140
2 6 1 1 1 1 4 2
4 5 1 1 1 1 1
3 5 1 1 1 1 1
2 6 2 1 1 2 4 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
3 5 1 1 1 1 1
4 5 1 1 1 1 1
6 5 1 1 1 1 1
2 5 1 1 1 1 1
6 5 1 1 1 1 1
3 5 1 1 1 1 1
7 6 1 1 1 1 1 2
1 5 1 1 1 1 1
3 5 1 1 1 1 1
2 5 1 1 1 1 1
3 5 1 1 1 1 1
2 5 2 3 2 1 1
4 6 1 1 1 1 1 1
4 6 1 1 1 1 1 1
5 5 1 1 1 1 1
5 5 1 1 1 1 2
2 5 1 1 1 1 1
2 6 1 1 1 1 1 1
1 5 1 1 1 1 1
3 5 1 1 1 1 1
1 5 2 1 2 1 1
1 5 1 2 1 1 1
4 5 1 1 1 1 1
10 5 1 1 1 1 1
1 5 1 1 1 1 1
6 5 1 1 1 1 1
2 5 2 1 1 1 1
5 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
7 5 1 1 1 1 1
3 5 1 1 1 1 1
3 5 1 1 1 1 1
2 5 1 1 1 1 1
1 5 1 1 1 1 1
6 5 1 1 1 1 1
1 5 1 1 1 1 1
2 5 1 1 1 1 1
5 5 1 1 1 1 1
11 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
46 1 1
46 1 1
18 5 1 1 1 1 1
8 5 1 1 1 1 1
5 5 1 1 1 1 1
5 5 2 2 2 2 2
3 5 3 3 3 3 3
6 5 3 3 3 3 3
1 5 2 2 2 2 2
3 5 2 2 2 2 2
19 5 2 2 2 2 2
2 5 1 1 1 1 1
5 5 1 1 1 1 1
8 5 2 2 2 2 2
13 5 1 1 1 1 1
8 5 2 2 2 2 2
10 5 1 1 1 1 1
8 5 3 3 3 3 3
6 5 2 2 2 2 2
4 5 1 1 1 1 1
9 5 1 1 1 1 1
2 5 4 4 4 4 4
5 5 1 1 1 1 1
6 5 1 1 1 1 1
2 5 1 1 1 1 1
5 5 1 1 1 1 1
6 5 1 1 1 1 1
8 5 1 1 1 1 1
7 5 1 1 1 1 1
2 2 1 1
3 2 1 1
3 2 1 1
2 2 1 1
2 5 1 1 1 1 1
7 5 1 1 1 1 1
13 5 1 1 1 1 1
14 5 1 1 1 1 1
3 5 1 1 1 1 1
1 5 1 1 1 1 1
6 5 1 2 2 2 2
3 5 1 1 1 1 1
6 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
2 5 1 1 1 1 1
5 5 1 1 1 1 1
1 5 1 1 1 1 2
2 5 1 1 1 1 1
2 5 1 1 1 1 1
6 5 1 1 1 1 1
3 5 1 1 1 1 1
2 5 1 1 1 1 1
1 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
1 5 1 2 1 1 1
1 5 1 1 1 1 1
5 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
16 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
6 5 1 1 1 1 1
2 5 1 1 1 1 1
7 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
10 5 1 1 1 1 1
2 5 1 1 1 1 1
2 5 1 1 1 1 1
3 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 1 1 1 1 1
1 5 2 2 1 2 2
3 2 1 1
Which of the following is not a property of the
binomial distribution?
The individual trials are dependent on each other.
There are exactly two possible outcomes for each trial.
The probability of success is the same for each trial.
The number of trials is fixed.
The mean, variance, and standard deviation of the
number of successes can be calculated knowing only
the number of trials and the probability of success
for a given trial.
All of the other answers is a property of the binomial
distribution.
A multiple choice exam has 15 questions. Each question has
4 possible answers. Only one answer is correct per question.
What is the probability that by just guessing, a student
will get at most 7 correct?
0.9827
0.0393
0.0173
0.0566
50--50
If the random variable $ X $ has a binomial distribution
with $ n = 10 $ and $ p = .5 $, what is the variance
of $ X $?
2.5
5
0.25
1.58
null hypothesis
Which of the following is true of a normal
distribution?
Each of the other answers is a property of a normal
distribution.
The graph is a symmetric, bell--shaped curve.
The graph is centered at the mean $ \mu $.
The total area under of the region between the curve and
the horizontal axis is one.
If the mean $ \mu $ and the variance $ {\sigma}^2 $ are
known, the standard normal table may be used to obtain
relevant probabilities for the corresponding normal
distribution.
The entire graph is always strictly above the horizontal axis.
The area of the region under the standard normal curve between
$ -2.0 $ and $ -1.0 $ is
0.1359
0.0228
0.4772
0.3413
None of the other answers is correct.
