
.
5
< X
(9)
bracketrightbig
.
(5 pts)
Pr
bracketleftbig
X
(2)
≤
θ
0
.
5
< X
(9)
bracketrightbig
=
8
summationdisplay
k
=2
parenleftbigg
10
k
parenrightbigg parenleftbigg
1
2
parenrightbigg
10
≈
0
.
9785
3. Describe the general principle that related a 90% confidence interval to a test of
hypothesis.
(5 pts)
The 90% confidence interval consists of all the
θ
0
values such that the test of
H
0
:
θ
=
θ
0
H
a
:
θ
negationslash
=
θ
0
would not reject
H
0
in a twosided
α
= 0
.
1 test of hypothesis.
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$435
$1,735
$822
$572
$308
$116
$786
$51
4. The Internal Revenue Service takes a sample of tax returns and records the refund
amount requested from the government ( A negative refund means that person owed
the government.) A sample of eight returns showed the refunds below:
Does this data indicate that the population median tax refund amount is positive?
Make no assumption about the nature of the underlying population in conducting this
test. Find the
p
−
value
.
(10 pts)
H
0
:
θ
0
.
5
= 0
H
a
:
θ
0
.
5
>
0
Under
H
0
,
p
=
Pr
(
X >
0) = 0
.
5. Thus,
H
0
:
p
= 0
.
5
H
a
:
p >
0
.
5
Test stat
B
=
8
summationdisplay
i
=1
I
(
X
i
>
0)
Under
H
0
,
B
∼
Binomial(
n
= 8
, p
= 0
.
5). Therefore,
p
−
value
=
Pr
(
B
≥
B
obs
) =
Pr
(
B
≥
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 Spring '08
 Staff
 Normal Distribution, white blood cell, self esteem scores, nofeedback group

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