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29-1 If 100 random samples of size n were drawn
79-1 find in the table
79-2 percentile of students' distribution
80-1 "students' distribution, 90 percent of the area"
83-1 T-distributions are spread out
84-2 A t-distribution with 30 d.f. is most similar
86-1 a true statement regarding the comparison of t-distributions
87-2 A t-distribution is used in estimating MU when
88-1 Suppose that we repeatedly draw a random sample
88-2 Obtain the values of the t-distribution
89-1 the value of Student's t that subdivides
92-1 standard normal distribution and a student's distribution are alike
96-1 Since percentiles of the t-distribution approximate those of
148-1 When (for what level of confidence) do we use Z =
148-2 (Z-score) corresponds to the sample mean
154-2 the 95% confidence interval for the appropriate z-score is
155-1 on which test was his performance better
156-1 then P(70.5 < XBAR < 71.3) is approximately
161-1 what is the probability that the mean competency
161-2 The middle 99% points for the distribution of the sample mean
162-1 P(WBAR > 195)
163-1 what are the middle 90%
175-1 Find (approximately) P[XBAR>2]
193-1 the students' personal libraries at this
201-1 whether or not the manufacturing process is operating satisfactorily
215-2 average (mean) height of 9 randomly selected setters
732-1 "Which of the following describes a ""statistical inference""?"
765-2 Standard error of a mean
777-1 MU
1342-1 A pocket calculator enthusiast claims that 80% of all incoming freshmen
1394-1 Establish a 90% confidence interval for the mean speed of cars
1398-1 is 10. The standard error of the mean is
1399-2 "As the sample size (n) increases, the mean of a random sample"
1399-3 In general the sampling distribution of the mean and the distribution
1400-2 Suppose that all possible distinct samples of size n > 30
1401-1 Suppose that all possible distinct samples of size n > 1
1418-1 90% confidence limits corresponding to a
1418-2 the lower confidence limit for a two sided
1419-1 t distribution is usually used instead of the standard normal
1420-2 90% confidence interval for the true percentage of children
1421-1 "MU, which of the following is most precise:"
1421-2 An approximate 95% confidence interval
1425-3 "Other things being equal, the larger the confidence coefficient"
1426-1 An investigation reveals that a confidence interval with
1428-1 percentages of fat in 1 pound packages
1429-1 Construct a 99% confidence interval
1429-2 "1, 2, 3, 4, 5. Construct a 95% symmetric confidence"
1430-2 What kind of table whould you use
1431-2 Determine 95% confidence limits corresponding to a
1433-1 The value you would find in the table to complete the information
1433-2 A 95% confidence interval for the mean
1434-1 interval for the population proportion favoring Senator Claghorn is:
1434-2 "t-statistic, student A uses a confidence coefficient of"
1435-1 What will be the effects of changes
1435-2 "To three decimals, what is the upper limit of"
1438-1 The size of a confidence interval for a mean is affected by
1439-1 The 93% confidence interval for MU is closest to:
1440-1 Consider the following table which gives the deodorants preferred
1441-2 then a 99% confidence interval for p is given by
1443-1 What is the probability that XBAR differs from MU by more than 1?
1443-2 A 95% confidence interval for MU is:
1445-1 The mean weight of a random sample of 16 dogs on the quad
1446-1 "XBAR = 31 and SUM(i = 1, 400)((X(i) - 31)**2) = 1596"
1447-1 "XBAR = 20 and SUM(i = 1, 16)((X(i) - 20)**2) = 960"
1448-2 Eighty percent of a sample of 400 people support candidate B.
1449-1 "8, 11, 9, 17, 12, 15 are a sample of size 6"
1454-1 found that the mean cost (i.e. XBAR) of hospital care for
1455-1 by adding to and subtracting from the sample mean a certain multiple
1455-2 A 90% confidence interval
1456-1 "Given that 99% confidence limits for MU are 42 and 58, "
1457-1 Suppose a psychologist wishes to know the mean IQ of students
1459-1 an approximate 95% confidence interval for MU?
1460-2 Find the 0.95 confidence interval estimate
1461-2 "A parameter is fixed (non-fluctuating), a confidence interval is"
1463-1 contained within approximately how many of these intervals?
1463-2 The sample mean of 225 scores on a math test is 75.
1464-1 the probability of MU falling within a confidence interval
1469-2 Is 113 an acceptable value for MU at a 95% confidence level?
1470-1 The Chamber of Commerce in Miami Beach wishes to estimate
1471-1 estimate the proportion of people in a population who
1471-2 Set up a 90% confidence interval for the area of the plot.
1472-1 John has done an experiment on gallons of water per second
1472-2 500 accounts receivable is selected
1473-1 A survey on consumer finances reports that 33 per cent
1473-3 Use a confidence level of 90%.
1474-2 confidence interval for the proportion of males
1478-1 nine patients on this new diet had observed cholesterol
1478-2 interval for the actual percentage of television viewers
1480-1 voters in Portsmouth indicate that 60%
1481-1 confidence interval for the true mean arm reach of tournament fighters.
1481-2 "Estimate the proportion of ""poverty"" families"
1482-1 Interpret the statistical meaning of this confidence interval.
1483-1 Would your interval have been narrower if SIGMA would have been
1484-1 respond to him based on part a above
1485-1 "their bills for the evening were [$12.50, $10.75, $14.28]"
1486-1 from 0 (no parenting skills deemed successful) to 100
1487-1 A sociologist conducted a study of assertion by
1488-1 "A sample of six male delinquents, aged 16, indicates"
1491-2 Explain why the confidence interval in (b) is approximate.
1495-1 Make a confidence interval statement concerning the proportion of
1499-3 Would you forecast a win for candidate A? Why?
1500-1 Which of the following confidence limits would appear most favorable
1501-1 determine if these limits are consistent with
1509-1 "If (5,8) is a 95% confidence interval for a MU, then the probability"
1509-2 A confidence interval for MU will generally be smaller if
1510-1 As the size of a confidence coefficient (one minus ALPHA)
1510-2 The sample mean lies at the center of the confidence interval for MU.
1511-2 A 95% confidence interval is twice as long as a 90% confidence
1513-3 With any sample size a high level of confidence in an interval
1522-2 What would be the midpoint of an interval estimate of p?
1523-1 the width of the interval can be narrowed by
1547-1 The confidence coefficient is the probability that an unknown parameter
1547-3 If confidence intervals are computed from repeated samples of the same
1548-2 MARGIN OF ERROR
1551-2 The purpose of using a sample and calculating a mean is to
1566-1 will approximate the normal curve if
1566-3 "If we think of how all possible XBAR's are distributed, the"
1568-1 The Central Limit Theorem tells us that:
1568-2 "As the size of the sample, n, increases towards the size of the popu"
1569-1 The Central Limit Theorem
1570-1 The sampling distribution of means of random samples of size n drawn
1572-3 10 hours using the Central Limit Theorem.
1576-1 "According to the Central Limit Theorem, how does a sampling distribu"
1577-1 Define what is meant by the sampling distribution of XBAR for size
1578-1 What is the approximate distribution of Z?
1580-2 The Central Limit Theorem is of most value when we sample from
1581-1 "As the sample size increases, the distribution of the sample mean"
1581-2 The Central Limit Theorem applies to the case of sampling from
1582-2 "the same size, drawn from a severely skewed population, will equal"
1583-2 "According to the Central Limit Theorem, the shape of "
1584-2 "The central limit theorem tells us that, if we take a large sample,"
1585-2 The sampling distribution of XBAR is approximately normal if and only
1691-1 "Other things being equal, a low level of confidence is desirable."
1730-1 ________________ is the process of drawing conclusions about population
1791-3 A proportion is a special case of a mean when you have
1815-1 "variance 36. For samples of size 9, the sampling distribution"
1820-2 "If the population distribution of scores (X) is normally distributed,"
1857-4 Suppose we have sampled (with replacement) from a finite population.
1858-3 A larger mean implies a larger standard deviation.
1859-1 non-normal population with SIGMA**2 = 1 and MU = 0 would have mean
1865-1 "The variance of the sample mean, SIGMA(XBAR)**2, is computed"
1869-1 "Does this imply that one is ""above average"" and the other is ""below"
1869-2 Give an example illustrating that the standard error of a sample
1891-2 Explain how SIGMA differs from SIGMA(XBAR).
1898-1 "For a large sample size, the estimated standard error of the mean is"
1898-2 The standard deviation of the original observations is generally
1901-1 Distributions of population statistics have standard deviations
1905-2 estimator of the variance of the population from which
1927-1 VAR(XBAR) = __________. (sometimes denoted (SIGMA(XBAR))**2).
1928-1 "If the variance of the mean S(XBAR)**2 = 5 and n = 5, what is S**2?"
1951-3 The distribution of XBAR will have a variance equal to the
1959-1 If the random variable X has a normal distribution with mean MU and
1960-2 "In a population with MU = 10 and SIGMA**2 = 64, the standard error"
1962-1 "MU = 5 and SIGMA = 20, the probability that the mean"
1963-2 If independent samples of size 6 are drawn over and over again and
1964-1 The standard error of the mean is another name for the standard
1964-3 "SIGMA(XBAR) changes in what way when n,"
1966-1 S(XBAR) is:
1966-3 "SIGMA(XBAR), is customarily called the:"
1967-1 Find the standard error of the mean for
1968-1 How could you estimate the value of the standard error
1968-2 "as before, but using samples of size 9n"
1969-1 The standard error of the mean
1969-2 Which of the following best describes the standard error of YBAR?
1971-1 Which investigator is apt to obtain a better estimate of the corres
1977-4 "As the sample size increases, the standard error of the mean remains"
1978-1 The standard deviation of the random sampling distribution of the mean
1978-3 "If the sample size is greater than one, the sampling distribution"
1979-3 The standard error of the sample mean increases with the sample
1981-1 The formula SIGMA(XBAR) = SIGMA/SQRT(n) requires the population
2074-2 The standard error of the median is an index of
2074-4 it must be a _________________ distribution.
2103-1 Let MU(X) and SIGMA(X)**2 be the mean and the variance of a
2857-1 A sampling distribution:
2859-1 Which is NOT a characteristic of a (random) sampling distribution
2860-3 We __________ have a complete sampling distribution displayed for us.
2862-4 How does a sampling distribution differ from the distribution of
2866-1 if the sample size is increased nine-fold
2867-3 A sampling distribution could be considered a population.
2889-1 If a population is very large an especially large sample is usually
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Q: Which one of the following statements is correct?
a. If L and U are the lower and upper limits of a 99% confidence
interval for MU, then MU varies between L and U with a
probability of .99.
b. If for a sample of 100 students from the registrar's office
it was found that 95% of these students had dean's list
averages, then random sampling from the entire student body
could not possibly have been performed.
c. Assuming the population variance is known, then if the sample size
is doubled, the variance of the distribution of the sample mean of
a variable would be halved.
d. If 100 random samples of size n were drawn and, if, for each,
a 99% confidence interval for MU was computed, then exactly
99 of these confidence intervals would contain within their
limits the true population mean.
e. For continuous variables the probabilities P(a < x < b) and
P(a <= x <= b) are always the same, whereas for discrete
variables these probabilities are always different.
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Q: A sample of size 16 is taken from a normal population. Then a 99%
confidence interval is set up with XBAR = 30 and s = 20. The value
you would find in the table to complete the information necessary
to obtain the interval would be:
(a) 2.602 (d) 2.947
(b) 2.326 (e) 2.921
(c) 2.576 (f) none of these.
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Q: The first percentile of students' distribution with 24 degrees of
freedom is:
(1) -2.80 (4) 2.49
(2) -2.50 (5) 2.50
(3) -2.49
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Q: For students' distribution, 90 percent of the area lies between t =
-1.89 and t = 1.89 if the degrees of freedom are:
(1) 2 (4) 7
(2) 3 (5) 8
(3) 6
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Q: A large normally distributed population has mean 16. Consider all
samples of size 7. The numbers
(XBAR - 16)/(S/SQRT(7))
are:
(a) normally distributed
(b) t distributed with 7 degrees of freedom
(c) t distributed with 6 degrees of freedom
(d) neither t nor normal
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Q: T-distributions are spread out __________(more or less) than a normal
distribution with MU = 0, SIGMA = 1.
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Q: A t-distribution with 30 d.f. is most similar to a _____ distribution.
a. normal distribution with mean = 1 and variance = 1
b. normal distribution with mean = 0 and variance = SIGMA**2
c. normal distribution with mean = 0 and variance = 29
d. normal distribution with mean = 0 and variance = 1
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Q: Which of the following is a true statement regarding the comparison of
t-distributions to the standard normal distribution?
a. The normal distribution is symmetrical whereas the t-distributions
are slightly skewed.
b. The proportion of area beyond a specific value of t is less than
the proportion of area beyond the corresponding value of z.
c. The greater the df, the more the t-distributions resemble the
standard normal distribution.
d. All of the above.
e. None of the above.
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Q: A t-distribution is used in estimating MU when __________ is unknown
but its use assumes that the sample data ____________________________.
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Q: Suppose that we repeatedly draw a random sample from a normally
distributed population with a known mean and calculate a value for
student's t for each sample.
a. We will calculate t = (sample mean - _____) / _____
b. If each sample consists of 7 elements, the t distribution
generation will have _____ degrees of freedom.
c. Over a large number of trials _____% of the values generated
will be greater than 1.440.
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Q: Obtain the values of the t-distribution with 8 degrees of freedom that
subdivide the area under the curve so that 5% is to the left of the
smaller value and 5% is to the right of the larger value. Sketch the
curve of this t-distribution and indicate the areas involved.
