Chapter 4 Inferences About Process Quality LEARNING OBJECTIVES After completing this chapter you should be able to: 1. E
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Chapter 4 Inferences About Process Quality LEARNING OBJECTIVES After completing this chapter you should be able to: 1. Explain the concept of random sampling 2. Explain the concept of a sampling distribution 3. Explain the general concept of estimating the parameters of a population or probability distribution 4. Know how to explain the precision with which a parameter is estimated 5. Construct and interpret confidence intervals on a single mean and on the difference in two means 6. Construct and interpret confidence intervals on a single variance or the ratio of two variances 7. Construct and interpret confidence intervals on a single proportion and on the difference in two proportions 8. Test hypotheses on a single mean and on the differences in two means 9. Test hypotheses on a single variance and on the ratio of two variances 10. Test hypotheses on a single proportion and on the difference in two proportions 11. Use the P-value approach for hypothesis testing 12. Understand how the analysis of variance (ANOVA) is used to test hypotheses about the equality of more than two means 13. Understand how to fit and interpret linear regression models
IMPORTANT TERMS AND CONCEPTS Alternative hypothesis Analysis of variance (ANOVA) Binomial distribution Checking assumptions for statistical inference procedures Chi-square distribution Confidence interval Confidence intervals on means, known variance(s)
Point estimator Poisson distribution Pooled estimator Power of a statistical test Random sample Regression model Residual analysis
4-2 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY Confidence intervals on means, unknown variance(s) Confidence intervals on proportions Confidence intervals on the variance of a normal distribution Confidence intervals on the variances of two normal distributions Critical region for a test statistic F-distribution Hypothesis testing Least squares estimator Linear statistical model Minimum variance estimator Null hypothesis P-value P-value approach Parameters of a distribution
Sampling distribution Scaled residuals Statistic t-distribution Test statistic Tests of hypotheses on means, known variance(s) Tests of hypotheses on means, unknown variance(s) Tests of hypotheses on proportions Tests of hypotheses on the variance of a normal distribution Tests of hypotheses on the variances of two normal distributions Type I error Type II error Unbiased estimator
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-3
EXERCISES Note: New exercises are noted with . 4.1. Suppose that you are testing the following hypotheses where the variance is known:
H0 : 100 H1 : 100 Find the P-value for the following values of the test statistic. (a) Z0 = 2.75 From Appendix II, (Z0) = (2.75) = 0.99702 For a two-sided test: P = 2[1 − (|Z0|)] = 2[1 – 0.99702] = 0.00596 (b) Z0 = 1.86 From Appendix II, (Z0) = (1.86) = 0.96856 For a two-sided test: P = 2[1 − (|Z0|)] = 2[1 – 0.96856] = 0.06289 (c) Z0 = −2.05 From Appendix II, (Z0) = (|−2.05|) = (2.05) = 0.97982 For a two-sided test: P = 2[1 - (|Z0|)] = 2[1 – 0.97982] = 0.04036 (d) Z0 = −1.86 Same answer as for (b) From Appendix II, (Z0) = (|−1.86|) = (1.86) = 0.96856 For a two-sided test: P = 2[1 − (|Z0|)] = 2[1 – 0.96856] = 0.06289
4-4 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.2. Suppose that you are testing the following hypotheses where the variance is known: H0 : 100
H1 : 100 Find the P-value for the following values of the test statistic. (a) Z0 = 2.60 From Appendix II, (Z0) = (2.60) = 0.99534 For a one-sided, upper-tail test: P = 1 − (Z0) = 1 – 0.99534 = 0.00466 (b) Z0 = 1.85 From Appendix II, (Z0) = (1.85) = 0.96784 For a one-sided, upper-tail test: P = 1 − (Z0) = 1 – 0. 96784 = 0.03216 (c) Z0 = 2.15 From Appendix II, (Z0) = (2.15) = 0.98422 For a one-sided, upper-tail test: P = 1 − (Z0) = 1 – 0. 98422 = 0.01578 (d) Z0 = 3.72 From Appendix II, (Z0) = (3.72) = 0.99990 For a one-sided, upper-tail test: P = 1 − (Z0) = 1 – 0. 99990 = 0.0001
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-5 4.3. Suppose that you are testing the following hypotheses where the variance is known: H0 : 100
H1 : 100 Find the P-value for the following values of the test statistic. (a) Z0 = −2.35 From Appendix II, (Z0) = (2.35) = 0.99061 For a one-sided, lower-tail test and by symmetry of the normal distribution: P = (Z0) = (−2.35) = 1 − (2.35) = 1 – 0.99061 = 0.00939 (b) Z0 = −1.99 From Appendix II, (Z0) = (1.99) = 0.97670 For a one-sided, lower-tail test and by symmetry of the normal distribution: P = (Z0) = (−1.99) = 1 − (1.99) = 1 – 0. 97670 = 0.02330 (c) Z0 = −2.18 From Appendix II, (Z0) = (2.18) = 0.98537 For a one-sided, lower-tail test and by symmetry of the normal distribution: P = (Z0) = (−2.18) = 1 − (2.18) = 1 – 0. 98537 = 0.01463 (d) Z0 = −1.85 From Appendix II, (Z0) = (1.85) = 0.96784 For a one-sided, lower-tail test and by symmetry of the normal distribution: P = (Z0) = (−1.85) = 1 − (1.85) = 1 – 0. 96784 = 0.03216
4-6 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.4. Suppose that you are testing the following hypotheses where the variance is unknown: H0 : 100
H1 : 100 The sample size is n = 10. Find bounds on the P-value for the following values of the test statistic. (a) t0 = 2.75 From Appendix IV and ν = n – 1 = 10 – 1 =9, 0.025 0.01 2.262 2.75 2.821 T , 9 For a two-sided test: 2(0.01) < P < 2(0.025), or 0.02 < P < 0.05 (b) t0 = 1.86 From Appendix IV and ν = 9, 0.05 0.025 1.833 1.86 2.262 T , 9 For a two-sided test: 2(0.025) < P < 2(0.05), or 0.05 < P < 0.10 (c) t0 = −2.05 From Appendix IV and ν = 9, 0.05 0.025 1.833 2.05 2.262 T , 9 For a two-sided test and by symmetry of the t distribution: 2(0.025) < P < 2(0.05), or 0.05 < P < 0.10 (d) t0 = −1.86 Same answer as for (b)
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-7 4.5. Suppose that you are testing the following hypotheses where the variance is unknown: H0 : 100
H1 : 100 The sample size is n = 12. Find bounds on the P-value for the following values of the test statistic. (a) t0 = 2.55 From Appendix IV and ν = n – 1 = 12 – 1 = 11, 0.025 0.01 2.20 2.55 2.718 T , 11 For a one-sided, upper-tail test: 0.01 < P < 0.025 (b) t0 = 1.87 From Appendix IV and ν = 11, 0.05 0.025 2.20 T , 11 1.796 1.87 For a one-sided, upper-tail test: 0.025 < P < 0.05 (c) t0 = 2.05 From Appendix IV and ν = 11, 0.05 0.025 2.20 T , 11 1.796 2.05 For a one-sided, upper-tail test: 0.025 < P < 0.05 (d) t0 = 2.80 From Appendix IV and ν = 11, 0.01 0.005 3.106 T , 11 2.718 2.80 For a one-sided, upper-tail test: 0.005 < P < 0.01
4-8 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.6. Suppose that you are testing the following hypotheses where the variance is unknown: H0 : 100
H1 : 100 The sample size is n = 20. Find bounds on the P-value for the following values of the test statistic. (a) t0 = −2.80 From Appendix IV and ν = n – 1 = 20 – 1 = 19, 0.01 0.005 T , 19 2.539 2.80 2.861 For a one-sided, lower-tail test and by symmetry of the t distribution: 0.005 < P < 0.01 (b) t0 = −1.75 From Appendix IV and ν = 19, 0.05 0.025 2.093 T , 19 1.729 1.75 For a one-sided, lower-tail test and by symmetry of the t distribution: 0.025 < P < 0.05 (c) t0 = −2.54 From Appendix IV and ν = 19, 0.01 0.005 2.861 T , 19 2.539 2.54 For a one-sided, lower-tail test and by symmetry of the t distribution: 0.005 < P < 0.01 (d) t0 = −2.05 From Appendix IV and ν = 19, 0.05 0.025 2.093 T , 19 1.729 2.05 For a one-sided, lower-tail test and by symmetry of the t distribution: 0.025 < P < 0.05
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-9 4.7. The inside diameters of bearings used in an aircraft landing gear assembly are known to have a standard deviation of = 0.002 cm. A random sample of 15 bearings has an average inside diameter of 8.2535 cm. (a) Test the hypothesis that the mean inside bearing diameter is 8.25 cm. Use a two-sided alternative and = 0.05. Since is known, use the standard normal distribution: x ~ N(, ), n = 15, x = 8.2535 cm, = 0.002 cm, 0 = 8.25, = 0.05 Test H0: = 8.25 vs. H1: 8.25. Reject H0 if |Z0|>Z/2. x 0 8.2535 8.25 Z0 6.78 (Equation 4.23) n 0.002 15 Z/2 = Z0.05/2 = Z0.025 = 1.96 (from Appendix II) Reject H0: = 8.25, and conclude that the mean bearing ID is not equal to 8.25 cm. (b) Find the P-value for this test. From Appendix II, find the corresponding cumulative standard normal, (Z0) for Z0 = 6.78, and calculate: P-value = 2[1 (Z0)] = 2[1 (6.78)] = 2[1 1.00000] = 0 (c) Construct a 95% two-sided confidence interval on the mean bearing diameter. Construct a 95% two-sided confidence interval on mean bearing diameter.
x Z /2
8.25 1.96 0.002
n 15 8.25 1.96 0.002 n x Z /2
8.249 8.251
MTB > Stat > Basic Statistics > 1-Sample Z One-Sample Z The assumed standard deviation = 0.002 N 15
Mean 8.23530
SE Mean 0.00052
95% CI (8.23429, 8.23631)
15
(Equation 4.29)
4-10 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.8. The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi. (a) Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi. Use = 0.05.