Let $ Z $ be standard normal. Then
$ P( Z > -1.65) = $
0.9505
0.0495
0.9010
0.0165
\$ 11,111.11
Let $ Z $ be standard normal. The value of $ z $
such that $ P(Z > z) = 0.0217 $ is approximately
2.02
$ -2.02 $
0.5382
0.7643
$ \mu $
Let $ Z $ be standard normal. The value of $ z $
such that $ P(-z < Z < z) = .8500 $ is approximately
1.44
1.03
1.04
1.00
$ {\sigma}^2 $
For the standard normal distribution, if the $ z $ value
is equal to 0, this implies that
the raw score is the same as the mean
the raw score is 0
the raw score does not exist
the raw score is extremely large
the raw score is not so raw
If $ X $ is a normally distributed random variable with a
mean of 15 and a variance of 9, then $ P(x < 18) $ is
0.8413
0.0
0.3413
0.1587
unknown, to say the least
If $ X $ is a normally distributed random variable
with a mean of 15 and a variance of 9, then
$ P(X = 18) $ is
0.0
0.8413
0.3413
0.1587
negative
If IQ scores are normally distributed with a mean of
100 and a standard deviation of 20, then the
probability of a person having an IQ score of at least
130 is
0.0668
0.1332
0.9332
does not exist
0.0
The manager of a video rental store believes that the total
annual revenue of the store can be approximated by a normal
distribution with a mean of \$250,000 and a standard
deviation of \$30,000. She knows she needs \$180,000
to break even. What is the probability that the store will
make a profit next year?
0.9901
0.0099
0.9599
0.0401
0.0
The sample statistic $ \overline{x} $ is a point
estimate of
the population mean
the sample median
the population mode
the population variance
nothing
The mean score of international students at a certain
university on the TOEFL English language exam is
normally distributed with a mean of 490 and a
standard deviation of 80. Suppose groups of 30
students are studied. The mean and standard deviation for
the distribution of sample means will be, respectively,
490, 14.61
490, 8/30
16.33, 80
490,213.33
binomially distributed, uniformly distributed
Let $ X $ be a random variable with mean 4.83 and standard
deviation 2.79. A random sample of size 6 is selected. The
standard error of the mean is
1.139
2.790
7.784
1.297
unknown, probably forever
A random sample of size 7 was taken from a population
of size 40. The observed values were
$$
3 \qquad 8 \qquad 13 \qquad 15 \qquad
17 \qquad 22 \qquad 27 .
$$
A point estimate for the population mean is
15
116
105
2.625
7/40
not random since the population only has size 40
which is not evenly divisble by 7
The Central Limit Theorem applies to
any probability distribution
only discrete distributions
only continuous distributions
only sem--discrete distributions
only normal distributions
When considering sampling distributions, if the population
from which we sample is normally distributed, then the
distribution of the sample means
is normally distributed
is binomially distributed
has a $ T $ distribution
has an unknown distribution
is not distributed fairly
As the sample size increases, the length of the
confidence interval for the population mean should
decrease
increase
stay the same
decrease and then increase
Who knows?
If we change the confidence level from 98\% to 95\%
when constructing a confidence interval for the mean,
we can expect the width of the interval to
decrease
increase
stay the same
decrease and then increase
Who knows?
A 90\% confidence interval for a population mean
indicates that
we are 90\% certain that the population mean will fall
within the interval
we are 90\% certain that the interval will contain all
possible sample means with the same sample size taken
from the given distribution
we are 90\% certain that the population mean will be the
same as the sample mean used in constructing the interval
we are 90\% confident there are no free lunches
we are never 90\% confident when it comes to statistics
A 99\% confidence interval is to be constructed for the
population mean from a random sample of size 22. If
the population standard deviation is unknown,
the table value to be used in the computation is
2.831
2.518
2.330
2.580
2.576
Who knows?
A 99\% confidence interval is to be constructed for the
population mean from a random sample of size 22. If
the population standard deviation is known,
the table value to be used in the computation is
2.576
2.831
2.518
2.330
2.580
Who knows?
The heights in inches of the students on a campus are assumed
to have a normal distribution with a standard deviation of
4 inches. A random sample of 49 students was taken with a mean
of 68 inches. A 95\% confidence interval for the population
mean $ \mu $ is
66.88 inches to 69.12 inches
67.08 inches to 68.94 inches
63.42 inches to 72.48 inches
64.24 inches to 71.76 inches
cannot be measured in inches
A random sample was taken from a normal population.
The mean of the sample is 30, an estimate for the
standard deviation of the mean is 5, and the
$t$--value from the table is 1.725. The lower
limit for the corresponding confidence interval is
21.375
28.071
26.673
20.357
unknown since the confidence level is not specified
directly
In hypothesis testing, the level of significance
($ \alpha $) is the probability of
rejecting a true null hypothesis
failing to reject a true null hypothesis
failing to reject a false null hypothesis
rejecting a false null hypothesis
constructing a confidence interval
The collection of all elements under study and about
which one is trying to draw conclusions is
a population
a random sample
a biased sample
the data set
unknown
stranger than anything
Statistical inference is
the drawing of conclusions from data
the presentation and summarization of data
a subset of the population
a quality which characterizes a population
silly, just silly
The major objective of statistical analysis is to
make $ \dots $ about a $ \dots $ based on
information obtained from a $ \dots $.
inferences, population, sample
inferences, sample, population
hypotheses, statistic, sample
hypotheses, parameter, population
wishes, better life, fortune cookie
A random variable is always
a variable whose value is determined by the outcome of
some chance experiment
a variable that assumes its values only at isolated points
a variable that, prior to the experiment, can conceivably
assume any value in some interval of real numbers
None of the other answers is correct.
At least two of the other answers is correct.
