252solnC 2/13/08
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C. POWER FUNCTIONS
1. A one-sided Test.
Text 9.109 on CD (9.102), C1
2. A Two-Sided Test.
C2
Exercise 9.109 (9.102 in 8th edition): A soft drink machine is supposed to dispense at leas 7oz of the drink
with a standard deviation of 0.2 oz. A sample of 16 cupfuls is taken. If the probability of a type one error is
set at 5%, compute the probability of a type two error and the power of the test if the actual mean is a) 6.9
oz. b) 6.8 oz.
H :   7
Solution: Given:  0
,   0.2, n  16,   .05,
H 1 :   7

0.2
so  x 

 0.05 . a) First, state the problem and find a critical value or values. Since this is
n
16
a one-sided test, the formula for a two-sided critical value x cv   0  z   x becomes xcv   0  z  x , so
2
that xcv  7  1.645 0.05   6.918 . So we will not reject H 0 if the sample mean x is greater than or equal
to 6.918. Make a diagram. Put 7 in the middle and show a 'reject' region below 6.918. Shade this area to
represent the significance level.
a) 1  6.9 . We will not reject the null hypothesis if the sample mean is at least 6.918, so our probability
of a type II error is the probability that the sample mean is below 6.918 when the population mean is 6.9.
6.918  6.9 

  Px  6.918   6.9  P  z 
 Pz  0.36   .5  .1406 .  .3594 . Power  1    .6406.
0.05 

6.918  6.8 

 Pz  2.36   .5  .4909  .0098 .
b) 1  6.8   Px  6.918   6.8  P  z 
0.05 

Power  1    .9909.
Problem C1: Assume that   4 and n  70 . Find the critical values, power function and operating
characteristic curve for:
 H 0 :   50
Use a significance level of 5 percent.

 H 1 :   50
Solution: a) First, state the problem and find a critical value or values.
 H 0 :   50

4
  4, n  70,   .05 so  x 

 0.47809 . Since this is a one-sided test, the

n
70
 H 1 :   50
formula for a two-sided critical value x cv   0  z   x becomes xcv   0  z  x , so that
2
xcv  50  1.645 0.47809   49.2135 . So we will not reject H 0 if the sample mean x is greater than or
equal to 49.2135. Make a diagram. Put 50 in the middle and show a 'reject' region below 49.2135. Shade
this area to represent the significance level.
b) Decide on what values of 1 to use to compute  , the probability of a type II error. The usual set
of values includes the mean from the null hypothesis, the critical value, a point about midway between
these values and two points, one further out beyond the critical value by a distance equal to the distance
between the null hypothesis mean and the critical value, and another halfway between this point and the
critical value. We thus choose 50, 49.2135, and 49.6, which is about halfway between them. Since the
difference between 50 and 49.2135 is about 0.8, the lowest value of 1 that we use is 48.4, and a point
about halfway between 48.4 and 49.2135 is 48.8.
252solnC 2/13/08
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c) Compute  for each value of 1 . Since a type II error is wrongly ‘accepting’ the null hypothesis, we
compute the probability that the sample mean will be above or equal to the critical value for each value of

x  1 
1 . Our computations are below. Note that, in general, for this one-sided hypothesis   P  z  cv
.
x 

49 .2135  50 

  Px  49 .2135   50   P  z 
1  50
  Pz  1.645   .95
.47809


power  1    .05  
49 .2135  49 .6 
  Pz  0.81  .2910  .5  .7910
.47809


power  1    .2090 Make a diagram. Use the diagram from part a) and superimpose
a second identical Normal curve centered at 1  49.6 . Shade the area to the right of 49.2135 under the
second curve in the opposite direction to your shading in a) to represent  , the operating characteristic.
1  49.6
  Px  49 .2135   49 .6  P  z 
1  49.2135
  Px  49 .2135   49 .2135   P  z 



