Download For large sample sizes, n, the sampling distribution of

For large sample sizes, n, the sampling distribution of the sample standard deviation S is
sometimes approximated with a Normal distribution having mean  and variance
2
.
2n
Prove that this approximation leads to the following large-sample confidence interval for
S
S
the population standard deviation  :
.
 
z / 2
z / 2
1
1
2n
2n
Proof. For large sample sizes, n, the sampling distribution of the sample standard
deviation S is sometimes approximated with a Normal distribution having mean  and
2
variance
. So,
2n
Z
S 
2

S 

2n
2n
follows roughly the standard normal distribution N(0,1). So,
P( z / 2  S 2n  z / 2 )  1  
i.e.,
P( z / 2 / 2n  S  z / 2 / 2n )  1  
i.e.,
P( z / 2 / 2n  S  1  z / 2 / 2n )  1  
i.e.,
P(1  z / 2 / 2n  S  1  z / 2 / 2n )  1  
…………………..(1)
Note: z / 2 denotes the Z-score with area  / 2 in the upper tail.
(1) is the same as
P(
1
Hence, (
1
S
,
z / 2
deviation  :
2n
S
 
z / 2
1
S
)  1
z / 2
1
2n
2n
S
) is a confidence interval for the population standard
z / 2
2n