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270 CHAPTER 6 17. (a) Yes. The value 3.5 is greater than the upper confidence bound of 3.45. Quantities greater than the upper confidence bound will have P-values less than 0.05. Therefore P < 0.05. (b) No, we would need to know the 99% upper confidence bound to determine whether P < 0.01. 18. (a) No, we would need to know the 99% lower confidence bound in order to determine whether P < 0.01. (b) Yes, since 50 is less than the 98% lower confidence bound, we know that P < 0.02. Therefore P < 0.05. 19. Yes, we can compute the P-value exactly. √ √ Since the 95% upper confidence bound is 3.45, we know that 3.40 + 1.645s/ n = 3.45. Therefore s/ n = 0.0304. The z-score is (3.40 − 3.50)/0.0304 = −3.29. The P-value is 0.0005, which is less than 0.01. 20. No, the confidence bound itself tells us only that P < 0.02. To determine the P-value more precisely, we would √ need to know the standard deviation of the sample mean. This is approximately equal to s/ n where s is the sample standard deviation and n is the sample size. We know that s = 5, but we do not know n. Section 6.3 1. X = 25, n = 300, pb = 25/300 = 0.083333. The null and alternate hypotheses are H0 : p ≤ 0.05 versus H1 : p > 0.05. p z = (0.083333 − 0.5)/ 0.05(1 − 0.05)/300 = 2.65. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.65, so P = 0.0040. The claim is rejected. 2. 3. No. n = 100 and p0 = 0.05, so np0 = 5, which is less than 10. X = 25, n = 400, pb = 25/400 = 0.0625. The null and alternate hypotheses are H0 : p ≤ 0.05 versus H1 : p > 0.05. p z = (0.0625 − 0.05)/ 0.05(1 − 0.05)/400 = 1.15. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 1.15, so P = 0.1251. The company cannot conclude that more than 5% of their subscribers would pay for the channel. Page 270 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271 SECTION 6.3 4. n = 304, pb = 0.62. The null and alternate hypotheses are H0 : p ≤ 0.5 versus H1 : p > 0.5. p z = (0.62 − 0.5)/ 0.5(1 − 0.5)/304 = 4.18. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 4.18, so P ≈ 0. It can be concluded that more than half of patients prefer a physician with high technical skills. 5. X = 330, n = 600, pb = 330/600 = 0.55. The null and alternate hypotheses are H0 : p ≤ 0.50 versus H1 : p > 0.50. p z = (0.55 − 0.50)/ 0.50(1 − 0.50)/600 = 2.45. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.45, so P = 0.0071. It can be concluded that more than 50% of all likely voters favor the proposal. 6. X = 18, n = 50, pb = 18/50 = 0.36. The null and alternate hypotheses are H0 : p ≥ 0.50 versus H1 : p < 0.50. p z = (0.36 − 0.50)/ 0.50(1 − 0.50)/50 = −1.98. Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.98, so P = 0.0239. It can be concluded that fewer than half of the incinerators meet the standard. 7. X = 470, n = 500, pb = 470/500 = 0.94. The null and alternate hypotheses are H0 : p ≥ 0.95 versus H1 : p < 0.95. p z = (0.94 − 0.95)/ 0.95(1 − 0.95)/500 = −1.03. Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.03, so P = 0.1515. The claim cannot be rejected. 8. X = 12, n = 300, pb = 12/300 = 0.04. The null and alternate hypotheses are H0 : p ≥ 0.08 versus H1 : p < 0.08. p z = (0.04 − 0.08)/ 0.08(1 − 0.08)/300 = −2.55. Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −2.55, so P = 0.0054. The machine can be qualified. 9. X = 15, n = 400, pb = 15/400 = 0.0375. The null and alternate hypotheses are H0 : p ≥ 0.05 versus H1 : p < 0.05. p z = (0.0375 − 0.05)/ 0.05(1 − 0.05)/400 = −1.15. Page 271 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 272 CHAPTER 6 Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −1.15, so P = 0.1251. It cannot be concluded that the new process has a lower defect rate. 10. X = 37, n = 50, pb = 37/50 = 0.74. The null and alternate hypotheses are H0 : p ≤ 0.60 versus H1 : p > 0.60. p z = (0.74 − 0.60)/ 0.60(1 − 0.60)/50 = 2.02. