Archives
978-0073401331 Chapter 1 Part 1
Solutions Manual to accompany STATISTICS FOR ENGINEERS AND SCIENTISTS, 4th ed. Prepared by William Navidi PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. […]
978-0073401331 Chapter 1 Part 2
11. (a) 40 50 60 70 80 90 100 Number Absent (b) Yes. The value 100 is an outlier. 12. The mean. The median, and the first and third quartiles are indicated directly on a boxplot, and the interquartile range […]
978-0073401331 Chapter 10
472 CHAPTER 10 Chapter 10 Section 10.1 1. (a) Count 2. (a) True (b) False. If no special causes are operating, then the variability is constant. It is still possible for much of the output to fail to conform to […]
978-0073401331 Chapter 2 Part 1
28 CHAPTER 2 (f) The set A∪Bcontains the outcomes that are either in A, in B, or in both. Therefore A∪B={CCC, DDD, SSS, CDS, CSD, DCS, DSC, SCD, SDC}. 5. (a) The outcomes are the sequences of candidates that end […]
978-0073401331 Chapter 2 Part 2
48 CHAPTER 2 (e) The small probability in part (a) indicates that some good bottles will be scrapped. This is not so serious. The important thing is that of the bottles that pass inspection, very few should have flaws. The […]
978-0073401331 Chapter 2 Part 3
68 CHAPTER 2 8. Let X1, …, X24 be the volumes of the 24 bottles. Let S=X1+···+X24 be the total weight. The average volume per bottle is X. (a) µS=PµXi= 24(20.01) = 480.24 (b) σS=qPσ2 Xi=p24(0.022) = 0.098 (c) µX=µXi= […]
978-0073401331 Chapter 2 Part 4
88 CHAPTER 2 (b) µM1M2=µR2+E1R+E2R+E1E2=µR2+µE1µR+µE2µR+µE1µE2=µR2 26. Cov(X, X) = µX·X−µXµX=µX2−(µX)2=σ2 X. 27. (a) Cov(aX, bY ) = µaX·bY −µaX µbY =µabXY −aµXbµY=abµXY −abµXµY =ab(µXY −µXµY) = abCov(X, Y ). (b) ρaX,bY = Cov(aX, bY )/(σaX σbY ) = abCov(X, Y […]
978-0073401331 Chapter 2 Part 5
SUPPLEMENTARY EXERCISES FOR CHAPTER 2 99 20. (a) µX= 0.02pX(0.02) + 0.04pX(0.04) + 0.06pX(0.06) + 0.08pX(0.08) = 0.02(0.22) + 0.04(0.19) + 0.06(0.29) + 0.08(0.30) = 0.0534 (b) µY= 100pY(100) + 150pY(150) + 200pY(200) = 100(0.14) + 150(0.36) + 200(0.50) = […]
978-0073401331 Chapter 3 Part 1
110 CHAPTER 3 7. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is 21.3 µg. (b) No, the bias cannot be estimated, since we do not know the true value. 8. (a) […]
978-0073401331 Chapter 3 Part 2
SECTION 3.4 129 k= ln(T0−Ta)/t −ln(T−Ta)/t = 0.1 ln(T0−Ta)−0.1 ln(T−Ta) = 0.0626 18. (a) a= 2.5, σa= 0.1, b= 0.05, σb= 0.01, w= 1.2, σw= 0.1, v=a+bw = 2.56 ∂v ∂a = 1 ∂v ∂b =w= 1.2 ∂v ∂w =b= […]
978-0073401331 Chapter 4 Part 1
148 CHAPTER 4 4. (a) If Xand Ycannot both be equal to 1, the possible values for Zare 0 and 1, so Zis a Bernoulli random variable. (b) pZ=P(Z= 1) = P(X= 1 or Y= 1) = P(X= 1) + […]
978-0073401331 Chapter 4 Part 2
168 CHAPTER 4 The uncertainty is σb +σ2 =√6.25 + 3.5 = 3.1. λ1−λ2= 5.5±3.1 18. If the mean decay rate is exactly 1 per second, then the mean number of events in 10 seconds is 10, so X∼Poisson(10). (a) […]
978-0073401331 Chapter 4 Part 3
188 CHAPTER 4 5. (a) ln I∼N(1,0.2), ln R∼N(4,0.1), and Iand Rare independent. Therefore ln V∼N(5,0.3). 6. The probability that a given circuit has a voltage less than 200 volts is P(V < 200) = 0.7054. Let Xbe the number […]
978-0073401331 Chapter 4 Part 4
6. 0 0.5 1 1.5 2 2.5 3 0.001 0.01 0.05 0.1 0.25 0.5 0.75 0.9 0.95 0.99 0.999 The logs of the PM data do appear to come from an approximately normal dis- tribution. 7. Yes. If the logs […]
978-0073401331 Chapter 4 Part 5
12. (a,b) Answers will vary. (c) µmedian = 0 exactly (simulation results will be approximate), σmedian ≈0.536. (d) Answers will vary. (e) µX= 0 exactly, σX= 1/√5 exactly (simulation results will be approximate). (f) Bias is exactly 0 in both […]
