1
Homework 7 (15.2 pts)
due Oct. 28
(1.2 pts.) 8.10 (p.354). For each of the given pairs of Pvalues and significance levels,
state whether H
0
should be rejected:
a) Pvalue = 0.084,
= 0.05
Fail to reject H
0
, 0.084 > 0.05
b) Pvalue = 0.003.,
= 0.001
Fail to reject H
0
, 0.003 > 0.001
c) Pvalue = 0.048,
= 0.05
Reject H
0
, 0.048 ≤ 0.05
d) Pvalue = 0.084,
= 0.10
Reject H
0
, 0.084 ≤ 0.10
e) Pvalue = 0.039,
= 0.01
Fail to reject H
0
, 0.039 > 0.01
f) Pvalue = 0.017,
= 0.10
Reject H
0
, 0.017 ≤ 0.10
(2 pts.) 8.12 (you do not need to do the 7 steps, p.354). New purchased automobile tires
of a certain type are supposed to be filled to a pressure of 34 psi. Let μ denote the true
average pressure. A test of H
0
: μ = 34 versus H
a
: μ ≠ 34 will be based on a large sample of
tires so that the z test statistic will have approximately a standard normal distribution.
Determine the value of z and the corresponding Pvalue in each of the following cases:
a) n = 50, x
̅
= 34.43, s = 1.06
°
=
± − ?
0
² ³´
µ
=
34.43
−
34
1.06
³
50
µ
= 2.868
Pvalue = 2P (Z > 2.868) = 2(1 – P(Z ≤ 2.87)) = 2(1 – 0.9979) = 2(0.0021) = 0.0042
b) n = 50, x
̅
= 33.57, s = 1.06
°
=
± − ?
0
² ³´
µ
=
33.57
−
34
1.06
³
50
µ
=
−
2.868
Pvalue = 2P (Z < 2.87) = 2(0.0021) = 0.0042
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c) n = 32, x
̅
= 33.25, s = 1.89
°
=
± − ?
0
² ³´
µ
=
33.25
−
34
1.89
³
32
µ
=
−
2.24
Pvalue = 2P (Z < 2.24) = 2(0.0125) = 0.0250
d) n = 36, x
̅
= 34.66, s = 2.53
°
=
± − ?
0
² ³´
µ
=
34.66
−
34
2.53
³
36
µ
= 1.565
Pvalue = 2P (Z > 1.565) = 2(1 – P(Z ≤ 1.57)) = 2(1 – 0.9418) = 2(0.0582) = 0.1164
(1.5 pt.) 8.16 (use the 7 steps, p.355). To obtain information on the corrosionresistance
properties of a certain type of steel conduit, 35 specimens are buried in soil for an
extended period. The maximum penetration (in mils) is then measured for each
specimen, yielding a sample mean penetration of 52.7 and a sample standard deviation
of 4.8. The conduits were manufactured with the specification that the true average
penetration be at most 50 mils. Does the sample data indicate that specifications have
not been met?
1. μ = the true average maximum penetration distance for corrosion of a certain type of steel
conduit.
2. H
0
: μ = 50
H
a
: μ > 50
3. α = 0.05
4.
°
=
± − ?
0
² ³´
µ
=
52.7
−
50
4.8
³
35
µ
= 3.328
5. P = P(Z > 3.33) = 1 – P(Z ≤ 3.33) = 1 – 0.9996 = 0.0004
6. Reject H
0
because 0.0004 > 0.05
7. The data does give strong support (Pvalue = 0.0004) to the claim that the true average
maximum penetration distance for corrosion of a certain type of steel conduit is greater than
50 mils.
(1.5 pts.) 8.18 (these do NOT use the 7 steps, p.367). The paint used to make lines on
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 Spring '08
 Staff
 Statistics, Normal Distribution, PValues, H0, Reject H0, oil workers

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