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124-2 Identify the critical value
158-1 use a _____ confidence interval for MU and hence
209-1 so that eight-ounce cups will overflow only 1% of the time
210-1 A company manufactures rope.
720-3 A type I error is always made when:
722-1 The level of significance is (check all that apply):
745-2 Level of confidence is another name for level of significance.
1280-1 What is the calculated value
1285-1 we would reject H(0) if:
1287-2 You will reject H(0) if:
1288-1 The level of significance of this test is approximately:
1290-1 one should _____ the H(0) since the value _____ lies _____
1291-1 "ALPHA = _____, one should _____ the H(0) since the value "
1298-1 A result was said to be statistically significant at the 5% level.
1298-2 The reasoning in rejecting a null hypothesis is __________.
1299-2 The critical value of a test statistic is determined from:
1301-1 significance level of the test is:
1301-2 Which of the following assumptions are needed to
1302-1 "In testing a hypothesis using a statistic Y, a critical region is"
1306-2 Find all values of Z = [XBAR - MU(0)]/[S/SQRT(n)]
1307-1 Which of the following statements do you KNOW is correct?
1308-1 What type of decision is reached when the calculated value of any
1309-1 "ended with a decision of ""reject H(O)""."
1309-2 what is the function of a critical value that is
1312-1 Whether or not causation may be inferred in a research study
1313-3 Your correct interpretation of the outcome is
1323-1 Test the null hypothesis that the new ball
1325-1 Test the appropriate hypothesis at the 5% level.
1336-4 What objection is there to using the rejection region:
1349-3 "If we would reject a null hypothesis at the 5% level, we would also"
1352-1 The decision to use a one-sided or two-sided test is usually made
1352-3 Significance at the ALPHA = .001 level means that the null
1354-2 Testing at a 5% level of significance means that you only have
1489-1 can we conclude that the experimental mean differs
1490-1 Determine a 0.90 confidence interval for the mean reaction time
1493-1 Construct a 95% confidence interval for MU.
1494-1 Do you think that it would be quite unreasonable for
1502-2 Set up a decision rule for H(0): MU => 25
1508-1 "In a sample of 25 physicians, the mean annual income of $47,000"
1567-2 Which of the following assumptions are needed to test
1602-2 "XBAR = 22, one should conclude that:"
1603-1 "If the P-value for your test statistic satisfies P > .25, then:"
1608-1 weight reduction program. After four months the statistics
1609-1 "The meaning of ""testing the hypothesis MU <= 6 versus MU > 6"
1610-3 "Statistical Significance means that, if an experiment were "
1617-1 H(0): MU >= 114 against H(1): MU < 114
1618-1 H(0): MU = 43 against H(1): MU =/= 43
1619-1 "With a computed t-statistic of 2.63, what conclusion should"
1627-1 A new diet for the reduction of cholesterol is introduced.
1628-2 This term she believes that her students are doing sig
1631-1 Past production units of a certain jet engine model showed the mean
1632-1 "Forty-nine American soldiers, observed at random, yield a mean weight"
1635-1 A standard intelligence examination has been given for several years
1646-1 she has collected the final grades of her classes and found
1654-1 "If a statistic is significant at the 5 percent level, then it must be"
1654-2 One can never prove the truth of a statistical (null) hypothesis. One can
1655-1 "If the population mean is known, it makes no sense to test"
1655-2 Since the P-value in a test of hypothesis is based on the specific
1655-3 The descriptive level of significance (P-value) is chosen by
1657-1 A hypothesis accepted at the ALPHA = .20 level of significance
1657-2 A small significance level indicates that the hypothesis
1657-3 If the results of an investigation show that one sleeping tablet
1661-4 "In hypothesis testing, a type I error is"
1662-1 "With which of the following terms is the ""level of significance"" most"
1662-2 better than the old one. The Type I Error is to conclude that:
1664-2 "Usually, one would like the critical region for a test to be _______"
1664-4 "If the number of observations (n) is increased to 2n,"
1665-1 if the population variance (SIGMA**2) is decreased to
1667-2 "If Z(critical) = 2.04, what is the p-value for your test?"
1668-2 For testing the hypothesis MU = 28 against
1669-1 Interpret the quality control procedure described above as a test of
1670-1 What would be the consequences of a Type I and Type II error?
1672-2 When an experimenter selects a particular level of risk (ALPHA) he
1677-1 What type of error might you have made in part a?
1686-3 A Type I error is committed when one accepts the null hypothesis
1688-1 The significance level is computed under the assumption that
1688-2 The risk of type II error does not depend upon the risk of type I
1691-2 "Other things being equal, a small level of significance is desirable."
1693-1 Level of confidence equals (1 - level of significance).
1694-1 "Although we speak of two types of error, in testing any"
1694-2 "Generally, a larger sample size implies a smaller level of"
1702-1 "In this situation, a type II error would be:"
1704-1 Type II error refers to:
1706-2 What is the probability of a Type II error when ALPHA = .05?
2137-1 rejected at the .05 level but not at the .025 level?
2139-1 The value(s) of the test statistic you would use is (are):
2140-1 Sixteen one-acre plots of wheat were harvested.
2143-1 we find a critical value of t equals 2.015. This means that
2143-2 "According to these data, the researcher can reject his H(0) with"
2147-1 "For each of the following sets of information, find and specify"
2151-2 have a higher average height than American males as a whole?
2789-1 the starting point of the region of rejection in terms of XBAR
2790-1 Should the teacher use the new method? Why?
2792-1 What is the smallest ALPHA-level that could be chosen before you
2796-1 The Crapi Cable Company #35 cable has a mean breaking strength of 1800
2799-1 Analyze the data at both levels after setting up appropriate
2800-1 at a .001 significance level. For a sample of 49
2803-1 "Since the observed Z-value is _______ than the table value, we would
2808-1 Test the company's claim using ALPHA = .01.
2811-2 What is the significance probability of the observed result?
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Q: It is desired to test the claim that a steady diet of wolfbane will
cause a lycanthrope (werewolf) to lose 10 lbs. over 5 months. A random
sample of 49 lycanthropes was taken yielding an average weight loss over
5 months of 12.5 pounds, with S = 7 lbs. Identify the critical value
suitable for conducting a two-tail test of the hypothesis at the
2% level.
a. 2.06
b. 2.58
c. 1.96
d. 1.65
e. 2.33
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Q: Molybdenum rods produced on a production line are supposed to average
2.2 inches in length. It is desired to check whether the process is in
control. Let X = length of such a rod. Assume X is approximately nor-
mally distributed with mean = MU and variance = SIGMA**2, where the mean
and the variance are unknown.
Suppose a sample of n = 400 rods is taken and yields a sample average
length of XBAR = 2 inches, and SUM((X - XBAR)**2) = 399.
To test H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%, one
would use a _____ confidence interval for MU and hence a table value
of _____.
a) 92%, 1.67
b) 92%, 1.41
c) 92%, 1.75
d) 96%, 2.06
e) 96%, 1.75
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Q: A soft drink machine can be regulated so that it discharges an average
of MU ounces per cup. If the ounces of fill are normally distributed
with standard deviation equal to .3 ounces, give the setting for MU
so that eight-ounce cups will overflow only 1% of the time.
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Q: A company manufactures rope. From a large number of tests over a long
period of time, they have found a mean breaking strength of 300 lbs.
and a standard deviation of 24 lbs. Assume that these values are
MU and SIGMA.
It is believed that by a newly developed process, the mean breaking
strength can be increased.
(a) Design a decision rule for rejecting the old process with an
ALPHA error of 0.01 if it is agreed to test 64 ropes.
(b) Under the decision rule adopted in (a), what is the probability
of accepting the old process when in fact the new process has
increased the mean breaking strength to 310 lbs.? Assume SIGMA
is still 24 lbs. Use a diagram to illustrate what you have done,
i.e., draw the reference distributions.
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Q: A type I error is always made when:
a. the null hypothesis is rejected when it is true
b. the null hypothesis is not rejected when it is false
c. the research hypothesis is rejected when it is true
d. the research hypothesis is not rejected when it is false
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Q: The level of significance is (check all that apply):
A. the probability of rejecting the null hypothesis when the null
hypothesis is true.
B. the magnitude of the sample size.
C. symbolized by the greek letter ALPHA.
D. none of the above.
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Q: True or False? If False, correct it.
Level of confidence is another name for level of significance.
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Q: It is desired to test the claim that a steady diet of wolfbane
will cause an 18-year-old lycanthrope werewolf to lose EXACTLY
10 lbs. over 5 months. A random sample of 49 lycanthropes was
taken, yielding an average weight loss over 5 months of 12.5 lbs.
with S = 7 lbs. Let ALPHA = .02. What is the calculated value
suitable for testing the above hypothesis?
a. 12.5
b. 7 * 2.5
c. 2.5
d. -2.5
e. .35
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Q: A sample of size 36 is taken from a population with unknown mean
MU and standard deviation SIGMA = 3.
In a test of H(0): MU = 5 versus H(1): MU =/= 5 at ALPHA = .01,
we would reject H(0) if:
(a) XBAR - 5 > 1.29 or 5 - XBAR > 1.29
(b) XBAR - 5 > 7.74 or 5 - XBAR > 7.74
(c) XBAR - 5 > 1.29 or 5 - XBAR < 7.74
(d) XBAR - 5 < 1.29 or 5 - XBAR < 1.29
(e) XBAR - 5 < 7.74 or 5 - XBAR < 7.74
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Q: The standard deviation of a large population is 20. To test:
H(0): MU = 4 vs. H(A): MU > 4
at level of significance .05, a sample of size 100 will be taken.
