Statistics for Engineers and Scientists 4th Edition

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. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree

14 CHAPTER 1 14 CHAPTER 1 14 CHAPTER 1 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 can be computed as the difference between the first and third quartiles. interquartile range can be

472 CHAPTER 10 Chapter 10 Section 10.1 1. (a) Count (b) Continuous(c) Binary(d) Continuous (b) Continuous(c) Binary(d) Continuous (b) Continuous (c) Binary (d) Continuous 2. (a) True (b) False. If no special causes are operating, then the variability is constant. It is still possible for much ofthe output to fail to conform to specifications, for example through incorrect calibration, or

28 CHAPTER 2 (f) The set ABcontains the outcomes that are either in A, in B, or in both. Therefore AB={CCC, DDD, SSS, CDS, CSD, DCS, DSC, SCD, SDC}. (g) Cccontains the outcomes that are not in C.ACccontains the outcomes that are in Abut not inC. Therefore ACc={DDD, SSS}.(h) Accontains the outcomes that

48 CHAPTER 2 =P(Fc|F lc)P(F lc)P(Fc|F lc)P(F lc) + [1 P(F|F l)]P(F l)=(0.99)(0.9998)(0.99)(0.9998) + (1 0.995)(0.0002)= 0.999999(d) ii. Given that a bottle passes inspection, the probability that is has no flaw is 0.999999. =P(Fc|F lc)P(F lc)P(Fc|F lc)P(F lc) + [1 P(F|F l)]P(F l)=(0.99)(0.9998)(0.99)(0.9998) + (1 0.995)(0.0002)= 0.999999(d) ii. Given that a bottle

68 CHAPTER 2 (b) M=X+1.5Y=p2X+ 1.522Y=p0.052+ 1.52(0.12) = 0.158 (b) M=X+1.5Y=p2X+ 1.522Y=p0.052+ 1.52(0.12) = 0.158 (b) M=X+1.5Y=p2 X+ 1.522 Y=p0.052+ 1.52(0.12) = 0.158 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) S=PXi= 24(20.01) = 480.24(b) S=qP2Xi=p24(0.022) = 0.098(c) X=Xi= 20.01(d)

88 CHAPTER 2 (b) M1M2=R2+E1R+E2R+E1E2=R2+E1R+E2R+E1E2=R2 (c) M1M2=R+E1R+E2= (R+E1)(R+E2) = RR=2R(d) Cov(M1, M2) = M1M2M1M2=R22R=2R(e) M1,M2=Cov(M1, M2)M1M2=2RM1M2=4(2.2361)(2.2361) = 0.8 (c) M1M2=R+E1R+E2= (R+E1)(R+E2) = RR=2R(d) Cov(M1, M2) = M1M2M1M2=R22R=2R(e) M1,M2=Cov(M1, M2)M1M2=2RM1M2=4(2.2361)(2.2361) = 0.8 (c) M1M2=R+E1R+E2= (R+E1)(R+E2) = RR=2 R (d) Cov(M1, M2) = M1M2M1M2=R22 R=2 R (e) M1,M2=Cov(M1, M2) M1M2 =2 R M1M2 =4 (2.2361)(2.2361) = 0.8 26. Cov(X, X) = XXXX=X2(X)2=2 X. 27. (a) Cov(aX, bY ) = aXbY

SUPPLEMENTARY EXERCISES FOR CHAPTER 2 99 P(Y= 200 |X= 0.08) = P(Y= 200 and X= 0.08)P(X= 0.08) =0.120.30 = 0.400.The additive concentration should be 0.06. P(Y= 200 |X= 0.08) = P(Y= 200 and X= 0.08)P(X= 0.08) =0.120.30 = 0.400.The additive concentration should be 0.06. P(Y= 200 |X= 0.08) = P(Y= 200 and X=

110 CHAPTER 3 (b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weightis 0. (b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weightis 0. (b) Yes, the bias can be

