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CHAPTER 8
Section 8-2
8-1
a.) The confidence level for
x
−
2.14
σ
/
n
≤
µ
≤
x
+
2.14
σ
/
n
is determined by the
x
−
2.49
σ
/
n
≤
µ
≤
x
+
2.49
σ
/
n
is determined by the
x
−
1.85
σ
/
n
≤
µ
≤
x
+
1.85
σ
/
n
is determined by the
by the value of z
0
which is 2.14. From Table II, we find
Φ(2.14)
= P(Z<2.14) = 0.9838 and the
confidence level is 100(1-.032354) = 96.76%.
b.) The confidence level for
by the value of z
0
which is 2.14. From Table II, we find
Φ(2.49)
= P(Z<2.49) = 0.9936 and the
confidence level is is 100(1-.012774) = 98.72%.
c.) The confidence level for
by the value of z
0
which is 2.14. From Table II, we find
Φ(1.85)
= P(Z<1.85) = 0.9678 and the
confidence level is 93.56%.
8-2
a.) A z
α
= 2.33 would give result in a 98% two-sided confidence interval.
b.) A z
α
= 1.29 would give result in a 80% two-sided confidence interval.
c.) A z
α
= 1.15 would give result in a 75% two-sided confidence interval.
a.) A z
α
= 1.29 would give result in a 90% one-sided confidence interval.
b.) A z
α
= 1.65 would give result in a 95% one-sided confidence interval.
c.) A z
α
= 2.33 would give result in a 99% one-sided confidence interval.
a.) 95% CI for
8-3
8-4
µ
,
n
=
10,
σ
=
20
x
=
1000,
z
=
1.96
x
−
z
σ
/
n
≤
µ
≤
x
+
z
σ
/
n
1000
−
1.96(20 / 10 )
≤
µ
≤
1000
+
1.96(20 / 10 )
b.) .95% CI for
987.6
≤
µ
≤
1012.4
µ
,
n
=
25,
σ
=
20
x
=
1000,
z
=
1.96
x
−
z
σ
/
n
≤
µ
≤
x
+
z
σ
/
n
1000
−
1.96(20 / 25 )
≤
µ
≤
1000
+
1.96(20 / 25 )
992.2
≤
µ
≤
1007.8
c.) 99% CI for
µ
,
n
=
10,
σ
=
20
x
=
1000,
z
=
2.58
x
−
z
σ
/
n
≤
µ
≤
x
+
z
σ
/
n
1000
−
2.58(20 / 10 )
≤
µ
≤
1000
+
2.58(20 / 10 )
d.) 99% CI for
983.7
≤
µ
≤
1016.3
µ
,
n
=
25,
σ
=
20
x
=
1000,
z
=
2.58
x
−
z
σ
/
n
≤
µ
≤
x
+
z
σ
/
n
1000
−
2.58(20 / 25 )
≤
µ
≤
1000
+
2.58(20 / 25 )
989.7
≤
µ
≤
1010.3
8-1
8-5
Find n for the length of the 95% CI to be 40. Z
a/2
= 1.96
1/2 length
=
(1.96)(20) /
n
=
20
39.2
=
20
n
39.2
n
=
20
2
=
3.84
Therefore,
n
= 4.
8-6
Interval (1):
3124.9
≤
µ
≤
3215.7
and Interval (2)::
3110.5
≤
µ
≤
3230.1
Interval (1): half-length =90.8/2=45.4 Interval (2): half-length =119.6/2=59.8
a.)
x
1
=
3124.9
+
45.4
=
3170.3
x
2
=
3110.5
+
59.8
=
3170.3
The sample means are the same.
b.) Interval (1):
3124.9
≤
µ
≤
3215.7
was calculated with 95% Confidence because it has a smaller
half-length, and therefore a smaller confidence interval. The 99% confidence level will make the interval
larger.
8-7
a.) The 99% CI on the mean calcium concentration would be longer.
b). No, that is not the correct interpretation of a confidence interval. The probability that
µ
is between
0.49 and 0.82 is either 0 or 1.
c). Yes, this is the correct interpretation of a confidence interval. The upper and lower limits of the
confidence limits are random variables.
