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the null hypothesis that the population means are
73. A sample size of 32 should suffice. equal.
¢. No, the values are positive and the standard deviation
78. a. Test Ho: = 2150 vs. Hy: > 2150 exceedsthe ean.
b. = (¥=2150)/(s//n)_¢. 1.33 d. 101 d. 95% CI: (10.0, 29.8)
e, Do not reject Ho at the .05 level.
IL. a. A.95% Cl for the true difference, fast food mean — not
77. Because t= .77 and the P-value is .23, there is no fast food mean is (219.6, 538.4)
evidence suggesting that coal increases the mean heat b. The one-tailed P-value is .014, so reject the null
flux. hypothesis of a 200-calorie difference at the .05
79. Conclude that activation time is too slow at the .05 level, Jevel, and conclude that- yes, there 1sstrong evidence.
but not at the .01 level. 13. 22. No.
81. A normal probability plot gives no reason to doubt the 45 p, It increases.
normality assumption. Because the sample mean is 9.815,
giving = 4.75 and a (upper tail) P-value of 00007, 17. Because z= 1.36, there is no reason to reject the
reject the null hypothesis at any reasonable level. The hypothesis of equal population means (p = .17).
true average flame time is too high.
19. Because z = .59, there is no reason to conclude that the
83. Assuming normality, calculate ¢ = 1.70, which gives a population mean is higher for the no-involvement group
two tailed P-value of .102. Do not reject the null (p = 28).
hypothesis Ho: ft = 1.75.
21. Because = —3.35 < -3.30=fooia2 yes, there is
85. The P-value for a lower tail test is .0014 (normal evidence that experts do hit harder.
approximation, .0005), so it is reasonable to reject the
idea that p = .75 and conclude that fewer than 75% of 23+ b-No _¢. Because |¢] = |~.38] < 2.228 = 1025.10, no,
mnschanios can identify the’ problem, there is no evidence of a difference.
87. Because 1 = 6.43, giving an upper tail P-value of 25+ Because the one-tailed P-value is 005 < .01, conclude at
0000002, conclude that the population mean time the O1-Ievelithat the difference ts.as tated:
exceeds 15 minutes, This could result in a type I error.
89. Because the P-value is .013 > .01, do not reject the null 27+ Yes, because ¢ = 2.08 with P-value = .046.
hypothesis at'the .01 level. 29 b. (127.6, 202.0) ¢. 131.8
--- Trang 840 ---
Chapter 10 827
31. Because ¢ = 1.82 with P-value .046 < .05, conclude at #start withX inC1, Y inc2
the .05 level that the difference exceeds 1. let k3 =N(c1)
let k4 =N(c2)
33. a. (F—¥) £ bayrmin-2 “Spat sample k3clc3;
b. (~.24, 3.64) replace:
¢. (~.34, 3.74), which is wider because of the loss of a sample k4c2c4;
degree of freedom replace:
35. a. The slender distribution appears to have a lower mean Lehi = mesh (Co) aean lea)
and lower variance. Spec eleoe?
b. With ¢= 1.88 and a P-value of .097, there is no eng
significant difference at the .05 level 71. a. Here is a macro that can be executed 999 times in
37. With ¢ = 2.19 and a two-tailed P-value of .031, there is a MINIEAB: ,
significant difference at the .05 level but not the .01 level. detartwithX incl; vin c2
2 let k3 =N(c1)
39. With ¢ = 3.89 and one-tailed P-value = .006, conclude let k4 =N(c2)
at the 1% level that true average movement is less for the sample k3clc3;
TightRope treatment. Normality is important, but the replace.
normal probability plot does not indicate a problem. sample k4c2c4;
replace.
41. a. The 95% confidence interval for the difference of let k2 = medi(c3)-medi(c4)
means is (,000046, .000446), which has only positive shack koce'ce
values. This omits 0 as a possibility, and says that the end
conventional mean is higher.
b. With 1 = 2.68 and P-value = .010, reject at the 05 73. a. (.593, 1.246)
level the hypothesis of equal means in favor of the b. Here is a macro that can be executed 999 times in
conventional mean being higher. MINITAB:
# start withX inC1, Y inC2
43. With = 1.87 and a P-value of .049, the difference is let k3 -N(c1)
(barely) significantly greater than 5 at the .05 level. let k4 =N(c2)
45.a.No b.—49.1 ©4911 SAMPLE ICS 1.82
replace.
47. 1 2 3 4 sample k4c2c4;
x 10 20 30 40 replace.
5 im 1 31 4 let k5 = stdev(c3) /stdev(c4)
stackk5c12¢12
end
49, a. Because Il = |-4.841 > 1.96, conclude that there is a
difference. Rural residents are more favorable to the 75. a. Because ¢ = —2.62 with a P-value of .018, conclude
increase. that the population means differ. At the 5% level,
b. 9967 blueberries are significantly better.
b. Here is a macro that can be executed repeatedly in
51. (016, 171) MIRTTAB:
53. Because 2 = 4.27 with P-value .000010, conclude that PStare with dats ine) Beoub yar ine?
the radiation is beneficial. detkt = N(eL)
Sample k3clc3.
55. a. Ho: ps = pa He: ps > Pr unstack c3 ¢4c5;
b. (X — Xo)in subs c2.
c. (Xy — X2)/VXa FX let k9 =mean(c4)-mean(c5)
d. With z = 2.67, P = .004, reject Hy at the .01 level. stack k9 c6 c6é
end
57. 769
71. a. Because f = 4.46 with a two-tailed P-value of .122,
22; Because 2 21d with B— OZ orelect Ha atthe (01 there is no evidence of unequal population variances.
level. Conclude that lefties are more accident-prone. by Hece: ie we mmacte: thavvan be es ected sepemedly
61. a..0175 hb. .1642 0200 d. 0448, MINITAB:
0035 let kl =n(C1)
Sample Klclc3.
63. No, because f = 1.814 < 6.72 = Foio2. unstack c3 4 c5;
subs c2.
65. Because f = 1.2219 with P = .505, there is no reason to letike= evdev(cdii/eedavicsi
question the equality of population variances. Been ke ce ce
67. 8.10 end
69. a. (158,735) 79, a. A MINITAB macro is given in #75(b).
b. Here is a macro that can be executed 999 times in gy. a. (11.85, -6.40)
MINITAB? b. See Exercise 57(a) in Chapter 8.