0 = 125; = 0.05 Test H0: = 125 vs. H1: > 125. Reject H0 if Z0 > Z. x 0 127 125 Z0 2.828 n 2 8 Z = Z0.05 = 1.645 Reject H0: = 125, and conclude that the mean tensile strength exceeds 125 psi. (b) What is the P-value for this test? P-value = 1 (Z0) = 1 (2.828) = 1 0.99766 = 0.00234 (c) Discuss why a one-sided alternative was chosen in part (a). In strength tests, we usually are interested in whether some minimum requirement is met, not simply that the mean does not equal the hypothesized value. A one-sided hypothesis test lets us do this. (d) Construct a 95% lower confidence interval on the mean tensile strength.
n 127 1.645 2 8 x Z
125.8
MTB > Stat > Basic Statistics > 1-Sample Z One-Sample Z Test of mu = 125 vs > 125 The assumed standard deviation = 2 N 8
Mean 127.000
SE Mean 0.707
95% Lower Bound 125.837
Z 2.83
P 0.002
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-11 4.9. The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7. (a) The manufacturer wants to be certain that the mean battery life exceeds 25 h. What conclusions can be drawn from these data (use = 0.05). Since is unknown, calculate the sample standard deviation S , and use the t-distribution, one-sided: x ~ N(, ), n = 10, x = 26.0, s = 1.62, 0 = 25, = 0.05 Test H0: = 25 vs. H1: > 25. Reject H0 if t0 > t.
t0 x 0 S
n 26.0 25 1.62
10 1.952 (Equation 4.33)
t, n1 = t0.05, 101 = 1.833 (from Appendix IV) Reject H0: = 25, and conclude that the mean life exceeds 25 h. MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-9 Test of mu = 25 vs > 25 Variable Ex 4-9
N 10
Mean 26.000
StDev 1.625
95% Lower Bound 25.058
SE Mean 0.514
T 1.95
P 0.042
(b) Construct a 90% two-sided confidence interval on mean life in the accelerated test. Construct a 90% two-sided confidence interval on mean life in the accelerated test. = 0.10 x t /2,n1 S
26.0 1.833 1.62
n x t /2,n1 S
n
10 26.0 1.833 1.62 25.06 26.94
10
(Equation 4.34)
4-12 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.9.(b) continued MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-9 Test of mu = 25 vs not = 25 Variable Ex 4-9
N 10
Mean 26.000
StDev 1.625
SE Mean 0.514
90% CI (25.058, 26.942)
T 1.95
P 0.083
(c) Construct a normal probability plot of the battery life data. What conclusions can you draw? MTB > Graph > Probability Plot > Single and ensure Distribution = Normal Probability Plot of Ex 4-9
The plotted points fall approximately along a straight line, so the assumption that battery life is normally distributed is appropriate.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-13 4.10. Using the data from Exercise 4.7, construct a 95% lower confidence interval on mean battery life. Why would the manufacturer be interested in a one-sided confidence interval? x ~ N(, ); n = 10; x = 26.0 h; s = 1.62 h; = 0.05; t, n1 = t0.05,9 = 1.833
x t ,n 1 S
26.0 1.833 1.62
10 n
25.06
The manufacturer might be interested in a lower confidence interval on mean battery life when establishing a warranty policy. 4.11. A new process has been developed for applying photoresist to 125-mm silicon wafers used in manufacturing integrated circuits. Ten wafers were tested, and the following photoresist thickness measurements ( in angstromx 1000) were observed: 13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002. (a) Test the hypothesis that mean thickness is 13.4 1000 Å. Use = 0.05 and assume a two-sided alternative. Since is unknown, calculate the sample standard deviation S , and use the t-distribution: x ~ N(, ), n = 10, x = 13.39618 1000 Å, s = 0.00391, 0 = 13.4 1000 Å, = 0.05 Test H0: = 13.4 vs. H1: 13.4. Reject H0 if |t0|>t/2.
t0 x 0 S
n 13.39618 13.4 0.00391
10 3.089 (Equation 4.33)
t/2, n1 = t0.025, 9 = 2.262 (from Appendix IV) Reject H0: = 13.4, and conclude that the mean thickness differs from 13.4 1000 Å. MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-11 Test of mu = 13.4 vs not = 13.4 Variable Ex 4-11
N 10
Mean 13.3962
StDev 0.0039
SE Mean 0.0012
95% CI (13.3934, 13.3990)
T -3.09
P 0.013
4-14 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.11. continued (b) Find a 99% two-sided confidence interval on mean photoresist thickness. Assume that thickness is normally distributed. = 0.01
x t /2,n1 S
13.39618 3.2498 0.00391
n 10 13.39618 3.2498 0.00391 n x t /2,n1 S
10
(Equation 4.34)
13.39216 13.40020
MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-11 Test of mu = 13.4 vs not = 13.4 Variable Ex 4-11
N 10
Mean 13.3962
StDev 0.0039
SE Mean 0.0012
99% CI (13.3922, 13.4002)
T -3.09
P 0.013
(c) Does the normality assumption seem reasonable for these data? MTB > Graph > Probability Plot > Single and ensure Distribution = Normal Probability Plot of Ex 4-11
The plotted points form a reverse-“S” shape, instead of a straight line, so the assumption that battery life is normally distributed is not appropriate.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-15 4.12. A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.06, 11.98, 11.97, 12.02, 12.05, and 11.99. (a) Suppose that the manufacturer wants to be sure that the mean net contents exceeds 12 oz. What conclusions can be drawn from the data (use = 0.01). x ~ N(, ), 0 = 12, = 0.01 n = 10, x = 12.017, s = 0.0298 Test H0: = 12 vs. H1: > 12. Reject H0 if t0 > t. x 0 12.017 12 t0 1.8040 S n 0.0298 10 t/2, n1 = t0.005, 9 = 3.250 Do not reject H0: = 12, and conclude that there is not enough evidence that the mean fill volume exceeds 12 oz. MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-12 Test of mu = 12 vs > 12 Variable Ex 4-12
N 10
Mean 12.0170
StDev 0.0298
SE Mean 0.0094
99% Lower Bound 11.9904
T 1.80
P 0.053
(b) Construct a 99% two-sided confidence interval on the mean fill volume.
= 0.01 t/2, n1 = t0.005, 9 = 3.250
x t /2,n1 S
12.017 3.250 0.0298
n 10 12.017 3.250 0.0298 n x t /2,n1 S
11.9864 12.0476
10
4-16 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.12. (b) continued MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-12 Test of mu = 12 vs not = 12 Variable Ex 4-12
N 10
Mean 12.0170
StDev 0.0298
SE Mean 0.0094
99% CI (11.9863, 12.0477)
T 1.80
P 0.105
(c) Does the assumption of normality seem appropriate for the fill volume data? MTB > Graph > Probability Plot > Single and ensure Distribution = Normal Probability Plot of Ex 4-12
The plotted points fall approximately along a straight line, so the assumption that fill volume is normally distributed is appropriate.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-17 4.13. Ferric chloride is used as a flux in some types of extraction metallurgy processes. This material is shipped in containers, and the container weight varies. It is important to obtain an accurate estimate of mean container weight. Suppose that from long experience a reliable value for the standard deviation of flux container weight is determined to be 4 lb. How large a sample would be required to construct a 95% two-sided confidence interval on the mean that has a total width of 1 lb?
= 4 lb, = 0.05, Z/2 = Z0.025 = 1.9600, total confidence interval width = 1 lb, find n
2 Z /2 2 1.9600 4
n total width n 1 n 246
4-18 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.14. The diameters of aluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001 in. A random sample of 20 rods has an average diameter of 0.5021 in. (a) Test the hypothesis that mean rod diameter is 0.5025 in. Assume a two-sided alternative and use = 0.05. x ~ N(, ), 0 = 0.5025, = 0.05 n = 20, x = 0.5021 in, = 0.0001 in Test H0: = 0.5025 vs. H1: 0.5025. Reject H0 if |Z0| > Z/2. x 0 0.5021 0.5025 Z0 17.89 n 0.0001 20 Z/2 = Z0.05/2 = Z0.025 = 1.96 Reject H0: = 0.5025, and conclude that the mean rod diameter differs from 0.5025. MTB > Stat > Basic Statistics > 1-Sample Z One-Sample Z Test of mu = 0.5025 vs not = 0.5025 The assumed standard deviation = 0.0001 N 20
Mean 0.502100
SE Mean 0.000022
95% CI (0.502056, 0.502144)
Z -17.89
P 0.000
(b) Find the P-value for this test. P-value = 2[1 (Z0)] = 2[1 (|−17.89|)] = 2[1 1] = 0 (c) Construct a 95% two-sided confidence interval on the mean rod diameter.
x Z /2
0.5021 1.960 0.0001
n 20 0.5021 1.960 0.0001 n x Z /2
0.50206 0.50214
20
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-19 4.15. The output voltage of a power supply is assumed to be normally distributed. Sixteen observations taken at random on voltage are as follows: 10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, and 10.85. (a) Test the hypothesis that the mean voltage equals 12 V against a two-sided alternative using = 0.05. x ~ N(, ), n = 16, x = 10.259 V, s = 0.999 V, 0 = 12, = 0.05 Test H0: = 12 vs. H1: 12. Reject H0 if |t0|>t/2.
t0 x 0 S
n 10.259 12 0.999
16 6.971 (Equation 4.33)
t/2, n1 = t0.025, 15 = 2.131 (from Appendix IV) Reject H0: = 12, and conclude that the mean output voltage differs from 12V. MTB > Stat > Basic Statistics > 1-Sample t One-Sample T: Ex 4-15 Test of mu = 12 vs not = 12 Variable Ex 4-15
N 16
Mean 10.259
StDev 0.999
SE Mean 0.250
95% CI (9.727, 10.792)
T -6.97
P 0.000
(b) Construct a 95% two-sided confidence interval on . Construct a 95% two-sided confidence interval on .
x t /2,n1 S
10.259 2.131 0.999
n 16 10.259 2.131 0.999 n x t /2,n1 S
16
(Equation 4.34)
9.727 10.792
(c) Test the hypothesis that 2 = 11 using = 0.05. Test H0: 2 = 1 vs. H1: 2 1. Reject H0 if 20> 2/2, n-1 or 20 < 21-/2, n-1. (16 1)0.9992 14.970 (Equation 4.38) 1 2/2, n1 = 20.025,161 = 27.488 (from Appendix III) 21/2, n1 = 20.975,161 = 6.262 (from Appendix III) Do not reject H0: 2 = 1, and conclude that there is insufficient evidence that the variance differs from 1.
