236 CHAPTER 5
5. (a) X= 50, s= 2, n= 100, z.025 = 1.96.
6. (a) X= 136.9, s= 22.6, n= 123, z.025 = 1.96.
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SECTION 5.1 237
7. (a) X= 178, s= 14, n= 120, z.025 = 1.96.
The confidence interval is 178 ±1.96(14/√120), or (175.495,180.505).
8. (a) X= 348.2, s= 5.1, n= 67, z.05 = 1.645.
The confidence interval is 348.2±1.645(5.1/√67), or (347.175,349.225).
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9. (a) X= 1.56, s= 0.1, n= 80, z.025 = 1.96.
The confidence interval is 1.56 ±1.96(0.1/√80), or (1.5381,1.5819).
10. (a) X= 85, s= 2, n= 60, z.025 = 1.96.
The confidence interval is 85 ±1.96(2/√60), or (84.494,85.506).
11. (a) X= 11.9, s= 1.1, n= 140, z.025 = 1.96.
The confidence interval is 11.9±1.96(1.1/√140), or (11.718,12.082).
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SECTION 5.1 239
12. (a) X= 50, s= 2, n= 100, z.05 = 1.645.
The lower confidence bound is 50 −1.645(2/√100) = 49.67.
13. (a) X= 136.9, s= 22.6, n= 123, z.02 = 2.05.
The lower confidence bound is 136.9−2.05(22.6/√123) = 132.72.
14. (a) X= 178, s= 14, n= 120, z.05 = 1.645.
The lower confidence bound is 178 −1.645(14/√120) = 175.898.
15. (a) X= 348.2, s= 5.1, n= 67, z.01 = 2.33.
The upper confidence bound is 348.2 + 2.33(5.1/√67) = 349.651.
16. (a) X= 1.56, s= 0.1, n= 80, z.10 = 1.28.
The upper confidence bound is 1.56 + 1.28(0.1/√80) = 1.574.
17. (a) X= 85, s= 2, n= 60, z.02 = 2.05.
The lower confidence bound is 85 −2.05(2/√60) = 84.471.
18. (a) X= 11.9, s= 1.1, n= 140, z.05 = 1.645.
The upper confidence bound is 11.9 + 1.645(1.1/√140) = 12.053.
19. With a sample size of 70, the standard deviation of Xis σ/√70. To make the interval half as wide,
the standard deviation of Xwill have to be σ/(2√70) = σ/√280. The sample size needs to be 280.
22. (a) True. This results from the expression X±1.96(s/√n), which is a 95% confidence interval.
23. (a) False. The confidence interval is for the population mean, not the sample mean. The sample mean is
known, so there is no need to construct a confidence interval for it.
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24. Let Xbe the number of 90% confidence intervals that fail to cover the true mean.
Section 5.2
1. (a) 28/70 = 0.4, or 40%.
(b) X= 28, n= 70, ˜p= (28 + 2)/(70 + 4) = 0.405405, z.025 = 1.96.
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The level is 0.9515, or 95.15%.
2. (a) X= 73, n= 100, ˜p= (73 + 2)/(100 + 4) = 0.72115, z.025 = 1.96.
The confidence interval is 0.72115 ±1.96p0.72115(1 −0.72115)/(100 + 4), or (0.635, 0.807).
(b) X= 73, n= 100, ˜p= (73 + 2)/(100 + 4) = 0.72115, z.005 = 2.58.
The confidence interval is 0.72115 ±2.58p0.72115(1 −0.72115)/(100 + 4), or (0.608, 0.835).
3. (a) X= 52, n= 70, ˜p= (52 + 2)/(70 + 4) = 0.72973, z.025 = 1.96.
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244 CHAPTER 5
(c) Let nbe the required sample size.
4. (a) X= 170, n= 444, ˜p= (170 + 2)/(444 + 4) = 0.38393, z.025 = 1.96.
The confidence interval is 0.38393 ±1.96p0.38393(1 −0.38393)/(444 + 4), or (0.3839, 0.4290).
(b) X= 170, n= 444, ˜p= (170 + 2)/(444 + 4) = 0.38393, z.005 = 2.58.
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SECTION 5.2 245
5. (a) X= 859, n= 10501, ˜p= (859 + 2)/(10501 + 4) = 0.081961, z.025 = 1.96.