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Q: What is the value of Student's t that subdivides the area under
any Student's t curve so that 50% of the area lies to the right
of that value?
What is the corresponding value for a standard normal distribution?
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Q: 1) State two ways in which a standard normal distribution and a
student's distribution are alike.
2) State at least one way in which they differ.
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Q: True or False? If False, correct it.
Since percentiles of the t-distribution approximate those of
the normal distribution if n is greater than 30, this is usually
considered a good sample size to use.
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Q: When (for what level of confidence) do we use Z = 1.645, for a
two-sided test or confidence interval?
a. 90%
b. 95%
c. 80%
d. 100%
e. 99%
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Q: A random sample of size 25 is taken from a population with mean 7 and
variance 4. The sample mean is calculated to be 8. What value of the
standard normal random variable (Z-score) corresponds to the sample
mean?
a. 25
b. 1.25
c. -1.25
d. +2.5
e. none of the above
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Q: Consider the following data:
1, 7, 3, 3, 6, 4
Assuming this data is drawn from a normal population with mean = MU
and variance = 6, the 95% confidence interval for the appropriate
z-score is _____ standard units long.
a) 5.14 b) 4.90 c) 4.58 d) 4.04 e) 3.92
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Q: The class average on the first midterm was 81 with a standard
deviation of 9. The class average on the second midterm was 78
with a standard deviation of 12.
If a student got a 93 on each test, on which test was his performance
better relative to the class?
A. First midterm B. Second midterm
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Q: The height of male college freshmen has a normal distribution with
mean 71 inches and standard deviation 3 inches. If 100 male college
freshmen are selected at random, and XBAR is the average of their
heights, then P(70.5 < XBAR < 71.3) is approximately:
a) .4525 b) .6732 c) .7938 d) .8413 e) none of these
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Q: Suppose that for a sample of 36 Family Nurse Practitioners (FNP's)
from several similar type hospital clinics, a competency score
ranging from 0 to 100 was derived based on performance at the clinic.
Suppose further that the population mean competency score for all
FNP's was 80 and the population variance was 100. For the sample of
36 FNP's, what is the probability that the mean competency score will
be between 75 and 80?
a. .4987 b. .1915 c. .5013 d. .2287 e. .5115
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Q: Suppose that for a sample of 36 Family Nurse Practitioners (FNP's)
from several similar type hospital clinics, a competency score
ranging from 0 to 100 was derived based on performance at the clinic.
Suppose further that the population mean competency score for all
FNP's was 80 and the population variance was 100.
The middle 99% points for the distribution of the sample mean
competency score described above is (rounded to two decimal places):
a. (54.24, 105.76) b. (76.12, 83.88) c. (56.74, 103.26)
d. (75.71, 84.29) e. None of these
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Q: Suppose that the weight (W) of male patients registered at a diet
clinic has the normal distribution with mean 190 and variance 100.
For a sample of size 25 from the clinic, which of the following
statements is equivalent to the statement:
P(WBAR > 195)
where WBAR denotes the mean weight of the sample?
a. P(Z < -2.5) d. P(Z > 2.5)
b. P(Z < 1) e. P(Z < 2.5)
c. P(Z > -1)
(Note: Z is a standard normal random variable.)
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Q: Suppose that the weight (W) of male patients registered at a diet
clinic has the normal distribution with mean 190 and variance 100.
For a sample of size 25 from the clinic, what are the middle 90%
points of the distribution of WBAR where WBAR denotes the mean weight
of the sample?
a. (186.08, 193.92) d. (186.71, 193.29)
b. (173.55, 206.45) e. None of these
c. (170.40, 209.6)
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Q: A random sample of size 100 is selected from a population with MU=0
and SIGMA=20. Find (approximately) P[XBAR>2].
a. .0228 b. .4207
c. .3085 d. .1587
e. None of the above.
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Q: Suppose the distribution of the students' personal libraries at this
university can be approximated by a normal distribution with mean equal
to 18.7 and variance equal to 1.08. If a random sample of 27 students
is polled, what is the probability that the average size of their lib-
raries will be at least 19.3 books?
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Q: A company manufactures cylinders that have a mean 2 inches in
diameter. The standard deviation of the diameters of the cylinders
is .10 inches. The diameters of a sample of 4 cylinders are
measured every hour. The sample mean is used to decide
whether or not the manufacturing process is operating satisfactorily.
The following decision rule is applied: If the mean diameter
for the sample of 4 cylinders is equal to 2.15 inches or more,
or equal to 1.85 inches or less, stop the process. If the
mean diameter is more than 1.85 inches and less then 2.15
inches, leave the process alone.
a. What is the probability of stopping the process if the
process average MU, remains at 2.00 inches?
b. What is the probability of stopping the process if the
process mean were to shift to MU = 2.10 inches?
c. What is the probability of leaving the process alone if
the process mean were to shift to MU = 2.15 inches?
To MU = 2.30 inches?
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Q: Suppose that height of English setter dogs is normally distributed
with a mean of 30 inches and a known variance of 9. What is the
probability that the average (mean) height of 9 randomly selected
setters will be greater than 31 inches?
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Q: Which of the following describes a "statistical inference"?
a. A true statement about a population made by measuring some sample
of that population.
b. A conjecture about a population made by measuring some sample of
that population.
c. A true statement about a sample made by measuring some population.
d. A conjecture about a sample made by measuring some population.
e. A true statement about a sample made by measuring the entire
population.
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Q: Define the following term and give an example of its use.
Your example should not be one given in class or in a handout.
Standard error of a mean
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Q: Define the following term and give an example of its use.
Your example should not be one given in class or in a handout.
MU
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Q: A pocket calculator enthusiast claims that 80% of all incoming freshmen
own a pocket calculator. To investigate this claim a random sample of
200 incoming freshmen have been interviewed. Of these 120 own
calculators.
a. Set 90% confidence limits for P1, the true proportion having
calculators. (Use ALPHA = .10.)
b. Test the enthusiast's claim.
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Q: A random sample of 64 cars passing a check point on a certain highway
showed a mean speed of 60.0 mph. The standard deviation of speeds is
known to be 15.0 mph.
a. Give a point estimate of the population mean speed on this highway.
b. Establish a 90% confidence interval for the mean speed of cars on
this highway. Interpret the meaning of this interval.
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Q: A sample of 3600 cases is drawn at random from an infinitely large
population. The standard deviation of the population is 10. The
standard error of the mean is
a. 2/15.
b. 1/6.
c. 4/5.
d. 10.
e. none of the above.
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Q: True or False? If false, correct it.
As the sample size (n) increases, the mean of a random sample is less
likely to be near the mean of the population.
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Q: True or False? If false, correct it.
In general the sampling distribution of the mean and the distribution
of the parent population have exactly the same shape.
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Q: Suppose we have sampled (with replacement) from a finite population.
Suppose that all possible distinct samples of size n > 30 have been
selected and that the mean and variance have been computed for each
sample.
Suppose here, that Variance = SUM((X - XBAR)**2/(n - 1)).
True or False? If False, correct it.
The distribution of XBAR will be symmetric.
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Q: Suppose we have sampled with replacement from a finite
population. Suppose that all possible distinct samples
of size n (where n > 1) have been selected and that the
mean has been computed for each sample.
True or false? If false, correct it.
The shape of the distribution of XBAR will be the same
as the shape of the population.
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Q: A sample of twenty-five observations is taken from a normal popu-
lation with variance 9. 90% confidence limits corresponding to a
sample mean of 30 are best represented by:
(a) 30 +/- 9.00
(b) 30 +/- .79
(c) 30 +/- 1.03
(d) 30 +/- .47
(e) 30 +/- .99
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Q: Suppose that a sample of size 16 from a normal distribution with mean
MU yielded a sample mean of 3.2 and a standard deviation of 4. For a
98% confidence level, the lower confidence limit for a two sided in-
terval for MU is:
(1) 0.6 (4) 2.555
(2) 0.62 (5) 5.8
(3) 2.5
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Q: To find confidence intervals for the mean of a normal distribution, the
t distribution is usually used instead of the standard normal distribu-
tion because:
(1) the mean of the population is not known
(2) the t distribution is more efficient
(3) the variance of the population is usually not known
(4) the standard error of the estimate is S/SQRT(n)
(5) the sample mean is known.
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Q: Out of 50 children drawn at random from a large population of
school children, all but 10 showed need of dental treatment.
Based on normal approximation, a 90% confidence interval for
the true percentage of children in that population who need
dental treatment is:
(a) (10.7%, 29.3%) (d) (65.4%, 94.6%)
(b) (77.1%, 82.9%) (e) None of these.
(c) (70.7%, 89.3%)
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Q: In repeated constructions of 95% confidence intervals for a
population mean, MU, which of the following is most precise:
a. MU falls in the interval approximately 95 times out of a
100.
b. the interval brackets the unknown MU approximately 95 times
out of a 100.
c. 95 out of a 100 populations will have their means in the
interval.
d. XBAR falls in the interval approximately 95 times out of
a 100.
e. The interval brackets XBAR approximately 95 times out of
a 100.
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Q: Suppose in a random sample of 9 men, the mean height is found to be
70 inches, and suppose it is known that the population standard
deviation is 3 inches. An approximate 95% confidence interval for the
population mean height is:
a. 64 to 76 inches d. 69 to 71 inches
b. 67 to 73 inches e. none of these
c. 68 to 72 inches
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Q: True or False? If False, correct it.
Other things being equal, the larger the confidence coefficient (i.e.
1 - ALPHA), the smaller the confidence interval.
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Q: True or False? If False, correct it.
An investigation reveals that a confidence interval with confidence
coefficient .95 for the mean extends from 11.2 to 17.5. This means
that in about 95 percent of all samples drawn by the same method, the
sample means will fall between 11.2 and 17.5.
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Q: Suppose we have a population of percentages of fat in 1 pound packages
of hamburger. Suppose these percentages are normally distributed with
mean 28 and standard deviation 4. The probability that the mean per-
centage of a sample of 16 packages is between 26.5 and 29.5 ounces is
closest to:
a. .433
b. .134
c. .866
d. .933
e. .067
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Q: The mean of a sample of 100 items, drawn from an infinite population
with variance 400, is XBAR = 5. Construct a 99% confidence interval
for MU, the mean of the population.
a. -.05 to 9.25 b. 2 to 7 c. -0.30 to 10.20
d. -0.5 to 11.2 e. -0.15 to 10.15
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Q: A sample of size 5 from a normal population consists of the numbers
1, 2, 3, 4, 5. Construct a 95% symmetric confidence interval for
the mean of the population.
a. 2.00 - 4.00 b. 1.04 - 4.96 c. 0.00 - 6.00 d. 1.52 - 4.48
e. 1.85 - 4.15 f. 1.90 - 4.10 g. 0.50 - 5.50 h. 0.95 - 5.05
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Q: In a survey study on the incidence of depression in a population
of psychiatric hospital administrators, scores on a depression
measure were obtained from 86 respondents. The observed mean
score was 62, with a standard deviation of 16. What kind of
table whould you use in constructing a 99% confidence interval
for the mean depression score?
a. Chi square
b. Kendall's tau
c. binomial probabilities
d. student's t
e. other (specify):
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Q: A sample of twenty-five observations is taken from a normal population
with variance 4. Determine 95% confidence limits corresponding to a
sample mean of 20.
(a) 20 +/- .4 (c) 20 +/- .32
(b) 20 +/- 8 (d) 20 +/- .8
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Q: A sample of size 26 is taken from a normal population. Then a 95%
confidence interval is set up with XBAR = 30 and s = 20. The value
you would find in the table to complete the information necessary to
obtain the interval would be:
a. 2.064
b. 2.060
c. 1.711
d. 1.708
e. 1.960
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Q: Nine men with a genetic condition that causes obesity entered a weight
reduction program. After four months the statistics of weight loss were
XBAR = 11.2, S = 9.0. A 95% confidence interval for the mean of the
population of which this is a sample (assuming normality and randomness)
is:
a. 11.2 +/- 1.96(3)
b. 11.2 +/- 1.86(3)
c. 11.2 +/- 2.262(3)
d. 11.2 +/- 2.306(3)
e. 11.2 +/- 2.306(9)/SQRT(8)
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Q: A political pollster is hired to estimate the proportion of voters in
favor of Senator Claghorn. He takes a sample of 400 and finds 56% of
the voters favor the Senator. A 95% confidence interval for the true
population proportion favoring Senator Claghorn is:
(a) .56 +/- 1.96SQRT((.56)(.44))
(b) .44 +/- 1.96SQRT((.56)(.44)/400)
(c) .56 +/- 1.96SQRT((.44)(.56)/400)
(d) .56 +/- 1.64SQRT((.56)(.44)/400)
(e) none of the above.
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Q: In finding confidence intervals for the mean of a normal population
by using a t-statistic, student A uses a confidence coefficient of
0.95 while student B uses 0.99. Which one of the following
statements is true about the length of the confidence intervals found
by A and B? (The length of the confidence interval is the upper
limit minus the lower limit).
(1) B's interval will always be smaller than A's interval
(2) B's interval will usually be smaller than A's interval
(3) B's interval will always be larger than A's interval
(4) B's interval will usually be larger than A's interval
(5) There is no way of knowing which of the intervals will be larger.