02 (n 1)S 2 02
4-20 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 14.15 continued (d) Construct a 95% two-sided confidence interval on . (n 1)S 2 2 /2,n1 2 (n 1)S 2 12 /2,n 1 (16 1)0.9992 27.488 2 (16 1)0.9992 6.262
(Equation 4.39) 0.545 2 2.391 0.738 1.546 Since the 95% confidence interval on contains the hypothesized value, 02 = 1, the null hypothesis, H0: 2 = 1, cannot be rejected. Using MINITAB: MTB > Stat > Basic Statistics > Graphical Summary Summary for Ex 4-15
(e) Construct a 95% upper confidence interval on . 2 0.05; 12 ,n1 0.95,15 7.2609 (from Appendix III)
2 (n 1)S 2 12 ,n1 2 (16 1)0.9992 7.2609 (Equation 4.40) 2 2.062 1.436
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-21 14.15 continued (f) Does the assumption of normality seem reasonable for the output voltage? MTB > Graph > Probability Plot > Single and ensure Distribution = Normal Probability Plot of Ex 4-15
Looking at the plot, the plotted points fall approximately along a straight line, so the assumption of a normal distribution for output voltage seems appropriate.
4-22 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.16. Two machines are used for filling glass bottles with a soft-drink beverage. The filling processes have known standard deviations 1 = 0.010 liter and 2 = 0.005 liter, respectively. A random sample of n1 = 25 bottles from machine 1 and n2 = 15 bottles from machine 2 results in average net contents of x1 = 2.03 liters and x2 = 2.07 liters. (a) Test the hypothesis that both machines fill to the same net contents, using = 0.05. What are your conclusions?
= 0.05, 0 = 0 Test H0: 1 – 2 = 0 versus H0: 1 – 2 0. Reject H0 if Z0 > Z/2 or Z0 < –Z/2. Z0
(x1 x2 ) 0
n1 n2 2 1
2 2
(2.03 2.07) 0 0.0102 25 0.0052 15
16.80
Z/2 = Z0.05/2 = Z0.025 = 1.96 Z/2 = 1.96 Reject H0: 1 – 2 = 0, and conclude that there is a difference in mean net contents between machine 1 and machine 2. (b) Find the P-value for this test. P-value = 2[1 (Z0)] = 2[1 (16.80)] = 2[1 1.00000] = 0 (c) Construct a 95% confidence interval on the difference in mean fill volume.
(x1 x2 ) Z /2 2 (2.03 2.07) 1.9600 0.010
12
n1
22
n2
(1 2 ) (x1 x2 ) Z /2
12
n1
22
n2
2
25
2 2 (1 2 ) (2.03 2.07) 1.9600 0.010 0.005 15 25 15 0.045 (1 2 ) 0.035
0.005
The confidence interval for the difference does not contain zero. We can conclude that the machines do not fill to the same volume.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-23 4.17. Two quality control technicians measured the surface finish of a metal part, obtaining the data in Table 4E.1. Assume that the measurements are normally distributed.
(a) Test the hypothesis that the mean surface finish measurements made by the two technicians are equal. Use = 0.05 and assume equal variances. Test H0: 1 2 = 0 vs. H1: 1 2 0. Reject H0 if |t0|>t/2, n1+n2-2. sP
(n1 1)s12 (n2 1)s22 (7 1)0.1152 (8 1)0.1252 0.1204 n1 n2 2 7 8 2
t0 x1 x2
s
P
(Equations 4.51 and 4.52)
1 n1 1 n2 1.383 1.376 0.1204 1 7 1 8 0.11
t/2, n1+n2-2 = t0.025, 13 = 2.160 (from Appendix IV) Do not reject H0, and conclude that there is not sufficient evidence of a difference between measurements obtained by the two technicians. MTB > Stat > Basic Statistics > 2-Sample t Two-Sample T-Test and CI: Ex4-17T1, Ex4-17T2 Two-sample T for Ex4-17T1 vs Ex4-17T2 Ex4-17T1 Ex4-17T2
N 7 8
Mean 1.383 1.376
StDev 0.115 0.125
SE Mean 0.043 0.044
Difference = mu (Ex4-17T1) - mu (Ex4-17T2) Estimate for difference: 0.0066 95% CI for difference: (-0.1280, 0.1412) T-Test of difference = 0 (vs not =): T-Value = 0.11 Both use Pooled StDev = 0.1204
P-Value = 0.917
DF = 13
4-24 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.17. continued (b) What are the practical implications of the test in part (a)? Discuss what practical conclusions you would draw if the null hypothesis were rejected. The practical implication of this test is that it does not matter which technician measures parts; the readings will be the same. If the null hypothesis had been rejected, we would have been concerned that the technicians obtained different measurements, and an investigation should be undertaken to understand why. (c) Assuming that the variances are equal, construct a 95% confidence interval on the mean difference in surface-finish measurements. n1 = 7, x1 = 1.383, s1 = 0.115; n2 = 8, x2 = 1.376, s2 = 0.125, sP = 0.120
= 0.05, t/2, n1+n22 = t0.025, 13 = 2.160 (x1 x2 ) t /2,n
s
1 n2 2 p
1 n1 1 n2 (1 2 ) (x1 x2 ) t /2,n
s
1 n2 2 p
1 n1 1 n2
(1.383 1.376) 2.1604(0.120) 1 7 1 8 (1 2 ) (1.383 1.376) 2.1604(0.120) 1 7 1 8 0.127 (1 2 ) 0.141 (Equation 4.56) The confidence interval for the difference contains zero. We can conclude that there is not sufficient evidence of a difference in measurements obtained by the two technicians. (d) Test the hypothesis that the variances of the measurements made by the two technicians are equal. Use What are the practical implications if the null hypothesis is rejected? Test H0 : 12 22 versus H1 : 12 22 . Reject H0 if F0 F /2,n
1 1,n2 1
or F0 F1 /2,n
1 1, n2 1
F0 S12 S22 0.1152 0.1252 0.8464
F /2,n
1 1,n2 1
F1 /2,n
F0.05/2,71,8 1 F0.025,6,7 5.119
1 1,n2 1
F10.05/2,71,81 F0.975,6,7 0.176
.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-25 4.17.(d) continued MTB > Stat > Basic Statistics > 2 Variances Test and CI for Two Variances: Ex4-17T1, Ex4-17T2 Method Null hypothesis Alternative hypothesis Significance level Statistics Variable N Ex4-17T1 7 Ex4-17T2 8
StDev 0.115 0.125
Sigma(Ex4-17T1) / Sigma(Ex4-17T2) = 1 Sigma(Ex4-17T1) / Sigma(Ex4-17T2) not = 1 Alpha = 0.05
Variance 0.013 0.016
Ratio of standard deviations = 0.920 Ratio of variances = 0.846 95% Confidence Intervals CI for Distribution CI for StDev Variance of Data Ratio Ratio Normal (0.406, 2.195) (0.165, 4.816) Continuous (0.389, 3.092) (0.151, 9.562) Tests Method F Test (normal) Levene's Test (any continuous)
DF1 6 1
DF2 7 13
Test Statistic 0.85 0.01
P-Value 0.854 0.920
Do not reject H0, and conclude that there is not sufficient evidence of a difference in variability between measurements obtained by the two technicians. If the null hypothesis is rejected, we would have been concerned about the difference in measurement variability between the technicians, and an investigation should be undertaken to understand why. (e) Construct a 95% confidence interval estimate of the ratio of the variances of technician measurement error. F1 /2,n
2 1,n1 1
F0.975,7,6 0.1954; F /2,n
2 1, n1 1
S12 12 S12 F F /2,n 1,n 1 1 /2,n2 1,n1 1 2 1 S22 22 S22
12 0.1152 0.1152 (0.1954) (5.6955) 0.1252 22 0.1252 0.165
12 4.821 22
F0.025,7,6 5.6955
4-26 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.17. continued (f) Construct a 95% confidence interval on the variance of measurement error for technician 2. n2 8; x2 1.376; S2 0.125
0.05; 2 /2,n
2 1
(n 1)S 2
2 /2,n1
2 0.025,7 16.0128; 12 /2,n
2
2 1
2 0.975,7 1.6899
(n 1)S 2
12 /2,n1
(8 1)0.1252 (8 1)0.1252 2 16.0128 1.6899 2 0.007 0.065
(g) Does the normality assumption seem reasonable for the data? MTB > Graph > Probability Plot > Multiple and ensure Distribution = Normal Probability Plot of Ex4-17T1, Ex4-17T2
The normality assumption seems reasonable for both of these readings.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-27 4.18. Suppose that x1 ~ N(1, 12) and x2 ~ N(2, 22), and that x1 and x2 are independent. Develop a procedure for constructing a 100(1 − )% confidence interval on 1 − 2, assuming that 12 and 22 are unknown and cannot be assumed equal. From Eqn. 4.54 and 4.55, for 12 22 and both unknown, the test statistic is t * 0
x1 x2 S12 n1 S22 n2
S
2 2 with degrees of freedom 2 2 S n S n n1 S22 n2
2 1
2 1
1
n1 1
2 2
2
n2 1
A 100(1-)% confidence interval on the difference in means would be:
(x1 x2 ) t /2, S12 n1 S22 n2 (1 2 ) (x1 x2 ) t /2, S12 n1 S22 n2 4.19. Two different hardening processes—(1) saltwater quenching and (2) oil quenching—are used on samples of a particular type of metal alloy. The results are shown in Table 4E.2. Assume that hardness is normally distributed.
Saltwater quench: n1 = 10, x1 = 147.6, s1 = 4.97; Oil quench: n2 = 10, x2 = 149.4, s2 = 5.46
4-28 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 14.19 continued (a) Test the hypothesis that the mean hardness for the saltwater quenching process equals the mean hardness for the oil quenching process. Use = 0.05 and assume equal variances. Test H0: 1 2 = 0 vs. H1: 1 2 0. Reject H0 if |t0|>t/2, n1+n2-2. sP
(n1 1)s12 (n2 1)s22 (10 1)4.972 (10 1)5.462 5.2217 n1 n2 2 10 10 2
t0 x1 x2
s
P
(Eqns 4.51 and 4.52)
1 n1 1 n2 147.60 149.40 5.2217 1 10 1 10 0.77
t/2, n1+n2-2 = t0.025, 18 = 2.101 (from Appendix IV) Do not reject H0, and conclude that there is not sufficient evidence of a difference between measurements produced by the two hardening processes. MTB > Stat > Basic Statistics > 2-Sample t Two-Sample T-Test and CI: Ex4-19SQ, Ex4-19OQ Two-sample T for Ex4-19SQ vs Ex4-19OQ Ex4-19SQ Ex4-19OQ
N 10 10
Mean 147.60 149.40
StDev 4.97 5.46
SE Mean 1.6 1.7
Difference = mu (Ex4-19SQ) - mu (Ex4-19OQ) Estimate for difference: -1.80 95% CI for difference: (-6.71, 3.11) T-Test of difference = 0 (vs not =): T-Value = -0.77 Both use Pooled StDev = 5.2217
P-Value = 0.451
DF = 18
(b) Assuming that the variances 12 and 22 are equal, construct a 95% confidence interval on the difference in mean hardness.