You will reject H(0) if:
(a) XBAR >= 7.3 (d) XBAR >= 7.8 or XBAR <= .2
(b) XBAR >= 7.3 or XBAR <= .8 (e) none of the above
(c) XBAR >= 7.8
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Q: To test H(0): MU = 20 vs. H(A): MU =/= 20, a sample of 400 will be
taken from a large population, whose standard deviation is 5. H(0) will
be rejected if XBAR >= 20.5 or XBAR <= 19.5. The level of significance
of this test is approximately:
a. .05
b. .02
c. .10
d. .20
e. .15
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Q: Molybdenum rods are produced by a production line setup. It is desir-
able to check whether the process is in control. Let X = length of such
a rod. Assume X is approximately normally distributed with mean = MU
and variance = SIGMA**2, where the mean and variance are unknown.
Take n = 400 sample rods, with sample average length XBAR = 2 inches,
and SUM((X - XBAR)**2) = 399.
In testing H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%,
one should _____ the H(0) since the value _____ lies _____ the
confidence interval.
a) not reject, 2.2, within
b) reject, 2, outside of
c) reject, 2.2, outside of
d) not reject, 2, within
e) either b or c
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Q: Molybdenum rods are produced by a production line setup. It is desired
to check whether the process is in control. Let X = length of such a
rod. Assume X is approximately normally distributed with mean = MU
and variance = SIGMA**2, where the mean and variance are unknown.
Take n = 400 sample rods, with sample average length XBAR = 2 inches
and SUM((X - XBAR)**2) = 399.
If one were testing H(0): MU = 1 vs. H(1): MU =/= 1 at level
ALPHA = _____, one should _____ the H(0) since the value 1 lies
_____ the confidence interval.
a) 16%, not reject (continue), within
b) 8%, not reject (continue), within
c) 4%, not reject (continue), within
d) 4%, not reject (continue), to the left of
e) 4%, reject, to the left of
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Q: A result was said to be statistically significant at the 5% level.
This means:
a. the null hypothesis is probably wrong
b. the result would be unexpected if the null hypothesis were
true
c. the null hypothesis is probably true
d. none of the above.
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Q: The reasoning in rejecting a null hypothesis is __________.
a. that a significant result usually occurs when the null hypo-
thesis is false.
b. that a significant result seldom occurs when the null hypo-
thesis is true.
c. some reason other than (a) or (b).
d. both (a) and (b).
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Q: The critical value of a test statistic is determined from:
a. calculations from the data.
b. calculations based on many actual repetitions of the same
experiment.
c. the sampling distribution of the statistic assuming H(A).
d. the sampling distribution of the statistic assuming H(O).
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Q: Suppose a t-test for the hypothesis that H(O): MU = 0 vs. H(A): MU
=/= 0 is carried out and we find t(obs.) = 1.8. The descriptive
significance level of the test is:
a. the Type I error probability of the test.
b. the probability of getting a t-value >= 1.8.
c. the probability of getting a t-value >= 1.8 or <= -1.8.
d. the Type II error probability of the test.
e. none of these.
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Q: Indicate which assumptions are needed to use the sample mean and normal
tables to test a hypothesis about a population mean, MU, and known
variance, SIGMA**2. Which of the following assumptions are needed to
use XBAR, the mean of the data, and normal tables to test a hypothesis
about MU?
I. the data are a random sample
II. the population distribution is normal
III. the sample size is large
a. I, II, and III
b. I and either II or III
c. II and III
d. only II
e. none of the above
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Q: In testing a hypothesis using a statistic Y, a critical region is
chosen to meet which of the following conditions:
I. the probability of Y falling in the critical region
when the null hypothesis is true is ALPHA
II. the probability of Y falling in the critical region when
the alternative hypothesis is true is greater than it not
falling in the critical region.
III. the sample size is large
a. I, II, and III
b. I and II only
c. I only
d. II only
e. none of the above
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Q: Suppose you are going to test H(O): MU = MU(0)
H(A): MU =/= MU(0)
using ALPHA =.05. Find all values of Z = [XBAR - MU(0)]/[S/SQRT(n)]
for which H(O) should be rejected.
a) Z < -1.96 b) Z < -1.645 or Z > 1.645
c) -1.96 < Z < 1.96 d) Z < -1.96 or Z > 1.96
e) -1.645 < Z < 1.645
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Q: Suppose a random sample of size 25 is selected from a population with
mean MU, the value of which is unknown. The sample statistics are
XBAR = 6.4, s = 10. Test
H(O): MU = 10
H(A): MU < 10 using ALPHA = .05.
Which of the following statements do you KNOW is correct?
a) A type 1 error has been committed.
b) H(O) is rejected.
c) H(O) is not rejected.
d) Statements (a) and (b) are correct.
e) None of the above.
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Q: What type of decision is reached when the calculated value of any test
statistic falls in the critical region when a false null hypothesis is
being tested?
a) A correct decision.
b) Type I error.
c) Type II error.
d) The type of decision can not be determined from the information
given.
e) None of the above are correct.
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Q: A home owner claims that the current market value of his house is at
least $40,000. Sixty real estate agents were asked independently to
estimate the house's value. The hypothesis test that followed ended
with a decision of "reject H(O)". Which of the following statements
accurately states the conclusion?
a) The home owner is right, the house is worth $40,000.
b) The home owner is right, the house is worth less than $40,000.
c) The home owner is wrong, the house is worth less than $40,000.
d) The home owner is wrong, the house is worth more than $40,000.
e) The home owner is wrong, he should not sell his home.
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Q: In hypothesis testing, what is the function of a critical value that is
taken from the tables?
a. It is equal to the calculated statistic from the observed data.
b. It is the point where the decision changes from reject to fail to
reject.
c. It is the center of the distribution of X's.
d. It is a point which is 1 standard deviation away from the mean.
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Q: Whether or not causation may be inferred in a research study
(a) is indicated by the magnitude of the test statistic employed.
(b) is given by the final p-value.
(c) must be decided by the investigator.
(d) depends upon the sample size employed in the study.
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Q: You have drawn a random sample of size n from a specified population and
you test the hypothesis that MU = 50. Suppose that you are unable to
reject the hypothesis. Your correct interpretation of the outcome is
that
a. in the population, MU = 50.
b. probability is high that in the population, MU = 50.
c. the sample data are not inconsistent with the hypothesis that in
the population, MU = 50.
d. all of the above are acceptable interpretations.
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Q: The FMA Company has designed a new type of 16 lb. bowling ball. The
company knows that the average man who bowls in a scratch league
with the company's old ball has a bowling average of 155. The
variance of these averages is 100. The company asks a random sample
of 100 men bowling in scratch leagues to bowl for five weeks with
their new ball. The mean of bowling averages for these men with the
new ball is 170. There is no reason to believe the variance is any
different with the new ball. Test the null hypothesis that the new
ball does not improve a bowler's average at the 5% level of
significance.
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Q: A reading coordinator in a large public school system suspects that
poor readers may test lower in IQ than children whose reading is satis-
factory. He draws a random sample of 30 fifth grade students who are
poor readers. Historically fifth grade students in the school system
have had an average IQ of 105. The sample of 30 has XBAR = 101.5 and
S(XBAR) = 1.42. Test the appropriate hypothesis at the 5% level.
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Q: What objection is there to using the rejection region:
[68 - [.125*(SIGMA/SQRT(N))]] < XBAR < [68 + [.125*(SIGMA/SQRT(N))]]
in testing the hypothesis H(O): MU = 68? HINT: What would be the
level of significance for this test?
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Q: True or False? If False, correct it.
If we would reject a null hypothesis at the 5% level, we would also
reject it at the 1% level.
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Q: True or False? If False, correct it.
The decision to use a one-sided or two-sided test is usually made
after the data is analyzed.
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Q: True or false? If false, explain why.
Significance at the ALPHA = .001 level means that the null hypo-
thesis is almost certainly false.
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Q: True or false? If false, explain why.
Testing at a 5% level of significance means that you only have a
5% chance of rejecting the null hypothesis.
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Q: The calculated nitrogen content of pure benzanilide is 7.10%. Five
repeat analyses of "representative" samples yielded values of 7.11%,
7.08%, 7.06%, 7.06%, and 7.04%. Using an ALPHA level of size 5%, can we
conclude that the experimental mean differs from the expected value?
Assume that the measured values are approximately normally distributed.
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Q: The following are the reaction times in seconds of five people to a
particular stimulus: 7,8,6,10,9
XBAR = 40/5 = 8
S(X)**2 = [(-1)**2 + (0)**2 + (-2)**2 + (1)**2]/4
= 10/4
= 2.5
a. Do these data present sufficient indication that the mean reac-
tion time of all people would be less than ten seconds? Test at
the 0.05 level of significance (ALPHA = .05).
b. Determine a 0.90 confidence interval for the mean reaction time
for all people to this stimulus.
c. State any assumptions that must be made in order to do parts (a)
and/or (b).
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Q: Suppose in a sample of 25 people, the mean height XBAR was observed
to be 70 inches. Suppose also SIGMA = 3.
A. Construct a 95% confidence interval for MU.
B. Would you reject the hypothesis H(0):MU = 71
versus H(1):MU =/= 71 on the basis of the observations,
when testing at level ALPHA = .05?
