STAT 500 — HW 2 — Solution
All parts of Q1 and Q2 are one point each except for 2b that is 2 points. Q3 is 4 points. Q4 is 1 point.
1. Volatility on stock exchanges
(a) No, this study is not a randomized experiment. We are observing the volatility in two major
stock exchanges. Treatments (exchange) were not randomly assigned to stocks.
(b) Yes, these data can be used to make inferences about the mean volatility of all stocks on the
NYSE during the week of 1216 Feb 2007. Because a random sample was selected from all stocks
listed on the exchange that week. By randomly choosing stocks, the sample is representative (on
average) of the whole exchange.
(c)
S
p
=
s
(
n
1

1)
S
2
1
+ (
n
2

1)
S
2
2
n
1
+
n
2

2
=
r
(40

1)
×
(2
.
23%)
2
+ (40

1)
×
(3
.
37%)
2
40 + 40

2
= 2
.
85%
df
= 40 + 40

2 = 78
(d) H
0
:
μ
1
=
μ
2
H
a
:
μ
1
6
=
μ
2
T
=
x
1

x
2
r
S
2
p
1
n
1
+
1
n
2
=
2
.
91

4
.
39
q
(2
.
85%)
2
(
1
40
+
1
40
)
=

2
.
31
df
= 78, so
p
value=0.024. Using a table, you should find p
<
0.05. There is evidence of difference
in mean volatility between the two exchanges.
Note: Different tables have different d.f. That shouldn’t change the conclusion.
(e) No, I do not agree.
Because the consultant’s argument requires a causal conclusion.
This
study is not a randomized experiment.
(f)
T
=
x
1

x
2
r
S
2
p
1
n
1
+
1
n
2
=
2
.
91

4
.
39
q
(2
.
85%)
2
(
1
20
+
1
20
)
=

1
.
64
df
= 20 + 20

2 = 38, so
p
value=0.11.
(g)
T
=
x
1

x
2
r
S
2
p
1
n
1
+
1
n
2
=
2
.
91

4
.
39
q
(2
.
85%)
2
(
1
100
+
1
100
)
=

3
.
66
df
= 100 + 100

2 = 198, so
p
value=0.0003.
(h) As the sample size increases, pvalue becomes smaller and we are more likely to reject the
null hypothesis.
This is reasonable because a larger sample gives more information about the
population which reduces the uncertainty about the mean volatility difference.
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 Fall '08
 Staff
 Mean, mean volatility

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