= 0.05, t/2, n1+n22 = t0.025, 18 = 2.1009 (x1 x2 ) t /2,n
S
1 n2 2 p
1 n1 1 n2 (1 2 ) (x1 x2 ) t /2,n
S
1 n2 2 p
1 n1 1 n2
(147.6 149.4) 2.1009(5.22) 1 10 1 10 (1 2 ) (147.6 149.4) 2.1009(5.22) 1 10 1 10 6.7 (1 2 ) 3.1 (Eqn 4.56) The confidence interval for the difference contains zero. We conclude that there is not sufficient evidence of a difference between measurements produced by the two hardening processes.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-29 (c) Construct a 95% confidence interval on the ratio 12 / 22. Does the assumption made earlier of equal variances seem reasonable? F1 /2,n
2 1,n1 1
F0.975,9,9 0.2484; F /2,n
2 1,n1 1
F0.025,9,9 4.0260
S12 12 S12 F F /2,n 1,n 1 1 /2,n2 1,n1 1 2 1 S22 22 S22
12 4.972 4.972 (0.2484) (4.0260) 5.462 22 5.462 0.21
12 3.34 22
MTB > Stat > Basic Statistics > 2 Variances Test and CI for Two Variances: Ex4-19SQ, Ex4-19OQ Method Null hypothesis Alternative hypothesis Significance level Statistics Variable N Ex4-19SQ 10 Ex4-19OQ 10
StDev 4.971 5.461
Sigma(Ex4-19SQ) / Sigma(Ex4-19OQ) = 1 Sigma(Ex4-19SQ) / Sigma(Ex4-19OQ) not = 1 Alpha = 0.05
Variance 24.711 29.822
Ratio of standard deviations = 0.910 Ratio of variances = 0.829 95% Confidence Intervals Distribution of Data Normal Continuous
CI for Variance Ratio (0.206, 3.336) (0.152, 4.652)
CI for StDev Ratio (0.454, 1.826) (0.390, 2.157)
Tests Method F Test (normal) Levene's Test (any continuous)
DF1 9 1
DF2 9 18
Test Statistic 0.83 0.08
P-Value 0.784 0.783
Since the confidence interval includes the ratio of 1, the assumption of equal variances seems reasonable.
4-30 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY (d) Does the assumption of normality seem appropriate for these data? MTB > Graph > Probability Plot > Multiple and ensure Distribution = Normal Probability Plot of Ex4-19SQ, Ex4-19OQ
The normal distribution assumptions for both the saltwater and oil quench methods seem reasonable. However, the slopes on the normal probability plots do not appear to be the same, so the assumption of equal variances does not seem reasonable.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-31 4.20. A random sample of 200 printed circuit boards contains 15 defective or nonconforming units. Estimate the process fraction nonconforming. (a) Test the hypothesis that the true fraction nonconforming in this process is 0.10. Use = 0.05. Find the P-value. p0 = 0.10, = 0.05. Test H0: p = 0.10 versus H1: p 0.10. Reject H0 if |Z0| > Z/2. np0 = 200(0.10) = 20 Since (x = 15) < (np0 = 20), use the normal approximation to the binomial for x < np0. (x 0.5) np0 (15 0.5) 20 Z0 1.0607 (Eqn 4.43) np0 (1 p0 ) 20(1 0.10) Z/2 = Z0.05/2 = Z0.025 = 1.96 Do not reject H0, and conclude that the sample process fraction nonconforming does not differ from 0.10. P-value = 2[1 |Z0|] = 2[1 |1.0607|] = 2[1 0.8556] = 0.2888 MTB > Stat > Basic Statistics > 1 Proportion Under Option, select to Use test and interval based on normal distribution Test and CI for One Proportion Test of p = 0.1 vs p not = 0.1 Sample 1
X 15
N 200
Sample p 0.075000
95% CI (0.038496, 0.111504)
Z-Value -1.18
P-Value 0.239
Using the normal approximation.
Note that MINITAB does not use the continuity correction factor (+/- 0.5 in Eqn 4.43) in computing the test statistic. (b) Construct a 95% two-sided confidence interval on the true fraction nonconforming in the production process.
= 0.05, Z/2 = Z0.05/2 = Z0.025 = 1.96 pˆ Z /2 pˆ(1 pˆ) n p pˆ Z /2 pˆ(1 pˆ) n 0.075 1.96 0.075(1 0.075) 200 p 0.075 1.96 0.075(1 0.075) 200 0.038 p 0.112
4-32 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.21. A random sample of 500 connecting rod pins contains 65 nonconforming units. Estimate the process fraction nonconforming. (a) Test the hypothesis that the true fraction defective in this process is 0.08. Use = 0.05. Test H0: p = 0.08 versus H1: p 0.08. Reject H0 if |Z0|>Z/2. np0 = 500(0.08) = 40 Since (x = 65)>(np0 = 40), use the normal approximation to the binomial for x>np0. (x 0.5) np0 (65 0.5) 40 Z0 4.0387 np0 (1 p0 ) 40(1 0.08) Z/2 = Z0.05/2 = Z0.025 = 1.96 Reject H0, and conclude the sample process fraction nonconforming differs from 0.08. MTB > Stat > Basic Statistics > 1 Proportion Under Option, select to Use test and interval based on normal distribution Test and CI for One Proportion Test of p = 0.08 vs p not = 0.08 Sample 1
X 65
N 500
Sample p 0.130000
95% CI (0.100522, 0.159478)
Z-Value 4.12
P-Value 0.000
Using the normal approximation.
Note that MINITAB does not use the continuity correction factor (+/- 0.5 in Eqn 4.43) in computing the test statistic. (b) Find the P-value for this test. Find the corresponding cumulative standard normal, (Z0) for Z0 = 4.0387, and calculate: P-value = 2[1 |Z0|] = 2[1 |4.0387|] = 2[1 0.99997] = 0.00006 (c) Construct a 95% upper confidence interval on the true process fraction nonconforming. = 0.05, Z = Z0.05 = 1.645 p pˆ Z pˆ(1 pˆ) n p 0.13 1.645 0.13(1 0.13) 500 p 0.155
The 95% upper confidence interval for the process fraction nonconforming is 0.155.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-33 4.22. Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not conform to the strength specifications, whereas of 300 forgings selected from process 2, 20 are nonconforming. (a) Estimate the fraction nonconforming for each process. n1 = 200, x1 = 10, pˆ1 = x1/n1 = 10/200 = 0.05 n2 = 300, x2 = 20, pˆ2 = x2/n2 = 20/300 = 0.067 (b) Test the hypothesis that the two processes have identical fractions nonconforming. Use = 0.05. Test H0: p1 = p2 versus H1: p1 p2. Reject H0 if Z0 > Z/2 or Z0 < –Z/2 pˆ1 pˆ2 0.05 0.067 x x 10 20 0.7842 pˆ 1 2 0.06 Z0 n1 n2 200 300 pˆ(1 pˆ) 1 n1 1 n2 0.06(1 0.06) 1 200 1 300 Z/2 = Z0.05/2 = Z0.025 = 1.96; Z/2 = 1.96 Do not reject H0. Conclude there is no strong evidence to indicate a difference between the fraction nonconforming for the two processes. MTB > Stat > Basic Statistics > 2 Proportions Test and CI for Two Proportions Sample 1 2
X 10 20
N 200 300
Sample p 0.050000 0.066667
Difference = p (1) - p (2) Estimate for difference: -0.0166667 90% CI for difference: (-0.0513613, 0.0180279) Test for difference = 0 (vs not = 0): Z = -0.77
P-Value = 0.442
Fisher's exact test: P-Value = 0.565
(c) Construct a 90% confidence interval on the difference in fraction nonconforming between the two processes. (pˆ1 pˆ2 ) Z /2 (0.050 0.067) 1.645
pˆ1 (1 pˆ1 )
0.05(1 0.05) 200
n1
pˆ2 (1 pˆ2 ) n2
0.067(1 0.067) 300
(p1 p2 ) (pˆ1 pˆ2 ) Z /2
pˆ1 (1 pˆ1 ) n1
( p1 p2 ) (0.05 0.067) 1.645
0.052 (p1 p2 ) 0.018
pˆ2 (1 pˆ2 ) n2
0.05(1 0.05) 200
0.067(1 0.067) 300
4-34 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.23. A new purification unit is installed in a chemical process. Before its installation, a random sample yielded the following data about the percentage of impurity: x12 9.85, s12 6.79 and n1 10 . After installation, a random sample resulted in x22 8.08, s22 6.18 and n2 8 . (a) Can you conclude that the two variances are equal? Use =0.05. Test H0 : 12 22 versus H1 : 12 22 , at 0.05 Reject H0 if F0 F /2,n1 1,n2 2 or F0 F1 /2,n1 1,n2 1 F /2,n1 1,n2 2 F0.025,9,7 4.8232; F1 /2,n1 1,n2 1 F0.975,9,7 0.2383 F0 S12 S22 6.79 6.18 1.0987 F0 1.0987 4.8232 and 0.2383, so do not reject H0
s12 6.79 2.606;
s22 6.18 2.486
MTB > Stat > Basic Statistics > 2 Variances Test and CI for Two Variances Method Null hypothesis Alternative hypothesis Significance level Statistics Sample N 1 10 2 8
StDev 2.606 2.486
Sigma(1) / Sigma(2) = 1 Sigma(1) / Sigma(2) not = 1 Alpha = 0.05
Variance 6.790 6.180
Ratio of standard deviations = 1.048 Ratio of variances = 1.099 95% Confidence Intervals Distribution of Data Normal
CI for StDev Ratio (0.477, 2.147)
CI for Variance Ratio (0.228, 4.611)
Tests Method F Test (normal)
DF1 9
DF2 7
Test Statistic 1.10
P-Value 0.922
The impurity variances before and after installation are the same.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-35 4.23. continued (b) Can you conclude that the new purification device has reduced the mean percentage of impurity? Use = 0.05. Test H0: 1 = 2 versus H1: 1>2, = 0.05. Reject H0 if t0>t,n1+n22. t,n1+n22 = t0.05, 10+82 = 1.746 SP
t0
n1 1 S12 n2 1 S22 n1 n2 2
x1 x2 SP 1 n1 1 n2
10 1 6.79 8 1 6.18
9.85 8.08 2.554 1 10 1 8
10 8 2
2.554
1.461
The mean impurity after installation of the new purification unit is not less than before. MTB > Stat > Basic Statistics > 2-Sample t Two-Sample T-Test and CI Sample 1 2
N 10 8
Mean 9.85 8.08
StDev 2.61 2.49
SE Mean 0.82 0.88
Difference = mu (1) - mu (2) Estimate for difference: 1.77 95% lower bound for difference: -0.35 T-Test of difference = 0 (vs >): T-Value = 1.46 Both use Pooled StDev = 2.5542
P-Value = 0.082
DF = 16
4-36 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.24. Two different types of glass bottles are suitable for use by a soft-drink beverage bottler. The internal pressure strength of the bottle is an important quality characteristic. It is known that 1 = 2 = 2.5 psi. From a random sample of n1 = n2 = 15 bottles, the mean pressure strengths are observed to be x1 = 175.9 psi and x2 = 181.2 psi. The company will not use bottle design 2 unless its pressure strength exceeds that of bottle design 1 by at least 5 psi. Based on the sample data, should they use bottle design 2 if we use = 0.05. What is the P-value for this test? Want to demonstrate that 2 is greater than 1 by at least 5 psi, so H1: 1 + 5 < 2. So test a difference 0 = 5, test H0: 1 2 = 5 versus H1: 1 2 < 5. Reject H0 if Z0 < Z .
