Applied Statistics and Probability for Engineers, 5th edition
October 22, 2011
Errata
Page 169, exercise 5-13, 5-14, 5-15, part (k) in each, change “Y 5 2” to “Y ≤ 2”
Page 236, exercise 71-8, change 2 to 2
Page 304, exercise 9-44(d), change “ = 3500” to “ = 3470”
Page 305, Under “Using Operating Characteristic Curves” the text should refer to Charts VIIa and VIIb,
not VIa and VIb.
Page 311, Should refer to Table III, not Table II, in second paragraph from bottom
Page 315, below equation (9-32), change the chart references from VIG and VIH to VIIg and VIIh,
respectively
Page 316, exercise 9-48, change “unknown variance ” to “unknown variance 2”
Page 316, exercise 9-49, change “known variance ” to “unknown variance 2”, and change the test
statistic from Z0 to T0
Page 317, exercise 9-50, change “known variance ” to “unknown variance 2”, and change the test
statistic from Z0 to T0
Page 320, equation, Equation 9-33 should be 9-36. Equation 9-34 should be 9-37
Page 320, box at the bottom of the page, in the first line replace Χ20 < −Χ2𝛼/2,𝑛−1 with Χ20 <
2
2
Χ1−𝛼/2,𝑛−1
and in the last line replace Χ20 < −Χ2𝛼,𝑛−1 with Χ20 < Χ1−𝛼,𝑛−1
Page 369, exercise 10-75, part (e), change “…answer in part (b)…” to “…answer in part (d)…”
Page 377, Example 10-10, change to 𝑑̅ = 0.2739 and t0 = 6.08
Page 383, Equations (10-27) and (10-28) are the same. Renumber H0: 12 = 22 as (10-26), H1: 12 ≠ 22 as
(10-27) and remove the equation number from the equation currently labeled as (10-27)
Page 384, last paragraph, replace “Table V contains...” with “Table VI contains…”
Page 386, The seven-step procedure at the top of the page should be renumbered. Replace Step 6 with
Step 5, Step 7 with Step 6, and Step 8 with Step 7
Page 439, exercise 11-78, replace “Clausis” with “Clausius”
Page 469, Table 12-8, replace NI Islanders with NY Islanders
Page 482, Figure 12-5, on the vertical axis replace x01 with x02
Page 508, exercise 12-99, change “do not use x5” to “do not use x5 or x6”
Page 533, exercise 13-16, change “types of chocolate” to “flow rates”
Page 538, line 3, change reference from Example 10-9 to Example 10-10
Page 5-38, Figure 13-9, treatments should be randomized within each block
Page 562, Table 14-4, MSE is in the wrong row, it should be in the Error row
Page 564, Table 14-6 Round-off errors, more accurate answers for f0 are 27.86, 59.70, 1.47, respectively
Page 568, Section 14-4, first paragraph, better notation is abc…(n)
Page 608, first paragraph, change reference from Table 14-24 to Table 14-26
Page 666, Example 15-3, denominator of PCR and PCRk should be 6(1.5) and 3(1.5), respectively. The
final answers are unchanged.
Answers to Exercises
4-87 c) 10,233
Applied Statistics and Probability for Engineers, 5th edition
5-77
8-43
9-61
10-55
10-57
10-73
10-75
10-89
11-23
11-41
11-43
11-75
12-9
12-21
12-23
12-35
12-55
12-87
12-99
13-43
14-1
14-15
14-59
15-39
15-47
15-59
15-63
15-71
15-101
October 22, 2011
d) 0.323
a) 13.85 b) 21.67
a) t0 = -1.291, fail to reject H0 at  = 0.05, 0.2 < p-value < 0.5 b) Yes c) power = 0.70 d) n > 100
e) 129.337 <  < 130.157
b) (0.323, 2.527)
a) (0.607, 1.463) b) (0.557, 1.559) c) 0.669 < 1/2
a) Confidence interval –2.622 < 1 – 2 < 0.902
d) (0.113, 3.673)
d) One-sided confidence interval V/M > 22.93 e) One-sided test, reject H0, f0 = 72.78 > 3.28
b) estimate of 2 = 1.8436
a) [7.464, 12.720] c) [91.698, 98.164] d) [83.801, 106.061]
a) Confidence interval 9.101 < 0 < 9.932
c) [0.660, 0.909]
a) y = 47.8 - 9.60x1 + 0.415x2 + 18.3x3 b) 12
a) add df Residual = 12, SS Residual = 307, R-Sq = 0.9867
a) f0 = 263.26 b) t0(beta1) = 19.32, t0(beta2) = -13.16
a) f0 = 191.09
a) -8.658 < 0 < 108.458; -0.08 < 2 < 0.059; -0.05 < 3 < 0.047; -0.006 < 7 <0;
-4.962 < 8 < 5.546; -7.811 < 9 < 0.101; 1.102 < 10 < -5.023
b) 25.557 < Y|x0 < 33.863
c) 50.855 < 0 < 71.147; -0.036 < 2 < -0.006; -0.006 < 7 < -0.001; -6.485 < 9 < -0.429
d) Intervals in part c) are narrower so that part c) may be a preferable model.
a) Min Cp = 5.1, Att, PctComp, Yds, YdsperAtt, TD, PctTD, PctInt
a) y* = - 0.908 + 5.48x1* + 1.13x2* - 3.92x3* - 1.14x4*
b) f0 = 109.02, p-value = 0.000; x1* t = 11.27, p-value = 0.000; x2* t = 14.59, p-value = 0.000;
x3* t = -6.98, p-value = 0.000; x4* t = -8.11, p-value = 0.000
b) 4
a) Add interaction hypotheses: H0 = 11 = 12 = … = 23, H1: at least one ij ≠ 0
a) Cleaning method = –5.593
b) Change to (1.22, –2.13, 1.62, –0.61)
a) 1.4 b) PCR = 0.95, PCRk = 0.64
b) points 5, 9, 12, 20 exceed the limits
a) 𝜎̂ = 6.928 b) 0.0401 c) 24.94
a) sigma estimate from s = 0.1736, estimate from moving range = 0.1695, compare to exercise
15-67
b) For h = 4, 2.01 < ARL < 2.57
a) Probability = 0.0197 b) 50.8