C. Would you reject the hypothesis H(0):MU = 72
versus the alternative H(1):MU =/= 72 on the basis of the
observations, when testing at level ALPHA = .05?
D. Would you reject the hypothesis H(0):MU = 69 versus
the (one-sided) alternative H(1):MU > 69 on the basis of your
observations, when testing at level ALPHA = .05?
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Q: A floor manager of a large department store is studying the buying
habits of the store's customers. Suppose he assumes that monthly
income of these customers is normally distributed with a standard
deviation of 500. If he draws a random sample of size N = 100 and
obtains a sample mean of YBAR = 800,
A 0.95 confidence interval for the true population mean is
702 < MU < 898.
Do you think that it would be quite unreasonable for
the true population mean to be $600? Explain.
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Q: Suppose that you are concerned with the claim that MU > 25. Assume that
you know that SIGMA = 1. Set up a decision rule for H(0): MU => 25 for
a sample of size 10 such that the type 1 error rate is 2%.
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Q: In a sample of 25 physicians, the mean annual income of $47,000 with a
variance of $360,000.
a. Estimate with 95% confidence the mean income for all physicians.
b. What assumptions are necessary to make this a valid estimate? Do
you feel the assumptions are reasonable in this case? Why or why
not?
c. Based on your answer in (a), would the null hypothesis that the true
mean is $50,000 be continued or rejected at the 5% significance
level? Why?
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Q: Which of the following assumptions are needed to test a hypothesis
about mean MU in a population with known variance SIGMA**2 using
the mean of the data X(1) ... X(N) and normal tables:
I. The data are a random sample.
II. The population distribution is normal.
III. The sample size is large.
a. I, II and III d. only II
b. I and either II or III e. none of these
c. II and III
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Q: In testing H(0): MU <= 20 vs. H(A): MU > 20 when ALPHA = .05,
n = 25, S**2 = 16 and XBAR = 22, one should conclude that:
a. MU > 20 with 5% chance of error
b. MU > 20 with 5 % confidence
c. MU = 20 with 95% confidence
d. MU = 20 with 95% chance of error
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Q: If the P-value for your test statistic satisfies P > .25, then:
(a) you would not reject H(O)
(b) you would reject H(O) for ALPHA = .05
(c) you would reject H(O) for ALPHA = .10
(d) your acceptance region has a lower limit of .25
(e) none of these
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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. The researcher wants
to test the hypothesis: The average four-month weight loss in
such a program is <= 6 pounds verses the alternative: > 6 pounds
at a 5% significance level. Given the data of our problem, we
a. reject the hypothesis.
b. reserve judgement about the hypothesis.
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Q: The meaning of "testing the hypothesis MU <= 6 versus MU > 6
at a 5% significance level" is:
a. if the population mean MU is > 6, the probability of deciding
wrongly is at least 95%.
b. if the population mean MU is <= 6, the probability of deciding
wrongly is at most 5%.
c. if the population mean MU is > 6, the probability of deciding
wrongly is at most 5%.
d. no matter what the population mean is, the probability of
deciding wrongly is at most 5%.
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Q: Statistical Significance means that, if an experiment were replicated
over and over again:
a) the same results would occur again with certainty
b) the same results would probably occur
c) the same results would probably not occur
d) the same results would certainly not occur again.
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Q: Suppose X(1) ... X(10) is a sample from a normal population with
mean = MU and variance = 22.5. The critical region for testing
H(0): MU >= 114 against H(1): MU < 114 at significance level
.05 is:
Reject H(0) if
a. XBAR < 112.00
b. XBAR < 111.54
c. XBAR < 111.25
d. XBAR < 111.06
e. none of these
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Q: If X(1),...,X(10) is a sample from a normal population with mean = MU
and variance = SIGMA**2, with SIGMA**2 unknown, and the sample standard
deviation (biased estimator) s = 7.5, then the critical region for
testing H(0): MU = 43 against H(1): MU =/= 43 at ALPHA = .01 is:
Reject H(0) if
a. XBAR > 50.02 or XBAR < 35.98
b. XBAR > 51.12 or XBAR < 34.88
c. XBAR > 50.56 or XBAR < 35.44
d. XBAR > 50.72 or XBAR < 35.28
e. none of these.
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Q: Blood samples from 40 patients are analyzed for glucose by two
different methods. With a computed t-statistic of 2.63, what
conclusion should you draw?
a. There is a significant difference in methods, p < .005.
b. There is a significant difference in methods, p < .01.
c. There is a significant difference in methods, p < .05.
d. There is no significant difference in the two methods.
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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.
This new method of cholesterol reduction was used on a sample from a
population with a mean MU cholesterol level of 225. Test the hypo-
thesis that the procedure used was effective (ALPHA = .05).
a. H(O): _______________.
b. H(A): _______________.
c. test statistic = _______________.
d. critical region = _______________.
e. Does the test statistic (c) fall in the critical region (d)?
f. Conclusion?
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Q: A teacher has been conducting the same course for many years. As part
of her check on her own teaching and on the general effectiveness of the
students in her classes from year to year, she has been giving the same
pop quiz during the 4th week of classes. Over the years the score has
averaged 13.5. This term she believes that her students are doing sig-
nificantly different than usual. If she can establish support of her
perception she will alter her teaching methods for this class. On the
basis of the test scores below should she conclude that there is support
for her perception and change her methods? On what basis do you offer
your consultation to her?
Scores: 20, 19, 6, 4, 3, 10, 13, 14, 16, 17,
10, 11, 8, 9, 11, 20, 18, 6, 4, 13.
XBAR = 11.6
S = 5.49
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Q: Past production units of a certain jet engine model showed the mean
military thrust to be 7600 pounds. The first ten production units
manufactured after a model change yielded military thrusts of 7620,
7680, 7570, 7700, 7650, 7720, 7600, 7540, 7670, and 7630. Is there
sufficient evidence (use ALPHA = 0.05) that the model change
resulted in a higher average military thrust?
Finding: YBAR = 7638
S(Y) = SQRT((583,420,000 - ((76,380)**2/10))/9) = 57.3
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Q: Forty-nine American soldiers, observed at random, yield a mean weight
of 160 pounds with a standard deviation (s) of 11 pounds. Are these
observations consistent with the assumption that the mean weight of
all American soldiers is 170 pounds?
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Q: A standard intelligence examination has been given for several years
with an average score of 80 and a standard deviation of 7. If 25
students taught with special emphasis on reading skill, obtain a mean
grade of 83 on the examination, is there reason to believe that the
special emphasis changes the result on the test? Use ALPHA = .05.
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Q: A teacher has just taken a course in statistics and decides to
put her new knowledge to work. During the last ten years she
has collected the final grades of her classes and found the mean
to be 73.6 with a standard deviation = 9. The final grades for
this years class are: 95, 97, 84, 64, 72, 88, 92, 61, 76, 74,
84, 85, 89, 75, 76, 64, 61, 72, 63, 74.
The teacher wishes to know if she should consider this years
class as significantly different than previous years. The
teacher did very well in her statistics class. What would you
believe her decision was? Why?
XBAR = 77.3; Hint: Do NOT calculate S in this problem.
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Q: True or False? If False, correct it.
If a statistic is significant at the 5 percent level, then it must be
significant at the 1 percent level also.
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Q: True or False? If False, correct it.
One can never prove the truth of a statistical (null) hypothesis. One can
only tend to discount it.
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Q: True or False? If False, correct it.
If the population mean is known, it makes no sense to test hypotheses
concerning the population mean.
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Q: True or False? If False, correct it.
Since the P-value in a test of hypothesis is based on the specific
observed value of a test statistic, it cannot be used in a two-sided
test.
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Q: True or False? If False, correct it.
The descriptive level of significance (P-value) is chosen by the
investigator before his experiment is conducted.
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Q: True or False? If False, correct it.
A hypothesis accepted at the ALPHA = .20 level of significance
is probably true.
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Q: True or False? If False, correct it.
A small significance level indicates that the hypothesis will
probably be continued.
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Q: TRUE OR FALSE. IF FALSE EXPLAIN WHY.
If the results of an investigation show that one sleeping tablet
works better than another at the 5% level of significance, the
conclusion would be similar if tested at the 10% level of significance.
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Q: In hypothesis testing, a type I error is
a. failing to reject the null hypothesis when it is false.
b. failing to reject the null hypothesis when it is true.
c. rejecting the null hypothesis when it is true.
d. rejecting the null hypothesis when it is false.
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Q: With which of the following terms is the "level of significance" most
closely associated?
a. one-tailed test of significance
b. two-tailed test of significance
c. type 1 error
d. type 2 error
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Q: Given the null hypothesis: that a new process is as good as or
better than the old one. The Type I Error is to conclude that:
(a) the old process is as good or better when it is not
(b) the old process is better when it is
(c) the old process is better when it is not
(d) the new process is as good or better when it is
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Q: Usually, one would like the critical region for a test to be _______
(long or short). The "size" of the critical region is determined by
_______ (ALPHA or BETA).
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Q: If the number of observations (n) is increased to 2n, the level of
significance (ALPHA) is:
a. increased b. unaffected c. decreased
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Q: When conducting a test for the population mean, if the population vari-
ance (SIGMA**2) is decreased to [(SIGMA**2)/2], then the level of sig-
nificance (ALPHA) is:
a. increased b. unaffected c. decreased
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Q: Suppose you are testing H(O): MU <= 21 vs. H(A): MU > 21, from a normal
distribution with SIGMA**2 known equal to 28 and n = 13. If Z(critical)
= 2.04, what is the p-value for your test?
a. .0207 d. .025
b. .4793 e. none of these
c. .0414
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Q: A population is known to have a variance of 16. When a sample of size
25 is taken, the sample variance is found to be 14, while the
sample mean is 30. For testing the hypothesis MU = 28 against the
alternative MU =/= 28 at the 0.10 level, the critical values are:
a. +/- 1.711
b. +/- 1.318
c. +/- 1.645
d. +/- 1.28
e. none of these
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Q: Past experience shows that, if a certain machine is adjusted properly, 5
percent of the items turned out by the machine are defective. Each day
the first 25 items produced by the machine are inspected for defects.