An example of a continuous random variable is
the length of a telephone conversation
the interest paid on six--month money market
certificates per week
the number of dots on the top face of a fair die
dress size
discrete
To test the theory that the country is pleased with the present
economic conditions, a booth is set up on Wall Street and persons
who pass by the booth are asked their opinion. The results
of this survey might be suspect because
the sample isn't random, therefore, probably not representative
people are reluctant to talk to strangers
the sample isn't large enough
there are no numbers in the raw data
well, just because
Use the following sample of the ages of forest rangers
in Virginia to estimate the true mean age of all
forest rangers in Virginia.
$$
\aligned
& 35 \qquad 38 \qquad 36 \\
& 37 \qquad 34 \qquad 33 \\
& 40 \qquad 35 \qquad 37
\endaligned
$$
36.111
35 or 37
34.000
35.500
34.222
The range of the sample $ 3, 1, 2, 8, 6, 5 $ is
7
2
6
8
not defined
Lorenzo Watts is a statistician for the Fuquay Tyle Utility
Company. Long years of experience have shown that at the main
power plant in Tyleville, the amount of actual work time the
company linemen put in during an eight--hour day has a mean
of four hours and a variance of 0.64. What is the standard
deviation of the hours worked?
0.8
0.4096
0.32
0.4
unknown
The reason the standard deviation is more easily interpreted
than the variance is
the standard deviation is expressed in the same units
as the original data
the standard deviation is a measure of average deviation
the standard deviation is normally distributed
the standard deviation takes the mean into account
the standard deviation is equal to the sample mean
Bernie Ooly is a telephone salesman for the Everlast
Cemetary Company. It is his experience that 10\% of
his calls lead to a sale, and that each call is
independent of all other calls. What is the probability that
Bernie will make exactly 3 sales in 10 calls?
0.0574
0.0128
0.9426
0.2431
0.9872
Five fair coins are tossed. The probability of obtaining
two or fewer heads is
0.5000
0.1875
0.3125
0.1563
can't happen since 5 is not evenly divisible by 2
If the random variable $ Z $ has a standard normal distribution,
then $ P(Z \le -1.74) $ is
0.0409
$ -0.0409 $
0.9591
$ -0.9591 $
$ \dfrac{\sqrt{2}}{2} $
If $ Z $ is standard normal, then the value of $ z $ such
that $ P(Z \ge z) = .516 $ is
$ -0.04 $
0.04
0.484
1.04
$ -0.484 $
Dr. I. C. Trout, a professional iththyologist, is studying
a population of catfish with his assistant, Etta Bass.
Trout and Bass have established that the mean weight for a
catfish is 3.2 pounds with a standard deviation of 0.8 pound.
Additionally, weights of catfish are normally distributed.
What is the probability that a catfish will weigh more
than five pounds?
0.0122
0.4878
0.9878
0.9556
can't happen; catfish don't come that big
Refer to the previous question. If a sample of
64 fish is taken from the population of catfish,
what would the standard error of the mean weight equal?
0.100
0.013
0.080
0.112
0.800
Given that the random variable $ X $ is normally distributed
with a mean of 3 and a standard deviation of 5, find
$ E(X) $.
3
5
5/3
3/2
$ \sigma $
As the size of the sample is increased, the variance of the
distribution of the sample mean
is decreased
is increased
will stay the same
disappears
approaches $ \mu $
A point estimator is
a statistic
an interval extimator
a random sample
a parameter
very big
Consider the following sample of data
$$
700 \qquad 500 \qquad 340 \qquad 280 \qquad 600 .
$$
What is a point estimate for the mean of the population
from which the data were taken?
484
500
605
340
unknown
Refer to the previous problem. The sample median is
500
484
605
340
unknown
As the confidence interval increases, the width
of the interval
increases
remains the same
decreases
waivers
ceases to exist
A random sample of five observations provided a mean of
0.32 and a standard deviation of 0.0043. In constructing a
95\% confidence interval for the mean of the population from
which the data were selected, the appropriate table value
to use in the formula is
2.776
2.571
3.132
1.96
not available
Audie O'Fyle is testing whether a certain brand of audio tape
provides less than one decibel of tape hiss. His hypotheses
are
$$
\aligned
H_0 &: \mu_0 = 1 \\
H_1 &: \mu < 1
\endaligned
$$
If Audie concludes that the tape does provide at least one
decibel of hiss, when this is not true, Audie has made a
Type I error
Type II error
Type III error
tragic, tragic mistake
hissless tape
Which of the following could be called a null hypothesis?
$ H_0 : \mu = 10 $
$ H_0 : \overline{x} = 3200 $
$ H_0 : \mu = 1.5 $
$ H_1 : \mu > 150 $
More than one of the other answers could be a null hypothesis.
Acceptance of a false null hypothesis is
a Type II error
a Type I error
a Type III error
a significance level
a correct decision
One of the most feared predators in the New River is the
Giant Guppy. Although it is known that the Giant Guppy
grows to a mean length of 21 feet, an RU student in marine
biology believed the guppies near Radford are smaller
due to unusual feeding habits (which we won't go into).
To test his claim, he decided
to tackle randomly chosen guppies, measure them, and,
of course, then set them free. Because this is a difficult, costly,
and dangerous, not to mention silly, operation, only three
guppies were sampled. Their measured lengths were 22, 20, and
17 feet.
Test the hypothesis that
Radford Guppies are smaller than the average Giant Guppy. Use a
significance level of 1\%.
