AP Statistics Chapter 12
Write-ups
Problem 12.7
Step 1: 1-proportion Z interval
p = the proportion of teams with televisions in their rooms.
Step 2: The problem tells us to treat the sample as if it were an SRS. The population of teenagers is
certainly larger than 10 x 1048.
æ 692 ö
np̂ = 1048 ç
= 692 ³ 10
è 1048 ÷ø
692 ö
æ
n (1 - p̂ ) = 1048 ç 1 = 356 ³ 10
è 1048 ÷ø
Step 3: For televisions:
p̂ (1 - p̂ )
p̂ ± z*
n
692 æ 356 ö
ç
÷
692
1048 è 1048 ø
± 1.96
1048
1048
(.632, .689)
Concerning Fox network: The rules of thumb are met similarly to that above.
189 æ 859 ö
ç
÷
189
1048 è 1048 ø
± 1.96
1048
1048
(.157, .204)
Step 4: We are 95% confident that the true proportion of teenagers with televisions in their rooms is
between 63 and 69%. We are 95% confident that the true proportion of teenagers who name Fox
network as their favorite network is between 16 and 20%.
In repeated random sampling this method captures the true proportion 95% of the time.
AP Statistics Chapter 12
Write-ups
Problem 12.9
Step 1: 1-proportion Z interval
p = the proportion of applicants who lied about degrees.
Step 2: The problem tells us to treat the sample as if it were an SRS. The population of applicants is
laid to be large..
æ 15 ö
np̂ = 83 ç ÷ = 15 ³ 10
è 83 ø
æ 68 ö
n (1- p̂ ) = 83 ç ÷ = 68 ³ 10
è 83 ø
Step 3:
p̂ ± z *
p̂ (1- p̂ )
n
15 æ 68 ö
ç ÷
15
83 è 83 ø
± 1.645
83
83
(.111,.250)
Step 4: We are 90% confident that the true proportion of applicants who lied about degrees is between
11 and 25%.
In repeated random sampling, this method captures the true proportion 90% of the time.
AP Statistics Chapter 12
Write-ups
Problem 12.16
Step 1: p is the proportion of Harley-Davidson motorcycles among stolen motorcycles.
H 0 : p = 0.14
H a : p > 0.14
Step 2: We can think of our sample of stolen motorcycles as an SRS of all stolen motorcycles, given in
problem.
æ 2490 ö
æ 6734 ö
np = 9224 ç
= 2490 > 10 and n(1- p) = 9224 ç
= 6734 > 10
÷
è 9224 ø
è 9224 ÷ø
The population of motorcycles is well over 92,240.
2490
- .14
= 9224
= 36
p0 (1- p)0
.14(.86)
9224
n
p̂ - p0
Step 3: Z =
Step 4:
Step 5: P-value: P(Z > 36) » 0
Step 6: Reject H0, a test statistic this large almost never occurs by chance alone. p » 0 < .05 = a .
Step 7: We have strong evidence that the proportion of Harley-Davidsons among stolen motorbikes is
much greater than the proportion of Harleys among all motorcycles. In other words, Harleys seem to
be stolen preferentially.
Problem 12.23
Step 1: 2-proportion Z-interval
p1 is the proportion of mice in breeding condition in high acorn location.
p2 is the proportion of mice in breeding condition in low acorn location.
Step 2: We have 2 independent simple random samples, from the problem and its description of the
experiment. We meet the large population requirements as there are certain to be more than 72(10)
and 17(10) mice in the two populations.
æ 54 ö
æ 18 ö
n1 pˆ1 = 72ç ÷ ³ 5 n1 (1- pˆ1) = 72ç ÷ ³ 5
è 72 ø
è 72 ø
æ 10 ö
æ7ö
n 2 pˆ 2 = 17ç ÷ ³ 5 n 2 (1- pˆ 2 ) = 17ç ÷ ³ 5
è 17 ø
è 17 ø
Step 3
pˆ1 - pˆ 2 ± z *
pˆ1 (1- pˆ1) pˆ 2 (1- pˆ 2 )
+
=
n1
n2
54 (1- 54 ) 10 (1- 10 )
72
72 + 17
17
: 54 - 10 ± 1.645
72 17
(-.0518,.3752)
72
17
Step 4: We are 90% confident that the true difference in proportions of mice in breeding condition
between areas with added acorns and areas without extra acorns is between -.0518 and .3752. We
estimate that there are between 5% fewer to 37% more mice in the extra acorn location.
