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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
7+0.152
52
2. X= 6.83, sX= 1.72, nX= 5, Y= 3.32, sY= 1.17, nY= 7.
The number of degrees of freedom is
5+1.172
72
3. X= 59.6, sX= 5.295701, nX= 10, Y= 50.9, sY= 5.321863, nY= 10.
The number of degrees of freedom is
10 +5.3218632
10 2
4. X= 28.714286, sX= 7.387248, nX= 7, Y= 17.714286, sY= 4.4614753, nY= 7.
5. X= 73.1, sX= 9.1, nX= 10, Y= 53.9, sY= 10.7, nY= 10.
6. X= 18.7, sX= 3.3, nX= 11, Y= 11.2, sY= 2.4, nY= 9.
The number of degrees of freedom is
11 +2.42
92
7. X= 33.8, sX= 0.5, nX= 4, Y= 10.7, sY= 3.3, nY= 8.
SECTION 5.6 257
8. X= 7.9, sX= 0.6, nX= 8, Y= 3.4, sY= 0.6, nY= 10.
9. X= 4.8, sX= 1.9, nX= 24, Y= 2.8, sY= 1.0, nY= 24.
10. X= 4.495714, sX= 0.177469, nX= 7, Y= 2.465000, sY= 0.507149, nY= 8.
The number of degrees of freedom is
7+0.5071492
82
11. X= 36.893077, sX= 3.221054, nX= 13, Y= 33.000000, sY= 2.544882, nY= 9.
258 CHAPTER 5
13 +2.5448822
92
12. X= 62.71429, sX= 3.8607, nX= 7, Y= 60.40000, sY= 5.46097, nY= 10.
13. X= 229.54286, sX= 14.16885, nX= 7, Y= 143.95556, sY= 59.75699, nY= 9.
14. X= 95, sX= 1, nX= 5, Y= 70, sY= 6, nY= 5.
The number of degrees of freedom is
5+62
52
Page 258
15. X= 4500.8, sX= 271.6, nX= 5, Y= 1299.8, sY= 329.8, nY= 5.
Section 5.7
1. D= 6.736667, sD= 6.045556, n= 9, t9−1,.025 = 2.306.
4. The differences are: 15, 21, 18, 20, 22, 24, 15, 19, 20, 20.
7. The differences are: 8.4,8.6,10.5,9.6,10.7,10.8,10.7,11.3,10.7.
9. (a) The differences are: 3.8,2.6,2.0,2.9,2.2,−0.2,0.5,1.3,1.3,2.1,4.8,1.5,3.4,1.4,1.1,1.9,−0.9,−0.3.
D= 1.74444, sD= 1.46095, n= 18, t18−1,.005 = 2.898.
10. (a) Expression (5.21) in Section 5.6 should be used, because the design involves two independent samples.
Page 260
SECTION 5.8 261
Section 5.8
1. (a) χ2
12,.025 = 23.337
2. s= 15, n= 25, α= 0.05. χ2
24,.025 = 39.364, χ2
24,.975 = 12.401.
5. s= 8, n= 18, α= 0.05. χ2
17,.025 = 30.191, χ2
17,.975 = 7.564.
8. s= 0.8377, n= 6, α= 0.01. χ2
5,.005 = 16.750, χ2
5,.995 = 0.412.
A 95% confidence interval for σ2is 5(0.83772)
16.750 ,5(0.83772)
0.412 , or (0.21, 8.52).
11. s= 40 and the number of degrees of freedom is k= 100.
12. s= 40 and the number of degrees of freedom is k= 100.
Section 5.9
1. (a) X= 101.4, s= 2.3, n= 25, t25−1,.025 = 2.064.
Page 262
2. (a) X= 89.7, s= 8.2, n= 20, t20−1,.005 = 2.861.
4. (a) X= 19.35, s= 0.577, n= 6, t6−1,.05 = 2.015.
5. (a) X= 86.56, s= 1.02127, n= 5, t5−1,.025 = 2.776.
Section 5.10
1. (a) X∗∼N(8.5,0.22), Y∗∼N(21.2,0.32)
(b) Answers will vary.
(c) Answers will vary.
Page 263
264 CHAPTER 5
(e) Answers will vary.
2. (a) X∗∼N(1.18,0.022), Y∗∼N(0.85,0.022)
(b) Answers will vary.
(c) Answers will vary.
(e) Answers will vary.
3. (a) Yes, Ais approximately normally distributed.
(b) Answers will vary.
(c) Answers will vary.
4. (a) Yes, Tis approximately normally distributed.
(b) Answers will vary.
(c) Answers will vary.
5. (a) N(0.27,0.402/349) and N(1.62,1.702/143). Since the values 0.27 and 1.62 are sample means, their
Page 264
SECTION 5.10 265
(c) Answers will vary.
6. (a) Using Method 1, the limits of the 95% confidence interval are the 2.5 and 97.5 percentiles of the
7. (a,b,c) Answers will vary.
