Solucionario Capitulo 10 Estadistica Walpole

Chapter 10
One- and Two-Sample Tests of
Hypotheses
10.1 (a) Conclude that fewer than 30% of the public are allergic to some cheese products
when, in fact, 30% or more are allergic.
(b) Conclude that at least 30% of the public are allergic to some cheese products
when, in fact, fewer than 30% are allergic.
10.2 (a) The training course is effective.
(b) The training course is effective.
10.3(a) The firm is not guilty.
(b) The firm is guilty.
10.4 (a) α = P (X ≤ 5 | p = 0.6) + P (X ≥ 13 | p = 0.6) = 0.0338 + (1 − 0.9729) = 0.0609.
(b) β = P (6 ≤ X ≤ 12 | p = 0.5) = 0.9963 − 0.1509 = 0.8454.
β = P (6 ≤ X ≤ 12 | p = 0.7) = 0.8732 − 0.0037 = 0.8695.

(c) This test procedure is not good for detecting differences of 0.1 in p.

10.5 (a) α = P (X < 110 | p = 0.6) + P (X > 130 | p =0.6) = P (Z < −1.52) + P (Z >
1.52) = 2(0.0643) = 0.1286.
(b) β = P (110 < X < 130 | p = 0.5) = P (1.34 < Z < 4.31) = 0.0901.
β = P (110 < X < 130 | p = 0.7) = P (−4.71 < Z < −1.47) = 0.0708.

(c) The probability of a Type I error is somewhat high for this procedure, although
Type II errors are reduced dramatically.

10.6 (a) α = P (X ≤ 3 | p = 0.6) = 0.0548.

(b) β = P (X > 3 | p = 0.3) =1 − 0.6496 = 0.3504.
β = P (X > 3 | p = 0.4) = 1 − 0.3823 = 0.6177.
β = P (X > 3 | p = 0.5) = 1 − 0.1719 = 0.8281.
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Chapter 10 One- and Two-Sample Tests of Hypotheses

10.7 (a) α = P (X ≤ 24 | p = 0.6) = P (Z < −1.59) = 0.0559.

(b) β = P (X > 24 | p = 0.3) = P (Z > 2.93) = 1 − 0.9983 = 0.0017.
β = P (X > 24 | p = 0.4) = P (Z > 1.30) = 1 − 0.9032 = 0.0968.
β = P (X > 24 | p= 0.5) = P (Z > −0.14) = 1 − 0.4443 = 0.5557.

10.8 (a) n = 12, p = 0.7, and α = P (X > 11) = 0.0712 + 0.0138 = 0.0850.
(b) n = 12, p = 0.9, and β = P (X ≤ 10) = 0.3410.
√
10.9 (a) n = 100, p = 0.7, µ = np = 70, and σ = npq =
Hence z = 82..5−70 = 0.3410. Therefore,
4 583

(100)(0.7)(0.3) = 4.583.

α = P (X > 82) = P (Z > 2.73) = 1 − 0.9968 = 0.0032.
(b) n = 100, p = 0.9, µ = np = 90,and σ =
z = 82.53−90 = −2.5. So,

√

npq =

(100)(0.9)(0.1) = 3. Hence

β = P (X ≤ 82) = P (X < −2.5) = 0.0062.
10.10 (a) n = 7, p = 0.4, α = P (X ≤ 2) = 0.4199.

(b) n = 7, p = 0.3, β = P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.6471 = 0.3529.
√
10.11 (a) n = 70, p = 0.4, µ = np = 28, and σ = npq = 4.099, with z = 23..5−28 = −1.10.
4 099
Then α = P (X < 24) = P (Z < −1.10) = 0.1357.
√
(b)n = 70, p = 0.3, µ = np = 21, and σ = npq = 3.834, with z = 23..5−21 = 0.65
3 834
Then β = P (X ≥ 24) = P (Z > 0.65) = 0.2578.
√
10.12 (a) n = 400, p = 0.6, µ = np = 240, and σ = npq = 9.798, with
z1 =

259.5 − 240
= 1.990,
9.978

and z2 =

220.5 − 240
= −1.990.
9.978

Hence,
α = 2P (Z < −1.990) = (2)(0.0233) = 0.0466.
(b) When p = 0.48, then µ = 192 and σ = 9.992, with
z1 =220.5 − 192
= 2.852,
9.992

and z2 =

259.5 − 192
= 6.755.
9.992

Therefore,
β = P (2.852 < Z < 6.755) = 1 − 0.9978 = 0.0022.
10.13 From Exercise 10.12(a) we have µ = 240 and σ = 9.798. We then obtain
z1 =

214.5 − 240
= −2.60,
9.978

and z2 =

265.5 − 240
= 2.60.
9.978

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Solutions for Exercises in Chapter 10

So
α = 2P (Z < −2.60) = (2)(0.0047) = 0.0094.Also, from Exercise 10.12(b) we have µ = 192 and σ = 9.992, with
z1 =

214.5 − 192
= 2.25,
9.992

and z2 =

265.5 − 192
= 7.36.
9.992

Therefore,
β = P (2.25 < Z < 7.36) = 1 − 0.9878 = 0.0122.
10.14 (a) n = 50, µ = 15, σ = 0.5, and σX =
¯
Hence, α = P (Z < −1.41) = 0.0793.

0
√.5
50

14.9−15
0.071

= 0.071, with z =

= −1.41.

(b) If µ = 14.8, z = 14.09.−14.8 = 1.41. So,β = P (Z > 1.41) = 0.0793.
071
If µ = 14.9, then z = 0 and β = P (Z > 0) = 0.5.

10.15 (a) µ = 200, n = 9, σ = 15 and σX =
¯
z1 =

15
3

= 5. So,

191 − 200
= −1.8,
5

and z2 =

with α = 2P (Z < −1.8) = (2)(0.0359) = 0.0718.
(b) If µ = 215, then z − 1 =

191−215
5

= −4.8 and z2 =

209 − 200
= 1.8,
5

209−215
5

= −1.2, with

β = P (−4.8 < Z < −1.2) = 0.1151 −...