Solutions to Suggested Problems
6.21
n=6
x̄ = 2.3
s = 6.39
t '
µ = 20 ambulance service claim of mean service time
23 & 20
' 1.15
6.39/ 6
df = 5
t < t 0.1 = 1.476
í Claim is reasonable.
6.22
n = 10
x̄ = 0.5060
s = 0.0040
t '
µ = 0.5000 value determines whether process is in control
0.5060 & 0.5000
' 4.743
0.0040/ 10
df = 5
t > t 0.001 = 3.250
í Process is out of control
6.24
n = 10
2
= 42.5
Find P ( 3.14 < s < 8.94). Let
9 (3.14)2
' 2.088
42.5
Find P ( 2.088 <
2
2
=
(n & 1) s 2
2
.
9 (8.94)2
' 16.92
42.5
< 16.92) = 1 - 0.05 - 0.01 = 0.94
7.4
n = 50
x̄ = 11,795
s = 14,054
= 0.05
maximum error (E) estimate:
E = | x̄ - µ| < z 0.05
14,054
' 1.960
50
2
14,054
' 3,896 with 95% confidence
50
7.5
Confidence interval: ( x̄ - E, x̄ + E) = (11,795 - 3,896, 11,795 + 3,896) with 95%
confidence.
7.6
n = 40
x̄ = 12.73
s = 2.06
a)
= 0.01
maximum error (E) estimate:
E = | x̄ - µ| < z 0.01
2
b)
2.06
' 2.576
40
2.06
' 0.84 with 99% confidence
40
= 0.02
E2 = | x̄ - µ| < z 0.02
2.06
2
40
' 2.326
2.06
' 0.76 with 98% confidence
40
Confidence interval: ( x̄ - E2 , x̄ + E2 ) = (12.73- 0.76, 12.73 +0.76) with 98%
confidence.
7.7
E = 0.5 min.
2.06
E = | x̄ - µ| < z
2
' 0.5
40
/2 = 1 - 0.9376 = 0.0624
'
z
2
0.5 40
' 1.535
2.06
= 0.1248
Confidence level = 1 - 0.1248 = 0.8752 that maximum error is 0.5 min.
7.9
n = 80
x̄ = 472.36
s = 62.35
62.35
E = | x̄ - µ| < z
2
' 10
'
z
80
2
/2 = 1 - 0.9243 = 0.0757
10 80
' 1.435
62.35
= 0.1514
Confidence level = 1 - 0.1514 = 0.8486 that maximum error is 10.
7.12
= 60
1 - = 0.9
E = 10
E = | x̄ - µ| < z 0.1
2
n $
7.24
60
# 10
1.645
n
# 10
n
1.645 (60)
10
n=8
60
n $ 97.4
E1 = | x̄ - µ| < t 0.05
2
0.54
8
' 2.365
0.54
' 0.45 with 98%
8
confidence
x̄ = 2.1
s = 0.54
Confidence interval ( x̄ - E1 , x̄ + E1 ) = (2.1 - 0.45,
2.1 + 0.45) with 95% confidence.
1 - = 0.95
df = 7
7.27
a)
Erroneously rejecting the null hypothesis that the dam was safe would cause
unneeded renovation and/or repairs.
b)
Erroneously accepting the null hypothesis that the dam was safe would cause loss
of property and/or life.
7.30
µ = 3.0000
= 0.0250
a)
H0 :
H1 :
µ = 3.0000
µ ú 3.0000
n = 30
H0 will be rejected if | x̄ - µ| > 0.0040
|z| =
|x̄&µ|
0.0040
'
' 1.012
0.0250/ 30
/ n
P ( |z| > 1.012 ) = 2 P ( z > 1.012 ) = 2 (1 - 0.8670) = 0.2660
b)
n = 30
Find P (x < 2.9960) + P(x > 3.0040) if µ = 3.0050 prevails
za =
2.9960&3.0050
' &1.972
0.0250 / 30
zb=
3.0040&3.0050
' &0.219
0.0250 / 30
Find P (z < z a ) + P (z > z b) = 0.02433 + 1 - 0.4125 = 0.6118
7.32
µ = 100
= 12
n = 40
= 0.01
Find x̄0 so that P (x > x̄0 ) = 0.01.