If three or fewer defects are found, production is continued without
interruption. If four or more items are found to be defective, produc-
tion is interrupted and an engineer is asked to adjust the machine.
After adjustments have been made, production is resumed. This proce-
dure can be viewed as a test of the hypothesis p = .05 against the
alternative p > .05, p being the probability that the machine turns
out a defective item. In test terminology, the engineer is asked to
make adjustments only when the hypothesis is rejected.
Interpret the quality control procedure described above as a test of
the indicated hypothesis. A Type I error results in:
a. a justified production stoppage to carry out machine adjustments.
b. an unnecessary interruption of production.
c. the continued production of an excess of defective items.
d. the continued production, without interruption, of items that
satisfy the accepted standard.
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Q: In assessing the weather prior to leaving our residences on a spring
morning, we make an informal test of hypothesis "The weather will be
fair today." Using the "best" information available to us, we complete
the test and dress accordingly. What would be the consequences of a
Type I and Type II error?
(1) Type I error: inconvenience in carrying needless rain equipment;
Type II error: clothes get soaked.
(2) Type I error: clothes get soaked; Type II error: inconvenience
in carrying needless rain equipment.
(3) Type I error: clothes get soaked; Type II error: no consequence
since Type II error cannot be made.
(4) Type I error: no consequence since a Type I error cannot be made;
Type II error: inconvenience in carrying needless rain equipment.
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Q: When an experimenter selects a particular level of risk (ALPHA) he
effectively is setting the probability of rejecting H(0) when in fact
it is true.
a. true
b. false
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Q: The mean weight of adult women in the U.S. is 140 lb. with a standard
deviation of 20 lb. You are willing to accept the report of the
standard deviation but not the mean. You selected 400 women at random
and measured their weight and found the average in this group to be
137 lb.
a. Test at ALPHA=.05 the hypothesis that the true weight is 140 lb.
b. What type of error might you have made in part a? Do you know the
probability of making such an error? If so, what is it? If not,
why not?
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Q: True or False? If False, explain why.
A Type I error is committed when one accepts the null hypothesis
when it is false.
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Q: True or False? If False, correct it.
The significance level is computed under the assumption that the
alternative hypothesis is true.
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Q: True or False? If False, correct it.
The risk of type II error does not depend upon the risk of type I
error.
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Q: True or False? If False, correct it.
Other things being equal, a small level of significance is desirable.
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Q: True or False? If False, correct it.
Level of confidence equals (1 - level of significance).
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Q: True or False? Explain your answer.
Although we speak of two types of error, in testing any
specified hypothesis we can make only one.
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Q: True or False? If False, correct it.
Generally, a larger sample size implies a smaller level of
significance.
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Q: Given the null hypothesis: that a process is producing no more than
the maximum allowable rate of defective items. In this situation, a
type II error would be:
(a) to conclude that the process is producing too many defectives
when it actually is not
(b) to conclude that the process is not producing too many defec-
tives when it actually is
(c) to conclude that the process is not producing too many defec-
tives when it is not
(d) to conlcude that the process is producing too many defectives
when it is.
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Q: Type II error refers to:
a. rejecting the null hypothesis when the alternative is true.
b. choosing the wrong decision rule.
c. not rejecting the null hypothesis when the alternative is
true.
d. incorrectly assuming the data are normally distributed.
e. none of these.
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Q: What is the probability of a Type II error when ALPHA = .05?
(1) .025 (2) .050
(3) .950 (4) .975
(5) Cannot be determined without more information
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Q: 16 observations are taken from a normal population in order to test
the null hypothesis H(O): MU = 10 against H(A): MU < 10. A
t-statistic is evaluated. In which case would H(O) be rejected at
the .05 level but not at the .025 level?
a. t = -2.12
b. t = -2.50
c. t = 2.12
d. t = 10 - 2.50 = 7.50
e. t = 10 - 2.12 = 7.88
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Q: A sample of size 26 is taken from a finite Normal population in which
the variance is known to be 25. It is desirable to test the hypothe-
sis that the mean is greater than 50, with ALPHA = .025. The critical
value(s) of the test statistic you would use is (are):
(a) -1.96, 1.96 (d) -1.645
(b) 1.96 (e) -2.060, 2.060
(c) 1.645 (f) 1.708
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Q: Sixteen one-acre plots of wheat were harvested. Their average yield
was found to be 701 bushels, and their standard deviation was 50 bu.
Since last year's crop yielded 680 bu./acre, we wish to test H(0):
MU <= 680 against H(A): MU > 680 at ALPHA = .05. Based on the above
data we should
a. not reject H(0). b. reject H(0).
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Q: Employing ALPHA = 0.05, one-tailed test, for 5 df, we find that the
critical value of t equals 2.015. This means that
(1) in this t-distribution, 5% of the area lies below t=-2.015.
(2) there is a 95% probability of obtaining a t-ratio less than
2.015.
(3) if our obtained t = 2.00, we cannot reject H(0) using the
.05 level.
(4) all of the above.
(5) none of the above.
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Q: The birthweight of babies is normally distribution with a mean of 3.8
kg. A researcher suspects that the weight of babies from mothers who
smoked a lot during pregnancy will be lower than the population mean.
To examine this, he takes a random sample of 26 babies from mothers who
smoked a lot. The mean birthweight in this group was 3.48 kg with a
standard deviation of .80 kg. (biased estimator of SIGMA**2). He takes
as H(0): MU >= 3.80 kg and as H(1): MU < 3.80 kg. According to these
data, the researcher can reject his H(0) with
a. 2.5 % probability of a type I error
b. 5 % probability of a type I error
c. 10% probability of a type I error
d. 95% probability of a type I error
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Q: For each of the following sets of information, find and specify the
appropriate critical region, test the null hypothesis, and draw a
conclusion.
a) H(0): MU = 16 n = 27
H(A): MU =/= 16 YBAR = 15.0
ALPHA = .02 s**2 = 3.0
b) H(0): MU <= 18.2 n = 250
H(A): MU > 18.2 YBAR = 18.7
ALPHA = .005 s**2 = 3.6
c) H(0): MU >= 113 n = 14
H(A): MU < 113 YBAR = 108
ALPHA = .01 s**2 = 56
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Q: Suppose that the heights of American males are normally distributed
with MU=71". A random sample of n=100 university students has XBAR=72.5,
S**2 =25.0. Can it be claimed at the ALPHA=.10 level of significance
that university students have a higher average height than American
males as a whole?
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Q: Suppose a test was taken by 36 students and the variance of the
distribution of scores was 100. It is desired to test H(O): MU>=80
against H(1): MU<80, using ALPHA=.05 and z as the test statistic.
(Assume the population of test scores is normally distributed.) What
( to the nearest tenth) is the starting point of the region of
rejection in terms of XBAR values?
a. 76.1
b. 76.7
c. 77.3
d. 77.7
e. None of these.
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Q: A fourth grade teacher wants to try a new teaching method which the
authors recommend should only be used with particularly bright
children. The authors offer a short test which they feel can be
used as a guide to decide whether the method should be used. They
believe that the average score for a class on this test should
significantly exceed 73 and indicate that the national standard
deviation for the test is 10. The teacher's students take the test
and exhibit the following scores: 91, 91, 94, 63, 61, 40, 73, 75, 83,
84, 70, 71, 88, 93, 92, 91, 87, 83, 74, 80.
Should the teacher use the new method? Why? (Use ALPHA = .05.)
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Q: In a Cancer Experiment, Guinness Beer was given in large quantities
to rats (fortunate things]) to see if it inhibited the growth of
tumors. From comparative, comprehensive studies it is known that the
"typical" tumor should weigh 1.35 grams. These 25 rats had a mean
weight of 1.31 grams with a standard deviation of 0.2 grams.
a. Has there been a significant inhibiting of growth? Use an ALPHA of
0.10. (Assume tumors are normally distributed.)
b. What is the smallest ALPHA-level that could be chosen before you
would change your conclusion?
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Q: The Crapi Cable Company #35 cable has a mean breaking strength of 1800
pounds with a standard deviation of 100 pounds. A new material is used
which, it is claimed, increases the breaking strength. To test this
claim a random sample of 50 cables, manufactured with the new material,
is tested. It is found that the sample has a mean breaking strength
of 1850 pounds. Test this claim using ALPHA = .01.
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Q: In the past a chemical fertilizer plant has produced an average of
1100 pounds of fertilizer per day. The record for the past year based
on 256 operating days shows the following:
XBAR = 1060 lbs/day
S = 320 lbs/day
where XBAR and S have the usual meaning. It is desired to test
whether or not the average daily production has dropped significantly
over the past year. Suppose that in this kind of operation, the
traditionally acceptable level of significance has been .05. But the
plant manager, in his report to his bosses, uses level of significance
.01. Analyze the data at both levels after setting up appropriate
hypotheses, and comment.
