that New Yorkers sleep an amount different than
8 hours per night on average. (e) Yes, as we re
jected
H
0
.
332
APPENDIX B.
END OF CHAPTER EXERCISE SOLUTIONS
4.7
t
?
19
is 1.73 for a onetail. We want the lower
tail, so set 1.73 equal to the T score, then solve
for ¯
x
: 56.91.
4.9
(a) For each observation in one data set,
there is exactly one speciallycorresponding ob
servation in the other data set for the same
geographic
location.
The
data
are
paired.
(b)
H
0
:
μ
diff
= 0 (There is no difference in av
erage daily high temperature between January
1, 1968 and January 1, 2008 in the continental
US.)
H
A
:
μ
diff
6
= 0 (Average daily high tem
perature in January 1, 1968 is different than the
average daily high temperature in January, 2008
in the continental US.) (c) Independence: loca
tions are random and represent less than 10% of
all possible locations in the US. We are not given
the distribution to check the skew.
In prac
tice, we would ask to see the data to check this
condition, but here we will move forward under
the assumption that it is not strongly skewed.
(d)
T
= 1
.
60,
df
= 51

1 = 50.
→
pvalue be
tween 0.1, 0.2 (two tails!). (e) Since the pvalue
> α
(since not given use 0.05), fail to reject
H
0
. The data do not provide strong evidence of
a temperature change in the continental US on
the two dates.
(f) Type 2, since we may have
incorrectly failed to reject
H
0
.
There may be
an increase, but we were unable to detect it.
(g) Yes, since we failed to reject
H
0
, which had
a null value of 0.
4.11
(a) (0.28, 2.48).
(b) We are 95% con
fident that the average daily high on January
1, 2008 in the continental US was 0.24 degrees
lower
to 2.44 degrees
higher
than the average
daily high on January 1, 1968. (c) No, since 0
is included in the interval.
4.13
(a) Each of the 36 mothers is related to
exactly one of the 36 fathers (and viceversa),
so there is a special correspondence between
the mothers and fathers.
(b)
H
0
:
μ
diff
= 0.
H
A
:
μ
diff
6
= 0. Independence: random sample
from less than 10% of population. The skew of
the differences is, at worst, slight.
T
= 2
.
72,
df
= 36

1 = 35
→
pvalue
≈
0
.
01.
Since p
value
<
0.05, reject
H
0
. The data provide strong
evidence that the average IQ scores of mothers
and fathers of gifted children are different, and
the data indicate that mothers’ scores are higher
than fathers’ scores for the parents of gifted chil
dren.
4.15
Independence: Random samples that are
less than 10% of the population.
In practice,
we’d ask for the data to check the skew (which
is not provided), but here we will move forward
under the assumption that the skew is not ex
treme (there is some leeway in the skew for such
large samples). Use
t
?
999
≈
1
.
65. 90% CI: (0.16,
5.84).
We are 90% confident that the average
score in 2008 was 0.16 to 5.84 points higher than
the average score in 2004.
4.17
(a)
H
0
:
μ
2008
=
μ
2004
→
μ
2004

μ
2008
=
0 (Average math score in 2008 is equal to aver
age math score in 2004.)
H
A
:
μ
2008
6
=
μ
2004
→
μ
2004

μ
2008
6
= 0 (Average math score in 2008 is
different than average math score in 2004.) Con
ditions necessary for inference were checked in
Exercise
4.15
.
T
=

1
.
74,
df
= 999
→
pvalue
between 0.05, 0.10.
Since the pvalue
< α
, re
ject
H
0
. The data provide strong evidence that
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 Spring '14
 Simple random sample, Level of measurement