0 = 5 Z = Z0.05 = 1.645 Z0
x1 x2 0
n1 n2 2 1
2 2
(175.9 181.2) (5) 2.52 15 2.52 15
0.3286
(Z0 = 0.3286) > 1.645, so do not reject H0. The mean strength of Design 2 does not exceed Design 1 by 5 psi. P-value = (Z0) = (0.3286) = 0.3712 MTB > Stat > Basic Statistics > 2-Sample t Two-Sample T-Test and CI Sample 1 2
N 15 15
Mean 175.90 181.20
StDev 2.50 2.50
SE Mean 0.65 0.65
Difference = mu (1) - mu (2) Estimate for difference: -5.300 95% upper bound for difference: -3.747 T-Test of difference = -5 (vs t/2, n1 + n2 2. d 1
n
n j 1
xMicrometer, j xVernier, j 1
n d dj j 1 j 1 2 Sd n 1 n
2 j
t0 d Sd
0.150 0.151 12
0.151 0.152 0.000417
2
n 0.0013112
n 0.000417 0.001311
12 1.10
t/2, n1 + n2 2 = t0.005,22 = 2.8188 (|t0| = 1.10) < 2.8188, so do not reject H0. There is no strong evidence to indicate that the two calipers differ in their mean measurements. MTB > Stat > Basic Statistics > Paired t Paired T-Test and CI: Ex4-25MC, Ex4-25VC Paired T for Ex4-25MC - Ex4-25VC Ex4-25MC Ex4-25VC Difference
N 12 12 12
Mean 0.151167 0.151583 -0.000417
StDev 0.000835 0.001621 0.001311
SE Mean 0.000241 0.000468 0.000379
95% CI for mean difference: (-0.001250, 0.000417) T-Test of mean difference = 0 (vs not = 0): T-Value = -1.10
P-Value = 0.295
4-38 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.26. The cooling system in a nuclear submarine consists of an assembly pipe through which a coolant is circulated. Specifications require that weld strength must meet or exceed 150 psi. (a) Suppose the designers decide to test the hypothesis H0: = 150 versus H1: > 150. Explain why this choice of alternative is preferable to H1: < 150 The alternative hypothesis H1: > 150 is preferable to H1: < 150 we desire a true mean weld strength greater than 150 psi. In order to achieve this result, H0 must be rejected in favor of the alternative H1, > 150. (b) A random sample of 20 welds results in x = 153.7 psi and s = 11.5 psi. What conclusions can you draw about the hypothesis in part (a)? Use = 0.05. Test H0: = 150 versus H1: > 150. Reject H0 if t0 > t, n 1. t, n 1 = t0.05,19 = 1.7291.
t0 x S
n 153.7 150 11.5
20 1.4389
(t0 = 1.4389) < 1.7291, so do not reject H0. There is insufficient evidence to indicate that the mean strength is greater than 150 psi. MTB > Stat > Basic Statistics > 1-Sample t One-Sample T Test of mu = 150 vs > 150 N 20
Mean 153.70
StDev 11.50
SE Mean 2.57
95% Lower Bound 149.25
T 1.44
P 0.083
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-39 4.27. An experiment was conducted to investigate the filling capability of packaging equipment at a winery in Newberg, Oregon. Twenty bottles of Pinot Gris were randomly selected and the fill volume (in ml) measured. Assume that fill volume has a normal distribution. The data are as follows: 753, 751, 752, 753, 753, 753, 752, 753, 754, 754, 752, 751, 752, 750, 753, 755, 753, 756, 751, and 750. (a) Do the data support the claim that the standard deviation of fill volume is less than 1 ml? Use = 0.05. n = 20, x = 752.6 ml, s = 1.5, = 0.05 Test H0: 2 = 1 versus H1: 2 < 1. Reject H0 if 20 < 21-, n-1.
02 (n 1)S2 02 (20 1)1.52 1 42.75 21-, n-1 = 20.95,19 = 10.1170 (from Appendix III) (20 = 42.75)>10.1170, so do not reject H0. The standard deviation of the fill volume is not less than 1ml. MTB > Stat > Basic Statistics > 1 Variance Test and CI for One Variance: Ex4-27 Method Null hypothesis Alternative hypothesis
Sigma = 1 Sigma < 1
The chi-square method is only for the normal distribution. ... Statistics Variable Ex4-27
N 20
StDev 1.54
Variance 2.37
95% One-Sided Confidence Intervals Upper Bound for Upper Bound Variable Method StDev for Variance Ex4-27 Chi-Square 2.11 4.44 ... Tests Variable Ex4-27 ...
Method Chi-Square
Test Statistic 44.95
DF 19
P-Value 0.999
4-40 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.27. continued (b) Find a 95% two-sided confidence interval on the standard deviation of fill volume. 2/2, n-1 = 20.025,19 = 32.85; 21-/2, n-1 = 20.975,19 = 8.91 (from Appendix III) (n 1)S 2 2 /2,n1 2 (n 1)S 2 12 /2,n1 (20 1)1.52 32.85 2 (20 1)1.52 8.91 1.30 2 4.80 1.14 2.19
MTB > Stat > Basic Statistics > 1 Variance Test and CI for One Variance: Ex4-27 Method Null hypothesis Alternative hypothesis
Sigma = 1 Sigma not = 1
The chi-square method is only for the normal distribution. ... Statistics Variable N Ex4-27 20
StDev 1.54
Variance 2.37
95% Confidence Intervals Variable Ex4-27 ...
Method Chi-Square
CI for StDev (1.17, 2.25)
Method Chi-Square
Test Statistic 44.95
CI for Variance (1.37, 5.05)
Tests Variable Ex4-27 ...
DF 19
P-Value 0.001
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-41 4.27.(b) continued Also, in MINITAB MTB > Stat > Basic Statistics > Graphical Summary Summary for Ex4-27
The confidence interval does not include unity; therefore, we cannot conclude that the standard deviation of fill volume is less than 1 ml.
4-42 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.27. continued (c) Does it seem reasonable to assume that fill volume has a normal distribution? MTB > Graph > Probability Plot > Single and ensure Distribution = Normal Probability Plot of Ex4-27
The plotted points do not fall approximately along a straight line, so the assumption that battery life is normally distributed is not appropriate.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-43 4.28. Suppose we wish to test the hypotheses H0 : 25
H1 : 25 where we know that 2 = 16.0. If the true mean is really 30, what sample size must be used to ensure that the probability of type II error is no greater than 0.10? Assume that = 0.05. What n is needed such that the Type II error, , is less than or equal to 0.10?
1 2 30 25 5
d 5
16 1.25
From Figure 4.7, the operating characteristic curve for two-sided at = 0.05, n = 7. Check:
Z /2 n Z /2 n 1.96 5 7 4 1.96 5 7 4 (1.3472) (5.2672) 0.0890 0.0000 0.0890 MTB > Stat > Power and Sample Size > 1-Sample Z Power and Sample Size 1-Sample Z Test Testing mean = null (versus not = null) Calculating power for mean = null + difference Alpha = 0.05 Assumed standard deviation = 4 Difference 5
Sample Size 7
Target Power 0.9
Actual Power 0.911046
Power Curve for 1-Sample Z Test
4-44 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.29. Consider the hypotheses H0 : 0
H1 : 0 where 2 is known. Derive a general expression for determining the sample size for detecting a true mean of 1 0 with probability 1 − if the type I error is . Let 1 = 0 + .
From equation. 4.46, Z /2 n Z /2 n
If >0, then Z /2 n is likely to be small compared with . So,
Z /2 n
( ) 1 Z /2 n Z Z /2 n n (Z /2 Z )
2
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-45 4.30. Sample size allocation. Suppose we are testing the hypotheses H0 : 1 2
H1 : 1 2 where 12 and 22 are known. Resources are limited, and consequently the total sample size n1 + n2 = N. How should we allocate the N observations between the two populations to obtain the most powerful test? Maximize: Z0
x1 x2
Subject to: n1 n2 N .
n1 22 n2 2 1
Since (x1 x2 ) is fixed, an equivalent statement is Minimize: L
12 n1
22 n2
12 n1
22 N n1
22 dL 1 2 dL 12 1 n1 1 N n1 22 dn1 n1 N n1 dn1 1n12 12 (1)(1) N n1 22 0 2
12 n12
22
N n1
2
0
n1 1 n2 2
Allocate N between n1 and n2 according to the ratio of the standard deviations.
4-46 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.31. Note: There is a typographical error in the hypotheses (should be 22) in early printings of the 7th edition; this solution is for the correct hypotheses. Develop a test for the hypotheses H0 : 1 22
H1 : 1 22 where 12 and 22 are known. Given x ~ N, n1 , x1 , n2 , x2 , x1 independent of x2 . Assume 1 = 22 and let Q (x1 x2 ) . E (Q) E (x1 2 x2 ) 1 22 0 var(Q) var(x1 2 x2 ) var(x1 ) var(2 x2 ) var( x1 ) 22 var( x2 ) Z0
x1 2 x2 Q0 SD(Q) 12 n1 4 22 n2
And, reject H0 if Z 0 Z /2
var(x1 ) var(x2 ) 4 n1 n2
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-47 4.32. Nonconformities occur in glass bottles according to a Poisson distribution . A random sample of 100 bottles contains a total of 11 nonconformities. (a) Develop a procedure for testing the hypothesis that the mean of a Poisson distribution equals a specified value 0. Hint: Use the normal approximation to the Poisson. Wish to test H0: = 0 versus H1: 0. Select random sample of n observations x1, x2, …, xn. Each xi ~ POI().
n
x i 1
i
~ POI(n ) .