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Q: In J.B. Nimble's occupation, he is concerned with the length of
candles. Assuming that candle lengths are normally distributed with
mean MU and variance 4.0 square inches, he is interested in testing
the hypothesis that MU <= 15.0 inches against the alternative that MU
> 15.0 inches at a .001 significance level. For a sample of 49
candles, he obtains a sample mean of 15.6 inches. Conclusions?
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Q: It is desired to test the claim that a steady diet of wolfbane will
cause a lycanthrope (werewolf) to lose 10 lbs. over 5 months. A random
sample of 49 lycanthropes was taken yielding an average weight loss over
5 months of 12.5 pounds, with S = 7 lbs. Use a two-tailed test and let
ALPHA = .02.
Since the observed Z-value is _______ than the table value, we would ___
H(0). (Regard Z as an adequate approximation to t.)
a. smaller, reject
b. greater, not reject
c. greater, reject
d. smaller, not reject
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Q: The Pfft Light Bulb Company claims that the mean life of its 2 watt
bulbs is 1300 hours. Suspecting that the claim is too high, Nalph
Rader gathered a random sample of 64 bulbs and tested each. He found
the average life to be 1295 hours with s = 20 hours. Test the com-
pany's claim using ALPHA = .01.
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Q: A study was conducted to see if students living off-campus had a
grade point average which differed significantly from the
university-wide average GPA of 2.65. Extensive records indicate that
the standard deviation of the GPA is 0.3. What is your conclusion,
if a random sample of 100 off-campus students had a mean GPA of 2.72?
(The researcher feels that it is reasonable to assume that the
population of GPA scores is normally distributed.) What is the
significance probability of the observed result?
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A: e. 2.33
t(critical,twotail,ALPHA=.02,df=48) == +/- 2.33
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A: c) 92%, 1.75
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A: Z(for ALPHA=.01) = 2.33
Z = (XBAR - MU)/SIGMA
2.33 = (8 - MU)/.3
8 - MU = (2.33) (.3)
MU = 8 - (.699)
MU = 7.301
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A: a. One tail test at ALPHA = .01, therefore Z = 2.33.
Z = (YBAR-MU)/(SIGMA/SQRT(n))
2.33 = (YBAR-300)/(24/SQRT(64))
YBAR = 307
Decision Rule: If the mean strength of 64 ropes tested is 307
lbs. or more, we reject the hypothesis of no im-
provement, i.e., we continue that the new process
is better.
b. If available, consult file of graphs and diagrams that could not
be computerized for reference distributions.
Z = (307-310)/(24/SQRT(64)) = 1.00
Area = 0.1587 or 15.87%
P(type II error) = 0.1587
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A: a. the null hypothesis is rejected when it is true.
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A: A. the probability that the observed experimental
difference is due to chance given the null hypothesis.
C. symbolized by the greek letter ALPHA.
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A: False.
Level of confidence = 1 - (level of significance).
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A: c. 2.5
t(calculated) = (12.5-10)/(7/SQRT(49))
= 2.5
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A: (a) XBAR - 5 > 1.29 or 5 - XBAR > 1.29
Half Confidence Interval = Z*SIGMA/SQRT(n)
= 2.576 * .5 = 1.29
Hence, the critical value at the 1% level is 1.29.
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A: (a) XBAR >= 7.3
Z = (XBAR- MU)/(S/SQRT(n))
1.645 = (XBAR - 4)/2
XBAR = 7.29
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A: a. .05
Half Confidence Interval = Z * S/SQRT(N)
.5 = Z * 5/20
2 = Z
Area beyond Z = .023
So for both tails the significance level used = .023 + .023 = .046
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A: c) reject, 2.2, outside of
S**2 = 399/399 = 1
S(XBAR) = SQRT(S**2/n) = .05
If you center the confidence interval on the sample mean the
confidence interval = 2 +/- (1.75)(.05)
= from 1.9125 to 2.0875
which does not contain the hypothesized value, 2.2.
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A: e) 4%, reject, to the left of
S**2 = 399/399 = 1; S(XBAR) = SQRT(S**2/n) = .05;
C.I. = 2 +/- Z(ALPHA/2) * .05;
Z(16%/2) = 1.41, Z(8%/2) = 1.75, Z(4%/2) = 2.05
C.I.(ALPHA=16%) = 2 +/- (1.41)(.05)
= from 1.93 to 2.07
C.I.(ALPHA= 8%) = 2 +/- (1.75)(.05)
= from 1.91 to 2.09
C.I.(ALPHA= 4%) = 2 +/- (2.05)(.05)
= from 1.90 to 2.10
1 is not included in any of the confidence intervals, so H(0) should
be rejected in all cases.
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A: b. the result would be unexpected if the null hypothesis were true.
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A: d. both (a) and (b).
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A: d. the sampling distribution of the statistic assuming H(O).
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A: c. the probability of getting a t-value >= 1.8 or <= -1.8.
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A: b. I and either II or III.
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A: b. I and II only
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A: d) Z < -1.96 or Z > 1.96
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A: b) H(O) is rejected.
t(calc) = [XBAR - MU]/[s/SQRT(n)]
= [6.4 - 10]/[10/5]
= [-3.6]/2
= -1.8
t(crit., df=24, ALPHA=.05, one-tailed) = -1.711
Since t(calc) < t(crit), reject H(O). However, since the true
value of MU is not known, we do not know if a type 1 error has
been committed.
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A: a) A correct decision.
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A: c) The home owner is wrong, the house is worth less than $40,000.
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A: b. It is the point where the decision changes from reject to fail to
reject.
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A: (c) must be decided by the investigator.
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A: c. the sample data are not inconsistent with the hypothesis that in
the population, MU = 50.
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A: H(O): MU <= 155
H(A): MU > 155
ALPHA = .05 MU = 155 SIGMA = 10
SIGMA(XBAR) = 10/SQRT(100) = 1
Z(calculated) = (XBAR - MU)/SIGMA(XBAR)
= (170 - 155)/1
= 15
Z(critical, ALPHA =.05, one-tailed) = 1.645
Since Z(calculated) > Z(critical), reject H(O). Conclude that the
new bowling ball does improve a bowler's average at the 5% level.
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A: H(O): MU >= 105
H(A): MU < 105
t(obtained) = (XBAR - MU)/S(XBAR)
= (101.5 - 105)/1.42
= -3.5/1.42
= -2.465
t(critical, ALPHA=.05, one-tail, df=29) = -1.699
Since t(obtained) < t(critical), reject H(O). Therefore, the sample
evidence is strong enough to suggest that poor readers test signifi-
cantly lower in IQ.
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A: The main objection is that your level of confidence would be
approximately equal to .10 while the significance level would be
approximately equal to .90. This situation is the reverse of most
testing situations.
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A: False - A result may be significant at the 5% level, but not at
the 1% level.
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A: False, the decision to use a one-sided or two-sided test should be
made before the data is collected, so that the experimenter is not
influenced by the data.
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A: True, if the results are significant at .001 level, it is very
unlikely (one in one thousand) that the null hypothesis is true.
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A: False, this statement is accurate only if the null hypothesis is true.
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A: H(O): MU = 7.10
H(A): MU =/= 7.10
YBAR = 7.07
S(Y) = 0.0265
t = (YBAR - MU)/S(YBAR) = (7.07 - 7.10)/(0.0265/SQRT(5))
= 2.53
t(critical, ALPHA=.05, df=4) = +/- 2.776
Since the calculated value of t is not in the critical region, continue
H(O) that the nitrogen content has a true value of 7.10%, i.e., the
0.03% difference is ascribable to random error.
or
YBAR +/- t*(S(Y)/SQRT(n))
YBAR +/- 2.776*(0.0265)/(SQRT(5))
P(7.037 <= MU <= 7.103) = 0.95
Continue H(O) that the nitrogen content has a true value of 7.10% at 95%
level since 7.10 lies within the 95% confidence interval.
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A: a. H(O): MU = 10
H(A): MU < 10
t(calculated) = (XBAR - MU)/(S(X)/SQRT(n)
= (8 - 10)/SQRT(2.5/5)
= -2/.707
= -2.8
t(critical, one-tail, ALPHA = .05, df = 4) = -2.13
Decision: reject H(O) and conclude that this data does present
sufficient indication, with confidence level = .95, that the mean
reaction time of all people would be less than ten seconds.
b. C.I. = XBAR +/- t(ALPHA = .10)*S(XBAR)
= 8 +/- (2.13 * .707)
= 8 +/- 1.51
6.49 < MU < 9.51
c. 1.) The sample of n=5 observations is a simple random sample
of times from the population or phenomena being studied.
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A: A. XBAR +/- Z*SIGMA/SQRT(n) = 70 +/- (1.96*3)/SQRT(25)
= 66.824 to 71.176
B. Not Reject H(0)
C. Reject H(0)
D. H(0): MU = 69 vs. H(1): MU > 69.
Decision Rule: If Z obtained > Z critical then reject H(0).
Z (Calculated) = (XBAR - MU)/(SIGMA(XBAR))
= (70-69)/.6 = 1.667
Z (Critical) = 1.645
1.667 > 1.645 => Reject H(0)
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A: Yes, based on the above confidence interval, we would reject
the hypothesis that MU = 700 (at ALPHA = .05).
