Introduction to Probability and Statistics for Engineers and Scientists

7.3Interval Estimates 247

Area = a/2

ta/2, n–1 t

Area = a/2

–ta/2, n–1

P{–ta/2,n–1< Tn–1 < ta/2,n–1} = 1 – a

FIGURE 7.2 t-density function.

EXAMPLE 7.3e Let us again consider Example 7.3a but let us now suppose that when the
valueμis transmitted at location A then the value received at location B is normal with
meanμand varianceσ^2 but withσ^2 being unknown. If 9 successive values are, as in
Example 7.3a, 5, 8.5, 12, 15, 7, 9, 7.5, 6.5, 10.5, compute a 95 percent confidence
interval forμ.

SOLUTION A simple calculation yields that

x= 9

and

s^2 =

∑

x^2 i−9(x)^2

8

=9.5

or

s=3.082

Hence, ast.025,8=2.306, a 95 percent confidence interval forμis

[

9 −2.306

(3.082)

3

,9+2.306

(3.082)

3

]

=(6.63, 11.37)

a larger interval than obtained in Example 7.3a. The reason why the interval just obtained
is larger than the one in Example 7.3a is twofold. The primary reason is that we have
a larger estimated variance than in Example 7.3a. That is, in Example 7.3a we assumed
thatσ^2 was known to equal 4, whereas in this example we assumed it to be unknown