Applied Statistics and Probability for Engineers

The z-values corresponding to 48.5 and 51.5 when   52 are

Therefore

Thus, if we are testing H 0 :   50 against H 1 :   50 with n10, and the true value of the

mean is   52, the probability that we will fail to reject the false null hypothesis is 0.2643. By

symmetry, if the true value of the mean is   48, the value of will also be 0.2643.

The probability of making a type II error increases rapidly as the true value of

approaches the hypothesized value. For example, see Fig. 9-4, where the true value of the

mean is   50.5 and the hypothesized value is H 0 :   50. The true value of is very close

to 50, and the value for is

As shown in Fig. 9-4, the z-values corresponding to 48.5 and 51.5 when 50.5 are

Therefore

0.8980 0.0057 0.8923

Thus, the type II error probability is much higher for the case where the true mean is 50.5

centimeters per second than for the case where the mean is 52 centimeters per second. Of course,

P 1

 2.53Z1.27 2 P 1 Z1.27 2

 P 1 Z

2.53 2

z 1 

48.5 50.5

0.79



2.53 and z 2 

51.5 50.5

0.79

1.27

P 1 48.5X51.5 when 50.5 2



0.2643 0.00000.2643

P 1

 4.43Z

0.63 2 P 1 Z

0.63 2

 P 1 Z

4.43 2

z 1 

48.5 

52

0.79



4.43 and z 2 

51.5 

52

0.79



0.63

9-1 HYPOTHESIS TESTING 283

Figure 9-3 The probability of type II error

when 52 and n10.

46

0.3

0.4

0.5

0.6

0.2

0.1

0

48 50 52 54 56

Under H 0 : = 50 Under H 1 : = 52

Probability density

μ μ

x–

46

0.3

0.4

0.5

0.6

0.2

0.1

0

48 50 52 54 56

Under H 0 : = 50

Under H 1 : = 50.5

Probability density

μ

μ

x–

Figure 9-4 The probability of type II error

when 50.5 and n10.

c09.qxd 5/15/02 8:02 PM Page 283 RK UL 9 RK UL 9:Desktop Folder: