Inference for Proportions
12.1 (a) Population: the 175 residents of Tonya's dorm; p is the proportion who like the food.
(b) j?
=
0.28.
12.2 (a) The population is the 2400 students at Glen's college, and p is the proportion who believe
tuition is too high. (b)
fi
=
0.76.
12.3 (a) The population is the 15,000 alumni, and
#I
is the proportion who support the president's
decision.
(b) b
=
0.38.
12.4 (a) Nothe
population is not large enough relative to the sample. (b) Yeswe
have an SRS,
the population is 48 times as large as the sample, and the success count (38) and failure count (12)
i
are both greater than 10. (c) Nothere
were only 5 or 6 "successes" in the sample.
12.5 (a) Nonp,
and n(l

0,) are less than 10 (they both equal 5). (b) Nothe
expected number
of failures is less than 10 (n(1

p,)
=
2). (c) Yeswe
have an SRS, the population is more than 10
times as large as the sample, and np,
=
n(l

=
10.
12.6 (a) SEb
=
d(0.54)(0.46)/1019
=
0.01561, so the 95
%
confidence interval is 0.54
r
(1.96)(0.01561)
=
0.51 to 0.57. The margin of error is about 3%, as stated.
(b) We weren't given sample sizes for each gender. (However, students who know enough
algebra can get a good estimate of those numbers by solving the system x
+
y
=
1019 and
0.65~
+
0.43~
=
550: approximately 508 men and 511 women.)
(c) The margin of error for women alone would be greater than 0.03 since the sample size is
smaller.
12.7 (a) The methods can be used here, since we assume we have an SRS from a large population,
and all relevant counts are more than 10. For TVs in rooms: b1
2
0.66 and SE5
=
d(0.66)(0.34)/1048
=
0.01463, so the 95% confidence interval is 0.66
?
(1.96)(0.01463)
=
0.631 to 0.689. For preferring Fox: bi
=
0.18 and SE4
=
d(0.18)(0.82)/1048
=
0.01187, so the 95% confidence interval is 0.18
5
(1.96)(0.01187)
=
0.157 to 0.203.
(b) In both cases, the margin of error for a 95% confidence interval ("19 cases out of 20") was
(no more than) 3
%
.
(c) We test H,: p
=
0.5 versus Ha: p
>
0.5. The test statistic is
z
=
(0.660.50)/
10.36,
which gives very strong evidence against
H, (P
<
0.0002); we conclude that more than half
of teenagers have TVs in their rooms. (Additionally, the interval from (a) does not include
i
0.50 or less.) With the TI83,
z
=
10.379 and P
=
1.577
x
12.8 (a)
=
.66, and since
4
=
132 and n(1

b)
=
68 are both greater than 10, the confidence
interval
based
on
z
can
be
used.
The 95% confidence
interval for p
is
.66
?
(1.96)V'((.66)(.34)/200)
=
.66
?
0.06565, or 0.59435 to 0.72565.