\vskip .05in
\item{ a. }
Null hypothesis:
\vskip .5in
\item{ b. }
Alternative hypothesis:
\vskip .5in
\item{ c. }
Test statistic:
\vskip .5in
\item{ d. }
Appropriate table:
\vskip .5in
\item{ e. }
Table value for test statistic:
\vskip .5in
\item{ f. }
Rejection region:
\vskip .5in
\item{ g. }
Data value of the test statistic:
\vskip .5in
\item{ h. }
Conclusion:
\vskip .5in
\item{ i. }
If your conclusion is wrong, to what type of error
are you subject?
\vskip .5in
\item{ j. } What are the practical consequences
of the error committed in part i)?
\vskip .5in
1
One of the most feared predators in the New River is the
Giant Guppy. Although it is known that the Giant Guppy
grows to a mean length of 16 feet, an RU student in marine
biology believed the guppies near Radford are larger
due to unusual feeding habits (which we won't go into).
To test his claim, he decided
to tackle randomly chosen guppies, measure them, and,
of course, then set them free. Because this is a difficult, costly,
and dangerous, not to mention silly, operation, only three
guppies were sampled. Their measured lengths were 21, 20, and
15 feet.
Test the hypothesis that
Radford Guppies are larger than the average Giant Guppy. Use a
significance level of 5\%.
\vskip .05in
\item{ a. }
Null hypothesis:
\vskip .5in
\item{ b. }
Alternative hypothesis:
\vskip .5in
\item{ c. }
Test statistic:
\vskip .5in
\item{ d. }
Appropriate table:
\vskip .5in
\item{ e. }
Table value for test statistic:
\vskip .5in
\item{ f. }
Rejection region:
\vskip .5in
\item{ g. }
Data value of the test statistic:
\vskip .5in
\item{ h. }
Conclusion:
\vskip .5in
\item{ i. }
If your conclusion is wrong, to what type of error
are you subject?
\vskip .5in
\item{ j. } What are the practical consequences
of the error committed in part i)?
\vskip .5in
1
The number of letters per word for words used in
the crossword puzzle in a typical issue of
the {\it Tartan} has an interesting distribution.
The following table shows $ x $ the number of letters
in a word and $ n(x) $ the number of words having this
number of letters in a recent issue.
$$
\aligned
\quad x &
\quad \, 2 \quad \, \, \, 3 \quad \, \, \, 4
\quad \, \, 5 \quad 6 \quad 7
\\
n(x) &
\quad 11 \quad 43 \quad 14 \quad 4 \quad 4 \quad 2
\endaligned
$$
Estimate the number of letters per word expected
in a {\it Tartan} crossword puzzle.
3.4
4.0
2.8
3.2
4.5
A sample of underweight babies was fed a special diet
and the following weight gains (lbs) were observed at
the end of three months:
$$
6.7 \quad 2.7 \quad 2.5 \quad 3.6 \quad
3.4 \quad 4.1 \quad 4.8 \quad 5.9 \quad 8.3 .
$$
The mean and standard deviation are:
4.67, 1.95
4.67, 3.82
3.82, 4.67
1.95, 4.67
4.67, 1.84
The heights in centimeters of 5 students are:
$$
165, \quad 175, \quad 176, \quad 159, \quad 170 .
$$
The sample median and sample mean are respectively:
170, 169
170, 170
169, 170
176, 169
176, 176
If most of the measurements in a large data set are of
approximately the same magnitude except for a few
measurements that are quite a bit larger, how would
the mean and median of the data set compare and what
shape would a histogram of the data set have?
the mean would be larger than the median and the
histogram would be skewed with a long right tail
the mean would be smaller than the median and the
histogram would be skewed with a long left tail
the mean would be larger than the median and the
histogram would be skewed with a long left tail
the mean would be smaller than the median and the
histogram would be skewed with a long right tail
the mean would be equal to the median and the
histogram would be symmetrical
In measuring the center of the data from a skewed
distribution, the median would be preferred over
the mean for most purposes because:
the mean may be too heavily influenced by the
larger observations and this gives too high
an indication of the center
the median is the most frequent number while
the mean is most
likely
the median is less than the mean and smaller
numbers are always appropriate for the
center
the mean measures the spread
in the
data
the median measures the arithmetic average
of the data excluding
outliers
A sample of 99 distances has a mean of 24 feet
and a median of 24.5 feet. Unfortunately, it
has just been discovered that an observation
which was erroneously recorded as ``30'' actually
had a value of ``35.'' If we make this correction
to the data, then:
the median remains the same,
but the mean is
increased
the mean remains the same,
but the median is
increased
the mean and median remain
the
same
the mean and median
are both
increased
we do not know how the mean and median are
affected without further calculations;
but the variance is increased
Which of the following is false?
The numbers 1, 5, 9 have a smaller standard
deviation than 101, 105, 109
The numbers 3, 3, 3 have a standard deviation
of 0
The numbers 3, 4, 5 have the same standard
deviation as 1003, 1004, 1005
The standard deviation is a measure of spread
around the center of the data
The standard deviation can only be computed
for interval or ratio scaled data
In a statistical test for the equality of a mean,
such as $ \text{ Null }: \, {\mu}_0 = 10 $ , if
$ \alpha = 0.05 $,
5\% of the time we will say that there is a real
difference when there is no difference
95\% of the time we will make an incorrect
inference
5\% of the time we will say that there is no real
difference when there is a difference
95\% of the time the null hypothesis will be
correct
5\% of the time we will make a correct
inference
DDT is an insecticide that accumulates up the
food chain. Predator birds can be contaminated
with quite high levels of the chemical by eating
many lightly contaminated prey. One effect of
DDT upon birds is to inhibit the production of
the enzyme carbonic anhydrate which controls
calcium metabolism. It is believed that this
causes egg shells to be thinner and weaker than
normal and makes the eggs more prone to breakage.