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Q: For a certain normally distributed population, the value of the standard
deviation is known, but the value of the mean is unknown. What will
be the effects of changes in sample size and in the confidence
coefficient on the length of the confidence interval estimate of the
population mean?
a. Increasing sample size increases the length, given a fixed
coefficient.
b. Increasing the confidence coefficient decreases the length
given a fixed sample size.
c. Increasing sample size decreases the length, given a fixed
coefficient.
d. None of the above
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Q: A random sample of 625 boxes taken from the output of a box making
machine was inspected for flaws. It was found that 500 of the boxes
were free from flaws. To three decimals, what is the upper limit of
the 0.99 confidence interval estimate of the proportion of
acceptable boxes being produced?
a. .8 + 1.96*SQRT(.16/625)
b. .8 + 2.576*(.16/625)
c. .8 + 1.96*(.16/625)
d. .8 + 2.576*SQRT(.16/625)
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Q: The size of a confidence interval for a mean is affected by changes in
which of the following?
a. The size of the sample
b. The confidence coefficient
c. The variance of the sample
d. b and c only
e. a, b, and c
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Q: The strength of elevator cables are to be measured. Let
X = strength of a cable, and assume X is normal with mean
MU and variance SIGMA**2, both unknown. A sample of 89
straps is taken, with results XBAR = 31 and S**2 = 89.
The 93% confidence interval for MU is closest to:
(a) 31 +/- (2.11)*SQRT(89) (b) 31 +/- (1.60)*(1.0)
(c) 31 +/- (1/SQRT(89)) (d) 31 +/- 1.81
(e) 31 +/- (1.80)*(89/SQRT(89))
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Q: Consider the following table which gives the deodorants preferred
by samples of lacrosse and soccer players.
Brand X Brand Y
------------------
lacrosse players | 20 | 30
--------------------------------------
soccer players | 10 | 40
--------------------------------------
1. Now let PI(1) be the proportion of all lacrosse players who
prefer Brand X. An 80% confidence interval for PI(1) would
use a table value closest to:
A. 1.28 B. 1.10 C. 1.50 D. 1.70 E. 1.90
2. An 80% confidence interval for PI(1) would then be closest to:
A. (1/5) +/- (1.28)(SQRT((4/25)(1/100)))
B. (2/5) +/- (1.28)(SQRT((6/25)(1/50)))
C. (2/3) +/- (1.1)(SQRT((2/9)(1/30)))
D. (1/5) +/- (1.5)(SQRT((4/25)(1/50)))
E. (2/5) +/- (1.7)(SQRT((6/25)(1/100)))
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Q: A poll of 100 students revealed that 50 were in favor of returning
to the semester system. If p is the proportion of all U.C.D. stu-
dents in favor of semesters, then a 99% confidence interval for p
is given by
(a) .5 +/- .098 (d) .5 +/- .135
(b) .5 +/- .112 (e) .5 +/- .141
(c) .5 +/- .129
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Q: A sample of size 36 is taken from a population with unknown mean
MU and standard deviation SIGMA = 3.
What is the probability that XBAR differs from MU by more than 1?
(a) .3413 (b) .9544 (c) .0228 (d) .4772 (e) .0456
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Q: A sample of size 36 is taken from a population with unknown mean
MU and standard deviation SIGMA = 3.
A 95% confidence interval for MU is:
(a) XBAR +/- 1.96(1/2) (d) MU +/- 1.64(XBAR)
(b) XBAR +/- 1.96 (e) MU +/- 1.96(SIGMA)
(c) XBAR +/- 1.64(1/4)
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Q: The mean weight of a random sample of 16 dogs on the quad is 58 lbs.
with a sample standard deviation of 8 lbs. A 95% confidence interval
for the mean weight of dogs on the quad is:
a) 58 +/- (1.96*2) b) 58 +/- (2.13*2) c) 58 +/- (1.96*4)
d) 58 +/- (1.75*2) e) 58 +/- (2.33*4)
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Q: A sample of size 400 is taken from a large population, yielding
XBAR = 31 and SUM(i = 1, 400)((X(i) - 31)**2) = 1596
The 95% confidence interval for the mean is:
(a) [30.88, 31.12] (d) [30.84, 31.16]
(b) [30.80, 31.19] (e) [30.76, 31.24]
(c) [30.75, 31.25]
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Q: A sample of size 16 is taken from a normally distributed population,
yielding:
XBAR = 20 and SUM(i = 1, 16)((X(i) - 20)**2) = 960
The 95% confidence interval for the mean of this population is:
(a) [15.8, 24.3] (d) [17.1, 22.9]
(b) [15.6, 24.4] (e) [16.4, 23.6]
(c) [16.2, 23.8]
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Q: Eighty percent of a sample of 400 people support candidate B.
The 95% confidence interval for the proportion of people who
support B is nearest:
A. [.768, .832] B. [.761, .839]
C. [.775, .825] D. [.790, .810]
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Q: If the observations: 8, 11, 9, 17, 12, 15 are a sample of size 6
from a normal population with a mean = MU and a variance = 24, then
a 95% confidence interval for MU has confidence limits
a. 12 +/- 3.92
b. 11.5 +/- 3.92
c. 12 +/- 5.142
d. 11.5 +/- 5.142
e. none of these
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Q: Suppose for a random sample of 16 hospitals in North Carolina, it was
found that the mean cost (i.e. XBAR) of hospital care for recovery
from acute myocardial infarction (MI) was $15000 with a sample
standard deviation of $1200. Then a 99% confidence interval for the
mean cost of recovery for all hospitals in North Carolina would have
the following limits (rounded to nearest whole numbers):
a. (14116, 15884) b. (14123,15876) c. (14225, 15775)
d. (14227, 15773) e. none of these
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Q: The limits of a 95% confidence interval for the mean MU of some
population, whose variance SIGMA**2 is assumed known, are found by
adding to and subtracting from the sample mean a certain multiple of
the standard error. For a sample size of 36, the multiplier described
in the previous sentence is:
a. 1.645
b. 1.6896
c. 2.0301
d. 1.96
e. None of these
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Q: Suppose that a sample of sixteen heights yielded a sample mean of
75.1 inches and a sample variance of 9.0. A 90% confidence interval
(rounded to two decimal places) for the true mean height of the
population has which of the following limits:
a. (73.79, 76.41) d. (71.16, 79.04)
b. (73.87, 76.33) e. None of these
c. (74.09, 76.11)
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Q: Given that 99% confidence limits for MU are 42 and 58, which of
of the following could be 95% confidence limits?
a. 43 and 57
b. 41 and 59
c. 41 and 57
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Q: Suppose a psychologist wishes to know the mean IQ of students in a
given high school. 25 students were sampled and the sample mean was
103. The sample standard deviation was 12. Which of the following
is a 90% confidence interval for the mean IQ for all students?
a. 98.9 - 107.1
b. 97.6 - 108.3
c. 98.0 - 108.0
d. 94.5 - 111.5
e. none of these
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Q: In a sample of size 400 we found XBAR=10 and S**2=100. Which of the
following is an approximate 95% confidence interval for MU?
a. (.2, 19.8) b. (-9.8, 9.8)
c. (8.04, 11.96) d. (9.02, 10.98)
e. (-.98, .98)
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Q: A sample of size 25 was taken from a normal population whose standard
deviation SIGMA is believed to be 20. The sample data produced a mean
XBAR of 57.5 Find the 0.95 confidence interval estimate for the popu-
lation mean MU.
a) 18.3 to 96.7
b) 49.66 to 65.34
c) 50.9 to 64.1
d) 53.5 to 61.5
e) 55.93 to 59.07
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Q: A parameter is fixed (non-fluctuating), a confidence interval is
a. also fixed.
b. equal to + or - one standard deviation.
c. variable from sample to sample.
d. either fixed or variable depending on sample size.
e. None of the above.
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Q: We know the mean MU of a population. Suppose 1,000 samples of size n
are drawn from this population. For each sample we compute a 90%
confidence interval for MU. We would expect the mean of the
population would NOT be contained within approximately how many of
these intervals?
a. 0
b. 10
c. 100
d. 900
e. Impossible to tell.
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Q: The sample mean of 225 scores on a math test is 75. Find the 95
percent confidence interval for the mean of the population, assuming
that SIGMA(X) = 7.
a. 72.5 - 77.5
b. 74.1 - 75.9
c. 73.2 - 76.8
d. 73.8 - 76.2
e. None of the above.
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Q: In a 95% confidence interval for the mean,
a. if many samples are drawn, YBAR will fall within the confidence
interval 95% of the time.
b. the probability of YBAR falling within a confidence interval
computed from one sample is .95.
c. the probability of MU falling within a confidence interval compu-
ted on one sample is .95.
d. if many samples are drawn, the computed confidence intervals will
contain MU 95% of the time.
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Q: The following values of "Y" represent a random sample from
some population -- 115, 110, 112, 113, 111, 107, 110, 106, 112.
(a) Construct a 95% confidence interval for the population mean.
(b) Is 113 an acceptable value for MU at a 95% confidence level?
(c) Construct a 99% confidence interval for the population mean.
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Q: The Chamber of Commerce in Miami Beach wishes to estimate the mean
expenditures per tourist per visit in that city. A random sample of
one-hundred tourists has been selected for investigation, and it has
been found that the mean expenditures of the sample was $800.00 per
tourist per visit. It is known that the standard deviation of
expenditures for all tourists is $120.00.
Construct a 95 percent confidence interval for the true mean expendi-
ture per tourist. Interpret your result.
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Q: It is wanted to estimate the proportion of people in a population who
have English ancestors. Make a 95% interval estimate assuming that a
random sample of 100 people has 21 of English descent in it.
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Q: A plot of land is surveyed by 25 student surveyors with the
following results:
YBAR = 7.25 acres [SUM(i=1,25)(Y(i)-YBAR)**2]/[25] = .01660
Set up a 90% confidence interval for the area of the plot.
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Q: John has done an experiment on gallons of water per second that flow in
a sewer main in the city. He makes 16 measurements of this flow and
finds that their average is 100 and their variance is 9. Find a 98%
confidence interval for the mean flow.
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Q: A random sample of 500 accounts receivable is selected from the 4,032
accounts that a firm has, and the sample mean is found to be
$242.30. The sample standard deviation is computed to be $3.20.
Set up a .99 confidence interval estimate of the population mean.
How do you interpret the meaning of this interval?
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Q: A survey on consumer finances reports that 33 per cent of a sample
of 2,600 spending units expected good times during the next 12 months.
Assume that a simple random sample was used in the study. Set up a
.95 confidence interval estimate of the population proportion of
spending units expecting good times.
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Q: Compute a confidence interval for the mean, given XBAR = 24, n = 25,
and SIGMA(XBAR) = 2. (Use a confidence level of 90%.)
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Q: Out of the random sample of 360 full time students on the River
Campus, 225 were male and 135 were females. Obtain a 99% con-
fidence interval for the proportion of males among the full time
student body.
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Q: A new diet for the reduction of cholesterol is introduced. In order to
test this procedure, nine patients on this new diet had observed choles-
terol levels of:
patient cholesterol patient cholesterol
1 240 6 220
2 290 7 190
3 220 8 230
4 250 9 200
5 260
XBAR = 210 S**2 = SUM(([X(i) - 210]**2)/8) = 950 S = 30.9
Assume cholesterol levels are normally distributed.
Construct a 99% confidence interval for MU with XBAR = 210.
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Q: In a random sample of 200 television viewers in a certain area, 95
had seen a certain controversial program. Construct a 0.99 confi-
dence interval for the actual percentage of television viewers in
that area who saw the program.
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Q: The results of an imaginary random sample of 30 registered
voters in Portsmouth indicate that 60% of the voters favor
Henry Kissinger for president.
Set 95% confidence limits for this sample result.
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Q: A sample of size 9 is used to estimate the mean arm reach of
fighters in a certain tournament. Arm reach (inches) for all fighters
in the tournament is known to be normally distributed with a variance
4 inches squared. If the sample mean is 35 inches, establish a 95%
confidence interval for the true mean arm reach of tournament fighters.
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Q: A random sample of 100 families in a large city showed 20 with annual
earnings that placed them in a "poverty" category. Estimate the pro-
portion of "poverty" families in the city using a 90% confidence
interval.
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Q: A light bulb manufacturer desires to determine the mean life time of
the light bulbs in the most recent batch produced. Available are 25
light bulbs to be tested. For these bulbs, the average lifetime is 72
hours (i.e. XBAR = 72). From previous experience it is known that
each batch produced bulbs according to a normal distribution with
standard deviation of 15 hours.
a. Construct a 99% confidence interval for the mean MU of the
batch.
b. Interpret the statistical meaning of this confidence interval.
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Q: a) Find a 95% confidence interval for MU from a sample of 9 where
XBAR = 12 and s = 1, and interpret the interval.
b) Would your interval have been narrower if SIGMA would have been
known to be 1?
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Q: Taking a random sample from its very extensive files, a water company
finds that the amount owed in 16 delinquent accounts have a mean of
$16.35 and a standard deviation of $4.56.
a. Use these values to construct a .98 confidence interval for the
average amount owed on all delinquent accounts.
b. If Mr. Blackwater, the company president, claims the delinquent
accounts have a population mean of $19.01, how could you quickly
respond to him based on part a above (also after explaining that
you were using a 2% ALPHA level)?
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Q: You are interested in the amount of discretionary funds available to
male patrons of a disco. A random sample of three patrons indicates
that their bills for the evening were [$12.50, $10.75, $14.28]. Esti-
mate with 90% confidence the mean amount spent by patrons at this disco.
