Diseño Experimental

Solutions from Montgomery, D. C. (2004) 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 σ = 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: µ = 150 H1: µ > 150

(b) Test these hypotheses using α = 0.05. What are your conclusions? n = 4, σ = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75
zo = y − µo = 148.75 − 150 −1.25 = = −0.8333 3 3 2 4

σ

n

Since z0.05 = 1.645, do not reject. (c) Find the P-value for the test in part (b). From thez-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 − zα 2

σ

n n 148.75 − (1.96 )(3 2) ≤ µ ≤ 148.75 + (1.96 )(3 2)

≤ µ ≤ y + zα 2

σ

145. 81 ≤ µ ≤ 151. 69

2-2 The viscosity of a liquid detergent is supposed to average 800 centistokes at 25°C. A random sample of 16batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes. (a) State the hypotheses that should be tested. H0: µ = 800 H1: µ ≠ 800

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

2-1

Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY
y − µo 812− 800 12 = = 1.92 25 25 4 16

zo =

σ

=

Since zα/2 = z0.025 = 1.96, do not reject.

n

(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 − zα 2

σ

812 − (1.96 )(25 4) ≤ µ ≤ 812 + (1.96 )(25 4 ) 812 − 12.25 ≤ µ ≤ 812 + 12.25 799.75 ≤ µ ≤ 824.25

n

≤ µ ≤ y + zα 2

σ
n2-3 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 σ = 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 µ. H0: µ = 0.255 H1: µ ≠ 0.255

(b) Test these hypotheses using α = 0.05. Whatare your conclusions? n = 10, σ = 0.0001, y = 0.2545
zo = y − µo

σ

=

n

0.2545 − 0.255 = −15.81 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 − zα 2

σ
n

≤ µ ≤ y + zα 2

σ
n

⎛ 0.0001 ⎞ ⎛ 0.0001 ⎞ 0.2545 − (1.96 ) ⎜⎟ ≤ µ ≤ 0.2545 + (1.96 ) ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠

0. 254438 ≤ µ ≤ 0. 254562

2-4 A normally distributed random variable has an unknown mean µ and a known variance σ2 = 9. Find the sample size required to construct a 95 percent confidence interval on the mean, that has total length of 1.0.

2-2

Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Since y ∼N(µ,9), a 95% two-sided confidence interval on µ is
y − zα 2

σ
n

≤ µ ≤ y + zα 2 3 n

σ
n 3 n

y − (196) .

≤ µ ≤ y + (196) .

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

(1.96)(3 n ) = 0.5 n = (1.96)(3 0.5) = 11.76 n = (11.76 )2 = 138.30 ≅ 139
2-5 The shelf life of a carbonated beverage is of interest. Ten bottles arerandomly 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 this claim. H0: µ = 120 H1: µ > 120

(b) Test these hypotheses using α = 0.01. What are your conclusions?
y = 131 S2 = 3438 / 9 = 382 S = 382 = 19.54

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