AP Stats Unit 11 Review
1. Old World cuckoos, famous as the voice in the “cuckoo clock,” are brood parasites. They lay
their eggs in the nests of other bird species. The “host”
Wren
Tree Pipit
Pied Wagtail
birds then raise the young cuckoo, their own young
19.85
21.05
21.05
perishing in the process. Biologists wondered
20.05
21.85
21.85
whether the size of the cuckoo eggs might vary with
20.25
22.05
21.85
the species being parasitized, especially as some of
20.85
22.45
21.85
the target birds are much smaller than the cuckoo. The
20.85
22.65
22.05
data are egg lengths of cuckoo eggs (in mm) laid in
20.85
23.25
22.45
the nests of Wrens, Tree Pipits, and Pied Wagtails.
These can be considered to be random samples of all
21.05
23.25
22.65
cuckoo eggs laid in the nests of the given species.
21.05
23.25
23.05
21.05
23.45
23.05
21.25
23.45
23.25
a. Is there a significant difference in the mean
21.45
23.65
23.45
cuckoo egg length of eggs laid in Wren and
22.05
23.85
24.05
Tree Pipit nests?
22.05
24.05
24.05
22.05
24.05
24.05
22.25
24.05
24.85
H 0 : m1 - m2 = 0 The mean lengths of cuckoo eggs laid in Wren and Tree Pipit nests are the same.
H a : m1 - m2 ¹ 0 The mean lengths of cuckoo eggs laid in Wren and Tree Pipit nests are not the same.
The samples are random and are collected independently. The normal probability plots are very
roughly linear, meeting or normality requirement.
Tree Pipit
t=
x1 - x2
2
2
s1 s2
+
n1 n2
= -6.467 df = 24.4
Wren
P(t < -6.467 or t > 6.467) = 1´10-6
Reject H0, p < 10-4 < a = .05 . A test statistic this extreme will occur by chance alone less than 1%
of the time.
We have strong evidence that the mean length of cuckoo eggs laid in Wren nests is different
from the mean of those laid in Tree Pipit nests.
b. Is there a significant difference in the mean cuckoo egg length of eggs laid in Tree Pipit
and Pied Wagtail nests?
H 0 : m1 - m2 = 0 The mean lengths of cuckoo eggs laid in Tree Pipit and Pied Wagtail nests are the same.
H a : m1 - m2 ¹ 0 The mean lengths of cuckoo eggs laid in Tree Pipit and Pied Wagtail nests are not the same.
The samples are random and are collected independently. The normal probability plots are very
roughly linear, meeting or normality requirement.
Tree Pipit
t=
x1 - x2
2
2
s1 s2
+
n1 n2
Pied Wagtail
= .4111 df = 25.08
P(t < -.4111 or t > .4111) = 0.68
Fail to reject H0, p < .68 > a = .05 . A test statistic
this extreme will occur by chance alone 68% of the time.
We lack strong evidence that the mean length of cuckoo eggs laid in Tree Pipit nests is different
from the mean of those laid in Pied Wagtail nests.
c. What is the mean length of cuckoo eggs laid in Wren nests? Make a 95% confidence
interval.
We will make a 1-sample t-interval for means.
Our assumptions have been met above.
s
df=13
n
(20.64,21.45)
x ±t
*
We are 95% confident that the true mean length of cuckoo eggs laid in Wren nests is between
20.6 and 21.4 mm. In repeated random sampling, this method captures the true mean 95% of the
time.
2. An experiment on the side effects of pain relievers assigned arthritis patients to one of several
over-the-counter pain medications. Of the 440 patients who took one brand of pain reliever, 31
suffered some “adverse symptom.” Does the experiment provide strong evidence that fewer than
10% of patients who take this medication have adverse symptoms? Test at a 1% significance
level.
H 0 : p = 0.10 The proportion of patients who suffer adverse symptoms when taking the medicine is 0.10.
H a : p < 0.10 The proportion of patients who suffer adverse symptoms when taking the medicine is <0.10
The data came from an experiment, and presumably they randomly assigned to treatment.
np = 440(.1) = 44 > 10
n(1- p) = 440(.9) = 396
10n = 4400 The may be a population of patients taking the medicine larger than 4400, but I'm not certain.
z=
p̂ - p0
p0 (1- p0 )
n
31
- .1
= 440
= -2.06
.1(.9)
440
P(Z < -2.06) = .0194 Fail to reject H0,
p = 0.019 > a = .01, a test statistic this small may occur by chance alone more than 1% of the
time.
We lack strong evidence that the true proportion of adverse symptoms is less than 10%.
3. In the 2001 regular baseball season, the World Series Champion Arizona Diamondbacks
played 81 games at home and 81 games away. They won 48 of their home games and 44 of the
games played away. We can consider these games as samples from potentially large populations
of games played at home and away. Most people think that it is easier to win at home than away.
Perform a significance test to determine whether this is true for the Arizona Diamondbacks.
H 0 : p1 - p2 = 0 The proportions of wins is the same for home and away games for the Arizona Diamondbacks.
H a : p1 - p2 > 0 The proportions of wins is greater for home than away games for the Arizona Diamondbacks.
Let 1= home, 2= away
We are given samples of large populations, in effect.
48 + 44
p̂ =
= .5679
81+ 81
n1 p̂ = 81(.5679) = 46 > 5
n1(1- p̂) = 81(.4320) = 35 > 5
Since the 2 samples are the same size the products are the same for n1 and n2..
z=
p̂1 - p̂2
p̂(1- p̂) p̂(1- p̂)
+
n1
n2
= 0.6344
P(z > 0.6344) = 0.2629
Fail to reject H0, p = .26 > .05 = a . A test statistic as large as this may occur by chance alone
26% of the time.
We lack evidence of a higher proportion of wins at home games for the Arizona Diamondbacks.
The composition of the earth’s atmosphere may have changed over time. One attempt to
discover the nature of the atmosphere long ago studies the gas trapped in bubbles inside ancient
amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles
within amber should be a sample of the atmosphere at the time the amber was formed.
Measurement on specimens of amber from the late Cretaceous era (75 to 95 million years ago)
gives these percentages of nitrogen:
4.
63.4
65.0
64.4
63.3
54.8
64.5
60.8
49.1
51.0
These values are quite different from the present 78.1% of nitrogen in the atmosphere, but are
these differences significant? Assume (this is not yet agreed on by experts) that these
observations are an SRS from the Cretaceous atmosphere.
H 0 : m = 78.1% The mean percentage of nitrogen in the atmosphere during the Cretaceous era is 78.1%.
H a : m ¹ 78.1% The mean percentage of nitrogen in the atmosphere during the Cretaceous era is not 78.1%
We are given data that is from a sample of gases trapped in the late Cretaceous era. While not
random, this may be representative. The sample size of 9 is small, but the data is not strongly
skewed. We rely on the robustness of the t-test and continue.
x - m0 59.58 - 78.1
=
= -8.877 df = 8
s
6.255
9
n
(No shading is visible on a graph of the t-distribution.)
t=
P(t < -8.87 or t > 8.87) = 2.04 ´10-5
Reject H0, p < 2 ´10-5 < a = .05 . A test statistic this extreme will occur by chance alone well less
than 1% of the time.
We have strong evidence that the mean percentage of nitrogen trapped in amber during the
Cretaceous era are different from the 78.1% today.