Solucionario montgomery

Solutions from Montgomery, D. C. (2001) 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 andy4=147. (a) State the hypotheses that you think should be tested in this experiment. H0: = 150 H1: > 150

(b) Test these hypotheses using n = 4,
zo y n

= 0.05. What are your conclusions?

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

148.75 150 3 4

1.25 3 2

0.8333

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 23 0.7995 0.7967 0.2014

(d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is
y n 148.75 1.96 3 2
2

y

z

z

2

n 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 viscosityis 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. (2001) Design and Analysis of Experiments, Wiley, NY y n 812 800 25 16 12 25 4 Since z = z0.025 = 1.96, do not reject.

zo

o1.92

/2

(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

2

n

y

z

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,
zo y n
o

= 0.0001, y = 0.2545
0.2545 0.255 0.0001 10 15.81

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
0.2545 1.96

z

2

n
10

y

z

2

n
1.96 0.0001 10

0.0001

0.2545

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 toconstruct 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 y z N( ,9), a 95% two-sided confidence interval on y z 3 n is

2

n

2

n 3 n
/2

y (196) .

y (196) .

If the total interval is to have width 1.0, then the half-interval is 0.5. Since z 1.96 3 nn 11.76 n
2

= z0.025 = 1.96,

0.5 11.76 138.30 139

1.96 3 0.5

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 108 124 124 106 115 138 163 159 134 139

(a) We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate hypotheses for investigating thisclaim. 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 = 3438 / 9 = 382 s 382 19. 54 to y s
o

131 120 19. 54 10

n

1. 78

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