978-0073401331 Chapter 5 Part 1
236 CHAPTER 5 5. (a) X= 50, s= 2, n= 100, z.025 = 1.96. The confidence interval is 50 ±1.96(2/√100), or (49.608,50.392). (b) X= 50, s= 2, n= 100, z.005 = 2.58. The confidence interval is 50 ±2.58(2/√100), or (49.484,50.516). […]
978-0073401331 Chapter 5 Part 2
or (0.0379, 0.266). 11. No, these are not simple random samples. Section 5.6 1. X= 1.5, sX= 0.25, nX= 7, Y= 1.0, sY= 0.15, nY= 5. The number of degrees of freedom is ν=0.252 (0.252/7)2 7−1+(0.152/5)2 5−1 = 9, rounded […]
978-0073401331 Chapter 6 Part 1
274 CHAPTER 6 is the area to the left of z=−2.98. 6. (a) X= 15.2, s= 1.8, n= 87. The null and alternate hypotheses are H0:µ= 15 versus H1:µ6= 15. z= (15.2−15)/(1.8/√87) = 1.04. Since the alternate hypothesis is of […]
978-0073401331 Chapter 6 Part 2
294 CHAPTER 6 Section 6.7 1. (a) X= 3.05, sX= 0.34157, nX= 4, Y= 1.8, sY= 0.90921, nY= 4. The number of degrees of freedom is ν=0.341572 (0.341572/4)2 4−1+(0.909212/4)2 4−1 = 3, rounded down to the nearest integer. t3= (3.05 […]
978-0073401331 Chapter 6 Part 3
8. The ranks of the combined samples are Value Rank Sample 1255 1.5X 1255 1.5X 1270 3 X 1280 4 X 1287 5 X 1291 6 X 1296 7 X 1301 8 Y 1302 9 X 1306 10 X 1310 […]
978-0073401331 Chapter 6 Part 4
SECTION 6.14 331 10. (a) One-tailed. The alternate hypothesis is of the form µ > µ0. (b) 4. The alternate mean is equal to the null plus the difference, and the difference is 1, so, if the null mean is […]
978-0073401331 Chapter 7 Part 1
5. The heights and weights for the men (dots) are on the whole greater than those for the women (xs). Therefore the scatterplot for the men is shifted up and to the right. The overall plot exhibits a higher correlation […]
978-0073401331 Chapter 7 Part 2
368 CHAPTER 7 11. The width of a confidence interval is proportional to sby=ss1 n+(x−x)2 Pn i=1(xi−x)2. 12. The mean temperature is (x) = 65.0. Since 60◦C is closest to (x), its confidence interval would be the shortest. Since 45◦C […]
978-0073401331 Chapter 7 Part 3
SUPPLEMENTARY EXERCISES FOR CHAPTER 7 381 (b) 0 10 20 30 40 50 60 −30 −20 −10 0 10 20 30 Order of Observations Residual No violations of the standard assumptions are in- dicated. 5. (a) x= 50, y= 47.909091, […]
978-0073401331 Chapter 8 Part 1
394 CHAPTER 8 3. 44 46 48 50 52 −2 −1 0 1 2 Fitted Values Residuals The linear model is reasonable. There is no obvious pattern to the plot. 4. (a) by=−0.83 + 0.017(20) + 0.0895(17) + 42.771(0.006) + […]
978-0073401331 Chapter 8 Part 2
412 CHAPTER 8 S = 0.33050 R-sq = 92.0% R-sq(adj) = 90.7% (c) 6 7 8 9 10 11 −1 −0.5 0 0.5 1 Fitted Value Residual There is a some suggestion of het- eroscedasticity, but it is hard to […]
978-0073401331 Chapter 9 Part 1
6. (a) Source DF SS MS F P Age 3 20.964 6.988 0.89095 0.453 Error 45 352.95 7.8433 Total 48 373.91 (b) No. F3,45 = 0.89095, P > 0.10 (P= 0.453). 7. (a) Source DF SS MS F P Group […]
978-0073401331 Chapter 9 Part 2
450 CHAPTER 9 18. (a) Source DF SS MS F P Wafer 2 262627.1 131313.6 24118.8 0.000 (b) With the balance powered up continuously, there are no apparent differences in measurements obtained by different operators. When one of the operators […]
978-0073401331 Chapter 9 Part 3
SUPPLEMENTARY EXERCISES FOR CHAPTER 9 461 (c) Variable Effect Variable Effect A7.50 AE 7.00 B19.75 BC 3.00 AD −4.00 (d) The probability plot shows that factors Band E appear to affect the yield. None of the interactions appear to be […]