SECTION 3.4 129 k= ln(T0Ta)/t ln(TTa)/t = 0.1 ln(T0Ta)0.1 ln(TTa) = 0.0626 kT =110(TTa)=0.00543478kT0=110(T0Ta)= 0.00290698kTa=110(TTa)110(T0Ta)= 0.00252781k=skT 22T+kT022T0+kTa22Ta= 0.0013k= 0.0626 0.0013 min1 kT =110(TTa)=0.00543478kT0=110(T0Ta)= 0.00290698kTa=110(TTa)110(T0Ta)= 0.00252781k=skT 22T+kT022T0+kTa22Ta= 0.0013k= 0.0626 0.0013 min1 k T =1 10(TTa)=0.00543478 k T0 =1 10(T0Ta)= 0.00290698 k Ta =1 10(TTa)1 10(T0Ta)= 0.00252781 k=sk T 2 2 T+k T02 2 T0+k Ta2 2 Ta= 0.0013 k= 0.0626 0.0013 min1 18. (a) a= 2.5, a= 0.1, b= 0.05, b= 0.01, w= 1.2, w= 0.1, v=a+bw =

148 CHAPTER 4 (f) No. If the surface has both discoloration and a crack, then X= 1, Y= 1, and Z= 1, but X+Y= 2. (f) No. If the surface has both discoloration and a crack, then X= 1, Y= 1, and Z= 1, but X+Y= 2. (f) No. If the surface has

168 CHAPTER 4 The uncertainty is b 1b2=q2b1b2 1b2=q2b1b2 1 b 2=q2 b 1 b 2 +2 =6.25 + 3.5 = 3.1. 12= 5.53.1 12= 5.53.1 12= 5.53.1 18. If the mean decay rate is exactly 1 per second, then the mean number of events in 10 seconds is 10, so XPoisson(10). (a) P(X1) = P(X= 0) + P(X= 1) = e10 1000! +e10 1011! = 4.994 104(b)

188 CHAPTER 4 5. (a) ln IN(1,0.2), ln RN(4,0.1), and Iand Rare independent. Therefore ln VN(5,0.3). Since ln Vis normal, Vis lognormal, with V= 5 and 2V= 0.3.(b) P(V < 200) = P(ln V < ln 200) = P(ln V < 5.298317).Now ln VN(5,0.3), so the z-score of 5.298317 is (5.298317 5)/0.3

208 CHAPTER 4 208 CHAPTER 4 208 CHAPTER 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 tocome from an approximately normal dis-tribution.7. Yes. If the logs of the PM data come from a normal population, then the PM data come from alognormal population, and vice versa. The

222 CHAPTER 4 222 CHAPTER 4 222 CHAPTER 4 12. (a,b) Answers will vary. (c) median = 0 exactly (simulation results will be approximate), median 0.536. (c) median = 0 exactly (simulation results will be approximate), median 0.536. (c) median = 0 exactly (simulation results will be approximate), median 0.536. (d) Answers will vary. (e) X= 0 exactly,

236 CHAPTER 5 (d) z.025 = 1.96. 1.96(311.7/n) = 50, so n= 150.(e) z.01 = 2.33. 2.33(311.7/n) = 50, so n= 211. (d) z.025 = 1.96. 1.96(311.7/n) = 50, so n= 150.(e) z.01 = 2.33. 2.33(311.7/n) = 50, so n= 211. (d) z.025 = 1.96. 1.96(311.7/n) = 50, so n= 150. (e) z.01 =

SECTION 5.6 255 SECTION 5.6 255 SECTION 5.6 255 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 =0.252(0.252/7)271+(0.152/5)251= 9, rounded down to the nearest integer.t9,.005 = 3.250, so the confidence interval is 1.51.03.250r0.2527+0.1525,or

274 CHAPTER 6 is the area to the left of z=2.98. Thus P= 0.0014.(b) If the mean number of sick days were 5.4, the probability of observing a sample mean less than or equalto the observed value of 4.5 would be 0.0014. Since 0.0014 is a small probability, we are convincedthat the

294 CHAPTER 6 (d) Since the alternate hypothesis is of the form pXpY6= 0, the P-value is the sum of the areas to theright of z= 1.79 and to the left of z=1.79.Thus P= 0.0367 + 0.0367 = 0.0734. (d) Since the alternate hypothesis is of the form pXpY6= 0, the P-value