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 2.02, so P = 0.0217. It can be concluded that more than 60% of the measurements are satisfactory. 11. X = 17, n = 75, pb = 17/75 = 0.22667. The null and alternate hypotheses are H0 : p ≥ 0.40 versus H1 : p < 0.40. p z = (0.22667 − 0.40)/ 0.40(1 − 0.40)/75 = −3.06. Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −3.06, so P = 0.0011. It can be concluded that less than 40% of the fuses have burnout amperages greater than 15 A. 12. (a) One-tailed. The alternate hypothesis is of the form p < p0 . (b) H0 : p ≥ 0.4 (c) Yes, P = 0.001, which is less than 0.02. (d) The sample proportion is 0.304167. Therefore the P-value for the null hypothesis p ≥ 0.45 will be less than the P-value for the null hypothesis p ≥ 0.40, which is 0.001. It follows that the null hypothesis p ≥ 0.45 can be rejected at the 2% level. (e) X = 73, n = 240, pb = 73/240 = 0.304167. The null and alternate hypotheses are H0 : p ≤ 0.25 versus H1 : p > 0.25. p z = (0.304167 − 0.25)/ 0.25(1 − 0.25)/240 = 1.94. Since the alternate hypothesis is of the form p > p0 , the P-value is the area to the right of z = 1.94. Thus P = 0.0262. (f) X = 73, n = 240, p̃ = (73 + 2)/(240 + 4) = 0.307377, z.05 = 1.645. p The confidence interval is 0.307377 ± 1.645 0.307377(1 − 0.307377)/(240 + 4), or (0.2588, 0.3560). Page 272 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273 SECTION 6.4 13. (a) Sample p = pb = 345/500 = 0.690. (b) The null and alternatephypotheses are H0 : p ≥ 0.7 versus H1 : µ < 0.7. n = 500. From part (a), pb = 0.690. z = (0.690 − 0.700)/ 0.7(1 − 0.7)/500 = −0.49. (c) Since the alternate hypothesis is of the form p < p0 , the P-value is the area to the left of z = −0.49. Thus P = 0.3121. Section 6.4 1. (a) X = 100.01, s = 0.0264575, n = 3. There are 3 − 1 = 2 degrees of freedom. The null and alternate hypotheses are H0 : µ = 100 versus H1 : µ 6= 100. √ t = (100.01 − 100)/(0.0264575/ 3) = 0.6547. Since the alternate hypothesis is of the form µ 6= µ0 , the P-value is the sum of the areas to the right of t = 0.6547 and to the left of t = −0.6547. From the t table, 0.50 < P < 0.80. A computer package gives P = 0.580. We conclude that the scale may well be calibrated correctly. (b) The t-test cannot be performed, because the sample standard deviation cannot be computed from a sample of size 1. 2. (a) Reject the assumptions: the value 221.03 is an outlier. (b) The assumptions appear to be met. (c) Reject the assumptions: the values are in increasing order, so this does not appear to be a simple random sample. 3. (a) H0 : µ ≤ 35 versus H1 : µ > 35 (b) X = 39, s = 4, n = 6. There are 6 − 1 = 5 degrees of freedom. From part (a), the null and alternate hypotheses are H0 : µ ≤ 35 versus H1 : µ > 35. √ t = (39 − 35)/(4/ 6) = 2.449. Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 2.449. From the t table, 0.025 < P < 0.050. A computer package gives P = 0.029. Page 273 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274 CHAPTER 6 (c) Yes, the P-value is small, so we conclude that µ > 35. 4. (a) X = 20.75, s = 3.93, n = 26. There are 26 − 1 = 25 degrees of freedom. The null and alternate hypotheses are H0 : µ ≤ 20 versus H1 : µ > 20. √ t = (20.75 − 20)/(3.93/ 25) = 0.973. Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.973. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.170. We cannot conclude that the mean concentration is greater than 20 mg/kg. (b) X = 20.75, s = 3.93, n = 26. There are 26 − 1 = 25 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 25 versus H1 : µ < 25. √ t = (20.75 − 25)/(3.93/ 25) = −5.514. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −5.514. From the t table, P < 0.0005. A computer package gives P = 4.96 × 10−6. We conclude that the mean concentration is less than 25 mg/kg. 5. (a) X = 61.3, s = 5.2, n = 12. There are 12 − 1 = 11 degrees of freedom. The null and alternate hypotheses are H0 : µ ≤ 60 versus H1 : µ > 60. √ t = (61.3 − 60)/(5.2/ 12) = 0.866. Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.866. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.202. We cannot conclude that the mean concentration is greater than 60 mg/L. (b) X = 61.3, s = 5.2, n = 12. There are 12 − 1 = 11 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 65 versus H1 : µ < 65. √ t = (61.3 − 65)/(5.2/ 12) = −2.465. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −2.465. From the t table, 0.01 < P < 0.025. A computer package gives P = 0.0157. We conclude that the mean concentration is less than 65 mg/L. 6. X = 90.3, s = 1.290349, n = 5. There are 5 − 1 = 4 degrees of freedom. The null and alternate hypotheses are H0 : µ ≤ 90 versus H1 : µ > 90. √ t = (90.3 − 90)/(1.290349/ 5) = 0.5199. Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 0.5199. From the t table, 0.25 < P < 0.40. A computer package gives P = 0.315. Page 274 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 275 SECTION 6.4 We cannot conclude that the mean octane rating is greater than 90%. 7. (a) 3.8 4 4.2 (b) Yes, the sample contains no outliers. X = 4.032857, s = 0.061244, n = 7. There are 7 − 1 = 6 degrees of freedom. The null and alternate hypotheses are H0 : µ = 4 versus H1 : µ 6= 4. √ t = (4.032857 − 4)/(0.061244/ 7) = 1.419. Since the alternate hypothesis is of the form µ 6= µ0 , the P-value is the sum of the areas to the right of t = 1.419 and to the left of t = −1.419. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.2056. It cannot be concluded that the mean thickness differs from 4 mils. (c) 3.9 4 4.1 4.2 4.3 (d) No, the sample contains an outlier. 8. (a) H0 : µ ≤ 85 versus H1 : µ > 85 (b) X = 90.55, s = 2.901551, n = 6. There are 6 − 1 = 5 degrees of freedom. The null and alternate hypotheses are H0 : µ ≤ 85 versus H1 : µ > 85. √ t = (90.55 − 85)/(2.901551/ 6) = 4.685. Since the alternate hypothesis is of the form µ > µ0 , the P-value is the area to the right of t = 4.685. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0027. (c) Yes, the P-value is small, so we conclude that µ > 85. 9. X = 1.88, s = 0.21, n = 18. There are 18 − 1 = 17 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 2 versus H1 : µ < 2. √ t = (1.88 − 2)/(0.21/ 18) = −2.424. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −2.424. From the t table, 0.01 < P < 0.025. A computer package gives P = 0.013. We can conclude that the mean warpage is less than 2 mm. Page 275 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276 10. CHAPTER 6 X = 7.22, s = 2.328519, n = 5. There are 5 − 1 = 4 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 8 versus H1 : µ < 8. √ t = (7.22 − 8)/(2.328519/ 5) = −0.7490. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −0.7490. From the t table, P ≈ 0.25. A computer package gives P = 0.248. We cannot conclude that the mean amount is less than 8%. 11. X = 1.25, s = 0.624500, n = 4. There are 4 − 1 = 3 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 2.5 versus H1 : µ < 2.5. √ t = (1.25 − 2.5)/(0.624500/ 4) = −4.003. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −4.003. From the t table, 0.01 < P < 0.025. A computer package gives P = 0.014. We can conclude that the mean amount absorbed is less than 2.5%. 