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A: H(0): MU > 25
H(1): MU <= 25
With ALPHA = .02, a one-tailed Z = 2.05
Critical value = MU - (Z) (SIGMA(XBAR))
= 25 - (2.05)(1/SQRT(10))
= 24.35
Decision rule: If the sample mean is greater than or equal to 24.35,
then the claim that MU > 25 is to be continued.
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A: a. XBAR - t*SQRT((S**2)/n) <= MU <= XBAR + t*SQRT((S**2)/n)
47000 - 2.06*SQRT(360000/25) <= MU <= 47000 + 2.06*SQRT(360000/25)
46750 <= MU <= 47250
b. Physician incomes are normally distributed. This is not likely to
be a valid assumption - incomes are usually skewed.
c. The null hypothesis would be rejected since the hypothesized value
of $50,000 is not included in the confidence interval found in (a.).
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A: b. I and either II or III
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A: a. MU > 20 with 5% chance of error
t = (22-20)/(4/5) = 2.5
Critical value for t is 1.711, with df = 24
Reject H(0) and conclude H(A) is true with .05 chance of error.
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A: (a) You would not reject H(O).
The probability of the data under H(O) is greater than .25,
therefore you would not reject H(O).
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A: b. reserve judgement about the hypothesis.
S(XBAR) = S/SQRT(n)
= 9.0/SQRT(9)
= 3.0
t(calc) = [XBAR - MU]/[S(XBAR)]
= [11.2 - 6.0]/[3.0]
= 1.73
t(crit, df=8, ALPHA=.05, one-tail) = 1.86
Since t(calc) < t(crit), continue (or do not reject) H(O).
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A: b. if the population mean MU is <= 6, the probability of deciding
wrongly is at most 5%.
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A: b) the same results would probably occur again.
Statistical tests are always performed at some given probability
level which gives the probability of occurrence of same results.
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A: b. XBAR < 111.54
SIGMA(XBAR) = SQRT(22.5/10)
= 1.5
Critical point = 114 - (1.645)(1.5)
= 111.5325
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A: b. XBAR > 51.12 or XBAR < 34.88
S/SQRT(n-1) = 7.5/3
= 2.5
Critical points = 43 +/- t(df=9, ALPHA=.01/2)*(2.5)
= 43 +/- (3.25)(2.5)
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A: b. There is a significant difference in methods, p < .01.
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A: a. H(O): MU = 225
b. H(A): MU < 225
c. t(calculated) = (210 - 225)/(30.9/SQRT(9))
= (-15)/(10.3)
= -1.456
d. t(critical, df = 8, ALPHA = .05, onetail) = -1.86
e. No
f. The result of this sample does not provide strong enough
evidence to support the claim that cholesterol level was
reduced.
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A: Using ALPHA = .05 and two-tailed t-test with df = 19,
we test H(0): MU = 13.5
H(1): MU =/= 13.5
t(calculated) = (11.6 - 13.5)/(5.49/SQRT(20)) = -1.55
t(critical) = +/- 2.093
We conclude that, at the .05 ALPHA level, she should continue H(0),
since the mean score of this class is not significantly different
from the usual. The basis for this conclusion rests on the assumption
that the scores have a normal distribution.
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A: Using ALPHA = .05 and a one-tailed t-test we test:
H(0): MU <= 7600
H(A): MU > 7600
t = (YBAR - MU)/(S(Y)/SQRT(N))
t = (7638 - 7600)/(57.3/SQRT(10))
t = 2.097
t(critical) = 1.83
Since t(calculated) is larger than t(critical) for a one-sided test at
ALPHA = .05, reject the null hypothesis. At the 95% confidence level,
the sample evidence indicates a detectable increase.
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A: Null hypothesis: MU = 170
Alternative: MU =/= 170
ALPHA = .05
Reject if t is not between -2.01 and 2.01.
t = (160 - 170)/(11/SQRT(49))
= -6.36
Data from sample is not consistent with the assumption.
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A: H(O): MU = 80
H(A): MU =/= 80
SIGMA(XBAR) = SIGMA/SQRT(n)
= 7/5
= 1.4
Using a Confidence Interval to test our hypothesis:
XBAR(CRIT) = MU +/- Z(CRIT)*SIGMA(XBAR)
= 80 +/- 1.96*1.4
= 77.256 to 82.744
Since 83 > 82.744 we reject H(O) and conclude that the result on the
test is significantly different from 80.
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A: MU = 73.6
SIGMA = 9
Test the hypothesis that this years class is a sample of a
population with MU = 73.6 and standard deviation = 9.
H(0): MU = 73.6
H(1): MU =/= 73.6
SIGMA(XBAR) = 9/SQRT(20) = 2.01
Z(calculated) = (77.3 - 73.6)/2.01
= 3.7/2.01
= l.84
Z(critical) = 1.96 with ALPHA = .05
At the 5% significance level, Z(calculated) is less than
Z(critical) so based on this evidence I would conclude that
this years class is not different from previous years classes.
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A: False. A statistic significant at the 5% level is not necessarily
significant at the 1% level, although the latter may be true.
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A: True. In statistical inference there is always some probability, how-
ever small, of drawing the wrong conclusion. The fact that a hypothesis
is consistent with a set of data does not mean that it is correct;
whereas, if it is not consistent with the data set it may be incorrect.
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A: True, the purpose of a hypothesis test is to draw a conclusion about
a population based on a sample. If the population is known, there is
nothing to hypothesize.
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A: False, if the assumed distribution is symmetric, the P-value can be
used in a two-sided test by doubling it.
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A: False, the descriptive level of significance (P-value) is dependent
on the sample. The ALPHA level is chosen before the experiment is
conducted.
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A: False.
The failure to reject does not imply the null hypothesis is true.
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A: False. A small significance level merely indicates that the probability
of a type I error is small.
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A: True, 95% confidence automatically implies more than
90% confidence.
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A: c. rejecting the null hypothesis when it is true.
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A: c. type 1 error
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A: (c) the old process is better when it is not.
Type I Error = rejecting null hypothesis when it is true
= rejecting [newer is as good or better than old]
when true
= continuing [old better than new] when it is not
= Prob(getting old betternew process is better
than old one)
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A: Usually, one would like the critical region for a test to be SHORT.
The "size" of the critical region is determined by ALPHA.
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A: b. unaffected
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A: b. unaffected.
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A: a. .0207
Area beyond Z corresponding to Z value of 2.04 = .0207.
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A: c. +/- 1.645
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A: b. an unnecessary interruption of production.
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A: (1) Type I error: inconvenience in carrying needless rain equipment;
Type II error: clothes get soaked.
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A: a. true
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A: a. H(0): MU = 140
H(A): MU =/= 140
ALPHA = .05
Test statistic: Z = (XBAR-MU)/(SIGMA/SQRT(n))
Assumptions: Random sample. Sampling distribution of mean is
normal.
Critical region: Z > 1.96 or Z < -1.96
Z = (137-140)/(20/SQRT(400)) = -3.00
Reject H(0)
Conclude true mean is not equal to 140.
b. Type I error: rejection of true H(0). Probability is .05 (ALPHA).
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A: False, a Type I error is committed when one rejects the null hypothesis
when it is true.
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A: False - The significance level is computed under the
assumption that the null hypothesis is true.
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A: False - The risk of type II error depends upon the risk of type I
error. As the risk of one type increases, the risk of the other type
decreases.
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A: True.
Level of significance is the probability of Type I error. We would
always like to have it small.
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A: True.
Level of confidence = (1 - ALPHA).
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A: True.
Reject H(0) Don't Reject H(0)
________________________________________________
NO ERROR
H(0) True TYPE1ERROR 1 - ALPHA
________________________________________________
H(0) False NO ERROR TYPE2ERROR
1 - BETA
________________________________________________
Therefore, either we can make a Type I error or Type II error
depending on whether our hypothesis is true or false, but it is
impossible to make both simultaneously.
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A: False, the level of significance is usually chosen before the
measurements are collected and is not dependent on the sample
size.
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A: (b) to conclude that the process is not producing too many defec-
tives when it actually is.
Type II error = P(do not reject H(O) when H(O) is false)
H(O) = process producing no more than k defectives
H(O) false means process producing more than k de-
fectives.
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A: c. not rejecting the null hypothesis when the alternative is
true.
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A: (5) Cannot be determined without more information.
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A: a. t = -2.12
With df = 16 - 1 = 15,
t(critical, ALPHA = .05) = -1.753
t(critical, ALPHA = .025) = -2.131
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A: (b) 1.96
This is a one-sided test and, since SIGMA is known,
Z(ALPHA = .025) = 1.96.
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A: a. not reject H(0).
t(calculated) = (701 - 680)/(50/SQRT(16))
= 21/(50/SQRT(16)) = 1.68
t(critical, one tail, ALPHA = .05, df = 15) = 1.753
Therefore t(calculated) < t(critical) and we should
not reject H(0) at this significance level.
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A: (4) all of the above.
