Starnes TPS 4e Chapter 8
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Gerard, Heather
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1 of 12
A random sample of 85 students in Chicago's high schools take a course designed to
improve SAT scores. Based on this sample, a 90% confidence interval for the mean
improvement μ in SAT scores for all Chicago high-school students taking this course is
found to be (72.3, 91.4). Which of the following statements is the correct interpretation of
this interval?
A. Ninety percent of the students in the sample improved their scores by between 72.3
and 91.4 points.
B. Ninety percent of the students in the population who take the course should improve
their scores by between 72.3 and 91.4 points.
C
C. Neither (A) nor (B) is correct.
1 out of 1
Correct. The confidence level gives the probability that the method
used to construct the interval will give an interval that captures μ
when used repeatedly. In other words, if you use the recipe for a
90% confidence interval many times for many random samples, in
the long run 90% of the intervals you get will contain the true value
of μ.
2 of 12
Agricultural researchers plant 100 plots with a new variety of corn. The average yield for
these plots is
= 130 bushels per acre. Assume that the yield per acre for the new corn
variety follows a Normal distribution with unknown mean μ and standard deviation σ = 10
bushels per acre. A 90% confidence interval for μ is
A. 130 ± 1.645.
B. 130 ± 1.96.
A
C. 130 ± 16.45.
1 out of 1
Correct. Recall that a level C confidence interval for the mean μ of a
Normal population with standard deviation σ is
, where is the sample mean, z* is the upper (1 - C)/2
standard Normal critical value, and n is the sample size. Here, =
130, n = 100, and z* = 1.645 (since we want a 90% confidence
interval and thus must use the upper (1 - 0.9)/2 = 0.05 standard
Normal critical value). We are told that σ = 10, so the interval is
3 of 12
A random sample of size n is collected from a Normal population with standard deviation σ.
Using these data, a confidence interval is computed for the mean of the population. Which
of the following actions would produce a new confidence interval with a smaller width
(smaller margin of error), assuming that the same data were used?
A. increasing the value of σ
B. using a lower confidence level
B
C. using a smaller sample size n
1 out of 1
Correct. The margin of error of a confidence interval is reduced by
using a lower level of confidence, increasing the sample size, or
reducing the value of σ.
4 of 12
To assess the accuracy of a kitchen scale, a standard weight, known to weigh 1 gram, is
weighed a total of n times, and the mean
of the n weight measurements is computed.
Suppose the scale readings are Normally distributed with unknown mean μ and standard
deviation σ = 0.01 gram. How large should n be so that a 90% confidence interval for μ has
a margin of error of E = ± 0.0001 gram?
A. 165
B. 27,061
B
C. 38,416
1 out of 1
Correct. Recall that a level C confidence interval for a Normal
population mean μ with standard deviation σ will have a specified
margin of error E when the sample size is
, where z* is the upper (1 - C)/2 standard Normal critical
value. Here, σ = 0.01,
z* = 1.645 (since we are interested in a 90% confidence interval
and thus must use the upper
(1 - 0.9)/2 = 0.05 standard Normal critical value), and E = 0.0001.
The resulting sample size is n = [(1.645)(0.01)/0.0001]2 = (164.5)2
= 27,060.25, which rounds up to 27,061.
5 of 12
Which of the following is not a property of a student's t distribution?
A. The density curve of the t distribution is symmetric about 0.
B. The density curve of the t distribution has thicker "tails" than the density curve of the
standard Normal distribution.
C. The density curve of the t distribution more closely resembles the density curve of
the standard Normal distribution when the number of degrees of freedom is small.
C
1 out of 1
Correct. The t curve actually more closely resembles the standard
Normal distribution when the number of degrees of freedom is
large.
6 of 12
The heights (in inches) of adult males in the United States are believed to be Normally
distributed with mean μ. The average height of a random sample of 25 American adult
males is found to be
= 69.72 inches, with a sample standard deviation of s = 4.15 inches.
A 90% confidence interval for μ is
A. 69.72 ± 1.42.
B. 69.72 ± 1.09.
A
C. 69.72 ± 1.37.
1 out of 1
Correct. Recall that a level C confidence interval for the mean μ of a
Normal population is
, where n is the sample size, is the sample mean, s is the
sample standard deviation, and t* is the upper (1 - C)/2 critical
value of the t distribution with n - 1 degrees of freedom. Here, n =
25, = 69.72, s = 4.15, and t * = 1.711 (the upper (1 - 0.9)/2 =
0.05 critical value of the t (25 - 1) = t(24) distribution). The 90%
confidence interval is therefore
.