12. (a) One-tailed. The alternate hypothesis is of the form µ > µ0 . (b) H0 : µ ≤ 5.5 (c) Yes, the P-value is 0.002, which is less than 0.01. √ (d) X = 5.92563, n = 5, s/ n = 0.07046. There are 5 − 1 = 4 degrees of freedom. The null and alternate hypotheses are H0 : µ ≥ 6.5 versus H1 : µ < 6.5. t = (5.92563 − 6.5)/0.07046 = −8.152. Since the alternate hypothesis is of the form µ < µ0 , the P-value is the area to the left of t = −8.152. From the t table, 0.001 < P < 0.0005. A computer package gives P = 0.00062. √ (e) X = 5.92563, n = 5, s/ n = 0.07046. There are 5 − 1 = 4 degrees of freedom. t4,.005 = 4.604. The confidence interval is 5.92563 ± 4.604(0.07046), or (5.6012, 6.2500). √ √ 13. (a) StDev = (SE Mean) N = 1.8389 11 = 6.0989. (b) t10,.025 = 2.228. The lower 95% confidence bound is 13.2874 − 2.228(1.8389) = 9.190. (c) t10,.025 = 2.228. The upper 95% confidence bound is 13.2874 + 2.228(1.8389) = 17.384. (d) t = (13.2874 − 16)/1.8389 = −1.475. Page 276 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 277 SECTION 6.5 Section 6.5 1. X = 8.5, sX = 1.9, nX = 58, Y = 11.9, sY = 3.6, nY = 58. The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0. p z = (8.5 − 11.9 − 0)/ 1.92 /58 + 3.62/58 = −6.36. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the left of z = −6.36. Thus P ≈ 0. We can conclude that the mean hospital stay is shorter for patients receiving C4A-rich plasma. 2. X = 20.95, sX = 14.5, nX = 482, Y = 22.79, sY = 15.6, nY = 614. The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0. p z = (20.95 − 22.79 − 0)/ 14.52 /482 + 15.62/614 = −2.02. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the left of z = −2.02. Thus P = 0.0217. We can conclude that the mean weight of male spotted flounder is greater than that of females. 3. X = 5.92, sX = 0.15, nX = 42, Y = 6.05, sY = 0.16, nY = 37. The null and alternate hypotheses are H0 : µX − µY = 0 versus H1 : µX − µY 6= 0. p z = (5.92−6.05−0)/ 0.152 /42 + 0.162/37 = −3.71. Since the alternate hypothesis is of the form µX − µY 6= ∆, the P-value is the sum of the areas to the right of z = 3.71 and to the left of z = −3.71. Thus P = 0.0001 + 0.0001 = 0.0002. We can conclude that the mean dielectric constant differs between the two types of asphalt. 4. X = 27.2, sX = 1.2, nX = 40, Y = 28.1, sY = 2.0, nY = 40. The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0. p z = (27.2 − 28.1 − 0)/ 1.22 /40 + 2.02/40 = −2.44. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the left of z = −2.44. Thus P = 0.0073. We can conclude that the mean mileage is greater with premium fuel. 5. X = 40, sX = 12, nX = 75, Y = 42, sY = 15, nY = 100. The null and alternate hypotheses are H0 : µX − µY > 0 versus H1 : µX − µY ≤ 0. p z = (40 − 42 − 0)/ 122/75 + 152/100 = −0.98. Since the alternate hypothesis is of the form µX − µY ≤ ∆, the P-value is the area to the left of z = −0.98. Thus P = 0.1635. Page 277 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 278 CHAPTER 6 We cannot conclude that the mean reduction from drug B is greater than the mean reduction from drug A. 6. (a) The 60 counts may not be independent if they are taken in consecutive time periods. (b) X = 73.8, sX = 5.2, nX = 60, Y = 76.1, sY = 4.1, nY = 60. The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0. p z = (73.8 − 76.1 − 0)/ 5.22 /60 + 4.12/60 = −2.69. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the left of z = −2.69. Thus P = 0.0036. We can conclude that machine 2 is faster. 