Chapter 7 Solutions
7.1. (a)
The standard error of the mean is
s
√
n
=
$570
√
15
.
=
$27
.
1109.
(b)
The degrees of freedom
are df
=
n
−
1
=
14.
7.2.
In each case, use df
=
n
−
1; if that number is not in Table D, drop to the
lower
degrees
of freedom.
(a)
For 95% con±dence and df
=
11, use
t
∗
=
2
.
201.
(b)
For 99% con±dence
and df
=
23, use
t
∗
=
2
.
807.
(c)
For 90% con±dence and df
=
199, we drop to df
=
100
and use
t
∗
=
1
.
660. (Software gives
t
∗
=
1
.
6525 for df
=
199.)
7.3.
For the mean monthly rent, the 95% con±dence interval is $570
±
(
2
.
145
)(
$105
/
√
15
)
=
$570
±
$58
.
15
=
$511.85 to $628.15.
7.4.
The margin of error for 90% con±dence would be smaller—and the interval would be
narrower—because we are taking a greater risk that the interval does
not
include the true
mean
µ
.
7.5. (a)
Yes,
t
=
2
.
35 is signi±cant when
n
=
15. This
can be determined either by comparing to the df
=
14
line in Table D (where we see that
t
>
2
.
264, the 2%
critical value) or by computing the twosided
P
value
(which is
P
=
0
.
0340).
(b)
No,
t
=
2
.
35 is not
signi±cant when
n
=
6, as can be seen by comparing
to the df
=
5 line in Table D (where we see that
t
<
2
.
571, the 2.5% critical value) or by computing the
twosided
P
value (which is
P
=
0
.
0656).
(c)
Student
sketches will likely be indistinguishable from Normal
distributions; careful students may try to show that the
t
(
5
)
distribution is shorter in the
center and heavier to the left and right than the
t
(
14
)
distribution (as is the case here), but in
reality, the difference is nearly imperceptible.
123
–1
–2
–3
0
2.35
–2.35
–1
–2
–3
0
2.35
–2.35
t
(14)
t
(5)
7.6.
For the hypotheses
H
0
:
µ
=
$550 vs.
H
a
:
µ>
$550, we ±nd
t
=
570
−
550
105
/
√
15
.
=
0
.
7377 with
df
=
14, for which
P
.
=
0
.
2364. We do not reject
H
0
;wedo not have suf±cient evidence to
conclude that the mean rent is greater than $550.
Note:
Computing a Pvalue is not really necessary; a t
(
14
)
distribution is similar
enough to a Normal distribution that we can quickly determine that t
.
=
0
.
74 gives no
reason to doubt the null hypothesis.
7.7.
Software will typically give a more accurate value for
t
∗
than that given in Table D, but
otherwise the details of this computation are the same as what is shown in the textbook:
df
=
7,
t
∗
=
2
.
3646, 5
±
t
∗
(
3
.
63
/
√
8
)
=
5
±
3
.
0348
=
1.9652 to 8.0348, or about 2.0 to 8.0
hours per month.
198
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199
7.8.
We wish to test
H
0
:
µ
=
0 vs.
H
a
:
µ
6=
0, where
µ
is the mean of (drink A rating) minus
(drink B rating). Compute the differences for each subject (
−
2, 1, 5, 11, and 8), then Fnd
the mean and standard deviation of these differences:
x
.
=
4
.
6 and
s
.
=
5
.
2249. Therefore,
t
=
4
.
6
−
0
5
.
2249
/
√
5
.
=
1
.
9686 with df
=
4, for which
P
=
0
.
1204. We do not have enough
evidence to conclude that there is a difference in preference.
7.9.
Using the mean and standard deviation from the previous exercise, the 95% conFdence
interval is 4
.
6
±
(
2
.
7764
)(
5
.
2249
/
√
5
)
=
4
.
6
±
6
.
4876
=
1.8876 to 11.0876.
7.10.
See also the solutions to Exercises 1.30, 1.64 and 1.144. The CO
2
data are sharply
rightskewed (clearly nonNormal). However, the robustness of the
t
procedures should make
them safe for this situation because the sample size is large (
n
=
48). The bigger question
is whether we can treat the data as an SRS; we have recorded CO
2
emissions for every
country with a population over 20 million, rather than a random sample.
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 Spring '07
 Guggenberger
 Normal Distribution, Standard Deviation

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