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Chapter9 825
59, a. (198,230) b. .048 2M. Test Ho: w= Svs. Hy wz S
c. A 90% prediction interval is (.149, .279) a, Do not reject Hy because fo5.1 = 2.179 > 11.61
b. Do not reject Hy because 199512 = 2.179 > |-1.6]
61. 246 ¢. Do not reject Ho because 005.24 = 2.797 > |-2.6]
63. a. A 95% confidence interval for the mean is (.163, d. Reject Ho because £995.24 = 2.797 < |-3.9]
-174). Yes, this interval is below the interval for 59(a). 3, Because 1 = 2.24 > 1.708 = fsas: teject Ho: t= 360.
b. (089, 326) Yes, this suggests contradiction of prior belief.
65. (0.1263, 0.3018) 25. Because |z| = 3.37 > 1.96, reject the null hypothesis.
67. a. yes b. (196.88, 222.62) It appears that this population exceeds the national
” average in 1Q.
69. &. V(p) = 07/222, 0, = 0/ (EP
d. Pat the 3's far from 0 to minimize op 27. a. no,t = ~.02b. 58
©. BP ty/an_18//Sx2, 029.93, 30.15) ¢. n= 20 total observations
73, a..00985 _b. .0578 29 Babaibs PSR TASS tong dank ec Hy
78. a. (— tors.n-1.6¥ — (8/ VA) orsn1.6
b & ate 025.01 (s/Va)tors0-13) 31. Because t = —1.24 > ~1.397 = —to, we do not have
ee evidence to question the prior belief.
TI a2" be nf". (n+ 12", 1 — (n+ 2",
(29.9, 39.3) with confidence level 9785 35. a. The distribution is fairly symmetric, without outliers.
b. Because 1 = 4.25 > 3.499 = f95, there is strong
79. a. P(A\MAy) = 95? b. P(A,MA2) > 90 evidence to say that the amount poured differs from
€. P(A\MAg) > 1 = a — 933 PAO... AD) the industry standard, and indeed bartenders tend to
Dla ay — a exceed the standard,
¢. Yes, the test in (b) depends on normality, and a normal
probability plot gives no reason to doubt the
Chapter 9 assumption.
d. .643, .185, 016
1. ayes b.no eno dyes eno Fe yes 37, a, Do not reject Ho: p = -10 in favor of Ha: p > 10
5. Ho: a = .05 vs. Ha: 6 < .05. Type Terror: Conclude that because 2 = 13375 Leds, Because; “the: null
the standard deviation is less than .05 mm when it is hypothesis is not rejected, there could be a type II
really equal to .05 mm. Type II error: Conclude that the ‘errOty
standard deviation is .05 mm when it is really less than be 49,27, © 362
05. 39. a. Do not reject Ho: p = .02 in favor of Hy: p < .02
7. A type Lerror here involves saying that the plant is not in because == —1.1 > ~1.645. There is no strong
compliance when in fact it is. A type II error occurs when evidence suggesting that the inventory be postponed.
we conclude that the plant is in compliance when in fact it b. .195. ¢. <.0000001.
isn't. A government regulator might regard the type II 44. a, Reject Hp because z — 3.08 > 2.58. b. 03
error as being more serious. 4 en eet tee
43. Using n = 25, the probability of 5 or more leaky faucets
9. a. Ry is 0980 if p = .10, and the probability of 4 or fewer leaky
b. A type I error involves saying that the two companies faucets is .0905 if p = .3. Thus, the rejection region is
are not equally favored when they are. A type II error 5 or more, « = .0980, and ff = .0908.
involves saying that the two companies are equally % ¢
favored when they are not. 45. a. reject’ b. reject €. donot reject d. reject
¢. binomial, n = 25, p = .5; 0433 e. do not reject
d. .3, 4881; .4, 8452; .6, 8452; .7, 4881
e. If only 6 favor the first company, then reject the null 47- a 0778 b. 1841. 0250 d. .0066 €. .5438
hypothesis and conclude that the first company is not 49, a, p — 0403 b.P =.0176 ¢.P =.1304
preferred. d. P = 6532 e.P = .0021 f.P = .000022
A, a iHlos = 10. vs. Hees 10) .0009 51. Based on the given data, there is no reason to believe
¢. 5319..0076 dic =2.58 ec = 1.96 that pregnant women differ from others in terms of true
£, ¥= 10.02, so do not reject Hy average serum receptor concentration.
g. Recalibrate if < —2.58 or 2 > 2.58
53. a. Because the P-value is .17, no modification is
13. b. 00043, 0000075, less than .01 indicated b. 957
15, a. .0301, b: 0030 ¢:.0040 55. Because ¢ = ~1.759 and the P-value = .089, which is
17. a. Because z = 2.56 > 2.33, reject Hy b..84 less than .10, reject Ho: = 3.0 against a two-tailed
alternative at the 10% level. However, the P-value
e, 142d. .0052 exceeds .05, so do not reject Hp at the 5% level. There
19. a, Because z = ~2.27 > —2.58, do not reject Hy
b. 22 ©. 22
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826 Chapter 10
is just a weak indication that the percentage is not equal to. 91. a. For the test of Hy: {t= flo vs. Hy: > flo at level 2,
3% (lower than 3%). reject Ho if 2Exi/pto > Loy
87 a Test He: ji = 10 va Fig WS 10. For the test of Hy: 11 = lo vs. Hy: ft < fup at level 2,
b. Because the P-value is .017 <.05, reject Ho, reject Ho if 2Exi/po < Zia.2n
suggesting that the pens do not meet specifications. For the test of Ho: = lo vs. Ha: # Ho at level 2,
¢, Because the P-value is .045 > .01, do not reject Ho, reject Hy if 2ExJ/ pl > 72 94 OF
suggesting there is no reason to say the lifetime is if 2x0 < 2/220
inadequate. b. Because Xx; = 737, the test statistic is 2Exi/jt0
d. Because the P-value is .0011, reject Ho. There is good = 19.65, which gives a P-value of .52. There is no
evidence showing that the pens do not meet reason to reject the null hypothesis.
specifications,
93. a. yes
61. a. 98, .85, 43, .004, 0000002
b. 40, .11, .0062, 0000003
¢. Because the null hypothesis willbe rejected with high Chapter 10
probability, even with only slight departure from the
null hypothesis, it is not very useful to do.aOl level 4. g, —4:it doesn’t b..0724, .269
est ¢. Although the CLT implies that the distribution will be
63. b. 36.61 ¢. yes approximately normal when the sample sizes are each
100, the distribution will not necessarily be normal
65. a. Dx > cb yes when the sample sizes are each 10.
67. Yes, the test is UMP for the alternative Hy : 0 >.5 3. Do not reject Hp because z = 1.76 < 2.33
because the tests for Hy : 0 =.5 vs. Hy : 0 = py all
have the same form for any po > .5. 5. a. H, says that the average calorie output for sufferers is
more than | cal/em?/min below that for non-sufferers.
69. b. .05 Reject Hy in favor of H, because z = —2.90 < 2.33
¢. .04345, .05826; Because .04345 < .05, the test is not b. 0019 819 d..66
unbiased.
4. 05114; not most powerful 7. Yes, because z = 1.83 > 1.645.
71. b. The value of the test statistic is 3.041, so the P-value is 9%» & —¥=6.2
081, compared to 089 for Exercise 55. b. 2 = 1.14, two-tailed P-value = .25, so do not reject