Back to this chapter's Contents
Q: A researcher assigns each of his interviewers a list of 7 families,
drawn randomly from a region, to be interviewed. Each interviewer
is instructed to administer a successful parenting scale SPS to each
parent in his sample. The SPS scores, X(i), are defined as ranging
from 0 (no parenting skills deemed successful) to 100 (successful
parenting skills consistently and skillfully applied). The first
interviewer returned with the following scores for his seven female
respondents (i.e., mothers). Based on this sample, estimate the
mean SPS for females in Region I with 90% confidence:
[X(i)] = [75, 62, 48, 50, 55, 62, 69]
S = 9.856
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Q: A sociologist conducted a study of assertion by having one of her top
students, after appropriate training, note the number of assertive acts
performed in a day by each of 10 randomly selected coeds, producing the
following sample of data, in acts per day: [5,3,10,6,4,9,5,5,7,5].
Estimate with 90% confidence the mean number of assertive acts per day
performed by the coeds.
SUM(X(i)) = 59
XBAR = 5.9
S = 2.18
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Q: A sample of six male delinquents, aged 16, indicates that they have the
following number of delinquent acts recorded on their police record:
[5, 3, 3, 4, 3, 5]. Estimate, with 90% confidence, the mean number of
such acts recorded on the records of the universe of subjects (i.e., 16
year-old delinquent males in Gotham City).
SUM(X(i)) = 23
S = 0.983
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Q: In a random sample of 100 flashlight batteries, the average useful life
was 22 hours and the sample standard deviation was 5 hours.
(a) Estimate MU, the average life of all batteries of the type
sampled.
(b) Determine an approximate 99% confidence interval for MU.
(c) Explain why the confidence interval in (b) is approximate.
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Q: In a recent survey of Democrats in the Congress, 30% favored Senator
Kennedy of Massachusetts as the party's top presidential candidate
at this time, while the remaining 70% was divided up among numerous
other potential candidates. The results were reported accurate to
within +/- 10% with 95% confidence.
Make a confidence interval statement concerning the proportion of
Democrats in the Congress who preferred Senator Kennedy at the time
of the survey.
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Q: A TV network reports that their survey based on 200 randomly
selected voters indicates:
.60 favor candidate A
.40 favor candidate B
Margin of Error (95%) for the difference = .14.
Would you forecast a win for candidate A? Why?
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Q: Suppose that you are interested in an election and the latest
poll indicates:
Your candidate: 36% of the vote
Leading opponent: 40% of the vote
Suppose also that the poll indicates no uncommitted voters.
Which of the following confidence limits would appear most favorable
to your candidate? Why?
a. difference = 4% +/- 1%, 1 - ALPHA = .99
b. difference = 4% +/- 10%, 1 - ALPHA = .99
c. difference = 4% +/- 10%, 1 - ALPHA = .80
Back to this chapter's Contents
Q: A brewery producing beer has a number of specifications for quality.
Among these standards is the requirement that the degree of hop like
flavor should be a value of 8.0.
The production of the brewery consists of a large number of batches.
It's possible for differences to arise between batches, so we will
regard each batch as a different population. We will consider the
hoppiness of each batch as a normally distributed variable with mean
and variance unknown.
From each batch you can remove 6 samples for hoppiness. For each
batch you are to:
a. set confidence limits for the batch (population) mean, MU;
b. determine if these limits are consistent with the require-
ment that hoppiness is a value of 8.0.
1. Outline the procedure to be followed in setting confidence limits
where the probability of the interval calculated including MU is:
a. 90%
b. 99%
2. Apply the procedure outlined to this set of sample values: 13,
11, 9, 14, 8, 11. Is this sample data consistent with the speci-
fication of hoppiness = 8.0 when the probability level used is:
a. 90%
b. 99%
3. Do these results suggest any weakness in the procedure used? If
so, what?
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Q: True or false? If false, explain why.
If (5,8) is a 95% confidence interval for a MU, then the probability
that MU lies in the interval is .95.
Back to this chapter's Contents
Q: True or false? If false, correct it.
A confidence interval for MU will generally be smaller if a confidence
coefficient equals .90 than if this coefficient equals .95.
Back to this chapter's Contents
Q: True or False? Explain.
As the size of a confidence coefficient (one minus ALPHA) is
increased, the width of the corresponding confidence interval
will tend to increase.
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Q: True or False? If False, correct it.
The sample mean lies at the center of the confidence
interval for MU.
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Q: True or False? If False, correct it.
A 95% confidence interval is twice as long as a 90% confidence
interval.
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Q: True or false? If false, explain why.
With any sample size a high level of confidence in an interval
estimate may be obtained.
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Q: We are interested in the proportion (p) of voters in a community
that would favor a school issue to be settled by a referendum. A
random sample of size 225 from the community indicated that 65% were in
favor. What would be the midpoint of an interval estimate of p?
a. .03
b. 1.96
c. p
d. Insufficient information to obtain correct result.
e. Sufficient information but correct result is not given.
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Q: In interval estimation of a population mean, the width of the
interval can be narrowed by
a. increasing n.
b. lessening the confidence level (e.g., .99 to .90).
c. reducing the magnitude of SIGMA(X).
d. all of the above.
e. none of the above.
Back to this chapter's Contents
Q: True or False? If false, correct it.
The confidence coefficient is the probability that an unknown parameter
has a certain value.
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Q: True or False? If False, correct it.
If confidence intervals are computed from repeated samples of the same
size, in the long run they will cover the unknown parameter in the same
percentage of cases as the confidence coefficient.
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Q: Define the following term and give an example of its use.
Your example should not be one given in class or in a handout.
MARGIN OF ERROR
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Q: The purpose of using a sample and calculating a mean is to
a. find the average for the sample
b. determine the dispersion of the sample
c. estimate the mean of the population
d. estimate sample size
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Q: The distribution of means of all possible samples of the same size (n)
drawn from a population will approximate the normal curve if
a. the n is large enough
b. the population is large
c. the population is symmetrical
d. the mean of each sample equals the mean of the population
e. none of the above is correct.
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Q: If we think of how all possible XBAR's are distributed, the mean and
variance of those observations are
a. MU(X) and SIGMA(X)
b. zero and SIGMA(X)
c. MU(X) and ((SIGMA(X))**2)/SQRT(n)
d. MU(X) and ((SIGMA(X))**2)/n
e. none of these.
Back to this chapter's Contents
Q: The Central Limit Theorem tells us that:
a) the shape of all sampling distributions of sample means are nor-
mally distributed.
b) the mean of the distribution of sample means is less than the mean
of the parent population.
c) the standard deviation of the distribution of sample means is the
same as the standard deviation of the population.
d) all of the above are true.
e) none of the above are true.
Back to this chapter's Contents
Q: As the size of the sample, n, increases towards the size of the popu-
lation, the value of the standard error of the mean, SIGMA(XBAR),
approaches:
a) zero
b) one
c) SIGMA
d) SIGMA/SQRT(n)
e) None of the above are correct.
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Q: The Central Limit Theorem
a. says that YBAR approaches MU(Y) as sample size increases.
b. says that s(Y) approaches SIGMA(Y) as sample size increases.
c. says that both a and b will occur.
d. refers to a matter other than those stated above.
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Q: The sampling distribution of means of random samples of size n drawn
from some population will approach normality
a. only if the parent population is normally distributed and if n
is large.
b. only if the parent population is normally distributed regardless
of the value of n.
c. if n is large regardless of the shape of the parent population
distribution.
d. regardless of the value of n and regardless of the shape of the
parent population distribution.
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Q: The mean of a random sample of size n = 36 is used to estimate the
sample mean of a very large population of means which has standard
deviation SIGMA = 25 hours. Find the probability that the sample mean
will be "off" either way by less than 10 hours using the Central
Limit Theorem.
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Q: According to the Central Limit Theorem, how does a sampling distribu-
tion of means change as sample size increases? Of what significance
is this change (are these changes) in telling you how accurately a
sample mean will estimate a population mean?
Back to this chapter's Contents
Q: The scientific rationale for statistical inferences about population
means relies on a set of three propositions which we have called the
Central Limit Theorem. These propositions give us information about,
respectively, the mean, standard deviation, and shape of a sampling
distribution of XBAR.
A. Define what is meant by the sampling distribution of XBAR for size
25 samples.
B. According to the Central Limit Theorem, what relationship will
necessarily exist between the mean of this sampling distribution
and the mean of the population?
C. According to the Central Limit Theorem, what relationship will
necessarily exist between the standard error of the mean
(SIGMA(XBAR)) and the standard deviation of the population?
D. Based on your answer in C, explain how a confidence interval
estimate of MU will be affected by:
1) an increase in the size of the sample selected
2) an increase in the variability of the population from
which the sample is selected.
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Q: If X(1), X(2),...X(n) are n(n > 30) independent and identically
distributed random variables and
Z = (XBAR - MU(XBAR))/SQRT(VAR(XBAR))
a.) What is the approximate distribution of Z?
b.) MU(Z) = _________________________.
c.) VAR(Z) = _______________________.
d.) Why is the central limit theorem of fundamental importance
to statistics?
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Q: True or false? If false, explain why.
The Central Limit Theorem is of most value when we sample from a
normal distribution.
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Q: True or False? If False, correct it.
As the sample size increases, the distribution of the sample mean
approaches a normal distribution.
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Q: True or False? If False, correct it.
The Central Limit Theorem applies to the case of sampling from a
normal distribution as well as other cases.
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Q: True or False? If False, correct it.
The mean of the distribution of means of all possible random samples
of the same size, drawn from a severely skewed population, will equal
the population mean.
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Q: Determine if the following is true or false and explain why.
According to the Central Limit Theorem, the shape of the sampling
distribution of XBAR (given that n = 30) will be normal, whether
or not the shape of the population is normal.
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Q: True or False? If false, explain why.
The central limit theorem tells us that, if we take a large sample,
the sample values will follow an approximate normal distribution.
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Q: True or False? If False, correct it.
The sampling distribution of XBAR is approximately normal if and only
if the population is normally distributed.
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Q: True or False? If False, correct it.
Other things being equal, a low level of confidence is desirable.
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Q: ________________ is the process of drawing conclusions about population
characteristics from the facts given by a sample. It is generalization
from the specific to the general.
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Q: True or false?
A proportion is a special case of a mean when you have a
dichotomous population.
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Q: Consider simple random sampling from a normal distribution with mean
18 and variance 36. For samples of size 9, the sampling distribution
for the sample mean has mean and variance equal to:
a. 2 and 12 respectively
b. 18 and 4 respectively
c. 18 and 12 respectively
d. 2 and 4 respectively
e. none of these
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Q: If the population distribution of scores (X) is normally distributed,
then the sampling distribution of XBAR will be distributed
a. normally only if n is large.
b. normally for any given sample size if the sample is randomly
selected.
c. normally if XBAR is large.
d. Not enough information is given to determine the characteristics
of the distribution.
e. There is sufficient information but the correct answer is not
given.
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Q: Suppose we have sampled (with replacement) from a finite population.
Suppose that all possible distinct samples of size n (where n > 1)
have been selected and that the mean and variance have been computed
for each sample. (Suppose here, that the variance =
(SUM((X-XBAR)**2))/(n-1).)
True or false? If false, correct it.
The distribution of XBAR will have a mean equal to MU.
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Q: True or False? If False, correct it.
A larger mean implies a larger standard deviation.
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Q: True or False? If False, correct it.
The means of two hundred random samples each of size 4 from a
non-normal population with SIGMA**2 = 1 and MU = 0 would have mean
approximately 0 and standard deviation approximately .5.
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Q: A population is such that MU = 10, and SIGMA(X)**2 = 9. A
simple random sample of size 10 is taken and XBAR is computed.
The variance of the sample mean, SIGMA(XBAR)**2, is computed
to be:
(a) 9 (b) .9 (c) .6 (d) .36
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Q: You are the parent of two children that have been given an IQ
test that purports to be a sampling of the intellect. One child
receives a score of 99 and the other receives 101. You know,
of course, that the IQ is based on 100 as average or 'normal'.
Does this imply that one is "above average" and the other is "below
average"? Explain.
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Q: Give an example illustrating that the standard error of a sample
difference ought to be reported along with the sample difference.
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Q: Explain how SIGMA differs from SIGMA(XBAR).
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Q: True or False? If False correct it.
For a large sample size, the estimated standard error of the mean is
generally larger than the sample standard deviation obtained from a
random sample of the same size.
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Q: True or False? If False, correct it.
The standard deviation of the original observations is generally
larger than the standard deviation of all possible sample means.
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Q: True or false? If false, explain why.
Distributions of population statistics have standard deviations
while distributions of sample statistics have standard errors.
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Q: What is the value of the usual (unbiased) estimator of the variance
of the population from which the random sample composed of the values
5, 10, and 3 came?
a. 85.33
b. 13.00
c. 8.67
d. 6.00
e. none of the above
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Q: Let X be a random variable with the following probability
distribution:
P(X = 0) = .4
P(X = 1) = .3
P(X = 2) = .2
P(X = 3) = .1
A sample of 5 values is taken from a population with the above
probability distribution. The sample is:
X(1) = 0 X(2) = 1 X(3) = 0
X(4) = 3 X(5) = 2
VAR(XBAR) = __________. (sometimes denoted (SIGMA(XBAR))**2).
Estimated variance of XBAR, ((S(XBAR))**2) = __________.
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Q: A. If the sample standard deviation S = 3 and n = 3, what is S(XBAR)**2?
B. If the variance of the mean S(XBAR)**2 = 5 and n = 5, what is S**2?
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Q: Suppose we have sampled (with replacement) from a finite
population. Suppose that all possible distinct samples
of size n (where n > 1) have been selected and that the
mean has been computed for each sample.
True or false? If false, correct it.
The distribution of XBAR will have a variance equal to the
population variance.
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Q: If the random variable X has a normal distribution with mean MU and
variance SIGMA**2, which one of the following statements gives the
most accurate information about the sampling distribution of the
sample mean XBAR?