AP Statistics Chapter 12
Write-ups
Problem 12.26b
Step 1: p1 is the proportion of strokes in the aspirin treated group.
p2 is the proportion of strokes in the aspirin and dipyridamole treated group.
H 0 : p1 - p2 = 0
H a : p1 - p2 ¹ 0
Step 2: Patients are randomly assigned to treatments, so we have independent random samples of
patients for each population. The population of stroke patients is much larger than 10 times both
samples combined.
n1 p̂1 = 1649 (.125) ³ 5 n1(1- p̂1 ) = 1649 (.875) ³ 5
n2 p̂2 = 1650 (.095) ³ 5 n2 (1- p̂2 ) = 1650 (.905) ³ 5
Step 3:
: Z=
p̂1 - p̂2
p̂(1- p̂) p̂(1- p̂)
+
n1
n2
= 2.73
Step 4:
p̂ =
206 +157
= .1100
1649 +1650
Step 5: P(Z < -2.73 or Z > 2.73) = .006
Step 6: Reject H0, a test statistic this extreme will occur by chance alone less than 1% of the time.
p = .006 < .01 = a
Step 7: We have strong evidence that the proportion of strokes among patients treated with aspirin is
different from the proportion of strokes when both aspirin and dipyridamole are given.
AP Statistics Chapter 12
Write-ups
Problem 12.29
Step 1: p1 is the proportion of HIV patients who develop AIDS in the AZT treatment group.
p2 is the proportion of HIV patients who develop AIDS in the placebo control group.
H 0 : p1 - p2 = 0
H a : p1 - p2 < 0
Step 2: Patients are randomly assigned to treatments, so we have independent random samples of
patients for each population. The population of HIV patients is much larger than 10 times both samples
combined.
æ 17 ö
æ
17 ö
n1 p̂1 = 435 ç
³ 5 n1(1- p̂1 ) = 435 ç1³5
÷
è 435 ø
è 435 ÷ø
æ 38 ö
æ
38 ö
n2 p̂2 = 435 ç
³ 5 n2 (1- p̂2 ) = 435 ç1³5
÷
è 435 ø
è 435 ÷ø
Step 3:
: Z=
p̂1 - p̂2
p̂(1- p̂) p̂(1- p̂)
+
n1
n2
= -2.93
Step 4:
17 + 38
= .063
435 + 435
p̂1 = .039
p̂2 = .087
p̂ =
Step 5: P(Z < -2.93) = .0017
Step 6: Reject H0, a test statistic this extreme will occur by chance alone only about 1 in a 1000 times.
p = .0017 < .05 = a
Step 7: We have strong evidence that the proportion of AIDS cases among patients treated with AZT
is less than the proportion of AIDS when only a placebo is given.
AP Statistics Chapter 12
Write-ups
Problem 12.30
Step 1: p1 is the proportion of students from urban/suburban backgrounds who succeeded in a chemical
engineering course. p2 is the proportion of students from rural/small town backgrounds who succeeded
in a chemical engineering course.
H 0 : p1 - p2 = 0
H a : p1 - p2 ¹ 0
Step 2:
We are uncertain that the samples are simple random samples, but they do appear to be independent.
The large population requirements are met, as there are certain to be more than 650 urban/suburban
and 550 rural/small town students at NC State University.
æ 30 ö
n2 p̂2 = 55 ç ÷ ³ 5
è 55 ø
n2 (1- p̂2 ) = 55(1- 30) ³ 5
æ 52 ö
n1 p̂1 = 65 ç ÷ ³ 5
è 65 ø
æ 52 ö
n1(1- p̂1 ) = 65 ç1- ÷ ³ 5
è 65 ø
Step 3:
52 + 30
p̂ =
= .6833
65 + 55
p̂1 = .8
p̂2 = .545
z=
p̂1 - p̂2
æ1 1ö
p̂(1- p̂)ç + ÷
è n1 n2 ø
=
52
65
- 30
55
= 2.986
æ 1
1ö
.6833(1- .6833)ç + ÷
è 65 55 ø
Step 4:
The shaded area is on each side, but too small to be visible.
Step 5:
P(z £ -2.986 or z ³ 2.986) = .002826
Step 6:
Reject H0, a value this extreme may occur by chance alone less than 1% of the time.
p = .003 < .05 = a
Step 7:
We have strong evidence that the proportion of students who succeed is different for urban/suburban
and rural/small town students at NC State University.