314 CHAPTER 6We can conclude that the mean recovery times differ. 314 CHAPTER 6We can conclude that the mean recovery times differ. 314 CHAPTER 6 We can conclude that the mean recovery times differ. 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

SECTION 6.14 331 (e) Greater than 0.6. The power for a sample size of 150 is 0.691332, and the power for a larger samplesize of 200 would be greater than this.(f) Greater than 0.65. The power against the alternative p= 0.4 is 0.691332, and the alternative p= 0.3is farther from the

348 CHAPTER 7 348 CHAPTER 7 348 CHAPTER 7 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 shiftedup and to the right. The overall plot exhibits ahigher correlation than either plot separately. Thecorrelation between heights and weights

368 CHAPTER 7 11. The width of a confidence interval is proportional to sby=ss1 n+(xx)2 Pn i=1(xix)2. Since s,n,x, and Pni=1(xix)2are the same for each confidence interval, the width of the confidenceinterval is an increasing function of the difference xx.x= 21.7548. The value 20 is closest to xand the value 15 is the farthest from

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, Pn i=1(xix)2= 11000, Pni=1(yiy)2= 9768.909091,Pni=1(xix)(yiy) = 10360, n= 11.b1=Pni=1(xix)(yiy)Pn0=yb1x= 0.818182. i=1(xix)2= 11000, Pni=1(yiy)2= 9768.909091,Pni=1(xix)(yiy) = 10360, n= 11.b1=Pni=1(xix)(yiy)Pn0=yb1x= 0.818182. i=1(xix)2= 11000, Pn i=1(yiy)2= 9768.909091, Pn i=1(xix)(yiy) = 10360, n= 11. b 1=Pn i=1(xix)(yiy) Pn 0=yb 1x=

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) + 0.027(10) 0.0043(17)(10) = 0.8271 seconds. (b) Yes, the predicted change is 0.017(2) = 0.034 seconds.(c) No, the predicted change depends on the value

412 CHAPTER 8 S = 0.33050 R-sq = 92.0% R-sq(adj) = 90.7% Analysis of VarianceSource DF SS MS F PRegression 4 31.22 7.8049 71.454 0.000Residual Error 25 2.7307 0.10923Total 29 33.95Comparing this model with the one in part (a), F1,24 =(2.7307 2.6462)/(5 4)2.6462/24 = 0.77, P > 0.10.A computer package gives P=

430 CHAPTER 9(b) Yes, F5,73 = 7.9115, P < 0.01 (P= 5.35 106). 430 CHAPTER 9(b) Yes, F5,73 = 7.9115, P < 0.01 (P= 5.35 106). 430 CHAPTER 9 (b) Yes, F5,73 = 7.9115, P < 0.01 (P= 5.35 106). 6. (a) Source DF SS MS F P Age 3 20.964 6.988 0.89095 0.453 Error 45

450 CHAPTER 9 18. (a) Source DF SS MS F P Wafer 2 262627.1 131313.6 24118.8 0.000 Operator 2 24.778 12.389 2.28 0.159Interaction 4 16.222 4.0556 0.74 0.585Error 9 49.000 5.4444Total 17 262717.1 Operator 2 24.778 12.389 2.28 0.159Interaction 4 16.222 4.0556 0.74 0.585Error 9 49.000 5.4444Total 17 262717.1 Operator 2 24.778 12.389 2.28 0.159 Interaction

SUPPLEMENTARY EXERCISES FOR CHAPTER 9 461 (c) Variable Effect Variable Effect A7.50 AE 7.00 B19.75 BC 3.00 C1.25 BD 1.75D0.00 BE 0.25E44.50 CD 2.25AB 5.25 CE 6.25AC 1.25 DE 3.50 C1.25 BD 1.75D0.00 BE 0.25E44.50 CD 2.25AB 5.25 CE 6.25AC 1.25 DE 3.50 C1.25 BD 1.75 D0.00 BE 0.25 E44.50 CD 2.25 AB 5.25 CE 6.25 AC 1.25 DE 3.50 AD