Chapter 15 Solutions
15.1. (a)
If
µ
=
12, the distribution is approximately
Normal with mean
µ
=
12 g
/
dl and standard de
viation
σ/
√
50
.
=
0
.
2263 g
/
dl.
(b)
A result like
x
=
11
.
3g
/
dl lies out toward the low tail of the
curve, while 11
.
8g
/
dl is fairly close to the middle.
If
µ
=
12 g
/
dl, observing a value similar to 11
.
/
dl would not be too surprising, but
11
.
/
dl is less likely, and it therefore provides some evidence that
µ<
12 g
/
dl.
12.23 12.45 12.68
11.77
11.55
11.32
12
11.3
11.8
15.2. (a)
If
µ
=
115, the distribution is approximately
Normal with mean
µ
=
115 and standard deviation
√
25
=
6.
(b)
The actual result lies out toward the
high tail of the curve, while 118.6 is fairly close to the
middle. If
µ
=
115, observing a value similar to 118.6
would not be too surprising, but 125.8 is less likely, and it therefore provides some evidence
that
µ>
115.
121
127
133
109
103
97
115
125.8
118.6
15.3.
H
0
:
µ
=
12 g
/
dl
;
H
a
:
12 g
/
dl.
15.4.
H
0
:
µ
=
115
;
H
a
:
115.
15.5.
H
0
:
µ
=
51 mpg
;
H
a
:
51 mpg.
15.6.
H
0
:
µ
=
20 min
;
H
a
:
µ
6=
20 min.
15.7.
Hypotheses should be stated in terms of
µ
, not
x
.
Note:
Students who think about this problem a bit more might also point out that
1000 g (2.2 lb) is a dangerously low birth weight; babies smaller than this are classifed as
extremely low birth weight (ELBW).
15.8.
For
x
=
0
.
3,
z
=
0
.
3
−
0
1
/
√
10
.
=
0
.
95.
15.9.
For
x
=
11
.
/
dl,
z
=
11
.
3
−
12
1
.
6
/
√
50
.
=−
3
.
09. For
x
=
11
.
/
dl,
z
=
11
.
8
−
12
1
.
6
/
√
50
.
0
.
88.
15.10.
For
x
=
118
.
6,
z
=
118
.
6
−
115
30
/
√
25
=
0
.
60. For
x
=
125
.
8,
z
=
125
.
8
−
115
30
/
√
25
=
1
.
80.
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Chapter 15
Tests of SigniFcance: The Basics
15.11.
The applet’s output for
n
=
18,
σ
=
60,
and
x
=
17 is shown on the right. (This
Normal curve has mean 0 and standard
deviation
σ/
√
18
.
=
14
.
14.)
15.12.
In Exercise 15.8, the test statistic
z
was found to be 0.95, so for the alternative
µ>
0,
the
P
value is
P
(
Z
≥
0
.
95
)
=
0
.
1711.
15.13.
For
x
=
11
.
3g
/
dl,
z
=−
3
.
09 and
P
=
P
(
Z
<
−
3
.
09
)
=
0
.
0010. For
x
=
11
.
8g
/
dl,
z
0
.
88 and
P
=
P
(
Z
<
−
0
.
88
)
=
0
.
1894. The ±rst
P
value tells us that values of
x
similar to 11.3 g
/
dl would rarely occur when
H
0
is true, while the second
P
value is larger,
indicating that results similar to this one give little reason to doubt
H
0
.
15.14.
For
x
=
118
.
6, the
P
value is
P
(
x
>
118
.
6
)
=
P
(
Z
>
0
.
6
)
=
0
.
2743. For
x
=
125
.
8,
P
=
P
(
x
>
125
.
8
)
.
=
P
(
Z
>
1
.
80
)
=
0
.
0359. A small
P
value (such as 0.0359) tells us
that values of
x
similar to 125.8 would rarely occur when
H
0
is true, while
P
=
0
.
2743
indicates that results similar to 118.6 give little reason to doubt
H
0
.
15.15.
For
x
=
11
.
/
dl, we found
z
0
.
88 and
P
=
0
.
1894; this is not signi±cant at
α
=
0
.
05 or
α
=
0
.
01.
15.16.
For
x
=
125
.
8, we found
z
=
1
.
80 and
P
=
0
.
0359; this is signi±cant at
α
=
0
.
05, but
not at
α
=
0
.
01.
15.17. (a)
The results observed in this study would rarely have occurred by chance if vitamin
C were ineffective.
(b)
P
<
0
.
01 means that results similar to those observed would occur
less than 1% of the time if vitamin C supplements had no effect.
15.18.
We test
H
0
:
µ
=
5mgversus
H
a
:
µ<
5mg. The test statistic is
z
=
4
.
62
−
5
0
.
92
/
√
45
.
2
.
77,
and the
P
value is
P
=
P
(
Z
<
−
2
.
77
)
=
0
.
0028. This is very strong evidence that
5mg
.
15.19. State:
Do student SATM scores improve on their second attempt?
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 Spring '09
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 Normal Distribution, Standard Deviation, Mean, Null hypothesis, mean femininity score

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