7 of 12
Do students tend to improve their SAT Mathematics (SAT-M) score the second time they
take the test? Four randomly sampled students who took the test twice received the
following scores:
Assume that the change in SAT-M score (= second score - first score) for the population of
all students taking the test twice is Normally distributed with mean μ d. A 95% confidence
interval for μ d is
A. 25.0 ± 39.60.
B. 25.0 ± 56.09.
C
C. 25.0 ± 64.29.
1 out of 1
Correct. To analyze the data, we first compute the differences (=
second score - first score) and compute the mean and standard
deviation of the differences. The difference values are -10, 80, 0,
and 30. For these data, the mean is = 25 and the standard
deviation is sd = 40.41. A level C confidence interval for the
population mean difference μ d is
, where n = the number of
differences and t* is the upper (1 - C)/2 critical value of the t
distribution with n - 1 degrees of freedom. Here, n = 4 and t* =
3.182 (the upper (1 - 0.95)/2 = 0.025 critical value of the t(4 - 1) =
t(3) distribution). The confidence interval is, therefore,
.
8 of 12
You are thinking of using a t procedure to construct a 95% confidence interval for the
unknown mean μ of a population for a random sample taken from the population. You
suspect that the distribution of the population is not Normal and may in fact be strongly
skewed. Which of the following statements is then correct?
A. You should not use the t procedure because the population does not have a Normal
distribution.
B. You can use the t procedure provided that your sample size is large (say, at least
40).
C. You can use the t procedure regardless of the sample size because t procedures are
robust against non-Normality.
B
1 out of 1
Correct. The t procedures are robust against non-Normality of the
population except in the case of outliers or strong skewness.
However, if the sample size is large (roughly 40 or more), you can
use t procedures even when the distribution is strongly skewed.
9 of 12
A radio talk show host is interested in the proportion p of adults in his listening area who
think that the drinking age should be lowered to 18. To find this proportion, he poses the
following question to his listeners: "Do you think that the drinking age should be reduced to
18 in light of the fact that 18-year-olds are eligible for military service?" He asks listeners to
phone in and vote "yes" or "no" depending upon their opinions. Of 200 people who phone
in, 140 answer "yes." The standard error of the sample proportion
those who phone in is
A. 0.84.
of "yes" votes among
B. 0.032.
B
C. 0.00105.
1 out of 1
Correct. The standard error of is given by
In this case, n = 200 and = 140/200 = 0.7, so
.
10 of 12
An inspector monitors large truckloads of potatoes to determine the proportion p of
potatoes with major defects before the potatoes are used to make potato chips. She intends
to compute a 95% confidence interval for p. To do so, she selects an SRS of 50 potatoes
from a truckload of more than 2000 potatoes. Suppose that only two of the 50 potatoes
sampled are found to have major defects. Which of the following assumptions for inference
about a proportion using a confidence interval are violated?
A. The sample size n is large enough that both the count of successes
of failures n(1 -
and the count
) are 10 or more.
B. The population is at least 10 times as large as the sample.
A
C. There appear to be no violations.
1 out of 1
Correct. In this case, n = 50 and = 2/50, so
than 10. This assumption is therefore violated.
= 2, which is less
11 of 12
One hundred rats with mothers that were exposed to high levels of tobacco smoke during
pregnancy were put through a simple maze. At the outset, the maze required the rats to
make a choice between going left and going right. Eighty of the rats went right when
running the maze for the first time. Assume that the 100 rats can be considered an SRS
from the population of all rats born to mothers who were exposed to high levels of tobacco
smoke during pregnancy. (Note that this assumption may or may not be reasonable, but
researchers often assume that lab rats are representative of large populations because they
are often bred to have uniform characteristics.) Let p be the proportion of rats in this
population that would go right when running the maze for the first time. A 90% confidence
interval for p is
A. 0.8 ± 0.040.
B. 0.8 ± 0.078.
C
C. 0.8 ± 0.066.
1 out of 1
Correct. Recall that the formula for an approximate level C
confidence interval for a population proportion p is
,
where = the sample proportion of successes, n = the sample size,
and z* = the upper (1 - C)/2 critical value of the standard Normal
distribution. Here, = 80/100 = 0.8, n = 100, and z* = 1.645
(because we are interested in a 90% confidence interval and,
therefore, must use the upper (1 - 0.9)/2 = 0.05 standard Normal
critical value). The confidence interval is thus
12 of 12
A noted psychic is tested for extrasensory perception. The psychic is presented with 400
cards, all face down, and asked to determine if each card is marked with one of four
symbols: a star, a cross, a circle, or a square. Let p represent the probability that the
psychic correctly identifies the symbol on the card in a random trial. How many trials would
you have to conduct to estimate p to within ± 0.01 ("plus or minus one percentage point")
with 95% confidence? (Use the guess 0.25 as the value for p. Note that this would be the
value of p if the psychic were merely guessing.)
A. 1351
B. 7203
C. 9604
B
1 out of 1
Correct. Recall that a level C confidence interval for a population
proportion p will have a margin of error approximately equal to a
specified value E when the sample size is
, where p*
is the guessed value for the true proportion and z* is the upper (1 C)/2 standard Normal critical value. In this case, E = 0.01, p* =
0.25, and z* = 1.96 (because we are estimating with 95%
confidence and must use the upper (1 - 0.95)/2 = 0.025 standard
Normal critical value). Therefore,