Significance Tests and Confidence Intervals for Means and Proportions
Decide on the type of procedure used for each test or interval. Answer each question fully using
appropriate statistical methods.
1.
Typically, only very brave students are willing to speak out in a college classroom. Student
participation may be especially difficult if the individual is from a different culture or country. A journal
article considered a numerical "speaking-up" scale, with possible values from 3 to 15 (low means a
student rarely speaks). For a random sample of 64 males from Asian countries where English is not the
official language, the sample mean and standard deviation were 8.75 and 2.57, respectively. Suppose that
the mean for the population of all males having English as their native language is 10.0. Does it appear
that the population mean for males from non-English-speaking Asian countries is less than 10.0?
Answer: 1 sample t-interval
H 0 : µ = 10 The mean "speaking-up" score is 10.
H a : µ < 10 The mean "speaking-up" score is less than 10.
We are given an SRS. Our sample is large (64), so by the Central Limit Theorem the normal approximation is
useful.
t=
x − µ 8.75 − 10
=
= −3.8911 df=63
s
2.57
n
64
P(t ≤ −3.8911) = .00012
Reject H0, a value this extreme may occur by chance alone less than 1% of the time. We have strong evidence
that males from Asian non-English-speaking countries have lower “speaking-up” scores than males who are native
English speakers.
2.
Researchers at the National Cancer Institute released the results of a study that examined the effect
of weed-killing herbicides on house pets. The following data is compatible with summary values given in
the report. Dogs, some of whom were from homes where the herbicide was used on a regular basis, were
examined for the presence of malignant lymphoma. Below are the data:
Group
Sample Size
# with Lymphoma
ˆ
p
Exposed
827
473
0.572
Unexposed
130
19
0.146
Find a 95% confidence interval for the difference by which the proportion of exposed dogs that develop
lymphoma exceed that for unexposed dogs.
Answer: 2-proportion Z-interval
We are uncertain of having SRSs. The samples can reasonably be expected to be independent.
Let p1=proportion of dogs exposed to herbicide that develop lymphoma.
Let p2=proportion of dogs not exposed to herbicide that develop lymphoma.
The populations of dogs exposed to herbicide and those not exposed to herbicide are each well over 10
times the sample sizes.
Significance Tests and Confidence Intervals for Means and Proportions
n1 p̂1 = 827 (.572 ) ≥ 5 n2 p̂2 = 130 (.146 ) ≥ 5
n1 1− p̂1 = 827 (1− .572 ) ≥ 5 n2 1− p̂2 = 130 (1− .146 ) ≥ 5
(
)
(
)
p̂ (1− p̂1) p̂2 (1− p̂2 )
p̂1 − p̂2 ± z* 1
+
n1
n2
.572 (1− .572 ) .146 (1− .146 )
.572 − .146 ±1.96
+
827
130
(.3563,.4953)
We are 95% confident that the difference in the proportions of dogs that develop lymphoma is between 35
and 50%, subtracting the proportion not exposed from the proportion exposed to herbicide. In repeated
random sampling this method captures the true difference in proportions 95% of the time.
3.
Workers at a factory ask their supervisor to provide music during their shift. The supervisor wanted
to know whether the music really helped improve the workers’ performance. Since all workers were
assigned to different rooms at random, the supervisor randomly selected one room. For one week, he
provided music in this room, and for one week he provided none. He flipped a coin to decide which week
got music. Afterward he recorded productivity of the workers, using average number of items assembled
per day:
Worker 1
With
29.38
Music
Without 24.55
Music
2
31.53
3
27.45
4
27.76
5
29.31
6
27.69
7
29.07
8
27.48
9
29.83
10
28.96
25.59
23.17
24.71
25.26
24.88
24.34
26.13
25.57
26.63
Does music improve the performance of workers, as measured by the mean number of items assembled
per day?
Answer: Matched pairs (1-sample t-test on the differences)
H 0 : µd = 0 The true mean increase in productivity is zero.
H a : µd > 0 The true mean increase in productivity is greater than zero.
t=
x − µ0 3.763 − 0
=
= 8.735
s
1.362
n
10
df = 9
P(t > 8.735) ≈ 0
Reject H0, a value this extreme will rarely occur by chance alone. We have strong
evidence that there is an increase in productivity with exposure to music.
Significance Tests and Confidence Intervals for Means and Proportions
4.
Elite distance runners are thinner than the rest of us. Here are data on skinfold thickness, which
indirectly measure body fat, for 20 elite runners and 95 ordinary men in the same age group. The data are
in millimeters and are given in the form “mean (standard deviation).”
Abdomen
Thigh
Runners
7.1 (1.0)
6.1 (1.8)
Others
20.6 (9.0)
17.4 (6.6)
Describe these differences between runners and more typical young men.
Answer: Confidence intervals give more information than tests. We will calculate a two sample tinterval for “abdomen” and the same for “thigh.” Let 1=runners 2=others
The problem does not tell us if these are SRSs. We can conclude that they are independent, as there is no
reason to think otherwise. The sample sizes together are large enough to justify the t-procedure.
For the abdomen skinfold thickness:
This interval is calculated on the calculator. Your degrees of freedom and exact interval will be
different if you did not use the calculator. The calculator method is more accurate.
s12 s22
x1 − x2 ± t *
+
n1 n2
1 92
7.1 − 20.6 ± 1.983
+
20 95 df = 103.6
(−15.48, −11.72)
We are 95% confident that the mean difference in abdominal skinfold thickness of runners and other men
is between -15.5 and –11.7 mm, suggesting that runners have less abdominal skinfold thickness. In
repeated random samples this method captures the true difference approximately 95% of the time.
For the thigh skinfold thickness:
This interval is calculated on the calculator.
1.8 2 6.6 2
+
20
95
(−12.86, −9.74)
6.1 − 17.4 ± 1.983
df = 106.4
We are 95% confident that the mean difference in thigh skinfold thickness of runners and other men is
between –12.86 and –9.74 mm. In repeated random samples this method captures the true mean
approximately 95% of the time. These data show that runners tend to have lower skinfold thicknesses
indicating less fat than nonrunners.