7. (a) X = 7.79, sX = 1.06, nX = 80, Y = 7.64, sY = 1.31, nY = 80. Here µ1 = µX and µ2 = µY . The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0. p z = (7.79 − 7.64 − 0)/ 1.062/80 + 1.312/80 = 0.80. Since the alternate hypothesis is of the form µX − µY > ∆, the P-value is the area to the right of z = 0.80. Thus P = 0.2119. We cannot conclude that the mean score on one-tailed questions is greater. (b) The null and alternate hypotheses are H0 : µX − µY = 0 versus H1 : µX − µY 6= 0. The z-score is computed as in part (a): z = 0.80. Since the alternate hypothesis is of the form µX − µY 6= ∆, the P-value is the sum of the areas to the right of z = 0.80 and to the left of z = 0.80. Thus P = 0.2119 + 0.2119 = 0.4238. We cannot conclude that the mean score on one-tailed questions differs from the mean score on two-tailed questions. 8. (a) X = 495.6, sX = 19.4, nX = 50, Y = 481.2, sY = 14.3, nY = 50. The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0. p z = (495.6−481.2−0)/ 19.42 /50 + 14.32/50 = 4.22. Since the alternate hypothesis is of the form µX − µY > ∆, the P-value is the area to the right of z = 4.22. Thus P ≈ 0. We can conclude that the mean speed for the new chips is greater. (b) X = 495.6, sX = 19.4, nX = 50, Y = 391.2, sY = 17.2, nY = 60. The null and alternate hypotheses are H0 : µX − µY ≤ 100 versus H1 : µX − µY > 100. p z = (495.6 − 391.2 − 100)/ 19.42/50 + 17.22/60 = 1.25. Since the alternate hypothesis is of the form µX − µY > ∆, the P-value is the area to the right of z = 1.25. Page 278 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 279 SECTION 6.5 Thus P = 0.1056. The data do not provide convincing evidence for this claim. 9. (a) X = 625, sX = 40, nX = 100, Y = 640, sY = 50, nY = 64. The null and alternate hypotheses are H0 : µX − µY ≥ 0 versus H1 : µX − µY < 0. p z = (625 − 640 − 0)/ 402 /100 + 502/64 = −2.02. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the right of z = −2.02. Thus P = 0.0217. We can conclude that the second method yields the greater mean daily production. (b) The null and alternate hypotheses are H0 : µX − µY ≥ −10 versus H1 : µX − µY < −10. p z = (625 − 640 + 10)/ 402 /100 + 502/64 = −0.67. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the right of z = −0.67. Thus P = 0.2514. We cannot conclude that the mean daily production for the second method exceeds that of the first by more than 10 tons. 10. X = 91.1, sX = 6.23, nX = 50, Y = 90.7, sY = 4.34, nY = 40. The null and alternate hypotheses are H0 : µX − µY ≤ 0 versus H1 : µX − µY > 0. p z = (91.1 − 90.7 − 0)/ 6.232/50 + 4.342/40 = 0.36. Since the alternate hypothesis is of the form µX − µY > ∆, the P-value is the area to the right of z = 0.36. Thus P = 0.3594. We cannot conclude that the mean hardness of welds cooled at 10◦ C is greater than that of welds cooled at 30◦ C. 11. X 1 = 4387, s1 = 252, n1 = 75, X 2 = 4260, s2 = 231, n2 = 75. The null and alternate hypotheses are H0 : µ1 − µ2 ≤ 0 versus H1 : µ1 − µ2 > 0. p z = (4387 − 4260 − 0)/ 2522/75 + 2312/75 = 3.22. Since the alternate hypothesis is of the form µX − µY > ∆, the P-value is the area to the right of z = 3.22. Thus P = 0.0006. We can conclude that new power supplies outlast old power supplies. 12. (a) One-tailed. The alternate hypothesis is of the form µX − µY < ∆. (b) H0 : µX − µY ≥ 0 (c) Yes, the P-value is equal to 0.047, which is less than 0.050. Page 279 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280 CHAPTER 6 p (d) z = (3.94−4.43−0)/ 2.652 /135 + 2.382/180 = −1.70. Since the alternate hypothesis is of the form µX − µY < ∆, the P-value is the area to the left of z = −1.70. Thus P = 0.0446, and the result is similar to that of the t test. √ 0.232 + 0.182 is used for z, then z = −1.68 and P = 0.0465. The result is still Note that if the denominator similar to that of the t test. (e) X = 3.94, sX = 2.65, nX = 135, Y = 4.43, sY = 2.38, nY = 180, z.005 = 2.58. p The confidence interval is 3.94 − 4.43 ± 2.58 2.652/135 + 2.382/180, or (−1.235, 0.255). √ Alternatively, the confidence interval can be computed as 3.94−4.43±2.58 0.232 + 0.182, or (−1.244, 0.264). √ √ 13. (a) (i) StDev = (SE Mean) N = 1.26 78 = 11.128. √ √ (ii) SE Mean = StDev/ N = 3.02/ 63 = 0.380484. √ (b) z = (23.3 − 20.63 − 0)/ 1.262 + 0.3804842 = 2.03. Since the alternate hypothesis is of the form µX − µY 6= ∆, the P-value is the sum of the areas to the right of z = 2.03 and to the left of z = −2.03. Thus P = 0.0212 + 0.0212 = 0.0424, and the result is similar to that of the t test. √ √ (c) X = 23.3, sX / nX = 1.26, Y = 20.63, sY / nY = 0.380484, z.01 = 2.33. √ The confidence interval is 23.3 − 20.63 ± 2.33 1.262 + 0.3804842, or (−0.3967, 5.7367). Section 6.6 1. (a) H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0 (b) X = 960, nX = 1000, pbX = 960/1000 = 0.960, Y = 582, nY = 600, pbY = 582/600 = 0.970, pb = (960 + 582)/(1000 + 600) = 0.96375. The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0. 0.960 − 0.970 z= p = −1.04. 0.96375(1 − 0.96375)(1/1000 + 1/600) Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −1.04. Thus P = 0.1492. (c) Since P = 0.1492, we cannot conclude that machine 2 is better. Therefore machine 1 should be used. 2. (a) H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0 Page 280 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 281 SECTION 6.6 (b) X = 150, nX = 180, pbX = 150/180 = 0.83333, Y = 233, nY = 270, pbY = 233/270 = 0.86296, pb = (150 + 233)/(180 + 270) = 0.85111. The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0. 0.83333 − 0.86296 − 0.86. z= p 0.85111(1 − 0.85111)(1/180 + 1/270) Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −0.86. Thus P = 0.1949. (c) No. Since P = 0.1949, we cannot conclude that the proportion from vendor B is greater than that from vendor A. 3. X = 133, nX = 400, pbX = 133/400 = 0.3325, Y = 50, nY = 100, pbY = 50/100 = 0.5, pb = (133 + 50)/(400 + 100) = 0.366. The null and alternate hypotheses are H0 : pX − pY = 0 versus H1 : pX − pY 6= 0. 0.3325 − 0.5 z= p = −3.11. 0.366(1 − 0.366)(1/400 + 1/100) Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of z = 3.11 and to the left of z = −3.11. Thus P = 0.0009 + 0.0009 = 0.0018. We can conclude that the response rates differ between public and private firms. 4. X = 48, nX = 60, pbX = 48/60 = 0.8, Y = 44, nY = 60, pbY = 44/60 = 0.73333, pb = (48 + 44)/(60 + 60) = 0.76667. The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0. 0.8 − 0.73333 = 0.86. z= p 0.76667(1 − 0.76667)(1/60 + 1/60) Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 0.86. Thus P = 0.1949. We cannot conclude that the model with one hidden layer has the greater success rate. 5. X = 57, nX = 100, pbX = 57/100 = 0.57, Y = 135, nY = 200, pbY = 135/200 = 0.675, pb = (57 + 135)/(100 + 200) = 0.64. The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0. 0.57 − 0.675 z= p = −1.79. 0.64(1 − 0.64)(1/100 + 1/200) Page 281 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282 CHAPTER 6 Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −1.79. Thus P = 0.0367. We can conclude that awareness of the benefit increased after the advertising campaign. 6. X = 19, nX = 161, pbX = 19/161 = 0.11801, Y = 22, nY = 95, pbY = 22/95 = 0.23158, pb = (19 + 22)/(161 + 95) = 0.16016. The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0. 0.11801 − 0.23158 = −2.39. z= p 0.16016(1 − 0.16016)(1/161 + 1/95) Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −2.39. Thus P = 0.0084. We can conclude that the larger particles are more likely to contain coliform bacteria. 7. X = 20, nX = 1200, pbX = 20/1200 = 0.016667, Y = 15, nY = 1500, pbY = 15/1500 = 0.01, pb = (20 + 15)/(1200 + 1500) = 0.012963. The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0. 0.016667 − 0.01 z= p = 1.52. 0.012963(1 − 0.012963)(1/1200 + 1/1500) Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 1.52. Thus P = 0.0643. The evidence suggests that heavy packaging reduces the proportion of damaged shipments, but may not be conclusive. 