49 .2135  49 .2135 
  Pz  0  .5000
.47809

power  1    .5000
1  48 .8
  Px  49 .2135   48 .8  P  z 


49 .2135  48 .8 
  Pz  0.86   .5  .3051  .1949
.47809

power  1    .8051
1  48.4
  Px  49 .2135   48 .4  P  z 

49 .2135  48 .4 
  Pz  1.70   .5  .4554  .0446
.47809


power  1    .9554
Make a diagram. Put 0 to 1 on your y axis and 48.4 to 50 on your x axis. Graph one line representing 
and one representing power. They should cross at 49.2135.
Problem C2: A hardware firm charges a flat rate for mailing of small tools based on an average weight of
20 oz. with a standard deviation of 3.60 oz. A consultant challenges this assumption and a sample of 100
packages is taken. Find critical values for a significance level of 1% and compute the power function and
operating characteristic curve.
Solution: a) First, state the problem and find a critical value or values.
 H 0 :   20

3.60
  3.60, n  100 ,   .01 so  x 

 0.360 . Since this is a two sided test, the

H
:


20
n
100
 1
formula for a critical value is x cv   0  z   x , so that xcv  20  2.576 0.360   20  0.927 . So we will
2
not reject H 0 if the sample mean x is between 19.073 and 20.927. Make a diagram. Put 20 in the middle
and show two ‘reject’ regions, one below 19.073 and one above 20.927. Shade these areas to represent the
significance level.
b) Decide on what values of 1 to use to compute  , the probability of a type II error. The usual set
of values includes the mean from the null hypothesis, the critical values, a point about midway between
these values and two points, one further out beyond the critical value by a distance equal to the distance
between the null hypothesis mean and the critical value, and another halfway between this point and the
critical value. We thus choose the null hypothesis mean, 20, and the two critical values 19.073 and
20.927.20.5 and 21.5 are about halfway between 20 and the critical values. Since the difference between 20
and the critical values is about 1.0, the lowest value of 1 that we use is 18.0 and the highest is 22.0. Points
about halfway between these numbers and the critical values are 18.5 and 21.5.
252solnC 2/13/08
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c) Compute  for each value of 1 . Since a type II error is wrongly ‘accepting’ the null hypothesis, we
compute the probability that the sample mean will be between the critical values for each value of 1 . Our
 x  1
x  1 
computations are below. Note that, in general, for a two-sided hypothesis   P  cv1
 z  cv 2
.
x 
 x
20 .927  20 
19 .073  201
1  20
  P19 .073  x  20 .927   20   P 
z
0
.
360
0.360 

 P2.575  z  2.575   2.4950   .99  1  
20 .927  20 .5 
19 .073  20 .5
z

0.360
0.360


 P 3.96  z  1.19   .5  .3830  .8830 Make a diagram. Use the diagram
  P19 .073  x  20 .927   20 .5  P 
1  20.5 or 19.5
from part a) and superimpose a second identical Normal curve centered at 1  20 .5 . Shade the
area between 19.073
and
20.927 under the second curve in the opposite direction to
your shading in a) to represent  , the operating characteristic.
1  20.927 or 19.073
  P19.073  x  20.927   20.927
20 .927  20 .927 
19 .073  20 .927
 P
z

0.360
0.360


 P 5.15  z  0.00   .5000
20 .927  21 .5 
19 .073  21 .5
z

0.360
0.360


  P19 .073  x  20 .927   19 .5  P 
1  21 .5 or 18 .5
 P 6.74  z  1.59   .5  .4441  .0559
20 .927  22 .0 
19 .073  22 .0
z

0.360
0.360


  P19 .073  x  20 .927   22 .0  P 
1  22.0 or 18.0
 P 8.13  z  2.98   .5  .4986  .0014
If we round these results, we get the following values for the operating characteristic and power:
22.0
21.5
20.9
20.5
20.0
19.5
19.0
18.5
1

power
.00
1.00
.06
.94
.50
.50
.88
.12
.99
.01
.88
.12
.50
.50
.06
.94
18.0
.00
1.00
Make a diagram. Put 0 to 1 on your y axis and 18 to 22 on your x axis. Graph one line representing 
and one representing power. They should cross at 19.073 and 20.927.