Using the normal approximation to the Poisson, if n is large, x = x/n = ~ N(, /n). Z0 x
0 / n . Reject H0: = 0 if |Z0| > Z/2
(b) Use the results of part (a) to test the hypothesis that the mean occurrence rate of nonconformities is = 0.15. Use = 0.01. x ~ Poi(), n = 100, x = 11, x = x/N = 11/100 = 0.110 Test H0: = 0.15 versus H1: 0.15, at = 0.01. Reject H0 if |Z0| > Z/2. Z/2 = Z0.005 = 2.5758 Z0 x 0
0 n 0.110 0.15
0.15 100 1.0328
(|Z0| = 1.0328) < 2.5758, so do not reject H0.
4-48 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.33. An inspector counts the surface-finish defects in dishwashers. A random sample of five dishwashers contains three such defects. Is there reason to conclude that the mean occurrence rate of surface-finish defects per dishwasher exceeds 0.5? Use the results of part (a) of Exercise 4.32 and assume that = 0.05. x ~ Poi(), n = 5, x = 3, x = x/N = 3/5 = 0.6 Test H0: = 0.5 versus H1: >0.5, at = 0.05. Reject H0 if Z0>Z. Z = Z0.05 = 1.645 Z0 x 0
0 n 0.6 0.5
0.5 5 0.3162
(Z0 = 0.3162) < 1.645, so do not reject H0. 4.34. An in-line tester is used to evaluate the electrical function of printed circuit boards. This machine counts the number of defects observed on each board. A random sample of 1,000 boards contains a total of 688 defects. Is it reasonable to conclude that the mean occurrence rate of defects is = 1. Use the results of part (a) of Exercise 4.26 and assume that = 0.05. x ~ Poi(), n = 1000, x = 688, x = x/N = 688/1000 = 0.688 Test H0: = 1 versus H1: 1, at = 0.05. Reject H0 if |Z0| > Z. Z/2 = Z0.025 = 1.96 Z0 x 0
0 n 0.688 1
(|Z0| = 9.8663) > 1.96, so reject H0.
1 1000 9.8663
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-49 4.35. An article in Solid State Technology (May 1987) describes an experiment to determine the effect of C2F6 flow rate on etch uniformity on a silicon wafer used in integrated-circuit manufacturing. Three flow rates are tested, and the resulting uniformity (in percent) is observed for six test units at each flow rate. The data are shown in Table 4E.4.
(a) Does C2F6 flow rate affect etch uniformity? Answer this question by using an analysis of variance with = 0.05. MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex4-35Obs versus Ex4-35Flow Source Ex4-35Flow Error Total S = 0.7132
Level 125 160 200
N 6 6 6
DF 2 15 17
SS 3.648 7.630 11.278
MS 1.824 0.509
R-Sq = 32.34%
Mean 3.3167 4.4167 3.9333
StDev 0.7600 0.5231 0.8214
F 3.59
P 0.053
R-Sq(adj) = 23.32%
Individual 95% CIs For Mean Based on Pooled StDev -----+---------+---------+---------+---(---------*----------) (----------*---------) (----------*---------) -----+---------+---------+---------+---3.00 3.60 4.20 4.80
Pooled StDev = 0.7132
(F0.05,2,15 = 3.6823)>(F0 = 3.59), so flow rate does not affect etch uniformity at a significance level = 0.05. However, the P-value = 0.053 is just slightly greater than 0.05, so there is some evidence that gas flow rate affects the etch uniformity.
4-50 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.35. continued (b) Construct a box plot of the etch uniformity data. Use this plot, together with the analysis of variance results, to determine which gas flow rate would be best in terms of etch uniformity (a small percentage is best).
Gas flow rate of 125 SCCM gives smallest mean percentage uniformity.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-51 4.35. continued (c) Plot the residuals versus predicted C2F6 flow. Interpret this plot.
Residuals are satisfactory, with no unusual patterns and approximately equal variance over the range of fitted values. (d) Does the normality assumption seem reasonable in this problem?
The normality assumption is reasonable.
4-52 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.36. Compare the mean etch uniformity values at each of the C2F6 flow rates from Exercise 4.33 with a scaled t distribution. Does this analysis indicate that there are differences in mean etch uniformity at the different flow rates? Which flows produce different results? Flow Rate 125 160 200
Mean Etch Uniformity 3.3% 4.4% 3.9%
scale factor MSE n 0.5087 6 0.3
Scaled t Distribution
(200)
(125)
3.0
3.3
3.6
3.9
(160)
4.2
4.5
Mean Etch Uniformity
4.8
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-53 4.36 continued Or, alternatively approximating with the normal distribution: MTB > Graph > Probability Distribution Plot > View Single Distribution Plot
The graphs do not indicate a large difference between the mean etch uniformity of the three different flow rates. The statistically significant difference between the mean uniformities can be seen by centering the t distribution between, say, 125 and 200, and noting that 160 would fall beyond the tail of the curve.
4-54 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.37. An article in the ACI Materials Journal (Vol. 84, 1987, pp. 213–216) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3-in.-diameter cylinder was used, and the number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in Table 4E.5.
(a) Is there any difference in compressive strength due to the rodding level? Answer this question by using the analysis of variance with = 0.05. MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex4-37Str versus Ex4-37Rod Source Ex4-37Rod Error Total S = 71.53
Level 10 15 20 25
N 3 3 3 3
DF 3 8 11
SS 28633 40933 69567
MS 9544 5117
R-Sq = 41.16%
Mean 1500.0 1586.7 1606.7 1500.0
StDev 52.0 77.7 107.9 10.0
F 1.87
P 0.214
R-Sq(adj) = 19.09%
Individual 95% CIs For Mean Based on Pooled StDev ----+---------+---------+---------+----(-----------*----------) (-----------*-----------) (-----------*-----------) (-----------*----------) ----+---------+---------+---------+----1440 1520 1600 1680
Pooled StDev = 71.5
ANOVA P-value = 0.214, so no difference due to rodding level at = 0.05.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-55 4.37. continued (b) Construct box plots of compressive strength by rodding level. Provide a practical interpretation of these plots. Boxplot of Ex4-37Str
Level 25 exhibits considerably less variability than the other three levels. (c) Construct a normal probability plot of the residuals from this experiment. Does the assumption of a normal distribution for compressive strength seem reasonable? Normplot of Residuals for Ex4-37Str
The normal distribution assumption for compressive strength is reasonable.
4-56 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.38. Compare the mean compressive strength at each rodding level from Exercise 4.37 with a scaled t distribution. What conclusions would you draw from this plot? Rodding Level 10 15 20 25
Mean Compressive Strength 1500 1587 1607 1500
scale factor MSE n 5117 3 41
S c a le d t D is tr ib u t io n
(10 , 25 )
(1 5 )
(2 0 )
1418 1459 1500 1541 1582 1623 M e a n C o m p re s s iv e S tr e n g t h There is no difference due to rodding level.
1664
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-57 4.39. An aluminum producer manufactures carbon anodes and bakes them in a ring furnace prior to use in the smelting operation. The baked density of the anode is an important quality characteristic, as it may affect anode life. One of the process engineers suspects that firing temperature in the ring furnace affects baked anode density. An experiment was run at four different temperature levels, and six anodes were baked at each temperature level. The data from the experiment are shown in Table 4E.6.
(a) Does firing temperature in the ring furnace affect mean baked anode density? MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex4-39Den versus Ex4-39T Source DF Ex4-39T 3 Error 20 Total 23 S = 0.3238
SS MS F P 0.457 0.152 1.45 0.258 2.097 0.105 2.553 R-Sq = 17.89% R-Sq(adj) = 5.57% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev --------+---------+---------+---------+500 6 41.700 0.141 (----------*----------) 525 6 41.583 0.194 (----------*----------) 550 6 41.450 0.339 (----------*----------) 575 6 41.333 0.497 (----------*----------) --------+---------+---------+---------+41.25 41.50 41.75 42.00 Pooled StDev = 0.324
ANOVA P-value = 0.258. At = 0.05, temperature level does not significantly affect mean baked anode density.
4-58 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.39. continued (b) Find the residuals for this experiment and plot them on a normal probability scale. Comment on the plot.
Normality assumption is reasonable. (c) What firing temperature would you recommend using?
Since statistically there is no evidence to indicate that the means are different, select the temperature with the smallest variance, 500C (see Boxplot), which probably also incurs the least cost (lowest temperature).
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-59 4.40. Plot the residuals from Exercise 4.36 against the firing temperatures. Is there any indication that variability in baked anode density depends on the firing temperature? What firing temperature would you recommend using?
As firing temperature increases, so does variability. More uniform anodes are produced at lower temperatures. Recommend 500C for smallest variability.
4-60 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.41. An article in Environmental International (Vol. 18, No. 4, 1992) describes an experiment in which the amount of radon released in showers was investigated. Radon-enriched water was used in the experiment, and six different orifice diameters were tested in showerheads. The data from the experiment are shown in Table 4E.7.
(a) Does the size of the orifice affect the mean percentage of radon released? Use the analysis of variance and = 0.05. MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex4-41Rad versus Ex4-41Dia Source Ex4-41Dia Error Total S = 2.711
Level 0.37 0.51 0.71 1.02 1.40 1.99
N 4 4 4 4 4 4
DF 5 18 23
SS 1133.38 132.25 1265.63
MS 226.68 7.35
R-Sq = 89.55%
Mean 82.750 77.000 75.000 71.750 65.000 62.750
StDev 2.062 2.309 1.826 3.304 3.559 2.754
F 30.85
P 0.000
R-Sq(adj) = 86.65%
Individual 95% CIs For Mean Based on Pooled StDev ----+---------+---------+---------+----(---*---) (---*---) (---*---) (----*---) (---*---) (---*---) ----+---------+---------+---------+----63.0 70.0 77.0 84.0
Pooled StDev = 2.711
Orifice size does affect mean % radon release, at = 0.05.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-61 4.41.(a) continued
Smallest % radon released at 1.99 and 1.4 orifice diameters.