Let z a =
x̄0 & µ
/ n
'
x̄0 & 100
12/ 40
P (z > z a) = 0.01 is equivalent to z a > z0.01 = 2.326. Hence
x̄0 > 100 + 2.326
12
40
= 100 + 4.41.
.
7.39
n = 45
x̄ = 76.7
µ = 73.2
= 8.6
= 0.01
✑
✒
✓
H0 :
H1 :
µ = 7.32
µ > 73.2
= 0.01
z=
x̄ & µ
Critical value: z = 2.326
/ n
✔
76.7 & 73.2
' 2.730
8.6/ 45
✕
Reject H0. The data provides sufficient evidence that mean aptitude score is more than
73.2.
7.40
n = 35
x̄ = 1.4707
s = 0.5235
µ = 1.3
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 1.3
µ > 1.3
= 0.05
z=
x̄ & µ
Critical value: z = 1.645
/ n
✔
1.4707 & 1.3
' 1.929
0.5325/ 35
✕
Reject H0. The data provides sufficient evidence that mean cost is more than 1.3
(thousands).
7.41
n = 64
x̄ = 1,038
s = 146
µ = 1,000
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 1000
µ > 1000
= 0.05
x̄ & µ
z=
Critical value: z = 1.645
/ n
✔
1,038 & 1,000
' 2.08
146/ 64
✕
Reject H0. The data provides sufficient evidence that mean number of acceptable pieces is
more than 1,000.
7.43
n = 60
x̄ = 33.8
s = 6.1
µ = 32.6
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 32.6
µ > 32.6
= 0.05
z=
x̄ & µ
Critical value: z = 1.645
/ n
✔
33.8 & 32.6
' 1.52
6.1/ 60
✕
Fail to reject H0. The data does not provide sufficient evidence that mean travel time is
more than 32.6 minutes.
7.44
n=6
x̄ = 58,392
s = 648
µ = 58,000
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 58,000
µ ú 58,000
= 0.05
t=
x̄ & µ
Critical value: t
/2
= 2.571
s/ n
df = 5
✔
58,392 & 58,000
' 1.481
648/ 5
✕
Fail to reject H0. The data does not provide sufficient evidence that mean compressive
strength is not than 58,000 psi.
7.48
n=5
x̄ = 14.4
s = 0.158
µ = 14.0
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 14.0
µ > 14.0
= 0.05
t=
x̄ & µ
Critical value: t = 2.132
s/ n
df = 4
✔
14.4 & 14.0
' 5.660
0.158/ 4
✕
Reject H0. The data provides sufficient evidence that mean tar content is more than 14.0.
7.49
n=5
x̄ = 14.7
s = 0.742
µ = 14.0
= 0.05
✑
✒
✓
H0 :
H1 :
µ = 14.0
µ > 14.0
= 0.05
t=
x̄ & µ
Critical value: t = 2.132
s/ n
df = 4
✔
14.7 & 14.0
' 2.109
0.742/ 4
✕
Fail to reject H0. The data does not provide sufficient evidence that mean tar content is
more than 14.0.
The increase in the standard deviation allows for more variability in the population data.
7.63
µ 1 = 0.249
1 = 0.003
µx
1 & x2
x1 & x2
µ 2 = 0.255
2 = 0.002
' µ 1 & µ 2 ' &0.006
'
2
1
%
2
2
' 0.003605
P( x1 - x2 < 0 ) = P( z <
0 & (&0.006)
) = P (z < 1.664) = 0.9519
0.003605
7.64
n1 = 71
x̄1 = 83.2
n2 = 75
x̄2 = 90.8
s1 = 19.3
s2 = 21.4
= 0.05
(a)
✑
✒
✓
H0 :
H1 :
µ1 = µ2
µ1 ú µ2
= 0.05
z=
(x̄1 & x̄2) & (0)
Critical value: -z
/2
= -1.960
x̄1 & x̄2
x̄1 & x̄2
✔
✕
2
1
'
n1
%
2
2
n2
'
19.32 21.42
%
' 3.369
71
75
(83.2 & 90.8) & 0
' &2.2558
3.369
Reject H0. The data provides sufficient evidence that the two population means are
different.