(1) XBAR is often close to SIGMA regardless of the sample size.
(2) As the sample size decreases, XBAR is closer to MU.
(3) As the sample size decreases, the variability of the distri-
bution of XBAR about MU increases.
(4) The errors (XBAR - MU) are both positive and negative regard-
less of the sample size.
(5) XBAR becomes less than SIGMA**2 as the sample size increases.
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Q: In a population with MU = 10 and SIGMA**2 = 64, the standard error
of XBAR for a sample size of 16 is:
a. -1.25
b. 0.5
c. 2
d. 4
e. 16
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Q: If random samples of size 25 are drawn from a normal population for
which MU = 5 and SIGMA = 20, the probability that the mean of a ran-
dom sample will be less than zero is:
(1) .1056 (4) .3944
(2) .2119 (5) .8944
(3) .2881
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Q: Suppose Y is a continuous random variable (now assumed to be normally
distributed) with MU = 138 and SIGMA**2 = 126. If independent samples
of size 6 are drawn over and over again and YBAR is calculated for
each sample, then:
a. the mean of YBAR is 138 and the standard deviation of YBAR is 21.
b. the mean of YBAR is 138 and the standard deviation of YBAR is
square root of 21.
c. the mean of YBAR is 21.
d. the variance of YBAR is 21.
e. b and d.
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Q: The standard error of the mean is another name for the standard
deviation of:
a. a sample
b. a population
c. the sampling distribution of any statistic
d. the sampling distribution of the mean
e. none of the above
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Q: SIGMA(XBAR) changes in what way when n, the sample size, increases?
a. it increases
b. it stays the same
c. it decreases
d. it is 0
e. it is 1
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Q: S(XBAR) is:
a. the standard error of the mean
b. the standard deviation of the sampling distribution of XBAR
c. an estimate of the standard error of the mean
d. the standard deviation of the sampling distribution of X
e. none of these
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Q: The standard deviation for the sampling distribution of sample means,
SIGMA(XBAR), is customarily called the:
a) coefficient of variation for sample means.
b) sampling error for means.
c) standard deviation for the mean.
d) standard error of the mean.
e) all of the above are correct.
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Q: Find the standard error of the mean for a distribution where samples of
size 36 are taken from a population with a mean MU = 100 and a standard
deviation SIGMA = 18.
a) 0.5 b) 2.0 c) 3.0 d) 18.0
e) None of the above are correct.
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Q: A sample of 100 cases is drawn, and values of YBAR and s(Y) are
computed on the sample. The population mean and standard deviation
are unknown. How could you estimate the value of the standard error
of the mean, SIGMA(YBAR), in this situation?
a. Use [SUM(Y(i)**2)]/[n-1]
b. Use [SIGMA(Y)**2]/[SQRT(n-1)]
c. Use [s(Y)]/[SQRT(n)]
d. None of the above.
e. Impossible to obtain an estimate from such data.
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Q: A "large number" of random samples, of n cases each, is drawn from a
certain population of scores. The mean of each sample was computed
and the "large number" of means thus obtained was organized into a
frequency distribution. The standard deviation of this distribution
was determined. The whole process was then repeated with the same
population as before, but using samples of size 9n, i.e., 9 times as
large as before. How will the standard deviation of this second
distribution of means compare with that of the first distribution?
a. It will be one-ninth as large.
b. It will be one-third as large.
c. Since "large numbers" are involved, it will be the same.
d. It will be three times as large.
e. It will be nine times as large.
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Q: The standard error of the mean
a. is theoretically determined.
b. behaves just like a standard deviation.
c. is affected by sample size.
d. is characterized by all of the above.
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Q: Which of the following best describes the standard error of YBAR?
a. [SIGMA(Y)**2]/[n]
b. [SIGMA(Y)]/[n]
c. SQRT([SIGMA(Y)]/[n])
d. [SIGMA(Y)]/[SQRT(n)]
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Q: Investigator X conducted a survey in which he randomly selected and
weighed 1000 persons in a population of 1 million to estimate mean
weight.
Investigator B conducted a similar survey except that he obtained data
on 500 persons in a population of 10,000.
Which investigator is apt to obtain a better estimate of the corres-
ponding population mean? Why?
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Q: True or False? If False, correct it.
As the sample size increases, the standard error of the mean remains
unchanged.
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Q: True or False? If false, correct it.
The standard deviation of the random sampling distribution of the mean
is equal to the population standard deviation.
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Q: True or False? If False, correct it.
If the sample size is greater than one, the sampling distribution of
the mean will always have a variance which is larger than the variance
of the associated parent population.
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Q: True or False? If False, correct it.
The standard error of the sample mean increases with the sample
size.
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Q: True or False? If False, correct it.
The formula SIGMA(XBAR) = SIGMA/SQRT(n) requires the population
to be normally distributed.
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Q: The standard error of the median is an index of
a. the variability of a sampling distribution of S(y)'s.
b. the variability of a sampling distribution of medians.
c. the central tendency of a sampling distribution of S(y)'s.
d. the central tendency of a sampling distribution of medians.
e. none of the above.
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Q: If the appropriate measure of variability for a distribution is the
standard error it must be a _________________ distribution.
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Q: Let MU(X) and SIGMA(X)**2 be the mean and the variance of a
population. If MU(XBAR) and SIGMA(XBAR)**2 are the mean and variance
of the sampling distribution of XBAR, then:
a. MU(X) > MU(XBAR) and SIGMA(X)**2 >= SIGMA(XBAR)**2
b. MU(X) = MU(XBAR) and SIGMA(X)**2 >= SIGMA(XBAR)**2
c. SIGMA(X)**2 >= SIGMA(XBAR)**2; one cannot predict the
relationship between MU(X) and MU(XBAR)
d. MU(X) < MU(XBAR) and SIGMA(X)**2 <= SIGMA(XBAR)**2
e. none of these
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Q: A sampling distribution:
a) is a distribution of all the various sample statistics that can be
found for one sample.
b) of the mean is a distribution of the means taken from all possible
samples of a given size n that could be taken from the population.
c) of any statistic has an approximately normal distribution.
d) is a histogram showing the distribution of the sample.
e) all of the above are correct.
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Q: Which is NOT a characteristic of a (random) sampling distribution of
means?
a. Its mean is the same as the mean of the population of scores.
b. Its standard deviation is greater than that of the population
of scores.
c. It tends to resemble the normal distribution irrespective of
the shape of the population of scores with sufficient n.
d. Its standard deviation changes with variation in sample size.
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Q: We __________ have a complete sampling distribution displayed for us.
a. always
b. frequently
c. seldom
d. never
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Q: How does a sampling distribution differ from the distribution of
a sample?
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Q: True or False? If False, correct it.
Assuming random sampling, if the sample size is increased nine-
fold, then the standard deviation of the sample mean is reduced
by one third.
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Q: True or false?
A sampling distribution could be considered a population.
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Q: True or False? Explain your answer.
If a population is very large an especially large sample is usually
taken or needed.
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A: c. Assuming the population variance is known, then if the sample size
is doubled, the variance of the distribution of the sample mean of
a variable would be halved.
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A: (d) 2.947
df = n - 1
= 16 - 1
= 15
t(ALPHA = .005, df = 15) = 2.947
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A: (3) -2.49.
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A: (4) 7
t value from t table at 7 df = 1.90.
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A: (c) (t distributed with 6 degrees of freedom)
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A: T-distributions are spread out MORE than a normal distribution
with MU = 0, SIGMA = 1.
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A: d. normal distribution with mean = 0 and variance = 1.
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A: c. The greater the df, the more the t-distributions resemble the
standard normal distribution.
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A: A t-distribution is used in estimating MU when SIGMA is unknown but
its use assumes that the sample data COMES FROM A NORMAL DISTRIBUTION.
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A: a. MU
s/SQRT(n)
b. 6
c. 10
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A: t = -1.86 and +1.86
If available, consult file of graphs and diagrams that could not be
computerized for appropriate sketch.
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A: Zero
Zero
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A: 1) Both center at zero and are symmetric.
2) The student t has more area in the tails. The variance of the
standard normal is 1 while the variance of the student t depends
on the degrees of freedom and is in general greater than 1.
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A: True
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A: a. 90%
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A: d. +2.5
Sample Mean = XBAR
E(XBAR) = 7 = MU
VAR(XBAR) = SIGMA**2/n = 4/25
SIGMA(XBAR) = 2/5
Then Z = (XBAR - MU)/SIGMA(XBAR)
= (8 - 7)/(2/5) = 2.5
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A: e) 3.92
Z(ALPHA = .05/2) = 1.96
2 * 1.96 = 3.92
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A: A. First midterm
Z = (X - MU)/SIGMA
Z(1) = (93 - 81)/9 Z(2) = (93 - 78)/12
= 12/9 = 15/12
= 1.33 = 1.25
Since Z(1) > Z(2), the student did better on the first midterm
relative to the class.
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A: c) .7938
P(70.5 < XBAR < 71.3) = P([(70.5-71)/.3] < Z < [(71.3-71)/.3])
= P((-.5/.3) < Z < (.3/.3))
= P(-1.67 < Z < 1)
= .4525 + .3413
= .7938
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A: a. .4987
stderr = 10/6 = SS below
P(75 < XBAR < 80) = P[(75-80)/SS < Z < (80-80)/SS]
= P(-3 < Z < 0)
= .4987
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A: d. (75.71, 84.29)
99% C.I. = MU +/- (Z*S(XBAR))
= 80 +/- (2.575*(10/6))
= from 75.71 to 84.29
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A: d. P(Z > 2.5)
Z = (195 - 190)/[SQRT((100)/25)]
= 2.5
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A: d. (186.71, 193.29)
Z = 1.645
SIGMA(XBAR) = SQRT(100/25) = 2
190 +/- (2)(1.645)
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A: d. .1587
P[XBAR>2] = P[Z > [[2-0]/[20/SQRT(100)]]]
= P[Z > [[2]/[2]]]
= P[Z > 1]
= .1587
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A: MU = 18.7
SIGMA**2 = 1.08
SIGMA = 1.04
SIGMA(XBAR) = SIGMA/SQRT(n)
= 1.04/SQRT(27)
= .200
P(X >= 19.3) = P(Z >= (19.3 - 18.7)/.200)
= P(Z >= 3)
= .0013
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A: a. Z = (1.85 - 2.00)/(.10/SQRT(4)) or (2.15 - 2.00)/(.10/SQRT(4))
= -.15/.05 or = +.15/.05
= -3 = +3
P(Z<-3 or Z>+3) = .0013 + .0013
= .0026
b. Z = (1.85 - 2.10)/.05 or (2.15 - 2.10)/.05
= -2.5/.05 .05/.05
= -5 1
P(Z<-5 or Z>1) = .00000 + .1587
= .1587
c. Using MU = 2.15:
Z = (1.85 - 2.15)/.05 or (2.15 - 2.15)/.05
= -.30/.05 0/.05
= -6 0
P(-6
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A: Z = (XBAR - MU)/(SIGMA/SQRT(n))
= (31 - 30)/(3/SQRT(9))
= 1/(3/3) = 1
Area beyond Z = .1587.
Therefore, the probability that the average of a random sample of 9
setters will exceed 31 inches is 0.1587 or 15.87%.
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A: b. A conjecture about a population made by measuring some sample of
that population.
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A: Definition: A measure of variation among values for sample means for a
particular sample size. The usual method used to calculate
an estimate of the standard error of a mean for sample size n
is to obtain an estimate of the population variance from the
sample elements (S**2), convert it to an estimated variance
of a mean for the desired sample size (Divide S**2 by n), and
take the square root. Notice that the same estimated popula-
tion variance can be used to calculate standard errors of a
mean for several different sample sizes. If the population
variance is known, the process is the same but you are no
longer estimating.
Example: Suppose that a random sample of size 10 has provided an
estimated population variance of 25 with 9 df. That esti-
mated variance provides the basis for calculating
A: Estimated standard error of a mean for a sample
size of 4 = 2.5
B: Estimated standard error of a mean for a sample
size of 16 = 1.25
C: Estimated standard error of a mean for a sample
size of 25 = 1.00
Symbols: S(YBAR), S(XBAR), SIGMA(YBAR), SIGMA(XBAR),...
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A: Definition: A Greek letter usually used to indicate the population mean.
Example: Suppose that we are concerned with the population of coins
carried by each student in this class on a certain day.
MU, the population mean = .78, would be the arithmatic
average of the amounts carried by all students in the class
that day.
Symbol: MU
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A: Using ALPHA = .10, Z = 1.645
S = SQRT(pq/n) = SQRT((.6)(.4)/200) = .0346
a. C.I. = .6 +/- 1.645*.0346
C.I. = .6 +/- .057
C.I. = .543 to .657 at 90% confidence
b. Since .8 is not included in our confidence interval we must
reject his claim at the 10% significance level.
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A: a. 60.0 miles per hour
b. C.I. = XBAR +/- [Z * SIGMA(XBAR)]
= 60.0 +/- [1.645 * (15/SQRT(64))]
= 60.0 +/- [3.08]
= from 56.92 to 63.08
This means that the method used for arriving at this interval will
produce intervals containing the mean 90% of the time. This parti-
cular interval could be one of those, or it could be one of the 5%
that fail to include the mean.
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A: b. 1/6.
Standard Error of Mean = Standard Deviation/SQRT(n)
= 10/SQRT(3600) = 1/6
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A: False - change "less" to "more".
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A: False - in general, the sampling distribution of the mean and the
distribution of the parent population have different shapes.
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A: True, the central limit theorem implies that the sampling distribution
of the mean will approach normality for large samples, even if the
original population distribution is not normal.