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A: b. 5% probability of a type I error
t = [XBAR - MU]/[S(XBAR)]
= [3.48-3.80]/[[.80)/[SQRT(25)]]
= -2.00, df = 25, one-tailed test
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A: a) t(critical, df=26, ALPHA=.02, two-tail) = +/-2.479
critical region: t's < -2.479 and t's > +2.479
t(calculated) = (15-16)/SQRT(3/27)
= -3
t(calculated) falls in the critical region, so based on this sample
evidence you should reject H(0).
b) t(critical, df=249, ALPHA=.005, one-tail)
== Z(critical, ALPHA=.005, onetail) = +2.576
critical region: Z's > +2.576
Z(calculated) = (18.7-18.2)/(SQRT(3.6/250))
= 4.166
Z(calculated) falls in the critical region so based on this sample
evidence you should reject H(0).
c) t(critical, df=13, ALPHA=.01, one-tail) = -2.65
critical region: t's < -2.65
t(calculated) = (108-113)/(SQRT(56/14))
= -2.5
t(calculated) does not fall in the critical region so based on this
sample evidence do not reject (continue) H(0).
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A: Using a one-tail t-test, we test:
H(0): MU<=71
H(1): MU>71
t(calculated) =(XBAR-MU)/(S/SQRT(n))
=1.5/(5/SQRT(100))
=3
t(ALPHA=.10,df=99)==1.296
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A: c. 77.3
Z(crit., ALPHA=.05, one-tail) = 1.645
rejection region = MU - Z(crit.) * SIGMA/SQRT(n)
= 80 - (1.645 * 10/SQRT(36))
= 77.258
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A: H(0): MU <= 73 SIGMA = 10
H(1): MU > 73 XBAR = 79.2
Z = (79.25 - 73)/(10/SQRT(20)) = 2.79
Z(critical) = 1.645; with ALPHA = .05 and using a onetail
decision rule.
Reject H(0) and conclude that the average score for this class is
significantly higher than 73. Therefore, she should use the new
method.
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A: a. H(O): Guinness Beer does not affect tumors.
H(A): Guinness Beer inhibits the growth of tumors.
i.e., H(O): MU => 1.35
H(A): MU < 1.35
t(calc.) = [XBAR-MU]/[S/SQRT(n)] with (n-1) df
= [1.31-1.35]/[0.2/SQRT(25)]
= [-0.04]/[0.04]
= -1.00
t(crit., df=24, ALPHA=.10, one-tail) = -1.318
Since t(calc.) > t(crit.), continue (do not reject) H(O). It would
thus appear that Guinness Beer does not affect the growth of tumors.
b. In part (a) we continued H(O), now we must reject H(O) for some
level of ALPHA. Keeping same degrees of freedom we notice:
ALPHA t(crit.)
----- --------
0.05 -1.711
0.10 -1.318
0.15 -1.059
0.20 -0.857
0.25 -0.685
Since t(calc.) = -1.000, the test is significant for ALPHA=0.20,
but not significant for ALPHA=0.15. The approximate significance
level obtained by interpolation is about 0.16.
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A: Hypothetical population: All Crapi #35 cables made with the new
material.
Sample: The 50 cables randomly selected.
H(O): MU <= 1800. The mean breaking strength of the new cable is
1800 lb.
H(A): MU > 1800. The mean breaking strength of the new cable is
more than 1800 lb.
MU(XBAR) = 1800 by H(O)
SIGMA(XBAR) = SIGMA/SQRT(n)
= 100/SQRT(50)
= 14.142
XBAR(crit) = MU(XBAR) + Z(crit)*SIGMA(XBAR)
= 1800 + (2.33)*(14.142)
= 1832.951
Since the sample mean breaking strength is 1850, which is greater than
1832.51, we must reject H(O) and conclude that the mean breaking
strength of the new cable is significantly more than 1800 lb.
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A: H(O): MU => 1100
H(A): MU < 1100
Since n = 256, use Z to approximate t.
S(XBAR) = 320/SQRT(256)
= 320/16
= 20
Z(calculated) = (1060 - 1100)/20
= -40/20
= -2
Z(critical, ALPHA=.05, one-tailed) = 1.645
Z(critical, ALPHA=.01, one-tailed) = 2.33
Therefore, H(0) is rejected at ALPHA=.05 but not at ALPHA=.01.
It appears that the manager is trying to pull a fast one on his
bosses by using ALPHA=.01 and saying production has not dropped.
However, if the traditional level of significance is used, ALPHA=.05,
there is evidence that indicates a drop in production.
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A: XBAR = 15.6
SIGMA**2 = 4
n = 49
H(O): MU <= 15.0
H(A): MU > 15.0
ALPHA = .001
Critical Region: Z > 3.090
Z(calculated) = (XBAR - MU)/(SIGMA/SQRT(n))
= 2.1
Conclusion: Mean candle lengths may be 15 inches.
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A: c. greater, reject
Z(calculated) = (12.5-10)/(7/SQRT(49))
= 2.5/1
= 2.5
Z(critical,ALPHA=.02,twotail) = +/- 2.33
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A: Hypothetical population: All Pfft 2 watt bulbs.
Sample: The 64 randomly selected bulbs.
H(O): MU => 1300. The mean life of 2 watt bulbs is 1300 hours.
H(A): MU < 1300. The mean life of 2 watt bulbs is less than 1300
hours.
MU(XBAR) = 1300 by H(O)
S(XBAR) = S/SQRT(n)
= 20/8
= 2.5
XBAR(crit) = MU(XBAR) + Z(crit)*S(XBAR)
= 1300 - 2.33*2.5
= 1294.18
Since 1295 is not less than 1294.18, we cannot reject H(O). There is
not enough evidence to conclude that the mean life of the 2 watt bulbs
is significantly less than 1300 hours.
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A: MU = 2.65
SIGMA = 0.3
N = 100
XBAR = 2.72
H(O): MU(off-campus) = 2.65
H(A): MU(off-campus) =/= 2.65
Z(calculated) = (XBAR - MU)/(SIGMA/SQRT(100))
= 2.33
P(Z > 2.33 or Z < -2.33) = .02
Based on this sample evidence, it appears that the null hypothesis
is incorrect and that off-campus students do have a different GPA.
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Multiple Choice
TSCORE TESTING TWOTAIL/T
STANDUNITS/NORMA PROBDISTRIBUTION PROBABILITY
CONCEPT STATISTICS TTEST
PARAMETRIC
T= 2 Comprehension
D= 2 General
***Statistical Table Necessary***
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TESTING SIMPLE/CI ZSCORE
CONCEPT STATISTICS CONFIDENCEINTERV
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PROBABILITY
T= 2 Comprehension
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Numerical Answer
ZSCORE
SIMPLE/CI ONETAIL/Z STANDUNITS/NORMA
PROBDISTRIBUTION PROBABILITY CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
ZTEST PARAMETRIC
T= 5 Application
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TYPE2ERROR TESTING ZSCORE
I650I CONCEPT STATISTICS
STANDUNITS/NORMA PROBDISTRIBUTION PROBABILITY
T=15 Application
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***Multiple Parts***
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BASICTERMS/STATS TYPE1ERROR
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T= 2 Comprehension
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TESTING TWOTAIL/T
CONCEPT STATISTICS TTEST
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T= 2 Computation
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Multiple Choice
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CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T=10 Computation
D= 4 General
***Statistical Table Necessary***
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T= 5 Computation
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***Statistical Table Necessary***
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ZSCORE CONCEPT STATISTICS
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PROBDISTRIBUTION PROBABILITY
T=10 Application
D= 4 General Natural Sciences
***Statistical Table Necessary***
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TESTING SIMPLE/CI
ZSCORE CONCEPT STATISTICS
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PROBDISTRIBUTION PROBABILITY
T=10 Application
D= 4 Natural Sciences General
***Statistical Table Necessary***
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T= 2 Comprehension
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Multiple Choice
TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 Business General
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Multiple Choice
TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
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Multiple Choice
TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
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Multiple Choice
TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 4 General
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Numerical Answer
TESTING ONETAIL/Z
CONCEPT STATISTICS ZTEST
PARAMETRIC
T=10 Computation
D= 5 General
***Statistical Table Necessary***
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Item is still being reviewed
Numerical Answer
TESTING ONETAIL/T
CONCEPT STATISTICS TTEST
PARAMETRIC
T=10 Computation
D= 4 General Psychology Social Sciences
***Statistical Table Necessary***
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Item is still being reviewed
Short Answer
TYPE1ERROR TESTING
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
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Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
TESTOFSIGNIFICAN TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
True/False
TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
True/False
TYPE1ERROR TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
True/False
TYPE1ERROR TESTING
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Short Answer
TESTOFSIGNIFICAN SIMPLE/CI TWOTAIL/T
CONCEPT STATISTICS CONFIDENCEINTERV
ESTIMATION TTEST PARAMETRIC
T= 5 Application Computation
D= 3 Natural Sciences
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by R. E. Lund - Montana State U.
Short Answer
ONETAIL/T STANDERROROFMEAN SIMPLE/CI
TESTOFSIGNIFICAN TTEST PARAMETRIC
STATISTICS DESCRSTAT/P CONFIDENCEINTERV
ESTIMATION CONCEPT
T= 6 Comprehension
D= 4 Psychology General
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Short Answer
TWOTAIL/T ONETAIL/T SIMPLE/CI
TESTING TTEST PARAMETRIC
STATISTICS CONFIDENCEINTERV ESTIMATION
CONCEPT
T=15 Application
D= 7 General
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by R. E. Lund - Montana State U.