(This is one of reasons why the condor in
California is near extinction.) An experiment
was conducted where 16 sparrow hawks were fed
a mixture of 3 ppm dieldrin and 15 ppm DDT (a
combination often found in contaminated prey).
The first egg laid by each bird was measured
and the mean shell thickness was found to be
0.19 mm with a standard deviation of 0.01 mm.
A normal egg shell has a mean thickness of
0.2 mm. The null and alternate hypotheses are:
$ \text{ Null }: {\mu}_0 = 0.2 \qquad
\text{ Alternative }: {\mu}_0 < 0.2 $
$ \text{ Null }: {\mu}_0 < 0.2 \qquad
\text{ Alternative }: {\mu}_0 = 0.2 $
$ \text{ Null }: {\mu}_0 = 0.2 \qquad
\text{ Alternative }: {\mu}_0 < 0.2 $
$ \text{ Null }: {\mu}_0 = 0.19 \qquad
\text{ Alternative }: {\mu}_0 = 0.2 $
$ \text{ Null }: {\mu}_0 = 0.2 \qquad
\text{ Alternative }: {\mu}_0 = 0.2 $
Same question. The value of the data test
statistic is:
$ -4.00 $
$ -1.00 $
0.01
1.96
1.75
It is important to detect a decrease in the
average thickness to .18 mm because then the
eggs are so fragile that few survive. What
sample size would be needed to be 80\% sure
of detecting this decrease at $ \alpha = 0.05 $?
27
8
128
34
101
The average time it takes for a person to
experience pain relief from aspirin is 25
minutes. A new ingredient is added to help
speed up relief. Let $ {\mu}_0 $ denote the
average time to obtain pain relief with
the new product. An experiment is conducted
to verify if the new product is better.
What are the null and alternative hypotheses?
$ \text{ Null }: {\mu}_0 = 25 \text{ vs }
\text{ Alternative }: {\mu}_0 < 25 $
$ \text{ Null }: {\mu}_0 = 25 \text{ vs }
\text{ Alternative }: {\mu}_0 = 25 $
$ \text{ Null }: {\mu}_0 < 25 \text{ vs }
\text{ Alternative }: {\mu}_0 = 25 $
$ \text{ Null }: {\mu}_0 < 25 \text{ vs }
\text{ Alternative }: {\mu}_0 > 25 $
$ \text{ Null }: {\mu}_0 = 25 \text{ vs }
\text{ Alternative }: {\mu}_0 > 25 $
In order to study the amounts owed to the
city, a city clerk takes a random sample
of 16 files from a cabinet containing a
large number of delinquent accounts and
finds the average amount owed to the city
to be \$230 with a sample standard deviation
of \$36. It has been claimed that the true
mean amount owed on accounts of this type
is greater than \$250. If it is appropriate
to assume that the amount owed is a normally
distributed random variable, the value of
the test statistic appropriate for testing
the claim is:
$ -2.22 $
$ -3.33 $
$ -1.96 $
$ -0.55 $
$ -2.1314 $
An appropriate 95\% confidence interval for
$ {\mu}_0 $ has been calculated as $(-0.73,1.92)$
based on $ n = 15 $ observations from a normally
distributed population with a a standard deviation
of 2. The hypotheses of interest are
$ \text{ Null }: {\mu}_0 = 0 \text{ vs }
\text{ Alternative }: {\mu}_0 \ne 0 $.
Based on this confidence interval,
we should not reject the null hypothesis at the
$ \alpha = 0.05 $ level of significance
we should reject the null hypothesis at the
$ \alpha = 0.05 $ level of significance
we should reject the null hypothesis at the
$ \alpha = 0.10 $ level of significance
we should not reject the null hypothesis at the
$ \alpha = 0.10 $ level of significance
we cannot perform the required test since
we do not know the value of the test statistic
The Roanoke Times claims that the time of travel
from downtown to the Roanoke Airport via bus has
an average of $ {\mu}_0 = 27 $ minutes. A person who
often takes this bus believes that $ {\mu}_0 $ is
greater than 27 minutes.
A sample of six ride--times taken to test the
hypothesis of interest gave $ \overline{x} = 27.5 $
minutes and a standard deviation $ s = 2.43 $ minutes.
The value of the test statistic for testing this
hypothesis is:
$ 0.504 $
$ -0.532 $
$ 0.460 $
$ -0.504 $
$ -0.460 $
A 95\% confidence interval for $ {\mu}_0 $ is
calculated to be $ (1.7 , 3.5) $.
It is now decided to test the hypothesis
$ \text{ Null }: {\mu}_0 = 0 \text{ vs }
\text{ Alternative }: {\mu}_0 \ne 0 $ at the
$ \alpha = 0.05 $ level, using the same
data as was used to construct the confidence
interval.
we would reject the null hypothesis at
level
$ \alpha = 0.05 $
we cannot test the hypothesis
without the
original data
we cannot test the hypothesis at the $ \alpha = 0.05 $
level since the $ \alpha = 0.05 $ test is connected
to the 97.5\% confidence interval.
we can only make the connection between hypothesis
testing and confidence intervals if the sample sizes
are large
we would accept the null hypothesis at
level
$ \alpha = 0.05 $.
