246 Chapter 7: Parameter Estimation
Because the estimatexis within 1.96(σ/
√
n)=.588/
√
nof any point in the interval, it
follows that we can be 95 percent certain thatxis within 0.1 ofμprovided that
.588
√
n
≤0.1
That is, provided that
√
n≥5.88
or
n≥34.57
That is, a sample size of 35 or larger will suffice. ■
Random Variable 7.5Approximate Confidence Interval for the Mean of a Bernoulli
Is Unknown
Suppose now thatX 1 ,...,Xnis a sample from a normal distribution with unknown mean
μand unknown varianceσ^2 , and that we wish to construct a 100(1−α) percent confidence
interval for√ μ. Sinceσis unknown, we can no longer base our interval on the fact that
n(X−μ)/σis a standard normal random variable. However, by lettingS^2 =
∑n
i= 1 (Xi−
X)^2 /(n−1) denote the sample variance, then from Corollary 6.5.2 it follows that
√
n
(X−μ)
S
is at-random variable withn−1 degrees of freedom. Hence, from the symmetry of the
t-density function (see Figure 7.2), we have that for anyα∈(0, 1/2),
P
{
−tα/2,n− 1 <
√
n
(X−μ)
S
<tα/2,n− 1
}
= 1 −α
or, equivalently,
P
{
X−tα/2,n− 1
S
√
n
<μ<X+tα/2,n− 1
S
√
n
}
= 1 −α
Thus, if it is observed thatX =xandS=s, then we can say that “with 100(1−α)
percent confidence”
μ∈
(
x−tα/2,n− 1
s
√
n
,x+tα/2,n− 1
s
√
n
)