Short Answer
SIMPLE/CI
CONFIDENCEINTERV ESTIMATION CONCEPT
STATISTICS
T= 3 Comprehension Computation
D= 4 Business General
***Statistical Table Necessary***
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Item is still being reviewed
Short Answer
ONETAIL/Z SIMPLE/CI
TYPE1ERROR STANDERROROFMEAN TESTING
I650I ZTEST PARAMETRIC
STATISTICS CONFIDENCEINTERV ESTIMATION
CONCEPT DESCRSTAT/P
T=10 Comprehension
D= 5 General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
SIMPLE/CI
TTEST TESTOFSIGNIFICAN CONFIDENCEINTERV
ESTIMATION CONCEPT STATISTICS
PARAMETRIC
T=10 Computation Comprehension
D= 4 General Biological Sciences Economics
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
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Multiple Choice
ZTEST ASSUMPTCUSTOMARY CENTRALLIMITTHM
TESTING PARAMETRIC STATISTICS
MISCELLANEOUS CONCEPT
T= 5 Comprehension
D= 4 General
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Item is still being reviewed
Multiple Choice ***Calculus Necessary***
TESTOFSIGNIFICAN ONETAIL/T
CONCEPT STATISTICS TTEST
PARAMETRIC
T= 5 Computation Comprehension
D= 4 General
***Statistical Table Necessary***
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Item is still being reviewed
Multiple Choice ***Calculus Necessary***
TESTOFSIGNIFICAN
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
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Item is still being reviewed
Multiple Choice
ONETAIL/T TESTOFSIGNIFICAN
TTEST PARAMETRIC STATISTICS
CONCEPT
T= 5 Application
D= 4 Biological Sciences General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TESTOFSIGNIFICAN ONETAIL/T
CONCEPT STATISTICS TTEST
PARAMETRIC
T= 2 Application
D= 2 General
Back to this chapter's Contents
Based upon item submitted by R. Shavelson - UCLA
Multiple Choice
TESTOFSIGNIFICAN TYPE1ERROR
BASICTERMS/STATS CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by F. J. Samaniego - UC Davis
Multiple Choice
TESTOFSIGNIFICAN STANDERROROFMEAN
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 5 Computation
D= 3 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TESTOFSIGNIFICAN STANDERROROFMEAN
CONCEPT STATISTICS DESCRSTAT/P
PARAMETRIC
T= 5 Application
D= 3 General
Back to this chapter's Contents
Based upon item submitted by G. Shavlik - Loma Linda Univ.
Multiple Choice
TWOTAIL/T TESTOFSIGNIFICAN
TTEST PARAMETRIC STATISTICS
CONCEPT
T= 2 Comprehension
D= 3 General Biological Sciences
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
ONETAIL/T TESTOFSIGNIFICAN
TTEST PARAMETRIC STATISTICS
CONCEPT
T=10 Application
D= 4 Biological Sciences General
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
TWOTAIL/T TESTOFSIGNIFICAN
STANDERROROFMEAN I650I TTEST
PARAMETRIC STATISTICS CONCEPT
DESCRSTAT/P
T= 5 Application
D= 5 Education General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
ONETAIL/T TESTOFSIGNIFICAN
STANDERROROFMEAN SIMPLEDATASET I650I
TTEST PARAMETRIC STATISTICS
CONCEPT DESCRSTAT/P
T= 5 Application
D= 4 Natural Sciences
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
TWOTAIL/Z TESTOFSIGNIFICAN
TESTING ZTEST PARAMETRIC
STATISTICS CONCEPT
T=10 Application
D= 4 Biological Sciences General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
TESTOFSIGNIFICAN TWOTAIL/Z
CONCEPT STATISTICS ZTEST
PARAMETRIC
T=10 Computation
D= 2 General Education
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
TWOTAIL/Z TESTOFSIGNIFICAN
STANDERROROFMEAN I650I ZTEST
PARAMETRIC STATISTICS CONCEPT
DESCRSTAT/P
T=10 Application
D= 4 Education General
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
True/False
TESTOFSIGNIFICAN
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
True/False
TESTOFSIGNIFICAN
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
True/False
TESTOFSIGNIFICAN SCOPEOFINFERENCE
CONCEPT STATISTICS
T= 2 Comprehension
D= 5 General
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True/False
TESTOFSIGNIFICAN
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by D. Kleinbaum - Univ of North Carolina
True/False
TESTOFSIGNIFICAN
CONCEPT STATISTICS
T= 2 Comprehension
D= 5 General
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True/False
TESTOFSIGNIFICAN TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 4 General
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Item is still being reviewed
True/False
TESTOFSIGNIFICAN TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by A. Bugbee - UNH
True/False
TESTOFSIGNIFICAN
TYPE1ERROR CONFIDENCEINTERV CONCEPT
STATISTICS ESTIMATION
T= 2 Comprehension
D= 1 General
Back to this chapter's Contents
Based upon item submitted by R. Pruzek - SUNY at Albany
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 1 General
Back to this chapter's Contents
Based upon item submitted by S. Sytsma - Ferris State
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
***Multiple Parts***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Based upon item submitted by S. Selvin - UC Berkeley
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TYPE1ERROR ONETAIL/Z
CONCEPT STATISTICS ZTEST
PARAMETRIC
T= 2 Computation Comprehension
D= 2 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TWOTAIL/Z TYPE1ERROR
ZTEST PARAMETRIC STATISTICS
CONCEPT
T= 2 Comprehension
D= 2 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 Natural Sciences Business General
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Item is still being reviewed
Multiple Choice
TYPE1ERROR TYPE2ERROR
CONCEPT STATISTICS
T= 2 Comprehension Application
D= 4 General
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Item is still being reviewed
Multiple Choice
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
Back to this chapter's Contents
Item is still being reviewed
Short Answer ***Calculus Necessary***
TWOTAIL/Z TYPE1ERROR
ZTEST PARAMETRIC STATISTICS
CONCEPT
T=10 Application Comprehension
D= 4 General Biological Sciences
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
True/False
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
True/False
TYPE1ERROR
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
TYPE1ERROR TYPE2ERROR
CONCEPT STATISTICS
T= 5 Comprehension
D= 3 General
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True/False
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
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Item is still being reviewed
True/False
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
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Item is still being reviewed
True/False
TYPE1ERROR TYPE2ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 4 General
Back to this chapter's Contents
Based upon item submitted by J. L. Mickey -UCLA
True/False
TYPE1ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
TYPE2ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
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Item is still being reviewed
Multiple Choice
TYPE2ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 3 General
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Item is still being reviewed
Multiple Choice
TYPE2ERROR
CONCEPT STATISTICS
T= 2 Comprehension
D= 2 General
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
ONETAIL/T
TTEST PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
ONETAIL/T
TTEST PARAMETRIC STATISTICS
T= 2 Computation
D= 2 General
Back to this chapter's Contents
Based upon item submitted by F. J. Samaniego - UC Davis
Multiple Choice
ONETAIL/T
TTEST PARAMETRIC STATISTICS
T= 5 Application
D= 3 General Biological Sciences
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
ONETAIL/T
TTEST PARAMETRIC STATISTICS
T= 2 Comprehension
D= 3 General
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Item is still being reviewed
Multiple Choice
ONETAIL/T
TTEST PARAMETRIC STATISTICS
T= 2 Computation
D= 3 Biological Sciences General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
ONETAIL/T TWOTAIL/T
TTEST PARAMETRIC STATISTICS
T=10 Computation Application
D= 5 General
***Calculator Necessary***
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Based upon item submitted by W. J. Hall - Univ. of Rochester
Short Answer
ONETAIL/T
STANDERROROFMEAN TESTOFSIGNIFICAN I650I
TTEST PARAMETRIC STATISTICS
DESCRSTAT/P CONCEPT
T= 5 Application
D= 5 Biological Sciences General General
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Multiple Choice
ONETAIL/Z
ZTEST PARAMETRIC STATISTICS
T= 5 Application
D= 4 General Education
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
ONETAIL/Z
STANDERROROFMEAN I650I ZTEST
PARAMETRIC STATISTICS DESCRSTAT/P
T=10 Application
D= 4 Education
***Calculator Necessary***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Numerical Answer
ONETAIL/Z
ZTEST PARAMETRIC STATISTICS
T=10 Application
D= 4 General Biological Sciences
***Multiple Parts***
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
ONETAIL/Z
ZTEST PARAMETRIC STATISTICS
T=10 Computation
D= 2 General Business
***Calculator Necessary***
***Statistical Table Necessary***
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Item is still being reviewed
Short Answer
ONETAIL/Z
COMMONPITFALLS ZTEST PARAMETRIC
STATISTICS MISCELLANEOUS
T=10 Application
D= 4 Business Natural Sciences General
***Statistical Table Necessary***
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Short Answer
ONETAIL/Z
MEAN ZTEST PARAMETRIC
STATISTICS DESCRSTAT/P
T=10 Application
D= 4 General
Back to this chapter's Contents
Based upon item submitted by F. J. Samaniego - UC Davis
Multiple Choice
TWOTAIL/Z
ZTEST PARAMETRIC STATISTICS
T= 5 Comprehension Computation
D= 3 General
***Statistical Table Necessary***
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Item is still being reviewed
Short Answer
TWOTAIL/Z
ZTEST PARAMETRIC STATISTICS
T=10 Computation
D= 2 General Business
***Statistical Table Necessary***
Back to this chapter's Contents
Item is still being reviewed
Short Answer
TWOTAIL/Z
MEAN ZTEST PARAMETRIC
STATISTICS DESCRSTAT/P
T=10 Application
D= 4 General Education
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