95% Two-sided CI on the breaking strength of yarn: where
x
= 98 ,
σ
= 2 , n=9 and z
0.025
= 1.96
x
−
z
0 . 025
σ
/
n
≤
µ
≤
x
+
z
0 . 025
σ
/
n
8-8
98
−
1 . 96 ( 2 ) / 9
≤
µ
≤
98
+
1 . 96 ( 2 ) /
96 . 7
≤
µ
≤
99 . 3
8-9
9
95% Two-sided CI on the true mean yield: where
x
= 90.480 ,
σ
= 3 , n=5 and z
0.025
= 1.96
x
−
z
0 . 025
σ
/
n
≤
µ
≤
x
+
z
0 . 025
σ
/
n
90 . 480
−
1 . 96 ( 3 ) / 5
≤
µ
≤
90 . 480
+
1 . 96 ( 3 ) /
87 . 85
≤
µ
≤
93 . 11
8-10
5
99% Two-sided CI on the diameter cable harness holes: where
x
=1.5045 ,
σ
= 0.01 , n=10 and
z
0.005
= 2.58
x
−
z
0 . 005
σ
/
n
≤
µ
≤
x
+
z
0 . 005
σ
/
n
1 . 5045
−
2 . 58 ( 0 . 01 ) / 10
≤
µ
≤
1 . 5045
+
2 . 58 ( 0 . 01 ) / 10
1 . 4963
≤
µ
≤
1 . 5127
8-2
8-11
a.) 99% Two-sided CI on the true mean piston ring diameter
For
α
= 0.01, z
α/2
= z
0.005
= 2.58 , and
x
= 74.036,
σ
= 0.001, n=15
σ
σ
x
−
z
0.005
≤ µ ≤
x
+
z
0.005
n
n
74.036
−
2.58
0.001
15
≤ µ ≤
74.036
+
2.58
0.001
15
74.0353
≤ µ ≤
74.0367
b.) 95% One-sided CI on the true mean piston ring diameter
For
α
= 0.05, z
α
= z
0.05
=1.65 and
x
= 74.036,
σ
= 0.001, n=15
≤
µ
n
0.001
74.036
−
1.65
≤
µ
15
x
−
z
0.05
74.0356
≤ µ
8-12
a.) 95% Two-sided CI on the true mean life of a 75-watt light bulb
For
α
= 0.05, z
α/2
= z
0.025
= 1.96 , and
x
= 1014,
σ
=25 , n=20
σ
x
−
z
0.025
1014
−
1.96
σ
n
25
≤
µ
≤
x
+
z
0.025
σ
n
25
20
20
1003
≤
µ
≤
1025
≤
µ
≤
1014
+
1.96
b.) 95% One-sided CI on the true mean piston ring diameter
For
α
= 0.05, z
α
= z
0.05
=1.65 and
x = 1014,
σ
=25 , n=20
x
−
z
0.05
1014
−
1.65
σ
n
≤
µ
≤
µ
25
20
1005
≤
µ
8-3
8-13
a) 95% two sided CI on the mean compressive strength
z
α/2
= z
0.025
= 1.96, and
x
= 3250,
σ
2
= 1000, n=12
σ
σ
x
−
z
0.025
≤ µ ≤
x
+
z
0.025
n
n
3250
−
1.96
31.62
12
3232.11
≤
µ
≤
3267.89
≤
µ
≤
3250
+
1.96
31.62
12
b.) 99% Two-sided CI on the true mean compressive strength
z
α/2
= z
0.005
= 2.58
x
−
z
0.005
3250
−
2.58
σ
n
≤
µ
≤
x
+
z
0.005
σ
n
31.62
12
31.62
12
3226.4
≤
µ
≤
3273.6
≤
µ
≤
3250
+
2.58
8-14
95% Confident that the error of estimating the true mean life of a 75-watt light bulb is less than 5 hours.
For
α
= 0.05, z
α/2
= z
0.025
= 1.96 , and
σ
=25 , E=5
z
σ
n
=
a
/2
E
8-15
2
1.96(25)
=
5
2
=
96.04
Always round up to the next number, therefore
n=97
Set the width to 6 hours with
σ
= 25, z
0.025
= 1.96 solve for n.
1/2 width
=
(1.96)( 25) /
n
=
3
49
=
3
n
49
n
=
3
2
=
266.78
Therefore, n=267.
8-16
99% Confident that the error of estimating the true compressive strength is less than 15 psi.
For
α
= 0.01, z
α/2
= z
0.005
= 2.58 , and
σ
=31.62 , E=15
z
σ
n
=
a
/2
E
Therefore,
n=30
2
2.58(31.62)
=
15
2
=
29.6
≅
30
8-4
8-17
To decrease the length of the CI by one half, the sample size must be increased by 4 times (2
2
).
z
α
/ 2
σ
/
n
=
0.5
l
Now, to decrease by half, divide both sides by 2.
(
z
α
/ 2
σ
/
n
) / 2
=
(
l
/ 2) / 2
(
z
α
/ 2
σ
/ 2
n
)
=
l
/ 4
(
z
α
/ 2
σ
/ 2
2
n
)
=
l
/ 4
Therefore, the sample size must be increased by 2
2
.
8-18
If n is doubled in Eq 8-7:
x
−
z
α
/ 2
σ
n
≤
µ
≤
x
+
z
α
/ 2
σ
n
z
α
/ 2
σ
2
n
=
z
α
/ 2
σ
1.414
n
=
z
α
/ 2
σ
1.414
n
=
z
α
/ 2
σ
1
1.414
n
The interval is reduced by 0.293 29.3%
If n is increased by a factor of 4 Eq 8-7:
z
α
/ 2
σ
4
n
=
z
α
/ 2
σ
2
n
=
z
α
/ 2
σ
2
n
=
1
z
α
/ 2
σ
2
n
The interval is reduced by 0.5 or �½.
Section 8-3
8-19
t
0.025,15
=
2.131
t
0.005, 25
=
2.787
t
0.05,10
=
1.812
t
0.001,30
=
3.385
t
0.10, 20
=
1.325
8-20
a.)
b.)
c.)
d.)
t
0.025,12
=
2.179
t
0.025, 24
=
2.064
t
0.005,13
=
3.012
t
0.0005,15
=
4.073
t
0.05,14
=
1.761
t
0.01,19
=
2.539
t
0.001, 24
=
3.467
8-21
a.)
b.)
c.)
8-5

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