X
n
should be to
μ
.
Mathematically, the central limit theorem says that
√
n
(
¯
X
n

μ
)
→
d
N
(0
, σ
2
).
We cannot know the exact value of
σ
2
merely by observing our sample
X
1
, . . . , X
n
.
But we can estimate it with the sample variance ˆ
σ
2
n
, defined as
ˆ
σ
2
n
=
1
n
n
X
i
=1
(
X
i

¯
X
n
)
2
.
Some people prefer to define the sample variance with a divisor of
n

1 instead of
n
, but this makes no difference to us, since we will be looking for limiting results
as
n
→ ∞
. With a little algebra, we can see that
ˆ
σ
2
n
=
1
n
n
X
i
=1
X
2
i

¯
X
2
n
.
16
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The law of large numbers tells us that
1
n
∑
n
i
=1
X
2
i
→
p
E
(
X
2
) as
n
→ ∞
. It also
tells us that
¯
X
n
→
p
E
(
X
), and so the continuous mapping theorem implies that

¯
X
2
n
→
p

E
(
X
)
2
. Therefore, applying part 1 of Slutsky’s theorem, we find that
ˆ
σ
2
n
→
p
E
(
X
2
)

E
(
X
)
2
= Var(
X
) =
σ
2
as
n
→ ∞
. In other words, the sample variance ˆ
σ
2
n
provides a
consistent
estimate
of the population variance
σ
2
.
Since the square root function is continuous, the
continuous mapping theorem implies that ˆ
σ
n
→
p
σ
.
Combining this result with
part 3 of Slutsky’s theorem and the fact that
√
n
(
¯
X
n

μ
)
→
d
N
(0
, σ
2
), we find
that
¯
X
n

μ
ˆ
σ
n
/
√
n
=
√
n
(
¯
X
n

μ
)
ˆ
σ
n
→
d
N
(0
, σ
2
)
σ
=
N
(0
,
1)
,
provided that
σ >
0 (which will be true unless every
X
i
takes the same value).
We have shown that the distribution of the ratio
¯
X
n

μ
ˆ
σ
n
/
√
n
approximates the
N
(0
,
1)
distribution in large samples. This fact can be used to test hypotheses about the
value of
μ
, or to form confidence intervals for
μ
.
Suppose we wish to test the
hypothesis that
μ
=
μ
0
.
Here,
μ
is the true expected value of
X
, which we do
not know, and
μ
0
is a conjectured value of
μ
.
We can calculate the
t
statistic
t
n
=
¯
X
n

μ
0
ˆ
σ
n
/
√
n
from our sample
X
1
, . . . , X
n
.
If the hypothesis
μ
=
μ
0
is true, then
our earlier argument shows that
t
n
has the
N
(0
,
1) distribution in large samples.
Therefore, it should lie between

1
.
96 and 1
.
96 with probability approximately
equal to 0
.
95.
On the other hand, if
μ
6
=
μ
0
, we have no reason to believe that
t
n
will lie in this range, and in fact it can be shown that it will be greater than
1.96 in absolute value with probability approaching one as
n
→ ∞
. Therefore, if
we calculate
t
n
and find that it does not lie between 1.96 and 1.96, we reject the
hypothesis that
μ
=
μ
0
. If our hypothesis
μ
=
μ
0
is true, there is an approximately
5% chance that we will incorrectly reject it. This is usually deemed an acceptable
rate of error for applied work in the social sciences.
The approximation
¯
X
n

μ
ˆ
σ
n
/
√
n
∼
N
(0
,
1) can also be used to form large sample
confidence intervals for the unknown quantity
μ
. A standard normal random vari
able lies between 1.96 and 1.96 with probability 0.95. Recalling the definition of
convergence in distribution, we can see that
lim
n
→∞
P

1
.
96
≤
¯
X
n

μ
ˆ
σ
n
/
√
n
≤
1
.
96
= 0
.
95
.
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 Spring '08
 Stohs
 Normal Distribution, Probability theory, probability density function

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