4-62 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.41. continued (b) Analyze the residuals from this experiment.
Residuals violate the assumption of normally distributed errors.
Variability in residuals does not appear to depend on the magnitude of predicted (or fitted) values. The assumption of equal variance at each factor level appears to be violated, with larger variances at the larger diameters (1.02, 1.40, 1.99).
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-63 4.42. An article in the Journal of the Electrochemical Society (Vol. 139, No. 2, 1992, pp. 524–532) describes an experiment to investigate the low-pressure vapor deposition of polysilicon. The experiment was carried out in a large-capacity reactor at SEMATECH in Austin, Texas. The reactor has several wafer positions, and four of these positions are selected at random. The response variable is film thickness uniformity. Three replicates of the experiment were run, and the data are shown in Table 4E.8.
(a) Is there a difference in the wafer positions? Use the analysis of variance and = 0.05. MTB > Stat > ANOVA > One-Way One-way ANOVA: Ex4-42Un versus Ex4-42Pos Source Ex4-42Pos Error Total S = 0.8076
DF 3 8 11
Level 1 2 3 4
N 3 3 3 3
SS 16.220 5.217 21.437
MS 5.407 0.652
R-Sq = 75.66%
Mean 4.3067 1.7733 1.9267 1.3167
StDev 1.4636 0.3853 0.4366 0.3570
Pooled StDev = 0.8076
F 8.29
P 0.008
R-Sq(adj) = 66.53%
Individual 95% CIs For Mean Based on Pooled StDev --------+---------+---------+---------+(------*------) (------*------) (------*------) (------*------) --------+---------+---------+---------+1.5 3.0 4.5 6.0
4-64 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.42.(a) continued
There is a statistically significant difference in wafer position, 1 is different from 2, 3, and 4. (b) Estimate the variability due to wafer positions.
ˆ2
MSfactor MSE 5.4066 0.6522 0.3962 n 12
(c) Estimate the random error component.
ˆ 2 MSE 0.6522 2 ˆuniformity ˆ2 ˆ 2 0.3962 0.6522 1.0484
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-65 4.42. continued (d) Analyze the residuals from this experiment and comment on model adequacy.
Normality assumption is probably not unreasonable, but there are two very unusual observations – the outliers at either end of the plot – therefore model adequacy is questionable. Both outlier residuals are
4-66 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY from wafer position 1. The variability in residuals does appear to depend on the magnitude of predicted (or fitted) values.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-67 4.43. The tensile strength of a paper product is related to the amount of hardwood in the pulp. Ten samples are produced in the pilot plant, and the data obtained are shown in Table 4E.9.
(a) Fit a linear regression model relating strength to percentage hardwood. MTB > Stat > Regression > Regression Regression Analysis: Ex4-43Str versus Ex4-43Per The regression equation is Ex4-43Str = 144 + 1.88 Ex4-43Per Predictor Constant Ex4-43Per
Coef 143.824 1.8786
S = 2.20320
SE Coef 2.522 0.1165
R-Sq = 97.0%
T 57.04 16.12
P 0.000 0.000
R-Sq(adj) = 96.6%
Analysis of Variance Source DF SS Regression 1 1262.1 Residual Error 8 38.8 Total 9 1300.9
MS 1262.1 4.9
F 260.00
P 0.000
(b) Test the model in part (a) for significance of regression. H0 : 1 0 vs H1 : 1 0 (Equation 4.97) MSR 1262.1 F0 260.00; F0.05,1,8 5.32 260.00 MSE 4.9 Therefore, the regression is significant. (c) Find a 95% confidence interval on the parameter 1. ˆ t se ˆ ˆ t se ˆ 1
/2,n 2
1
1
1
/2,n 2
1
1.8786 2.306 0.1165 1 1.8786 2.306 0.1165 1.610 1 2.147
(Equation 4.113)
4-68 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.44. A plant distills liquid air to produce oxygen, nitrogen, and argon. The percentage of impurity in the oxygen is thought to be linearly related to the amount of impurities in the air as measured by the “pollution count” in parts per million (ppm). A sample of plant operating data is shown below:
(a) Fit a linear regression model to the data. MTB > Stat > Regression > Regression Regression Analysis: Ex4-44Pur versus Ex4-44Poll The regression equation is Ex4-44Pur = 96.9 - 3.10 Ex4-44Poll Predictor Constant Ex4-44Poll
Coef 96.8574 -3.0955
S = 0.457540
SE Coef 0.4308 0.3131
R-Sq = 87.5%
T 224.81 -9.89
P 0.000 0.000
R-Sq(adj) = 86.6%
Analysis of Variance Source Regression Residual Error Total
DF 1 14 15
SS 20.469 2.931 23.400
MS 20.469 0.209
F 97.78
P 0.000
The regression model relating purity (y) to pollution count ( x1 ) is given by yˆ 96.9 3.1x1 . (b) Test for significance of regression. H0 : 1 0 versus H1 : 1 0 Since the P-value for the F statistic is very small (approximately 0), we reject the null hypothesis and conclude that the regression model is significant. (c) Find a 95% confidence interval on 1. ˆ1 t.025,8 se ˆ1 1 ˆ1 t.025,8 se ˆ1
3.19 2.16(0.31) 1 3.19 2.16(0.31) 3.86 1 2.52
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-69 4.45. Plot the residuals from Problem 4.43 and comment on model adequacy.
The assumption of equal variance at each factor level appears to be satisfied. The normality assumption appears to be satisfied. Variability in residuals does not appear to depend on the magnitude of predicted (or fitted) values.
4-70 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.46. Plot the residuals from Problem 4.44 and comment on model adequacy.
Based on the residual plots the normality and constant variance assumptions seems to be satisfied.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-71 4.47. The brake horsepower developed by an automobile engine on a dynamometer is thought to be a function of the engine speed in revolutions per minute (rpm), the road octane number of the fuel, and the engine compression. An experiment is run in the laboratory and the data are drawn in Table 4E.10:
(a) Fit a multiple regression model to these data. MTB > Stat > Regression > Regression Regression Analysis: Ex4-47HP versus Ex4-47RPM, Ex4-47Oct, Ex4-47Com The regression equation is Ex4-47HP = - 266 + 0.0107 Ex4-47RPM + 3.13 Ex4-47Oct + 1.87 Ex4-47Com Predictor Constant Ex4-47RPM Ex4-47Oct Ex4-47Com
Coef -266.03 0.010713 3.1348 1.8674
S = 8.81239
SE Coef 92.67 0.004483 0.8444 0.5345
R-Sq = 80.7%
Analysis of Variance Source DF SS Regression 3 2589.73 Residual Error 8 621.27 Total 11 3211.00 Source Ex4-47RPM Ex4-47Oct Ex4-47Com
DF 1 1 1
Seq SS 509.35 1132.56 947.83
T -2.87 2.39 3.71 3.49
P 0.021 0.044 0.006 0.008
R-Sq(adj) = 73.4% MS 863.24 77.66
F 11.12
P 0.003
4-72 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.47. continued (b) Test for significance of regression. What conclusions can you draw? H0 : 1 2 3 0 vs H1 : at least one i 0 MSR 863.24 F0 11.12 (Equation 4.97) MSE 77.66 F0.05,3,8 4.07 11.12
Since the P-value for the F statistic is very small (approximately 0), we reject the null hypothesis and conclude that at least one of the three variables – rpm ( ), road octane number of the fuel ( ) and the engine compression ( ) – has a nonzero regression coefficient. (c) Based on t-tests, do you need all three regressor variables in the model? The P-values corresponding to the variables rpm, road octane number and compression are 0.044, 0.006 and 0.008, respectively. Since these are all less than a significance level of 0.05, each of the three variables contributes significantly to the regression model.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-73 4.48. Analyze the residuals from the regression model in Exercise 4.47. Comment on model adequacy.
4-74 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY
Based on the residual plots, there do not seem to be any violations of the necessary assumptions for multiple linear regression.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-75 4.49. Table 4E.11 contains the data from a patient satisfaction survey for a group of 25 randomly selected patients at a hospital. In addition to satisfaction, data were collected on patient age and an index that measured the severity of illness.
4-76 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.49. continued (a) Fit a linear regression model relating satisfaction to patient age. MTB > Stat > Regression > Regression Regression Analysis: Ex4-49Sat versus Ex4-49Age The regression equation is Ex4-49Sat = 131 - 1.29 Ex4-49Age Predictor Constant Ex4-49Age S = 9.37523
Coef 131.096 -1.2898
SE Coef 6.832 0.1292
R-Sq = 81.2%
Analysis of Variance Source DF SS Regression 1 8756.7 Residual Error 23 2021.6 Total 24 10778.2 Unusual Observations Obs Ex4-49Age Ex4-49Sat 9 27.0 75.00 21 70.0 59.00
T 19.19 -9.98
P 0.000 0.000
R-Sq(adj) = 80.4% MS 8756.7 87.9
Fit 96.27 40.81
F 99.63
SE Fit 3.61 3.11
P 0.000
Residual -21.27 18.19
St Resid -2.46R 2.06R
R denotes an observation with a large standardized residual.
The regression model relating satisfaction (y) to patient age ( x1 ) is given by yˆ 131 1.29*Age (b) Test for significance of regression. H0: β1 = 0 versus H1: β1 ≠ 0 Since the P-value for the F statistic is very small (approximately 0), we reject the null hypothesis at a significance level of = 0.05 and conclude that the regression model is significant. (c) What portion of the total variability is accounted for by the regressor variable age? 81.2% (R2) of the total variability is accounted for by Age.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-77 4.50. Analyze the residuals from the regression model on the patient satisfaction data from Exercise 4.49. Comment on the adequacy of the regression model. MTB > Stat > Regression > Regression
Plot points on a normal probability plot of the residuals tend to fall along a straight line, indicating that the assumption of normally distributed model errors is valid. Plots of the residuals versus predicted Satisfaction and Age are relatively uniform over the range of values and do not indicate any issues with the assumptions of constant variance. These plots do not reveal any model inadequacy or unusual problem with the assumptions.