6. X= 28, n= 70, ˜p= (28 + 2)/(70 + 4) = 0.405405, z.05 = 1.645.
The lower confidence bound is 0.405405 −1.645p0.405405(1 −0.405405)/(70 + 4), or 0.312.
9. (a) X= 30, n= 400, ˜p= (30 + 2)/(400 + 4) = 0.079208, z.025 = 1.96.
The confidence interval is 0.079208 ±1.96p0.079208(1 −0.079208)/(400 + 4), or (0.0529, 0.1055).
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246 CHAPTER 5
10. q= (1 −p)2. From part (a), we are 95% confident that 0.0529 < p < 0.1055.
11. (a) X= 89, n= 710, ˜p= (89 + 2)/(710 + 4) = 0.12745, z.05 = 1.645.
12. (a) X= 63, n= 150, ˜p= (63 + 2)/(150 + 4) = 0.42208, z.025 = 1.96.
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SECTION 5.2 247
13. (a) Let nbe the required sample size.
Then nsatisfies the equation 0.05 = 1.96p˜p(1 −˜p)/(n+ 4).
14. (a) Let nbe the required sample size. Then nsatisfies the equation 0.05 = 2.33p˜p(1 −˜p)/(n+ 4).
(b) X= 30, n= 200, ˜p= (30 + 2)/(200 + 4) = 0.15686, z.01 = 2.33.
15. (a) X= 293, n= 335, ˜p= (293 + 2)/(335 + 4) = 0.87021, z.05 = 1.645.
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248 CHAPTER 5
Section 5.3
1. (a) 1.796
3. (a) 95%
Page 248
4. False. The population must be approximately normal.
8. (a) X= 3410.14, s= 1.018, n= 8, t8−1,.025 = 2.365.
9. (a)
1.3 1.305 1.31 1.315 1.32 1.325
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10. X= 13, s= 2, n= 15, t15−1,.025 = 2.977.
14. (a) N −1 = 10 −1 = 9.
15. (a) SE Mean is StDev/√N, so 0.52640 = StDev/√20, so StDev = 2.3541.
16. The value 85 comes from a normal population with standard deviation σ= 8. Since σis known, a
95% confidence interval is based on z.025 = 1.96. The confidence interval is 85 ±1.96(8), or (69.32,
100.68).
17. (a) X= 21.7, s= 9.4, n= 5, t5−1,.025 = 2.776.
Section 5.4
1. X= 620, sX= 20, nX= 80, Y= 750, sY= 30, nY= 95, z.025 = 1.96.
The confidence interval is 750 −620 ±1.96p202/80 + 302/95, or (122.54, 137.46).
6. X= 14, sX= 7, nX= 43, Y= 25, sY= 9, nY= 78, z.025 = 1.96.
The confidence interval is 25 −14 ±1.96p72/43 + 92/78, or (8.107, 13.893).
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9. X= 242, sX= 20, nX= 47, Y= 220, sY= 31, nY= 42, z.025 = 1.96.
The confidence interval is 242 −220 ±1.96p202/47 + 312/42, or (11.018, 32.982).
difference in hardness.
12. The standard deviation of the difference between the means is pσ2
X/nX+σ2
Y/nY. Estimate σX≈
sX= 6.23 and σY≈sY= 4.34.
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SECTION 5.5 253
Section 5.5
1. X= 20, nX= 100, ˜pX= (20 + 1)/(100 + 2) = 0.205882,
Y= 10, nY= 150, ˜pY= (10 + 1)/(150 + 2) = 0.072368, z.05 = 1.645.
3. (a) X= 841, nX= 5320, ˜pX= (841 + 1)/(5320 + 2) = 0.158211,
Y= 134, nY= 1120, ˜pY= (134 + 1)/(1120 + 2) = 0.120321, z.01 = 2.33.
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254 CHAPTER 5
4. X= 43, nX= 50, ˜pX= (43 + 1)/(50 + 2) = 0.846154,
Y= 25, nY= 40, ˜pY= (25 + 1)/(40 + 2) = 0.619048, z.005 = 2.58.
6. X= 112, nX= 143, ˜pX= (112 + 1)/(143 + 2) = 0.77931,
Y= 182, nY= 349, ˜pY= (182 + 1)/(349 + 2) = 0.52137, z.025 = 1.96.
9. X= 30, nX= 164, ˜pX= (30 + 1)/(164 + 2) = 0.186747,
Y= 19, nY= 185, ˜pY= (19 + 1)/(185 + 2) = 0.106952, z.005 = 2.58.
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