Solutions from Montgomery, D. C. Design and Analysis of Experiments, Wiley, NY

Solutions from Montgomery, D. C. Design and Analysis of Experiments, Wiley, NY


Chapter 2
Simple Comparative Experiments
Solutions

2-1 The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that
the standard deviation of breaking strength is  = 3 psi. A random sample of four specimens is tested. The
results are y1=145, y2=153, y3=150 and y4=147.

(a) State the hypotheses that you think should be tested in this experiment.

H0:  = 150 H1:  > 150

(b) Test these hypotheses using  = 0.05. What are your conclusions?

n = 4,  = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75

y − o 148.75 −150 −1.25
zo = = = = −0.8333
 3 3
4 2

Since z0.05 = 1.645, do not reject.

(c) Find the P-value for the test in part (b).

From the z-table: P  1− 0.7967 + (2 3)(0.7995 − 0.7967)= 0.2014


(d) Construct a 95 percent confidence interval on the mean breaking strength.

The 95% confidence interval is

 
y − z    y + z
2 2




148.75 − (1.96)(3 2)    148.75 + (1.96)(3 2)

145.81    151.69


2-2 The viscosity of a liquid detergent is supposed to average 800 centistokes at 25C. A random sample
of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard
deviation of viscosity is  = 25 centistokes.

(a) State the hypotheses that should be tested.

H0:  = 800 H1:   800

(b) Test these hypotheses using  = 0.05. What are your conclusions?




2-1

, Solutions from Montgomery, D. C. Design and Analysis of Experiments, Wiley, NY


y − o 812 − 800 12 Since z/2 = z0.025 = 1.96, do not reject.
z = = = = 1.92
o
 25 25
n 16 4

(c) What is the P-value for the test? P = 2(0.0274) = 0.0549

(d) Find a 95 percent confidence interval on the mean.

The 95% confidence interval is
 
y − z    y + z
2
n 2
n

812 − (1.96)(25 4)    812 + (1.96)(25 4)
812 −12.25    812 + 12.25
799.75    824.25


2-3 The diameters of steel shafts produced by a certain manufacturing process should have a mean
diameter of 0.255 inches. The diameter is known to have a standard deviation of  = 0.0001 inch. A random
sample of 10 shafts has an average diameter of 0.2545 inches.

(a) Set up the appropriate hypotheses on the mean .

H0:  = 0.255 H1:   0.255

(b) Test these hypotheses using  = 0.05. What are your conclusions?

n = 10,  = 0.0001, y = 0.2545

y − o 0.2545 − 0.255
z = = = −15.81
o
 0.0001
10
Since z0.025 = 1.96, reject H0.


(c) Find the P-value for this test. P=2.6547x10-56

(d) Construct a 95 percent confidence interval on the mean shaft diameter.

The 95% confidence interval is
 
y − z    y + z
2
n 2
n
 0.0001   0.0001 
0.2545 − (1.96)    0.2545 + (1.96)
   
 10   10 

0.254438    0.254562


2-4 A normally distributed random variable has an unknown mean  and a known variance 2 = 9. Find
the sample size required to construct a 95 percent confidence interval on the mean, that has total width of 1.0.

2-2

, Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments, Wiley, NY


Since y  N(,9), a 95% two-sided confidence interval on  is

 
y − z    y + z

2 2
n n
3 3
y − (1.96)    y + (1.96)
n
n

If the total interval is to have width 1.0, then the half-interval is 0.5. Since z /2 = z0.025 = 1.96,

(1.96)(3 n ) = 0.5
n = (1.96)(3 0.5) = 11.76
n = (11.76) = 138.30  139
2




2-5 The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested,
and the following results are obtained:

Days
108 138
124 163
124 159
106 134
115 139

(a) We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate
hypotheses for investigating this claim.

H0:  = 120 H1:  > 120

(b) Test these hypotheses using  = 0.01. What are your conclusions?

y = 131
s2 = [ (108 - 131)2 + (124 - 131)2 + (124 - 131)2 + (106 - 131)2 + (115 - 131)2 + (138 - 131)2
+ (163 - 131)2 + (159 - 131)2 + (134 - 131)2 + ( 139 - 131)2 ] / (10 - 1)

s2 = = 382
s = 382 = 19.54

y − o 131− 120
to = = = 1.78
s n 19.54 10


since t0.01,9 = 2.821; do not reject H0

Minitab Output
T-Test of the Mean

Test of mu = 120.00 vs mu > 120.00

Variable N Mean StDev SE Mean T P
Shelf Life 10 131.00 19.54 6.18 1.78 0.054

T Confidence Intervals
2-3

, Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments, Wiley, NY


Variable N Mean StDev SE Mean 99.0 % CI
Shelf Life 10 19.54 6.18 ( 110.91, 151.09)



(c) Find the P-value for the test in part (b). P=0.054

(d) Construct a 99 percent confidence interval on the mean shelf life.
s s
The 95% confidence interval is y − t  2 ,n −1    y + t  2 ,n −1
n n


 1954   1954 
131− (3.250)    131+ (3.250)
   
   

110.91    151.09



2-6 Consider the shelf life data in Problem 2-5. Can shelf life be described or modeled adequately by a
normal distribution? What effect would violation of this assumption have on the test procedure you used in
solving Problem 2-5?

A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of the
normality assumption. If shelf life is not normally distributed, then the impact of this on the t-test in problem
2-5 is not too serious unless the departure from normality is severe.

Normal Probability Plot for Shelf Life
ML Estimates



99 ML Estimates
Mean 131
95
StDev 18.5418
90


Goodness of Fit
80
70 AD* 1.292
Percent




60
50
40
30
20

10
5


1




86 96 106 116 126 136 146 156 166 176



Data




2-7 The time to repair an electronic instrument is a normally distributed random variable measured in
hours. The repair time for 16 such instruments chosen at random are as follows:

Hours
159 280 101 212
224 379 179 264
222 362 168 250
2-4