8. (a) 95% exactly (simulation results will be approximate).
9. (a) Coverage probability for traditional interval is ≈0.89, mean length is ≈0.585.
266 CHAPTER 5
10. (a) General coverage is ≈0.95, pooled coverage is exactly 0.95 (simulation results will be approximate).
Supplementary Exercises for Chapter 5
1. The differences are 21,18,5,13,−2,10.
2. X= 27.833333, sX= 8.328665, nX= 6, Y= 17.500000, sY= 4.370355, nY= 6.
or (1.252, 19.415).
3. X= 1919, nX= 1985, ˜pX= (1919 + 1)/(1985 + 2) = 0.966281,
4. (a) X= 4.20, s= 0.10, n= 87, z.01 = 2.33.
The confidence interval is 4.20 ±2.33(0.10/√87), or (4.175,4.225).
5. X= 6.1, sX= 0.7, nX= 125, Y= 5.8, sY= 1.0, nY= 75, z.05 = 1.645.
6. The standard deviation for the difference between the means is pσ2
X/nX+σ2
Y/nY. Estimate σX≈
sX= 0.7 and σY≈sY= 1.0.
7. (a) X= 13, n= 87, ˜p= (13 + 2)/(87 + 4) = 0.16484, z.025 = 1.96.
Page 267
268 CHAPTER 5
(b) Let nbe the required sample size.
8. X= 72.5, s= 5.8, n= 10, t10−1,.005 = 3.250.
9. The higher the level, the wider the confidence interval. Therefore the narrowest interval, (4.20, 5.83),
10. Let nbe the required sample size.
11. X= 7.909091, sX= 0.359039, nX= 11, Y= 8.00000, sY= 0.154919, nY= 6.
12. (a) X= 1.69, s= 0.25, n= 80, z.025 = 1.96.
Page 268
13. Let nbe the required sample size.
15. (a) False. This a specific confidence interval that has already been computed. The notion of probability
does not apply.
16. (a) True. A 95% confidence interval is found by adding and subtracting 1.96(650/√400) from the mean
of 370.
Page 269
17. (a) X= 37, and the uncertainty is σX=s/√n= 0.1. A 95% confidence interval is 37 ±1.96(0.1),
or (36.804, 37.196).
or (36.774, 37.226).
18. (a) X= 1500, s= 5, n= 10, t10−1,.025 = 2.262.
The confidence interval is 1500 ±2.262(5/√10), or (1496.4, 1503.6).
19. (a) Since Xis normally distributed with mean nλ, it follows that for a proportion 1 −αof all possible
samples, −zα/2σX< X −nλ < zα/2σX.
it without permission.
SUPPLEMENTARY EXERCISES FOR CHAPTER 5 271
20. (a) n= 1, b
λ= 64/1 = 64, σb
λ=p64/1 = 8. A 95% confidence interval for λis therefore 64 ±1.96(8),
or (48.32, 79.68).
21. (a) µ= 1.6, σµ= 0.05, h= 15, σh= 1.0, τ= 25, στ= 1.0. V=τ h/µ = 234.375
∂V
∂µ =−τh/µ2=−146.484375, ∂V
∂h =τ/µ = 15.625, ∂V
∂τ =h/µ = 9.375
22. (a) X= 2.3367, s= 0.093737, n6, t6−1,.025 = 2.571.
23. (a) X= 8.95, s= 0.5682, n6, t8−1,.01 = 2.998.
272 CHAPTER 5
16.013 ,r7(6.08862)
1.690 !, or (4.026, 12.391).
25. It is not appropriate. The sample contains an outlier, which suggests that the population is not
26. (a) Using Method 1, the limits of the 95% confidence interval are the 2.5 and 97.5 percentiles of the
(b) Using Method 2, the 95% confidence interval is (2X−X.975,2X−X.025 ).
27. (a) Answers may vary somewhat from the 0.5 and 99.5 percentiles in Exercise 26 parts (c) and (d).
(b) Answers may vary somewhat from the answer to Exercise 26 part (c).
(c) Answers may vary somewhat from the answer to Exercise 26 part (d).
Page 272
SECTION 6.1 273
Chapter 6
Section 6.1
1. (a) X= 783, s= 120, n= 73. The null and alternate hypotheses are H0:µ≤750 versus H1:µ > 750.
2. (a) X= 1.102, s= 0.02, n= 65. The null and alternate hypotheses are H0:µ≤1.1 versus H1:µ > 1.1.
3. (a) X= 231.7, s= 2.19, n= 66. The null and alternate hypotheses are H0:µ= 232 versus H1:µ6= 232.
4. (a) X= 2.6, s= 0.3, n= 50. The null and alternate hypotheses are H0:µ≤2.5 versus H1:µ > 2.5.
5. (a) X= 4.5, s= 2.7, n= 80. The null and alternate hypotheses are H0:µ≥5.4 versus H1:µ < 5.4.
Page 273
