Solution Manual/Test Bank For Design and Analysis of Experiments, Wiley, NY

Solutions from Montgomery, D. C. (2008) Design and Analysis of Experiments, Wiley, NY



Chapter 2
Simple Comparative Experiments
Solutions

2.1. The Minitab output for a random sample of data is shown below. Some of the quantities are missing.
Compute the values of the missing quantities.

Variable N Mean SE Mean Std. Dev. Sum

Y 15 ? 0.159 ? 399.851


Mean = 26.657 Std. Dev. = 0.616

2.2. The Minitab output for a random sample of data is shown below. Some of the quantities are missing.
Compute the values of the missing quantities.

Variable N Mean SE Mean Std. Dev. Variance Minimum Maximum

Y 11 19.96 ? 3.12 ? 15.94 27.16


SE Mean = 0.941 Variance = 9.73

2.3. Consider the Minitab output shown below.

One-Sample Z
Test of mu = 30 vs not = 30
The assumed standard deviation = 1.2
N Mean SE Mean 95% CI Z P

16 31.2000 0.3000 (30.6120, 31.7880) ? ?



(a) Fill in the missing values in the output. What conclusion would you draw?

Z=4 P = 0.00006

(b) Is this a one-sided or two-sided test?

Two-sided.

(c) Use the output and the normal table to find a 99 percent CI on the mean.

CI = 30.42725, 31.97275

(d) What is the P-value if the alternative hypothesis is H 1: µ > 30?

P-value = 0.00003




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, Solutions from Montgomery, D. C. (2008) Design and Analysis of Experiments, Wiley, NY


2.4. Suppose that we are testing H0: µ = µ 0 versus H1: µ > µ0. Calcualte the P-value for the following
observed values of the test statistic:

(a) Z0 = 2.45 P-value = 0.00714

(b) Z0 = -1.53 P-value = 0.93699

(c) Z0 = 2.15 P-value = 0.01578

(d) Z0 = 1.95 P-value = 0.02559

(e) Z0 = -0.25 P-value = 0.59871


2.5. Suppose that we are testing H0: µ = µ 0 versus H1: µ ≠ µ0. Calcualte the P-value for the following
observed values of the test statistic:

(a) Z0 = 2.25 P-value = 0.02445

(b) Z0 = 1.55 P-value = 0.12114

(c) Z0 = 2.10 P-value = 0.03573

(d) Z0 = 1.95 P-value = 0.05118

(e) Z0 = -0.10 P-value = 0.92034


2.6. Suppose that we are testing H0: µ = µ0 versus H 1: µ ≠ µ0 with a sample size of n = 9. Calculate
bounds on the P-value for the following observed values of the test statistic:

(a) t0 = 2.48 Table P-value = 0.02, 0.05 Cumputer P-value = 0.03811

(b) t0 = -3.95 Table P-value = 0.002, 0.005 Cumputer P-value = 0.00424

(c) t0 = 2.69 Table P-value = 0.02, 0.05 Cumputer P-value = 0.02750

(d) t0 = 1.88 Table P-value = 0.05, 0.10 Cumputer P-value = 0.09691

(e) t0 = -1.25 Table P-value = 0.20, 0.50 Cumputer P-value = 0.24663


2.7. Suppose that we are testing H0: µ = µ 0 versus H1: µ > µ0 with a sample size of n = 13. Calculate
bounds on the P-value for the following observed values of the test statistic:

(a) t0 = 2.35 Table P-value = 0.01, 0.025 Cumputer P-value = 0.01836

(b) t0 = 3.55 Table P-value = 0.001, 0.0025 Cumputer P-value = 0.00200

(c) t0 = 2.00 Table P-value = 0.025, 0.005 Cumputer P-value = 0.03433

(d) t0 = 1.55 Table P-value = 0.05, 0.10 Cumputer P-value = 0.07355




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, Solutions from Montgomery, D. C. (2008) Design and Analysis of Experiments, Wiley, NY


2.8. Consider the Minitab output shown below.

One-Sample T: Y
Test of mu = 25 vs > 25
Variable N Mean Std. Dev. SE Mean 95% Lower Bound T P

Y 12 25.6818 ? 0.3360 ? ? 0.034


(a) How many degrees of freedom are there on the t-test statistic?

11

(b) Fill in the missing information.

Std. Dev. = 1.1639 95% Lower Bound = 2.0292


2.9. Consider the Minitab output shown below.

Two-Sample T-Test and CI: Y1, Y2
Two-sample T for Y1 vs Y2
N Mean Std. Dev. SE Mean

Y1 20 50.19 1.71 0.38

Y2 20 52.52 2.48 0.55

Difference = mu (X1) – mu (X2)

Estimate for difference: -2.33341

95% CI for difference: (-3.69547, -0.97135)

T-Test of diffence = 0 (vs not = ) : T-Value = -3.47

P-Value = 0.01 DF = 38

Both use Pooled Std. Dev. = 2.1277


(a) Can the null hypothesis be rejected at the 0.05 level? Why?

Yes, the P-Value of 0.001 is much less than 0.05.

(b) Is this a one-sided or two-sided test?

Two-sided.

(c) If the hypothesis had been H0: µ 1 - µ 2 = 2 versus H1: µ 1 - µ 2 ≠ 2 would you reject the null hypothesis
at the 0.05 level?

Yes.

(d) If the hypothesis had been H0: µ 1 - µ 2 = 2 versus H1: µ 1 - µ 2 < 2 would you reject the null hypothesis
at the 0.05 level? Can you answer this question without doing any additional calculations? Why?

Yes. No additional calculations are required because the test becomes more significant with the
change from -2.33341 to -4.33341.


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, Solutions from Montgomery, D. C. (2008) Design and Analysis of Experiments, Wiley, NY



(e) Use the output and the t table to find a 95 percent upper confidence bound on the difference in
means?

95% upper confidence bound = -1.21.

(f) What is the P-value if the alternatie hypotheses are H0: µ 1 - µ 2 = 2 versus H1: µ1 - µ2 ≠ 2?

P-value = 1.4E-07.


2.10. Consider the Minitab output shown below.

One-Sample T: Y
Test of mu = 91 vs. not = 91
Variable N Mean Std. Dev. SE Mean 95% CI T P

Y 25 92.5805 ? 0.4675 (91.6160, ? ) 3.38 0.002


(a) Fill in the missing values in the output. Can the null hypothesis be rejected at the 0.05 level? Why?

Std. Dev. = 2.3365 UCI = 93.5450
Yes, the null hypothesis can be rejected at the 0.05 level because the P-value is much lower at 0.002.

(b) Is this a one-sided or two-sided test?

Two-sided.

(c) If the hypothesis had been H0: µ = 90 versus H 1: µ ≠ 90 would you reject the null hypothesis at the
0.05 level?

Yes.

(d) Use the output and the t table to find a 99 percent two-sided CI on the mean.

CI = 91.2735, 93.8875

(e) What is the P-value if the alternative hypothesis is H 1: µ > 91?

P-value = 0.001.


2.11. 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 12 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 = 12,  = 0.0001, y = 0.2545



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