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A: False.
The Central Limit Theorem states that the sampling distribution
of the mean will be normal in shape for large samples, even
if the original population distribution is not.
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A: (e) 30 +/- .99
Since SIGMA = 3, SQRT(n) = 5, and Z(critical) for 5% in each
tail = 1.645:
SIGMA/SQRT(n) = .6 and
(.6)*(1.645) = .987
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A: (1) 0.6
Lower confidence limit = XBAR - [t * (S/SQRT(n))]
= 3.2 - [2.6 * (4/4)]
= 0.6
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A: (3) The variance of the population is usually not known.
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A: (c) (70.7%, 89.3%)
Confidence interval is:
80.0 +/- (100*1.645*SQRT(.80*(1 - .80)/50))
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A: b. the interval brackets the unknown MU approximately 95 times out
of a 100.
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A: c. 68 to 72 inches
C.I. = XBAR +/- Z * Standard Error
= 70 +/- 1.96 * SIGMA/SQRT(n)
= 70 +/- 1.96 * 3/SQRT(9)
= 68.04 to 71.96 inches or approximately 68 to 72 inches.
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A: False, the confidence interval increases as the confidence coefficient
increases, other things being equal.
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A: False, it means that in about 95 percent of all samples drawn by the
same method, the true mean will be included in the confidence interval.
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A: c. .866
Z = (X - MU)/(SIGMA/SQRT(n))
Z = (2.95-28)/(4/SQRT(16)) = 1.50
Area between 26.5 and 29.5 is .866.
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A: e. -0.15 to 10.15
SIGMA(XBAR) = 20/10 = 2
C.I. = XBAR +/- Z(.005)SIGMA(XBAR)
= 5 +/- 2.576(2)
= 5 +/- 5.15
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A: b. 1.04 - 4.96
Noting n = 5, we find: XBAR = 3, S**2 = 2.5, S = SQRT(2.5) = 1.581
Using t(ALPHA = .05, df = 4) = 2.776:
C.I. = XBAR +/- (t)*(S/SQRT(n))
= 3 +/- (2.776)*(1.581/SQRT(5))
= 3 +/- 1.96
= 1.04 to 4.96
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A: d. student's t--this assumes depression scores are normally
distributed and student's t is the most common way of making
statements about 1 or 2 means.
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A: (d) 20 +/- .8
SIGMA(XBAR) = SIGMA/SQRT(n)
= 2/5
= .40
(SIGMA(XBAR))*Z(crit) = (.40)( 1.96)
= .784
== .8
C.I. = 20 +/- .8
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A: b. 2.060
Since all statistics involved are sample statistics, use the
t distribution with df = 25, and it is found that 2.5% of the
area lies in each tail in a two-tailed test when t = 2.060.
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A: d. 11.2 +/- 2.306(3)
C.I. = XBAR +/- t(C)(S/SQRT(n))
= 11.2 +/- 2.306(9/SQRT(9))
= 11.2 +/- 2.306(3)
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A: (c) .56 +/- 1.96SQRT((.44)(.56)/400)
SIGMA = SQRT((P)(Q)/n) = SQRT((.56)(.49)/400).
Population Proportion = P +/- Z(C)(S) = .56 +/- 1.96((.44)(.56)/
400)
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A: (3) B's interval will always be larger than A's interval.
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A: c. Increasing sample size decreases the length, given a fixed
coefficient.
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A: d. .8 + 2.576*SQRT(.16/625)
C.I. = p +/- Z(ALPHA/2)*SQRT(pq/n)
p = 500/625 = .8, q = 125/625 = .2
C.I. = .8 +/- Z(.005)*SQRT(.8*.2/625)
Upper limit = .8 + 2.576*SQRT(.16/625)
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A: e. a, b, and c
Confidence Interval = XBAR +/- (Z)*(Standard Deviation/SQRT(n))
Z value is dependent on the confidence coefficient, and the
standard error is SQRT(Variance).
Therefore, a, b, and c affect the size of the confidence in-
terval.
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A: (d) 31 +/- 1.81
Since the df are so high, we can use the normal table even though SIGMA
is unknown.
93% confidence interval = XBAR +/- Z*S/SQRT(n)
Area beyond Z = .035
Therefore, Z = 1.81
C.I. = 31 +/- (1.81)(SQRT(89)/SQRT(89))
= 31 +/- 1.81
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A: 1. A. 1.28
2. B. (2/5) +/- (1.28)(SQRT((6/25)(1/50)))
PI(1) = 20/50 = 2/5
S.D. = SQRT(pq/n) = SQRT((2/5)(3/5)/50)
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A: (c) .5 +/- .129
S = SQRT((.5)(.5)/100)
= .05
C.I. = .5 +/- (2.576)(.05)
= .5 +/- (.1288)
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A: (e) .0456
Half confidence interval = (SIGMA/SQRT(n))*Z
= (3/6)*2
Z = 2
Area beyond Z = .0228
Total Probability = 2(.0228) = .0456, because it is the proba-
bility of getting the difference of -1 as well as +1.
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A: (a) XBAR +/- 1.96(1/2)
C.I. = XBAR +/- Z*SIGMA/SQRT(n)
= XBAR +/- 1.96*(3/6)
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A: b) 58 +/- (2.13*2)
C.I. = MU +/- (t*S/SQRT(N))
= 58 +/- (2.13 * 8/4)
= 58 +/- 4.26
Note: t(df=15, ALPHA=.05, two-tailed) = 2.13
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A: (b) [30.80, 31.19]
C.I. = XBAR +/- Z*S/SQRT(n)
= 31 +/- 1.96*2/SQRT(400)
= 30.80, 31.19
note:
VARIANCE = 1596/(n - 1) = 1596/399 = 4
Therefore, S will be = SQRT(4) = 2
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A: (a) [15.8, 24.3]
C.I. = XBAR +/- t*S/SQRT(n)
= 20 +/- 2.13*8/4
= [15.8, 24.3]
Variance = 960/15 = 64
Standard Deviation = SQRT(64) = 8
Standard Error Of Mean = 8/SQRT(16) = 2
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A: B. [.761, .839]
C.I. = PHAT +/- Z(ALPHA/2) * SQRT((PHAT*(1-PHAT))/n)
= .8 +/- 1.96 * SQRT((.8*.2)/400)
= .8 +/- .0392
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A: a. 12 +/- 3.92
XBAR = (8 + 11 + 9 + 17 + 12 + 15)/6
= 12
SIGMA(XBAR) = SQRT(24/6)
= 2
C.I. = XBAR +/- Z(ALPHA/2) * (SIGMA(XBAR))
= 12 +/- (1.96)(2)
= 12 +/- (3.92)
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A: a. (14116, 15884)
C.I. = 15,000 +/- (2.947*1200)/SQRT(16)
= 15,000 +/- 884.1
= from 14116 to 15884
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A: d. 1.96 (Using normal table)
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A: a. (73.79, 76.41)
75.1 +/- (1.75)[SQRT(9)/SQRT(16)]
75.1 +/- 1.31
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A: a. 43 and 57.
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A: a. 98.9 - 107.1
S(XBAR) = S/SQRT(n) = 12/5 = 2.4
t(critical, df=24, two-tailed, ALPHA=.1) = 1.711
C.I. = XBAR +/- [t * S(XBAR)]
= 103 +/- [1.711 * (12/5)]
= 103 +/- [4.1064]
= from 98.89 to 107.1064
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A: d. (9.02, 10.98)
Since SIGMA is unknown, use the t distribution. However because
n=400, t is approximated by the Z distribution.
95% C.I. = MU +/- Z(.025)*[S/SQRT(n)]
= 10 +/- (1.96)(10/SQRT(400))
= 10 +/- (1.96)(.5)
= 10 +/- 0.98
= from 9.02 to 10.98
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A: 2) 49.66 to 65.34
CI = XBAR +/- [Z*SIGMA(XBAR)]
= 57.5 +/- [1.96*(20/SQRT(25))]
= 57.5 +/- [7.84]
= from 49.66 to 65.34
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A: c. variable from sample to sample.
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A: c. 100
.10 * 1000 = 100
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A: b. 74.1 - 75.9
C.I. = 75 +/- (1.96) * (7/SQRT(225))
= 75 +/- .91467
= (74.1, 75.9)
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A: d. if many samples are drawn, the computed confidence intervals will
contain MU 95% of the time.
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A: XBAR = 110.67
S(X)**2 = 7.1
S(X) = 2.667
df = 8
(a) 110.67 +/- 2.306 * 2.667/SQRT(9)
108.61 <= MU <= 112.72 at 95% confidence
(b) No, it is outside the interval in part a.
(c) 110.67 +/- 3.355 (2.667/3)
107.69 <= MU <= 113.65 at 99% confidence
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A: CI = XBAR +/- ((Z(.025))*SIGMA)/SQRT(n)
= 800 +/- ((1.96)(120))/SQRT(100)
= 800 +/- 23.52
= from 776.48 to 823.52
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A: S = SQRT((.21)(.79)/100)
= .0407
Z(ALPHA/2 = .025) = 1.96
C.I. = .21 +/- (1.96)(.0407)
= .21 +/- (.0798)
= from .13 to .29
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A: S(YBAR) = SQRT([n/(n-1)][.01660/n])
= .0263
YBAR - t(crit)*S(YBAR) < MU < YBAR + t(crit)*S(YBAR)
7.25 - 1.711 * .0263 < MU < 7.25 + 1.711 * .0263
7.205 < MU < 7.295
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A: C.I. = XBAR +/- [t(df=15,ALPHA=.01)*S/SQRT(n)]
C.I. = 100 +/- [2.602*(3/SQRT(16))]
C.I. = 100 +/- 1.95
= from 98.05 to 101.95
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A: Using t with ALPHA = .01 and df = 499,
C.I. = XBAR +/- (t) (S/SQRT(n))
= 242.30 +/- (2.576) (3.20/SQRT(500))
= 241.93 to 242.67.
99% of the time that this procedure is used to calculate an interval,
the resulting interval will contain MU. This interval may or may not
include MU.
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A: PHAT = .33
n = 2600
Z(ALPHA=.025) = 1.96
Stand. error of proportion = SQRT((PHAT(1-PHAT))/n)
= SQRT((.33*.67)/2600)
= .009
C.I. = .33 +/- (1.96 * .009)
= from .312 to .348
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A: C.I. = XBAR +/- (Z)*(SIGMA(XBAR))
= 24 +/- (1.645) (2)
= 24 +/- 3.29
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A: PHAT = 225/360 = .625 or 62.5%
Standard error of proportion = SQRT((PHAT(1-PHAT))/n)
S(PHAT) = SQRT(.625*.375/360)
= .026
C.I. = PHAT +/- Z(ALPHA = .01) * S(PHAT)
= 62.5% +/- (2.58)*(02.6)%
= 62.5% +/- 6.7%
= from 55.8 to 69.2%
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A: S(XBAR) = S/SQRT(n)
= 30.9/3
= 10.3
C.I. = XBAR +/- ([t(ALPHA = .005, df = 8)]*S(XBAR))
= 210 +/- (3.355*10.3)
= from 175.4 to 244.6
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A: .475 +/- 2.58 SQRT((.475*.525)/200) = .475 +/- .091
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A: SIGMA = SQRT(pq/n) and t with ALPHA = .05 and df = 29 equals 2.042
C.I. = .6 +/- 2.042 * SQRT ((.6*(1-.6))/30)
C.I. = .6 +/- 2.042 * SQRT (.008)
C.I. = .6 +/- .1826
C.I. = .417 - .783 at 95% confidence
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A: C.I. = XBAR +/- (Z) (SIGMA/SQRT(n))
= 35 +/- (1.96) (SQRT(4/9))
= 35 +/- 1.31
= 33.69 to 36.31
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A: 90% confidence interval is from .1342 to .2658.
C.I. = PHAT +/- [Z(ALPHA(2))*SQRT((PHAT*QHAT)/n)]
= .2 +/- [1.645 * (.2*.8/100)]
= .1342 to .2658
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A: a. Confidence interval = XBAR +/- [Z * S(XBAR)]
= 72 +/- (2.576 * [15/SQRT(25)])
= 72 +/- (2.576 * 3)
= from 64.272 to 79.728
b. Based on the sample statistic, we are 99% confident that the in-
terval from 64.27 to 79.73 covers the mean. This implies that if all
possible samples of size n (= 25) were taken from the population,
and a 99% confidence interval were calculated from each sample, 99%
of all these intervals would contain the true population mean.
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A: a) C.I. = XBAR +/- (t(ALPHA=.05,df=8) * [s/SQRT(n)])
= 12 +/- (2.306 * [1/SQRT(9)])
= 12 +/- .769
= from 11.231 to 12.769
This procedure for producing confidence limits will yield a confi-
dence interval that contains MU 95% of the time. A probability
statement cannot be made about the calculated interval (11.231,
12.769) since it either contains MU or does not contain MU.
b) Yes, because a Z value would then be used.
Z(.025) = 1.96 < 2.306
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A: a. C.I. = XBAR +/- [t*S(XBAR)]; t(ALPHA=.01, one-tail, df=15) = 2.602
= 16.35 +/- [2.602*(4.56/SQRT(16))]
= from 13.384 to 19.316
b. According to the confidence interval found in part a, Mr.
Blackwater's estimate of $19.01 is a possible estimate for
the population mean. It should be pointed out that we are 98%
confident that such a confidence interval would contain the
population mean.