We want to test $ \text{ Null }: {\mu}_0 = 1.5
\text{ vs } \text{ alternative }: {\mu}_0 \ne 1.5
\text{ at } \alpha = .05 $.
A 95\% confidence interval for $ {\mu}_0 $ calculated
from a given random sample is $ (1.4, 3.6) $.
Based on this finding we:
fail to
reject the null hypothesis
reject the null
hypothesis
cannot make any decision at all because the value
of the test statistic is not available
cannot make any decision at all because the
distribution of the population is unknown
cannot make any decision at all because $ (1.4,3.6) $
is only a 95\% confidence interval for $ {\mu}_0 $
A random sample of 100 observations is to be drawn
from a population with a mean of 40 and a standard
deviation of 25. The probability that the mean of
the sample will exceed 45 is:
0.0228
0.4772
0.4207
0.0793
not possible to compute, based on the information provided
Cans of salmon have a nominal net weight of 250 grams.
However, due to variation in the canning process,
the actual net weight has an approximate normal
distribution with a mean of 255 grams and a standard
deviation of 10 grams. According to Consumer Affairs,
a sample of 16 tins should have less than a 5\%
chance that the mean weight is less than 250 grams.
What is the actual probability that a sample of
16 tins will have a mean weight less than 250 grams?
.0228
.1915
.3085
.4772
.0500
The Central Limit Theorem is important in
Statistics because:
it enables reasonably accurate probabilities to be
determined for events involving the sample average
when the sample size is large regardless of the
distribution of the variable
it tells us that large samples
do not need to
be
selected
it guarantees that, when it applies,
the samples that are drawn are always
randomly
selected
it tells us that if several samples have produced
sample averages which seem to be different than
expected, the next sample average will likely be
close to its expected value
it is the basis for much of the theory that has
been developed in the area of discrete random
variables and their probability
distributions
A random sample of 15 people is taken from a
population in which 40\% favor a particular
political stand. What is the probability that
exactly 6 individuals in the sample favor
this political stand?
0.2066
0.4000
0.5000
0.4000
0.0041
It has been estimated that about 30\% of frozen
chickens contain enough salmonella bacteria to
cause illness if improperly cooked. A consumer
purchases 12 frozen chickens. What is the
probability that the consumer will have more
than 6 contaminated chickens?
.039
.961
.118
.882
.079
Which of the following is not an assumption of
the Binomial Distribution?
the probability of success is equal to .5 in all trials
all trials must be identical
all trials must be independent
each trial must be classified as a success or a failure
the number of successes in the trials is counted
The probability that a certain machine will produce a
defective item is 0.20. If a random sample of 6 items
is taken from the output of this machine, what is the
probability that there will be 5 or more defectives
in the sample?
.0016
.0001
.0154
.0015
.2458
Seventeen people have been exposed to a particular
disease. Each one independently has a 40\% chance
of contracting the disease. A hospital has the
capacity to handle 10 cases of the disease. What
is the probability that the hospital's capacity
will be exceeded?
.035
.965
.989
.011
.736
There are 10 patients on the Neo-Natal Ward of a
local hospital who are monitored by 2 staff
members. If the probability (at any one time) of
a patient requiring emergency attention by a
staff member is .3, assuming the patients to be
behave independently, what is the probability at
any one time that there will not be sufficient
staff to attend all emergencies?
.6172
.3828
.3000
.0900
.9100
Newsweek in 1989 reported that 60\% of young children
have blood lead levels that could impair their
neurological development. Assuming that a class in
a school is a random sample from the population of
all children at risk, the probability that at least
5 children out of 10 in a sample taken from a school
may have a blood level that may impair development is:
about .84
about .25
about .20
about .16
about .64
The number of children in a family is an example
of a continuous random variable.
False
True
The Central Limit Theorem guarantees that the population
we sample from is normal whenever the sample size is
sufficiently large.
False
True
As the degrees of freedom increases to infinity, the
$ T $ distribution approaches the normal distribution
with mean one and variance two.
False
True
Confidence intervals for $ {\mu}_x $ are valid only if the
sample is taken from a normal distribution.
False
True
As the sample size increases, the variance of the distribution
of $ \overline{X} $
decreases
increases
remains the same
flucuates back and forth
flits hither and yon
The Council on Alcohol and Drug Abuse wants to
determine the percentage of young people in the
United States in the 35--40 year old age group
who have a serious problem with alcohol.
A sample of 1000 people in this group yielded 250
showing signs of alcohol abuse. A point estimate
for the desired proportion is
0.250
250
2.50
0.500
not enough information is given to answer this question
Ashley and Shannon were late to Statistics class
because they had stopped to pick some poppies.
In order to not get yelled at by their other
class mates for interrupting class, they collected
a sample of 23 poppies, counted the number of petals
on each, and brought the following information
to class:
$$
\overline{X} \, = 9.16 \, \, \text{ and }
\, \, S \, = \, 1.3 .
$$
What is a point estimate of the average number of
petals on a poppy leaf?