4-78 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.51. Reconsider the patient satisfaction data in Table 4E.11. Fit a multiple regression model using both patient age and severity as the regressors. (a) Test for significance of regression. MTB > Stat > Regression > Regression Regression Analysis: Ex4-49Sat versus Ex4-49Age, Ex4-49Sev The regression equation is Ex4-49Sat = 143 - 1.03 Ex4-49Age - 0.556 Ex4-49Sev Predictor Constant Ex4-49Age Ex4-49Sev
Coef 143.472 -1.0311 -0.5560
S = 7.11767
SE Coef 5.955 0.1156 0.1314
R-Sq = 89.7%
Analysis of Variance Source DF SS Regression 2 9663.7 Residual Error 22 1114.5 Total 24 10778.2 Source Ex4-49Age Ex4-49Sev
DF 1 1
T 24.09 -8.92 -4.23
P 0.000 0.000 0.000
R-Sq(adj) = 88.7% MS 4831.8 50.7
F 95.38
P 0.000
Seq SS 8756.7 907.0
Unusual Observations Obs Ex4-49Age Ex4-49Sat 9 27.0 75.00
Fit 92.28
SE Fit 2.90
Residual -17.28
St Resid -2.66R
R denotes an observation with a large standardized residual.
The regression model relating satisfaction (y) to patient age and severity of illness is given by yˆ 143 1.03 Age 0.556 Severity Test H0: β1 = β2 = 0 versus H1: at least one βi ≠ 0. Since the P-value for the F statistic is very small (approximately 0), we reject the null hypothesis at a significance level of = 0.05 and conclude that at least one of the two variables – has a nonzero regression coefficient.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-79 4.51. continued (b) Test for the individual contribution of the two regressors. Are both regressor variables needed in the model? Test H0: βAge = 0 versus H1: βAge ≠ 0. Since the P-value for the t statistic is very small (0.000), reject the null hypothesis at a significance level of = 0.05 and conclude that Age contributes significantly to the model. Test H0: βSeverity = 0 versus H1: βSeverity ≠ 0. Since the P-value for the t statistic is very small (0.000), reject the null hypothesis at a significance level of = 0.05 and conclude that Severity contributes significantly to the model. (c) Has adding severity to the model improved the quality of the model fit? Explain your answer. Adding Severity to the model has improved measures of model quality, including R2 (increasing R2, adjusted R2, and R2-prediction) and decreasing MS-Error (from 87.9 to 50.7).
4-80 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.52. Analyze the residuals from the multiple regression model on the patient satisfaction data from Exercise 4.51. Comment on the adequacy of the regression model. MTB > Stat > Regression > Regression
Plot points on a normal probability plot of the residuals tend to fall along a straight line, indicating that the assumption of normally distributed model errors is valid. Plots of the residuals versus predicted Satisfaction, Age and Severity are relatively uniform over the range of values and do not indicate any issues with the assumptions of constant variance. These plots do not reveal any model inadequacy or unusual problem with the assumptions.
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-81 4.53. Consider the Minitab output below.
(a) Fill in the missing values. What conclusions would you draw?
31.400 30 4.1667 0.336 P 2Pr Z 4.1667 0.00003 Z
The small P-value would lead to concluding that the mean is not 30. (b) Is this a one-sided or two-sided test? This is a two-sided test (given in the first line of the output). This is why the P-value above was double the probability instead of just simply the probability. (c) Use the output and a normal table to find a 95% CI on the mean. x Z /2 31.4 1.96
n 1.3
x Z /2
n
31.4 1.96
15 30.7421 32.0579
1.3 15
(Equation 4.29)
This answer agrees with the given CI. (d) How was the SE mean calculated? SE mean=
1.3 15
0.336 .
(e) What is the P-value if the alternative hypothesis is H1: > 30? P-value=0.00003/2=0.000015
4-82 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.54. Suppose that you are testing H0 : 1 2 versus H1 : 1 2 with a n1 = n2 = 15. Use the table of the t distribution percentage points to find lower and upper bounds on the P-value for the following observed values of the test statistic: Degrees of freedom, = n1 + n2 – 2 = 15 + 15 – 2 = 28 (a) t0 = 2.19 0.01 < P-value < 0.025 (b) t0 = 3.50 0.0005 < P-value < 0.001 (c) t0 = 2.88 0.0025 < P-value < 0.005 (d) t0 = 1.42 0.05 < P-value < 0.10
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-83 4.55. Suppose that you are testing H0 : 1 2 versus H1 : 1 2 with a n1 = n2 = 10. Use the table of the t distribution percentage points of find lower and upper bounds on the P-value of the following observed values of the test statistic: Degrees of freedom, = n1 + n2 – 2 = 10 + 10 – 2 = 18 (a) t0 2.48 t (18) Prob 2.101 0.025 2.48 p-val 2.552 0.01 Since this is a two-sided test, we double the probabilities. 0.02 < P-value < 0.05. (b) t0 2.41
Notice that we can find the probabilities associated with the absolute value of the test
statistic, since the t distribution is symmetric about 0. t (18) Prob 2.101 0.025 2.48 p-val 2.552 0.01 Since this is a two-sided test, we double the probabilities. 0.02 < P-value < 0.05. (c) t0 2.98 t (18) Prob 2.878 0.005 2.98 p-val 3.197 0.0025 Since this is a two-sided test, we double the probabilities. 0.005 < P-value < 0.01. (d) t0 1.89 t (18) Prob 1.734 0.05 1.89 p-val 2.101 0.025 Since this is a two-sided test, we double the probabilities. 0.05 < P-value < 0.10.
4-84 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.56. Consider the Minitab output below.
(a) Fill in the missing values. Can the null hypothesis be rejected at the 0.05 level? Why?
94.58 95 0.626 . Since the P-value is greater than 0.05, the null hypothesis cannot 0.671 be rejected at the 0.05 level. StDev=3.00; T
(b) Is this a one-sided or two-sided test? Since the hypotheses are mu=95 vs not =95, this is a two-sided test. (c) How many degrees of freedom are there on the t-test statistic? Since N=20, there are N-1=19 df. (d) Use the output and a normal table to find a 95% CI on the mean. A 95% CI on the mean is given by (93.176, 95.984). (e) Suppose that the hypotheses had been H0 : 90 versus H1 : 90 . What conclusions would you have drawn?
94.58 90 6.83 with a corresponding P-value of approximately 0. Thus for this set of hypotheses 0.671 we would reject the null hypothesis and conclude that the mean is greater than 90. T
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-85 4.57. Consider the Minitab output below.
(a) Is this a one-sided or two-sided test? This is a two-sided test (given in the first line of the output). (b) Can the null hypothesis be rejected at the 0.05 level? Since the given P-value of 0.313 is larger than 0.05, we cannot reject the null hypothesis at this level. (c) Construct an approximate 90% CI for p.
pˆ Z /2 0.3267 1.645
pˆ 1 pˆ n
0.3267 1 0.3267 300
p pˆ Z /2
pˆ 1 pˆ n
p 0.3267 1.645
0.3267 1 0.3267 300
(Equation 4.44)
0.2822 p 0.3712
(d) What is the P-value if the alternative hypothesis is H1: p > 0.3? The P-value is half the P-value for the two sided alternative, so P = 0.131/2 = 0.156.
4-86 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.58. Consider the Minitab output below.
(a) Fill in the missing values. Estimate for difference=50.2-51.98= -1.78; DF=N1+N2-2=28. (b) Can the null hypothesis be rejected at the 0.05 level? Why? Since the P-value=0.019 is less than 0.05, the null hypothesis can be rejected at the 0.05 level. (c) Use the output and the t-table to find a 99% CI on the difference in means. A 99% CI on the difference in means is given by x1 x2 t0.005,28 sp 50.2 51.98 2.763(1.9602)
1 1 1 1 1 2 x1 x2 t0.005,28 sp n1 n2 n1 n2 1 1 1 1 1 2 50.2 51.98 2.763(1.9602) 15 15 15 15 3.76 1 2 0.198
(d) Suppose that the alternative hypothesis was H1: 1 = 2 versus H1: 1 > 2. What is the P-value? What conclusions would you draw?
50.2 51.98
2.49 , 1 1 1.9602 15 15 P-value = P(T>-2.49 with 28 df) = 0.99. Since the P-value is larger than 0.05, we fail to reject the null hypothesis. There is insufficient evidence to show that 1 2 . T-Value =
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-87 4.59. Consider the Minitab output below.
(a) Fill in the missing values. N2=301/0.602=500; Estimate for difference=185/300-301/500=0.0146667; Z: x x 185 301 pˆ 1 2 0.6075 n1 n2 300 500 Z
pˆ1 pˆ2 1 1 pˆ 1 pˆ n1 n2
0.616667 0.602 1 1 0.6075(1 0.6075) 300 500
0.411 .
(b) Is this a one-sided or two-sided test? Since the hypotheses are difference = 0 (vs not =0), this is a two-sided test. (c) What is the P-value if the alternative hypothesis is H1: p1 = p2 versus H0: p1 > p2? P-value = 0.68/2=0.34. (d) Construct an approximate 90% CI for the difference in the two proportions. A 90% CI for the difference in the two proportions is given by pˆ1 pˆ2 Z /2
pˆ1 1 pˆ1 n1
0.616667 0.602 1.645
pˆ2 1 pˆ2 n2
p1 p2 pˆ1 pˆ2 Z /2
pˆ1 1 pˆ1 n1
pˆ2 1 pˆ2 n2
0.616667(1 0.616667) 0.602(1 0.602) p1 p2 300 500
0.616667 0.602 1.645
0.616667(1 0.616667) 0.602(1 0.602) 300 500
0.0439 p1 p2 0.0732
4-88 CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4.60. Consider a one-way or single-factor ANOVA with four treatments and five replications. Use the table of the F distribution percentage points to find lower and upper bounds on the P-value for the following observed values of the test statistic. ν1 = a − 1 = 4 – 1 = 3, ν2 = a (n – 1) = 4 (5 – 1) = 16 F, 3, 16
0.25 1.51
0.10 2.46
(a) F0 = 7.50 P-value < 0.01 (b) F0 = 4.25 0.01 < P-value < 0.025 (c) F0 = 6.28 P-value < 0.01 (d) F0 = 1.70 0.10 < P-value < 0.25
0.05 3.24
0.025 4.08
0.01 5.29
CHAPTER 4 INFERENCES ABOUT PROCESS QUALITY 4-89 4.61. Consider the Minitab ANOVA output below. Fill in the blanks. You may give bounds on the P-value. What conclusions can you draw based on the information in this display?
MSFactor = SSFactor/3 = 18.30 DF Error = DF Total – DF Factor = 12 MSError = SSError/12 = 1.6475 F = MSFactor/MSError = 11.11 P-value < 0.01 Since the P-value is smaller than 0.01, there is strong evidence against the null hypothesis. We conclude that the means are not the same for all factor levels, and that changing the level of the factor will significantly affect the response.