text stringlengths 0 6.73k |
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Xyn |
37. a..9615 b..0617 bs bat |
39. a. Sinn + D4 b..25, n(n + 1)Qn + 1/24 19876 be GOS |
41, 10:52.74 15. a. 0=SDX2/(2n) b. 74.505 |
43. AB, 17. b.4/9 |
45. b. My) = ML — F/2n)} 19. a. p=21—.30=.20 .p = (1002 —9)/70 |
47. Because 7; is the sum of v independent random variables, 31. a.15 byes. .4437 |
each distributed as 7, the Central Limit Theorem applies. . |
23, a. 0 = (2 1)/(1-3) =3 |
53. 2.3.2. 10.04, the square of the answer to (a) b. = [-n/Zin(x)] —1= 3.12 |
ST: Dalla 2)s ae 2 3 25. p=r/(r+x)=.15 This is the number of successes |
b. 2v3(v1 + v2 —2)/[vi(v2 — 2)°(v2 —4)}, v2 > 4 over the number of trials, the same as the result in |
61. 0.432 Exercise 21, It is not the same as the estimate of |
Exercise 17. |
65. a. The approximate value, .0228, is smaller because of 50. RASA. dl HC |
skewness in the chi-squared distribution 2a =, OX be = 5 UX? |
b. This approximation gives the answer 03237, agreeing 9g, ) — 5~x2/(2n) — 74.505, the same as in Exercise 15 |
with the software answer to this number of decimals. ‘ |
Ea |
67. No, the sum of the percentiles is not the same as the b. 4 20In(2) = 10.16 |
percentile of the sum, except that they are the same for 34, j= — In(p)/24 = .0120 |
the 50th percentile. For all other percentiles, the |
percentile of the sum is closer to the SOth percentile 33. No, statistician A does not have more information, |
than is the sum of the percentiles |
69. a. 2360, 73.70 b..9713 |
37. WS max(x1, X2, Xn) SO < min@r, x2, ..-n)) |
71. 9685 |
39. a. 2X(n — X/[n(n — 1) |
73. 9093 Independence is questionable because con- was ~ |
sumption one day might be related to consumption the 41. a Xb. (XK —e)/V1— 1/n) |
next day. |
--- Trang 837 --- |
824 Chapter 8 |
43. a.V(0)=P/[n(n+2)] db. Pn 33. a. (38.081, 38.439) —_b, (100.55, 101.19), yes |
¢. The variance in (a) is below the bound of (b), but the | _ |
theorem does not apply because the domain is a 35+ @ Assuming normality, a 95% lower confidence bound |
finclionét tie'parsmeter is 8.11. When the bound is calculated from repeated |
independent samples, roughly 95% of such bounds |
45. a. Xb. Nu, o7/n) should be below the population mean, |
c. Yes, the variance is equal to the Cramér-Rao bound b. A 95% lower prediction bound is 7.03. When the |
d. The answer in (b) shows that the asymptotic bound is calculated from repeated independent |
distribution of the theorem is actually exact here. samples, roughly 95% of such bounds should be |
‘ below the value of an independent observation, |
47. a. 2a” |
b. The answer in (a) is different from the answer, 37. a.378.85 b.413.09 _e. (333.88, 407.50) |
1/(20*), to 46(a), so the information does depend on |
dive paratneteriestioti, 39, 95% prediction interval: (.0498, .0772) |
49, 1 = 6/ (616 — ty —...~ ts) = 6/(01 +202 +... + 6x6) = 0436, 41+ B+ (169.36, 179.37) |
Wie Sheek hn Beeb b. (134.30, 214.43), which includes 152 |
¢. The second interval is much wider, because it allows |
53. 1.275, s = 1.462 for the variability of a single observation. |
6 . d. The normal probability plot gives no reason to doubt |
55. b. no, E(a*) = 07/2, so 26° is unbiased normality. This is especially important for part (b), but |
59, 416, 448 the large sample size implies that normality is not so |
critical for (a). |
61. d(X) = (—1)*, 6(200) = 1, 6199) = -1 |
45. 2.18307 b.3.940 95d. .10 |
63. b. B =X xiy;/3D17 = 30.040, the estimated minutes |
per item; a =F S(yi — pri)? = 16.912; 7s 34, 5:60), |
258.= 751 49. a. (7.91, 12.00) |
b. Because of an outlier, normality is questionable for |
this data set. |
Chapter 8 ¢. In MINITAB, put the data in Cl and execute the |
following macro 999 times |
1. 299.5% b.85% 6.297 9 dL LIS Let k3 =N(c1) |
sample k3clc3; |
3. a. Anarrower interval has a lower probability _b. No, replace, |
Ass not random let k1 =mean(c3) |
¢. No, the interval refers to 1, not individual observations giecme cscs |
d. No, a probability of 95 does not guarantee 95 aha |
successes in 100 trials |
51. a. (26.61, 32.94) |
3 52218) G28) eS de b. Because of outliers, the weight gains do not seem |
7. Increase n by a factor of 4. Decrease the width by a factor normally distributed. |
of 5. ¢. In MINITAB, see Exercise 49(c). |
9. a. (¥— 1.6450/ fi, 00); (4.57, 00) 53.2. G8.46, 38.84) |
b. G22 0/V/n,00) b. Although the normal probability plot is not perfectly |
a: straight, there is not enough deviation to reject |
c. (=00,8 + 24+ @/ Vi); (—90, 59.7) normality. |
11. 950; .8724 (normal approximation), .8731 (binomial) ¢ In MINITAB, see Exercise 49(c). |
13. a. (.99, 1.07) b. 158 SBuri ass LOTS, 205345) |
b. Because of an outlier, normality is questionable for |
15. a. 80% — b.98% —€. 75% this data set. |
¢. In MINITAB, see Exercise 49(c). |
17. .06, which is positive, suggesting that the population |
mean change is positive 57. a. In MINITAB, put the data in Cl and execute the |
following macro 999 times |
19. (513, .615) Tet k3 oN(61) |
21. 218 sample k3cl¢3; |
replace. |
23. (.439, .814) let k1 = stdev(c3) |
stackklc5c5 |
25. a.381 —b.339 end |
29, 01341 b.1.753. 1.708. 1.684 b. Assuming normality, a 95% confidence interval for |
©2704 cis (3.541, 6.578), but the interval is inappropriate |
because the normality assumption is clearly not |
31. a.2.228 2131 2.947 4.604 satisfied. |
e.2.492 £.2.715 |
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