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A: n = 3
SUM(X(i)) = 37.53
XBAR = [37.53]/[3]
= 12.51
S**2 = [[(12.50-12.51)**2]+[(10.75-12.51)**2]+[(14.28-12.51)**2]]/[2]
= [[0.0001] +[3.0976] + [3.1329]]/[2]
= 3.1153
S = 1.7650
S(XBAR) = [1.7650]/[SQRT(3)]
= 1.0190
90% C.I. = XBAR +/- [t(ALPHA=0.05, df=2) * S(XBAR)]
= 12.51 +/- [2.92 * 1.019]
= 12.51 +/- 2.98
= from $9.53 to $15.49
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A: n = 7
SUM(X(i)) = 421
XBAR = 421/7
= 60.143
S(XBAR) = [9.856]/[SQRT(7)]
= 3.725
90% C.I. = XBAR +/- [t(ALPHA=.05, df=6) * S(XBAR)]
= 60.143 +/- [(1.943) * (3.725)]
= 60.143 +/- [7.238]
= from 52.905 to 67.381
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A: n = 10
90% C.I. = 5.9 +/- [t(df=9,ALPHA=.05,one-tail)*(2.18/SQRT(10)]
= 5.9 +/- [1.833 * 0.69]
= 5.9 +/- [1.266]
= from 4.63 to 7.166
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A: n = 6
XBAR = 3.83
90% C.I. = 3.83 +/- [t(df=5,ALPHA=.05,one-tail)*(0.983/SQRT(6)]
= 3.83 +/- [2.015 * 0.401]
= 3.83 +/- [0.809]
= from 3.02 to 4.64
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A: (a) MU(HAT) = 22
(b) With t(ALPHA = .01, df = 99) = 2.63
22 +/- 2.63 * 5/SQRT(100)
22 +/- 1.315
20.685 to 23.315
(c) There are several reasons why this confidence interval is approxi-
mate, among them is that the term average is not clearly defined
and that S approximates SIGMA. Also, the population distribution
may not be normal.
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A: P = 3/10
C.I. for this proportion = 3/10 +/- 10%
= 3/10 +/- 1/10
= 2/10 to 4/10
Therefore, between 20% and 40% of the Democrats favored Senator Kennedy.
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A: Confidence Interval for the difference:
= .20 +/- .14
= .06 <= the true difference <= .34 at 95% confidence.
Since this interval does not include zero, I would forecast
a win for candidate A.
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A: The most favorable confidence limit for my candidate would be
c., 4% +/- 10%, 1-ALPHA=.80. This limit indicates a larger
standard deviation than the other two, since t(ALPHA=.20) is
smaller than t(ALPHA=.05). Since my candidate is behind
the most favorable situation is that which indicates the greatest
uncertainty that the difference is anything more than random
sampling variation.
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A: 1. a. To set 90% confidence limits including MU, we need XBAR, the
sample standard deviation, s, the sample size, n, and a t
value for ALPHA = .1.
Use the formula:
XBAR +/- (t, ALPHA/2)(s/SQRT(n)) with t(ALPHA/2) = 2.015.
b. Use the same procedure as in 1a, but use t(ALPHA/2) = 4.032.
2. XBAR = 11
Standard deviation = 2.28
n = 6
a. 11 +/- (2.015)(2.28/SQRT(6))
= 11 +/- (2.015)(.93)
= 11 +/- 1.876
= 9.124 to 12.876
This is inconsistent with the specification of hoppiness = 8.0.
b. 11 +/- (4.032)(2.28/SQRT(6))
= 11 +/- (4.032)(.93)
= 11 +/- 3.753
= 7.247 to 14.753
This is consistent with the specification of hoppiness = 8.0.
3. Ths basic weakness seems to be in using a procedure that produces a
confidence interval consistent with a wide range of values, which
makes it difficult to detect departures from MU = 8. This situation
is exaggerated when the 99% level is used. In addition, the sample
size used seems small relative to the variability measured.
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A: False.
If (5,8) is a 95% confidence interval for MU then we are 95%
"confident" that the interval contains the value of MU. However,
since MU is a constant it either does (P=1.00) or does not (P=0.00)
lie in the interval, so it does not make sense to say that P=.95.
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A: True - the more confidence one needs to place in the interval containing
the true mean, the more leeway one must provide.
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A: True
CONFIDENCE INTERVAL = Mean +/- Z*Standard Error
As you increase (1 - ALPHA) the Z value increases, so the width of the
confidence interval will also increase.
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A: True
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A: False. If confidence intervals are based on the same sample
observations, and are for the same unknown parameter,
then, for example, under normality assumption, a 90%
CI for MU is = XBAR +/- 1.645 S(XBAR)
A 95% CI for MU is: XBAR +/- 1.96 S(XBAR)
Hence the statement is not true.
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A: True.
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A: e. Sufficient information but correct result is not given.
midpoint of interval = PHAT = .65
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A: d. all of the above.
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A: False - the confidence coefficient is the probability that an unknown
parameter will be within the given interval.
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A: True.
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A: Definition: The product of a probability dependent multiplier
times the standard error of a random variable. If
the standard error is known the multiplier usually comes
from the standard normal table. If the standard error
is estimated the multiplier usually comes from the
student's t.
Example: Suppose that the significance level is .05 and the
population variance is known to be 16. Suppose further
that a sample of size 4 has been drawn. Then the 95%
margin of error for the population mean is 1.96 times 2
or 3.92. (The population standard error of a mean is the
square root of the variance divided by the sample size,
here: SQRT(16/4).
Symbol: Such as MOE = Z(.05) * S(XBAR)
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A: c. estimate the mean of the population
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A: a. the n is large enough
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A: d. MU(X) and ((SIGMA(X))**2)/n
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A: e) none of the above are true.
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A: a) zero
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A: d. refers to a matter other than those stated above.
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A: c. if n is large regardless of the shape of the parent population
distribution.
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A: Confidence Interval = XBAR +/- Z*SIGMA(XBAR)
= XBAR +/- 10
10 = Z * SIGMA(XBAR)
= Z * (25/6)
Z = 2.4
Probability = 2 * (.4918) = .9836
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A: The variance of the sampling distribution of means decreases as sample
size increases. Therefore, one's estimate becomes more accurate as
sample size increases.
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A: A. The sampling distribution of XBAR for a sample size of 25 is the
frequency distribution of XBARs that results from finding the mean
of every distinct sample of 25 from the population. The Central
Limit Theorem states, among other things, that the shape of this
sampling distribution of the means will approach that of a normal
distribution as larger samples are used, regardless of the shape
of the population distribution. The figure below is an example
of a sampling distribution of XBAR where MU is the population
mean and SIGMA is the population standard deviation.
(If available, consult file of graphs and diagrams that could not
be computerized for appropriate figure.)
B. They will be the same.
C. SIGMA(XBAR) = SIGMA/SQRT(n)
D. 1) Confidence interval size will decrease.
2) Confidence interval size will increase.
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A: a.) Z is approximately normally distributed with MEAN = 0
and SIGMA = 1.
b.) MU(Z) = 0
c.) VAR(Z) = 1
d.) It is of fundamental importance because it specifies that
the sampling distribution of the mean will be normal in
shape for large samples, even if the original population
distribution is not.
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A: False.
The Central Limit Theorem holds equally whether we sample from a
normal distribution or not.
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A: True, the Central Limit Theorem justifies approximating the distribution
of XBAR with a normal distribution when n is sufficiently large.
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A: True, the Central Limit Theorem applies to sampling from any
distribution.
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A: True, the mean of the sampling distribution of means is exactly
equal to the population mean regardless of sample size and type
of distribution.
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A: True, the normal approximation to the probability distribution of XBAR
is reasonably good in most cases for sample sizes of 30 or more, pro-
vided we have a random sample from X(i).
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A: False.
The central limit theorem says that, under the given condition, the
sample means, obtained upon repeated sampling from a population,
will be normally distributed.
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A: False. The sampling distribution of XBAR will be normal in shape
for large samples, (n>30), even if the original population distri-
bution is not.
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A: False.
Other things being equal, a HIGH level of confidence is desirable.
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A: STATISTICAL INFERENCE is the process of drawing conclusions about popula-
tion characteristics from the facts given by a sample. It is generali-
from the specific to the general.
Other possible answers might be: INFERENCE, INDUCTIVE INFERENCE,
INFERENTIAL STATISTICS
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A: True. Let X = 1 if it has the desired attribute and X = 0
if it does not. Then the proportion, p, is: p = (SUM(X))/n.
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A: b. 18 and 4 respectively
Sampling distribution for the sample mean has a mean equal to
the population mean and variance equal to SIGMA**2/n.
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A: b. normally for any given sample size if the sample is randomly
selected.
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A: True
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A: False, size of mean is independent of size of standard deviation.
Standard deviation depends on the variability among individual ob-
servations.
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A: True.
Mean of sampling distribution = population mean.
Standard deviation of sampling distribution = SIGMA/SQRT(n)
= 1/SQRT(4)
= .5
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A: (b) .9
Since SIGMA(XBAR)**2 = SIGMA(X)**2/n.
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A: This question cannot be answered accurately without knowing the
standard deviation for IQ scores. Then these differences from
the mean or average can be evaluated relative to the standard
deviation as to whether there is a significant difference between
them.
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A: The difference in mean income between two towns of 10,000 people,
when based on samples of 5 people, will be much more variable from
sampling situation to situation than a difference based on samples of
100 people. The standard error allows us to take this into account.
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A: SIGMA is the standard deviation of the population, while SIGMA(XBAR)
is the standard error of the mean, which is the standard deviation of
the sampling distribution of means. The relationship between the
two is:
SIGMA(XBAR) = SIGMA/SQRT(n);
where n is the size of the sample.
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A: False. For large n, S(XBAR) should be smaller than the standard
deviation because we know S(XBAR) = S/SQRT(n).
Therefore, S(XBAR) will always be smaller than S, since n is larger
than 1.
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A: True. The standard deviation of all possible sample means should equal
SIGMA/SQRT(n), where SIGMA is the standard deviation of the original
observations, and n is the sample size.
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A: True.
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A: b. 13.00
XBAR = (5 + 10 + 3)/3 = 6
S(X)**2 = [(5 - 6)**2 + (10 - 6)**2 + (3 - 6)**2]/(3 - 1)
= (1 + 16 + 9)/2
= 26/2 = 13
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A: MU = E(X) = (0*.4) + (1*.3) + (2*.2) + (3*.1) = 1.0
E(X**2) = (0*.4) + (1*.3) + (4*.2) + (9*.1) = 2.0
VAR(X) = E(X**2) - (E(X))**2 = 2.0 - 1.0 = 1.0
VAR(XBAR) = (VAR(X))/n = 1.0/5 = 0.2
SUM(X) = 6
SUM(X**2) = 14
S(X)**2 = [SUM(X**2) - ((SUM(X))**2)/n]/(n - 1)
= [14 - (36/5)]/4
= 1.7
S(XBAR)**2 = (S(X)**2)/n = 1.7/5
= 0.34
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A: A. 3
B. 25
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A: False.
The distribution of XBAR will have a variance equal to the
population variance divided by n.
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A: (3) As the sample size decreases, the variability of the distribution
of XBAR about MU increases.
Variability = (XBAR - MU)/(SIGMA/SQRT(n))
Therefore, variability increases as n decreases because n is in
the denominator.
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A: c. 2
SIGMA(XBAR) = SIGMA/SQRT(n)
= 8/4
= 2
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A: (1) .1056
Z = (XBAR - MU)/(SIGMA/SQRT(n))
Z = (0 - 5)/(20/5) = -1.25
AREA = .5 - .3944 = .1056
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A: (e.) b and d
According to the Central Limit Theorem:
MU(XBAR) = MU = 138
SIGMA(XBAR)**2 = SIGMA**2/n = 126/6 = 21
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A: (d.) the sampling distribution of the mean.
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A: c. it decreases
SIGMA(XBAR) = SIGMA/SQRT(n)
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A: c. an estimate of the standard error of the mean
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A: d) standard error of the mean.
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A: c) 3.0
SIGMA(XBAR) = SIGMA/SQRT(n)
= 18/SQRT(36)
= 3
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A: c. Use [s(Y)]/[SQRT(n)]
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A: b. It will be one-third as large.
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A: d. is characterized by all of the above.
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A: d. [SIGMA(Y)]/[SQRT(n)]
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A: Investigator X. It is the size of the sample rather than the population
that matters, when dealing with such large populations and samples.
SIGMA(sample) = SIGMA(population)/SQRT(n)
If we regard SIGMA(population) as a known constant, then
1/SQRT(500) = .044 and 1/SQRT(1000) = .031
These "multiplication factors" will be introduced, leading to a larger
standard error with a smaller sample. I.E., the bigger the sample,
the larger our denominator when we take the square root, and the
larger the denominator, the smaller the value of the standard error
of the mean.
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A: False, as the sample size increases, the standard error of the mean
decreases. [SIGMA(XBAR) = SIGMA/SQRT(n)]
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A: False - the standard deviation of the random sampling distribution of
the mean is equal to the population standard deviation divided
by SQRT(n).
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A: False, the sampling distribution of the mean will always have a variance
which is less than the variance of the associated parent population.
Variance of sampling distribution of mean = S**2/n
Variance of parent population = S**2
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A: False, the standard error of the sample mean decreases as the
sample size increases.
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A: False.
The only requirement for this formula is that sample members are
independently and identically distributed.
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A: b. the variability of a sampling distribution of medians.
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A: If the appropriate measure of variability for a distribution is the
standard error it must be a sampling distribution.
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A: b. MU(X) = MU(XBAR) and SIGMA(X)**2 >= SIGMA(XBAR)**2
Mean of a sampling distribution is always equal to mean of parent
distribution and variance of sampling distribution is equal to
variance of parent distrbution divided by n.