9.16
1.3
$ {1.3}^2 $
$ 9.16 \, {}_{-}^{+} \, 0.271 $
none of the other answers is even vaguely familiar to me
A machine is producing metal pieces that are cylindrical
in shape. A sample of pieces is taken and the diameters,
measured in inches, are recorded as follows.
$$
1.01 \qquad \qquad 1.04 \qquad \qquad 0.99
$$
$$
0.97 \qquad \qquad 0.99 \qquad \qquad 1.01
$$
$$
1.03 \qquad \qquad 0.98 \qquad \qquad 1.03
$$
A point estimate for the mean diameter of metal pieces
from this machine is
1.0056 inches
2.0056 inches
0.0005 inches
0.0224 inches
Inches? What's an inch?
A sample of size 16 gives
$ \overline{X} = 7864 $ and $ S^2 = 552 $.
A point estimate of the standard error of the mean is
5.87
138.0
552.0
34.5
none of the other answers is correct
The standard error of the mean is
the standard deviation of the distribution of the sample mean
an error often made on Statistics exams
an error or mistake in the calculation of the sample mean
the arithmetic mean of the distribution of the sample mean
does not exist unless the population is at least approximately normal
A 95\% confidence interval for a population mean
was calculated from a random sample of 50
observations, and turned out to be
$ 24 \, {}_{-}^{+} \, 7 $ ({\it i.e.}, 18 to 32). For
practical purposes we can afford to draw only one sample
and conclude
We are 95\% confident that $ \mu $ lies between 18 and 32.
We are confident that about 95\% of the population
means are between 18 and 32.
We are confident that 95\% of our observations in
our sample are between 18 and 32.
We are 95\% confident that the sample mean will fall
between 18 and 32.
We are confident of nothing; the only certainties are
death and taxes.
To construct a 93.56\% confidence interval on $ \mu $
when $ \sigma $ is known and the sampled population is normal,
the table value needed for substitution in the formula is
1.85
0.0644
$ -1.52 $
1.52
No such confidence interval exists.
Use the following information to answer the next
three questions.
The travel time for a business woman between Radford and Roanoke
is uniformly distributed between 40 and 90 minutes.
What is the probability she will finish her trip in
80 minutes or less?
0.8
0.02
0.2
1.0
none of the other answers is correct
The probability her trip will take longer than 60 minutes is
0.6
0.4
0.02
1.00
none of the other answers is correct
The probability her trip will take exactly 50 minutes is
0.0
1.0
0.02
0.06
none of the other answers is correct
All of the following statements about the normal
distribution are true except
the variance is always 1
mean = median = mode
the graph of the distribution is a bell--shaped curve
the mean may be unequal to 0
the distribution is symmetric about $ \mu $
Let $ X $ denote the score obtained by a particular
gymnast on her floor routine. Assume that $ X $ is
normally distributed with mean 9.0 and standard deviation
0.25. The probability that the score for a given performance
falls between 8.75 and 9.25 is approximately equal to
0.68
0.95
0.99
0.75
\$2.50
The value of a standard normal random variable tells us
the number of standard deviations from the mean
the mean of any normal distribution
the variance of any normal distribution
whether the random variable is discrete or continuous
that for every light on Broadway, there's a broken heart
in Topeka
Suppose $ Z $ is standard normal. Find
$ p(Z = 2.6) $.
0.0
0.9953
0.0047
$ -.0047 $
$ \dfrac{{\sigma}_x}{\sqrt{n}} $
Suppose $ Z $ is standard normal. Find the value of
$ L $ for which $ p(Z \ge L) = 0.1251 $.
1.15
$ -1.15 $
0.5517
0.8749
not enough information is given to answer this question
Use the following information to answer the next
three questions. Let $ X $ denote the amount of radiation
that can be absorbed by an individual before death ensues.
Assume that $ X $ is normally distributed with a mean of
500 roentgens and a standard deviation of 150 roentgens.
Find $ p(X \le 560) $.
0.6554
0.3446
0.5
0.4
Roentgens? What's a roentgen?
How many standard deviations from the mean is an
individual who can absorb 560 roentgens of radiation
before death?
0.4
1.0
0.2
$ -0.4 $
Roentgens? What's a roentgen?
Above what dosage level will only 2.5\% of those
exposed survive?
794
975
1.96
650
Roentgens? What's a roentgen?
How are statistics and parameters related?
A sample statistic estimates a population parameter.
usually by birth, but sometimes by marriage
A population statistic estimates a sample parameter.
A population parameter estimates a sample statistic.
A sample parameter estimates a population statistic.
For the standard normal probability distribution, the area
to the left of the mean is
0.5
greater than 0.5
$ -0.5 $
1
none of the other answers is correct
If a $ X $ value is to the left of the mean,
the corresponding $ Z $ value is
negative
positive
zero
non--existent
none of the other answers is correct
Suppose $ Z $ is standard normal. Find
$ p(-1.50 \le Z \le 1.90) $.
0.9045
0.0381
$ -0.0381 $
0.4
none of the other answers is correct
Suppose $ Z $ is standard normal.
Find $ p(-2.0 \le Z \le -1.0) $.
0.1359
0.8185
0.1469
1.0000
none of the other answers is correct
Suppose $ Z $ is standard normal. Find
$ p( 2.0 \le Z \le 2.5) $.
0.0166
0.9710
0.5000
4.500
none of the other answers is correct
Suppose $ Z $ is standard normal. Find
$ p(-2.54 \le Z \le 2.54) $.
