Introduction to Probability and Statistics for Engineers and Scientists

310 Chapter 8:Hypothesis Testing

A computation gives that

X=37.2833, S=2.7319

and so the value of the test statistic is

T=

√

12(37.2833−40)

2.7319

=−3.4448

Since this is less than−t.05,11=−1.796, the null hypothesis is rejected at the 5 percent
level of significance. Indeed, thep-value of the test data is

p-value=P{T 11 <−3.4448}=P{T 11 >3.4448}=.0028

indicating that the manufacturer’s claim would be rejected at any significance level greater
than .003. ■

The preceding could also have been obtained by using Program 8.3.2, as illustrated in
Figure 8.4.

The value of the t-statistic is −3.4448

The p-value is 0.0028

The p-value of the One-sample t-Test

Start

Quit

One-Sided

Two-Sided

Is greater than m 0

Is less than m 0

Enter the value of m 0 : 40

Data value = 36

Sample size = 12

Add This Point To List

Remove Selected Point From List

Data Values

Clear List

This program computes the p-value when testing that a normal

population whose variance is unknown has mean equal to m 0

35.8

37

41

36.8

37.2

33

36

Is the alternative hypothesis Is the alternative that the mean

??

FIGURE 8.4