8. X = 260, nX = 300, pbX = 260/300 = 0.86667, Y = 370, nY = 400, pbY = 370/400 = 0.925, pb = (260 + 370)/(300 + 400) = 0.9. The null and alternate hypotheses are H0 : pX − pY ≥ 0 versus H1 : pX − pY < 0. 0.86667 − 0.925 = −2.55. z= p 0.9(1 − 0.9)(1/300 + 1/400) Since the alternate hypothesis is of the form pX − pY < 0, the P-value is the area to the left of z = −2.55. Thus P = 0.0054. We can conclude that the proportion of items rated acceptable or better is greater on Wednesday than on Monday. 9. X = 22, nX = 41, pbX = 22/41 = 0.53659, Y = 18, nY = 31, pbY = 18/31 = 0.58065, Page 282 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 283 SECTION 6.6 pb = (22 + 18)/(41 + 31) = 0.55556. The null and alternate hypotheses are H0 : pX − pY = 0 versus H1 : pX − pY 6= 0. 0.53659 − 0.58065 = −0.37. z= p 0.55556(1 − 0.55556)(1/41 + 1/31) Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of z = 0.37 and to the left of z = −0.37. Thus P = 0.3557 + 0.3557 = 0.7114. We cannot conclude that the proportion of wells that meet the standards differs between the two areas. 10. X = 76, nX = 100, pbX = 76/100 = 0.76, Y = 128, nY = 200, pbY = 128/200 = 0.64, pb = (76 + 128)/(100 + 200) = 0.68. The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0. 0.76 − 0.64 z= p = 2.10. 0.68(1 − 0.68)(1/100 + 1/200) Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.10. Thus P = 0.0179. We can conclude that drug A is more effective than drug B. 11. X = 285, nX = 500, pbX = 285/500 = 0.57, Y = 305, nY = 600, pbY = 305/600 = 0.50833, pb = (285 + 305)/(500 + 600) = 0.53636. The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0. 0.57 − 0.50833 = 2.04. z= p 0.53636(1 − 0.53636)(1/500 + 1/600) Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.04. Thus P = 0.0207. We can conclude that the proportion of voters favoring the proposal is greater in county A than in county B. 12. (a) X = 4243, nX = 82, 486, pbX = 4243/82, 486 = 0.05144, Y = 10, 701, nY = 219, 170, pbY = 10, 701/219, 170 = 0.04883, pb = (4243 + 10, 701)/(82, 486 + 219, 170) = 0.04954. The null and alternate hypotheses are H0 : pX − pY ≤ 0 versus H1 : pX − pY > 0. 0.05144 − 0.04883 z= p = 2.95. 0.04954(1 − 0.04954)(1/82, 486 + 1/219, 170) Since the alternate hypothesis is of the form pX − pY > 0, the P-value is the area to the right of z = 2.95. Page 283 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284 CHAPTER 6 Thus P = 0.0016. We can conclude that accidents involving drivers aged 15–24 are more likely to occur in driveways than accidents involving drivers aged 25–64. (b) No, the observed difference is only 0.2%. Although this is statistically significant, it is not practically significant. 13. No, because the two samples are not independent. 14. (a) Two-tailed. The alternate hypothesis is of the form p1 − p2 6= 0. (b) H0 : p1 − p2 = 0 (c) Yes, P = 0.024, which is less than 0.05. 15. (a) 101/153 = 0.660131. (b) 90(0.544444) = 49. (c) X1 = 101, n1 = 153, pb1 = 101/153 = 0.660131, X2 = 49, n2 = 90, pb2 = 49/90 = 0.544444, pb = (101 + 49)/(153 + 90) = 0.617284. 0.660131 − 0.544444 z= p = 1.79. 0.617284(1 − 0.617284)(1/153 + 1/90) (d) Since the alternate hypothesis is of the form pX − pY 6= 0, the P-value is the sum of the areas to the right of z = 1.79 and to the left of z = −1.79. Thus P = 0.0367 + 0.0367 = 0.0734. 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 2 0.341572 0.909212 + 4 4 ν= = 3, rounded down to the nearest integer. 2 2 (0.909212/4)2 (0.34157 /4) + 4−1 4−1 Page 284 c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual PROPRIETARY MATERIAL. may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.