(b)
'
d=
2
1
%
2
2
|µ 1 & µ 2|
'
'
19.32 % 21.42 ' 28.81
|0 & (&12)|
' 0.416
28.81
From Table 8 (c) the probability of failing to reject the null hypothesis when µ 1 - µ 2 = -12
prevails is 0.
7.65
n1 = 60
x̄1 = 292.5
n2 = 60
x̄2 = 266.1
s1 = 15.6
s2 = 18.2
= 0.01
✑
✒
✓
H0 :
H1 :
µ 1 - µ 2 = 20
µ 1 - µ 2 > 20
= 0.01
z=
(x̄1 & x̄2) & (20)
Critical value: z = 2.326
x̄1 & x̄2
x̄1 & x̄2
✔
✕
'
2
1
n1
%
2
2
n2
'
15.62 18.22
%
' 3.095
71
75
(292.5 & 266.1) & 20
' 2.068
3.095
Fail to reject H0. The data does not provide sufficient evidence that the mean weekly
salaries for men are $20 more than the mean weekly salaries for women.
7.67
n1 = 8
x̄1 = 9.67
n2 = 10
x̄2 = 7.43
s1 = 1.81
s2 = 1.48
1
=
2
=
= 0.05
✑
✒
✓
H0 :
H1 :
µ 1 - µ 2 = 1.5
µ 1 - µ 2 > 1.5
= 0.05
t=
(x̄1 & x̄2) & (1.5)
sp
Critical value: t = 1.746
1
1
%
n1 n2
df = 16
2
2
sp
✔
'
n1 % n2 & 2
(9.67 & 7.43) & 1.5
1.633
✕
2
(n1 & 1) s1 % (n2&1) s2
'
7(1.81)2 % 9(1.48)2
' 2.665
16
' 0.9556
1
1
%
8 10
Fail to reject H0. The data does not provide sufficient evidence that the mean deniers from
the two spinning machines differ by more than 1.5.
7.68
n1 = 10
x̄1 = 70
n2 = 10
x̄2 = 74
s1 = 3.37
s2 = 5.40
=
1
2
=
= 0.05
✑
✒
✓
H0 :
H1 :
µ1 = µ2
µ1 < µ2
= 0.05
t=
(x̄1 & x̄2) & (0)
Critical value: -t = -1.734
1
1
%
n1 n2
sp
df = 18
2
2
sp
✔
'
✕
n1 % n2 & 2
(70 & 74) & 0
4.50
2
(n1 & 1) s1 % (n2&1) s2
'
9(3.37)2 % 9(5.40)2
' 20.258
18
' &1.9876
1
1
%
10 10
Reject H0. The data provides sufficient evidence that the training under Method B is more
effective.
7.69
n1 = 9
x̄1 = 58
n2 = 6
x̄2 = 51.8
s1 = 10.4
s2 = 12.7
1
=
2
=
= 0.01
✑
✒
✓
H0 :
H1 :
µ1 = µ2
µ1 ú µ2
= 0.01
t=
(x̄1 & x̄2) & (1.5)
sp
Critical value: t
/2
= 3.012
1
1
%
n1 n2
df = 13
2
2
sp
✔
'
✕
n1 % n2 & 2
(58 & 51.8) & 0
11.34
2
(n1 & 1) s1 % (n2&1) s2
'
8(10.4)2 % 5(12.7)2
' 128.594
13
' 1.037
1 1
%
9 6
Fail to reject H0. The data does not provide sufficient evidence that the means from the
two populations are different.
7.71
n = 10
d̄ = -0.02
sd = 0.0287
= 0.05
✑
✒
✓
d̄ = 0
d̄ ú 0
H0 :
H1 :
= 0.05
t=
d̄ & 0
Critical value: -t
/2
= -2.262
sd / n
df = 9
✔
&0.02 & 0
' &2.2059
0.0287/ 10
✕
Fail to reject H0. The data does not provide sufficient evidence that the two scales weigh
differently.