0.9890
0.4945
0.0000
0.5400
none of the other answers is correct
Suppose $ Z $ is standard normal. Find
$ p(2.32 \le Z \le 3.05) $.
0.0091
0.4989
0.9887
0.9909
none of the other answers is correct
Suppose $ Z $ is standard normal. If the area between
zero and $ L $ is 0.4115, then $ L $ is
1.35
2.70
1.00
0.2077
none of the other answers is correct
Suppose $ Z $ is standard normal. If the area to the right
of $ L $ is 0.8413, then $ L $ is
$ -1.0 $
1.0
2.0
$ -2.0 $
none of the other answers is correct
Suppose $ Z $ is standard normal. If the area to the
right of $ L $ is 0.0668, then $ L $ is
1.50
0.17
2.00
1.00
none of the other answers is correct
A standard normal distribution is a normal distribution
with a mean of 0 and a standard deviation of 1
which is the standard by which all future normal
distributions should be judged
with a mean of 1 and a standard deviation of 0
with any mean and any standard deviation
with a mean of 0 and any standard deviation
A normal probability distribution
is a continuous probability distribution
is a discrete probability distribution
can be either continuous or discrete
must have a mean of 0
none of the other answers is correct
Use the following information to answer the next three questions.
The life expectancy of a particular brand of tire is normally
distributed with a mean of 40,000 miles and a standard
deviation of 5,000 miles. What is the probability that a randomly
selected tire will have a life of at least 30,000 miles?
0.9772
0.4772
0.0228
none of the other answers is correct
Mile? What's a mile?
What is the probability that a randomly selected
tire will have a life of at least 47,500 miles?
0.0668
0.4332
0.9332
50--50
none of the other answers is correct
What percentage of the tires will have a life
between 34,000 and 46,000 miles?
76.98\%
38.49\%
50--50
3\%
none of the other answers is correct
Use the following information to answer the next
four questions.
A volunteer ambulance service handles 0 to 5
service calls on any given day. The actual
probability distribution for the number of
service calls is as follows.
$$
x: \qquad 0 \qquad 1 \qquad 2
\qquad 3 \qquad 4 \qquad 5
$$
$$
f(x): \quad .10 \quad .15 \quad .30
\quad .20 \quad .15 \quad .10
$$
What is the expected number of service calls
per day?
2.45
2.50
6.25
4.0
none of the other answers is correct
What is the standard deviation of the number of
service calls per day?
1.43
2.05
2.45
2.92
none of the other answers is correct
What is the probability of at least three service
calls in a given day?
0.45
0.20
0.35
0.75
none of the other answers is correct
What is the probability of fewer than two service
calls in a given day?
0.25
0.30
0.55
0.75
none of the other answers is correct
Use the following information to answer the next
two questions.
Twenty--nine percent of lawyers and judges are women
({\it Statistical Abstract of the United States}, 1997).
In a jurisdiction with 30 lawyers and judges, what is the
expected number of women?
8.7
15.0
2.9
17.4
none of the other answers is correct
In the same jurisdiction, what is the standard
deviation of the number of women?
2.5
1 or 2
6.2
more than 10
none of the other answers is correct
Use the following information to answer the next
three questions. According to the WTA and the ATP
Tour, forty percent of women tennis players use
Wilson rackets. Suppose ten women players
are chosen at random. What is the
probability exactly one of them uses a Wilson
racket?
0.1209
0.1673
.0464
0.5000
none of the other answers is correct
What is the probability exactly two use Wilson rackets?
0.2150
0.1673
0.3823
0.16
50--50
What is the probability none of them use a Wilson racket?
0.0060
0.0404
0.9940
50--50
Wilson rackets? What a Wilson racket?
Use the following information to answer the next
three questions. Many utility companies have begun
to promote energy conservation by offering discount
rates to consumers who keep their usage below certain
established subsidy standards. A recent EPA reports
indicates that 70\% of the consumers of one area
have reduced their usage enough to qualify for
discounted rates. Suppose six consumers from the
region are chosen randomly. What is the probability
all six will qualify for discounted rates?
0.1176
1.0000
0.8824
0.0102
none of the other answers is correct
What is the probability at least four qualify
for discounted rates?
0.7443
0.2557
0.5798
0.4202
none of the other answers is correct
What is the probability fewer than two qualify for
discounted rates?
0.0109
0.0705
0.0007
0.7443
none of the other answers is correct
Use your TI calculator to construct a frequency histogram for this set of
data. Let it determine the number of classes and class boundaries.
What is the number of classes it uses?
7
5
3
9
1
What is the frequency of the third class?
7
8
4
5
6
What are the boundaries of the second class?
18.03--18.87
17.20--18.03
18.87--19.70
17--22
There is no well to tell.
Which class is the modal class?
3
7
5
1
There is no modal class.
Which of the following {\it best} describes the shape of the histogram?
Somewhat symmetric with a well-defined mean
Skewed left
Extremely variable
Bimodal
It makes no sense to talk about shape.
Which of these is not a reason that Statistics is of interest?
Statistics prove with 100\% certainty what we want to know
about a population.
It is often hard to express our requirements without the
use of numbers.
Numerical quantification can add specificity to vague statements.
Most statistical studies are performed using samples of data
from the population of interest.
Statistics allows us to make inferences about important population
parameters.
One of the aims of inferential statistics is to
make inferences about samples from a population
by analyzing the entire population.
False
True