Introduction to Probability and Statistics for Engineers and Scientists

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

)