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A: b) of the mean is a distribution of the means taken from all possible
samples of a given size n that could be taken from the population.
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A: b. Its standard deviation is greater than that of the population
of scores.
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A: c. We SELDOM have a complete sampling distribution displayed for us.
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A: The sampling distribution is the distribution of estimates for some
particular statistic found from taking many samples. The process is
to take a sample, calculate the estimate, draw another sample,
calculate its estimate, and repeat this a large number of times. The
sampling distribution is formed by getting a frequency diagram of
these estimates. In contrast, the distribution of the sample is a
frequency diagram of the observations occurring within one sample.
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A: True.
Standard deviation of the sample mean = (SIGMA)/SQRT(n)
and:
(SIGMA)/SQRT(9*n) = (1/3)*[(SIGMA)/SQRT(n)]
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A: True.
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A: False, sample size depends on economic considerations and the varia-
bility in the population. A more variable population needs a bigger
sample, while a less variable population requires a smaller sample.
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T= 5 Computation
D= 3 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Application
D= 5 Biological Sciences Economics
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 3 General
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Application
D= 5 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI MEAN
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI STANDERROROFMEAN MEAN
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T=10 Computation
D= 3 General Psychology
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
MEAN STANDERROROFMEAN CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
DESCRSTAT/P PARAMETRIC
T= 2 Computation
D= 2 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Computation
D= 3 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 3 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 3 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Application
D= 4 General Education
***Calculator Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by CIID - UNH
Numerical Answer
SIMPLE/CI
MEAN CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 5 Application Comprehension
D= 3 General
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
1 Page 1470
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Application
D= 4 Economics General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
PROPORTION SIMPLE/CI
DESCRSTAT/P PARAMETRIC STATISTICS
CONFIDENCEINTERV ESTIMATION CONCEPT
T= 5 Computation
D= 2 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Computation
D= 1 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by C. R. Buncher - Cincinnati
Numerical Answer
SIMPLE/CI
TDISTRIBUTION CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS PROBDISTRIBUTION
PROBABILITY
T= 5 Comprehension Computation
D= 3 General Biological Sciences Natural Sciences
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by G. Miaoulis - Wright State Univ.
Numerical Answer
SIMPLE/CI STANDERROROFMEAN
MEAN I650I CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
DESCRSTAT/P PARAMETRIC
T= 5 Application Comprehension
D= 4 General Business
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
PROPORTION SIMPLE/CI
STANDERROR/OTHER DESCRSTAT/P PARAMETRIC
STATISTICS CONFIDENCEINTERV ESTIMATION
CONCEPT
T= 5 Computation
D= 3 Business General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Numerical Answer
SIMPLE/CI
STANDERROROFMEAN I650I CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
DESCRSTAT/P PARAMETRIC
T= 5 Computation Comprehension
D= 4 General
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Numerical Answer
PROPORTION SIMPLE/CI
DESCRSTAT/P PARAMETRIC STATISTICS
CONFIDENCEINTERV ESTIMATION CONCEPT
T=10 Comprehension Computation
D= 4 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by S. Selvin - UC Berkeley
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Computation
D= 3 Biological Sciences General
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by K. Gorowara - Wright State U.
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Application
D= 6 General Business
***Calculator Necessary***
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Numerical Answer
SIMPLE/CI PROPORTION
I650I CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 5 Comprehension
D= 4 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Numerical Answer
SIMPLE/CI STANDERROROFMEAN
MEAN I650I CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
DESCRSTAT/P PARAMETRIC
T= 5 Computation
D= 4 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
PROPORTION SIMPLE/CI
DESCRSTAT/P PARAMETRIC STATISTICS
CONFIDENCEINTERV ESTIMATION CONCEPT
T= 5 Computation
D= 4 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI MEAN
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T=10 Computation Comprehension
D= 4 General
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Comprehension Computation
D= 2 General
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Computation Comprehension
D= 4 General Business Economics
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI MEAN
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T= 5 Computation
D= 3 General Social Sciences Sociology
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI MEAN
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T= 5 Computation
D= 3 General Psychology Social Sciences
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Computation
D= 2 Sociology General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
SIMPLE/CI
MEAN CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 5 Computation
D= 3 General Sociology Social Sciences
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by R. F. Tate - U. of Oregon
Short Answer
SIMPLE/CI
STANDERROROFMEAN MEAN I650I
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T= 5 Comprehension Computation
D= 5 General
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by L. J. Tashman - U. of Vermont
Short Answer
SIMPLE/CI PROPORTION
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS DESCRSTAT/P PARAMETRIC
T=10 Computation
D= 4 General
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Short Answer
PROPORTION SIMPLE/CI
I650I DESCRSTAT/P PARAMETRIC
STATISTICS CONFIDENCEINTERV ESTIMATION
CONCEPT
T= 5 Comprehension Computation
D= 4 General Social Sciences
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Short Answer
SIMPLE/CI PROPORTION
TYPE1ERROR I650I CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
DESCRSTAT/P PARAMETRIC
T= 5 Comprehension
D= 4 General Social Sciences
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Short Answer
SIMPLE/CI STANDERROROFMEAN
TDISTRIBUTION MEAN SIMPLEDATASET
I650I CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC PROBDISTRIBUTION PROBABILITY
MISCELLANEOUS I650/TEMPORARY
T=10 Application
D= 6 General Business
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
True/False
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Computation
D= 6 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
SIMPLE/CI
TYPE1ERROR I650I CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
SIMPLE/CI TYPE1ERROR
TYPE2ERROR CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
SIMPLE/CI ESTIMATION/OTHER
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
True/False
SIMPLE/CI SAMPLESIZE
TYPE1ERROR CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS SAMPLING
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
PROPORTION OTHER/CI ESTIMATION/OTHER
OTHER/Z DESCRSTAT/P PARAMETRIC
STATISTICS CONFIDENCEINTERV ESTIMATION
CONCEPT ZTEST
T= 5 Computation
D= 5 General Education Social Sciences
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
OTHER/CI
MEAN CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 4 General Education
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
OTHER/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
OTHER/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Definition
OTHER/CI
I650I CONFIDENCEINTERV ESTIMATION
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
ESTIMATION/OTHER SAMPLE
ESTIMATION CONCEPT STATISTICS
SAMPLING
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
CENTRALLIMITTHM
SAMPLESIZE SAMPLINGDISTRIB CONCEPT
STATISTICS SAMPLING
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Multiple Choice
CENTRALLIMITTHM VARIANCE/OTHER
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
CENTRALLIMITTHM STANDERROROFMEAN
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Comprehension
D= 6 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
CENTRALLIMITTHM SAMPLINGDISTRIB
MEAN CONCEPT STATISTICS
SAMPLING DESCRSTAT/P PARAMETRIC
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 5 Application
D= 7 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
SAMPLINGDISTRIB SAMPLESIZE CENTRALLIMITTHM
SAMPLING STATISTICS CONCEPT
T= 5 Comprehension
D= 4 General
***Multiple Parts***
Back to this chapter's Contents
Based upon item submitted by L. J. Tashman - U. of Vermont
Short Answer
CENTRALLIMITTHM SAMPLINGDISTRIB STANDERROROFMEAN
I650I CONCEPT STATISTICS
SAMPLING DESCRSTAT/P PARAMETRIC
T=10 Comprehension
D= 4 General
***Multiple Parts***
Back to this chapter's Contents
Based upon item submitted by S. Selvin - UC Berkeley
Short Answer
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
***Multiple Parts***
Back to this chapter's Contents
Based upon item submitted by H. B. Christensen - BYU
True/False
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Computation
D= 1 General
Back to this chapter's Contents
Based upon item submitted by H. B. Christensen - BYU
True/False
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
True/False
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
MEAN CENTRALLIMITTHM STANDERROROFMEAN
SAMPDIST/C SAMPLESIZE DESCRSTAT/P
PARAMETRIC STATISTICS CONCEPT
ESTIMATION SAMPLING
T= 5 Comprehension
D= 4 General
Back to this chapter's Contents
Item is still being reviewed
True/False
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
True/False
CENTRALLIMITTHM
CONCEPT STATISTICS
T= 2 Computation
D= 1 General
Back to this chapter's Contents
Item is still being reviewed
True/False
SAMPLINGDISTRIB CENTRALLIMITTHM
SAMPLING STATISTICS CONCEPT
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
True/False
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by W. Federer - Cornell
Fill-in
SCOPEOFINFERENCE SAMPLE
CONCEPT STATISTICS SAMPLING
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
True/False
PROPORTION MEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB MEAN VARIANCE
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Computation Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB MEAN
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 3 General Education
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
MEAN SAMPLINGDISTRIB
DESCRSTAT/P PARAMETRIC STATISTICS
SAMPLING
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
True/False
MEAN STANDARDDEVIATIO
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
True/False
SAMPLINGDISTRIB STANDERROROFMEAN MEAN
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension Application
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
VARIABILITY/P VARIANCE/OTHER
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Computation Comprehension
D= 4 General
Back to this chapter's Contents
Based upon item submitted by A. Bugbee - UNH
Short Answer
VARIABILITY/P STANDARDDEVIATIO
I650I DESCRSTAT/P PARAMETRIC
STATISTICS
T= 5 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Short Answer
VARIABILITY/P STANDERROR/OTHER
I650I DESCRSTAT/P PARAMETRIC
STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Short Answer
STANDARDDEVIATIO STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
STANDERROROFMEAN STANDARDDEVIATIO
SAMPLESIZE SAMPLINGDISTRIB DESCRSTAT/P
PARAMETRIC STATISTICS SAMPLING
T= 5 Comprehension
D= 4 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
STANDARDERROR STANDARDDEVIATIO SAMPLINGDISTRIB
PARAMETRIC STATISTICS DESCRSTAT/P
SAMPLING
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
True/False
STANDARDDEVIATIO STANDERROR/OTHER
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
VARIANCE
DESCRSTAT/P PARAMETRIC STATISTICS
T= 5 Computation
D= 2 General
Back to this chapter's Contents
Based upon item submitted by S. Selvin - UC Berkeley
Numerical Answer
VARIANCE STANDERROR/OTHER
OTHER/RV DESCRSTAT/P PARAMETRIC
STATISTICS RANDOMVARIABLES PROBABILITY
T=10 Computation Comprehension
D= 6 General
***Calculator Necessary***
***Multiple Parts***
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Numerical Answer
VARIANCE VARIANCE/OTHER
STANDERROROFMEAN I650I DESCRSTAT/P
PARAMETRIC STATISTICS
T= 5 Comprehension
D= 3 General
***Multiple Parts***
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
VARIANCE SAMPLINGDISTRIB
DESCRSTAT/P PARAMETRIC STATISTICS
SAMPLING
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by B. Weir - N. C. State & Massey Univ.
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Computation Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Computation Comprehension
D= 2 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by R. F. Tate - U. of Oregon
Multiple Choice
STANDERROROFMEAN SAMPLINGDISTRIB
CENTRALLIMITTHM I650I DESCRSTAT/P
PARAMETRIC STATISTICS SAMPLING
CONCEPT
T= 2 Computation Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
Multiple Choice
STANDERROROFMEAN
SAMPLINGDISTRIB I650I DESCRSTAT/P
PARAMETRIC STATISTICS SAMPLING
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Multiple Choice
STANDERROROFMEAN SAMPLESIZE
DESCRSTAT/P PARAMETRIC STATISTICS
SAMPLING
T= 2 Comprehension
D= 1 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Computation
D= 2 General
***Calculator Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 6 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB STANDERROROFMEAN
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 4 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 5 General Education
Back to this chapter's Contents
Based upon item submitted by J. Warren - UNH
Essay ***Calculus Necessary***
STANDERROROFMEAN
I650I DESCRSTAT/P PARAMETRIC
STATISTICS
T= 5 Comprehension
D= 5 General General
Back to this chapter's Contents
Item is still being reviewed
True/False
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
True/False
STANDERROROFMEAN VARIANCE/OTHER
SAMPDIST/C CENTRALLIMITTHM DESCRSTAT/P
PARAMETRIC STATISTICS ESTIMATION
CONCEPT
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
SAMPLESIZE STANDERROROFMEAN
SAMPLINGERROR SAMPLING STATISTICS
DESCRSTAT/P PARAMETRIC
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
True/False
STANDERROROFMEAN
DESCRSTAT/P PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
STANDERROR/OTHER
MEDIAN DESCRSTAT/P PARAMETRIC
STATISTICS DESCRSTAT/NP NONPARAMETRIC
T= 2 Comprehension
D= 4 General Education
Back to this chapter's Contents
Item is still being reviewed
Fill-in
SAMPLINGDISTRIB STANDERROR/OTHER
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB VARIANCE/OTHER
SAMPLING STATISTICS DESCRSTAT/P
PARAMETRIC
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB
SAMPLING STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB
SAMPLING STATISTICS
T= 2 Comprehension
D= 3 General Education
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
SAMPLINGDISTRIB
SAMPLING STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Short Answer
SAMPLE SAMPLINGDISTRIB
SAMPLING STATISTICS
T= 5 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
True/False
SAMPLESIZE SAMPLINGDISTRIB
STANDERROROFMEAN SAMPLING STATISTICS
DESCRSTAT/P PARAMETRIC
T= 2 Computation Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
True/False
SAMPLINGDISTRIB
SAMPLING STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Item is still being reviewed
True/False
SAMPLESIZE
VARIABILITY/P SAMPLING STATISTICS
DESCRSTAT/P PARAMETRIC
T= 2 Comprehension
D= 4 General
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