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3
rd Quarterly Exam ~ Z-Scores, Confidence Interval and Hypothesis
Part 1 – Review
VI. DISTRIBUTION (Cont.) Z-Scores
MC VI-1.) [CBAPSTATSPracticeProblem] Some descriptive statistics for a set of test scores are shown
below. For this test, a certain student has a standardized score of z = –1.2. What score did this student
receive on the test?
MC VI-2.) [APSTATSMC2014-22] A distribution of scores is approximately normal with a mean of 78
and a standard deviation of 8.6. Which of the following equations can be used to find the score x above
which 33 percent of the scores fall?
MC VI-3.) [APSTATSMC2012-7] The weight of adult male grizzly bears living in the wild in the
continental US is approximately normally distributed with a mean of 500 pounds and a standard deviation
of 50 pounds. The weight of a adult female grizzly bears is approximately with a mean of 300 pounds and a
standard deviation of 40 pounds. Approximately what would be the weight of a female grizzly bear with
the same standardized score (z-score) as a male grizzly bear with a weight of 530 pounds?
MC VI-4.) [2012APSTATSMC2012-25] The commuting time for a student to travel from home to a
college campus is normally distributed with a mean of 30 minutes and a standard deviation of 5 minutes. If
the student leaves home at 8:25 A.M., what is the probability that the student will arrive at the college
campus later than 9 A.M.?
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MC VI-5.) [CBAPSTATSPracticeProblem] A summer resort rents rowboats to customers but does not
allow more than four people to a boat. Each boat is designed to hold no more than 800 pounds. Suppose the
distribution of adult males who rent boats, including their clothes and gear, is normal with a mean of 190
pounds and standard deviation of 10 pounds. If the weights of individual passengers are independent, what
is the probability that a group of four adult male passengers will exceed the acceptable weight limit of 800
pounds?
(A) 0.023
(B) 0.046
(C) 0.159
(D) 0.317
(E) 0.977
FRQ VI-1.) [APSTATSFRQ2012-2] A charity fundraiser has a Spin the Pointer game that uses a spinner
like the one illustrated in the figure below.
A donation of $2 is required to play the game. For each $2 donation, a player spins the pointer once and
receives the amount of money indicated in the sector where the pointer lands on the wheel. The spinner has
an equal probability of landing in each of the 10 sectors.
a.) Let X represent the net contribution to the charity when one person plays the games once. Complete the
table for the probability distribution of X.
x $2 $1 -$8
( )p x
b.) What is the expected value of the net contribution to the charity for one play of the game?
c.) The charity would like to receive a net contribution of $500 from this game. What is the fewest number
of times the game must be played for the expected value of the net contribution to be at least $500?
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d.) Based on last year’s event, the charity anticipates that the Spin the Pointer game will be played 1000
times. The charity would like to know the probability of obtaining a net contribution of at least $500 in
1000 plays of the game. The mean and standard deviation of the net contribution to the charity in 1000
plays of the game are $700 and $92.79, respectively. Use the normal distribution to approximate the
probability that the charity would obtain a net contribution of at least $500 in 1000 plays of the game.
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VII. CONFIDENCE INTERVAL
MC VII-1.) [CBAPStatsPracticeQuestion] Courtney has constructed a cricket out of paper and rubber
bands. According to the instructions for making the cricket, when it jumps it will land on its feet half of the
time and on its back the other half of the time. In the 50 jumps, Courtney’s cricket landed on its feet 35
times. In the next 10 jumps, it landed on its feet only twice. Based on this experience, Courtney can
conclude that
(A) the cricket was due to land on its feet less than half the time during the final 10 jumps, since it had
handed too often on its feet during the first 50 jumps.
(B) a confidence interval for estimating the cricket’s true probability of landing on its feet is wider after the
final 10 jumps than it was before the final 10 jumps.
(C) a confidence interval for estimating the cricket’s true probability of landing on its feet after the final 10
jumps is exactly the same as it was before the final 10 jumps.
(D) a confidence interval for estimating the cricket’s true probability of landing on its feet is more narrow
after the final 10 jumps than it was before the final 10 jumps.
(E) a confidence interval for estimating the cricket’s true probability of landing on its feet based on the
initial 50 jumps does not include 0.2, so there must be a defect in the cricket’s construction to account for
the poor showing in the final 10 jumps.
MC VII-2.) [APSTATSMC2012-24] A random sample of 432 voters revealed that 100 are in favor of a
certain bond issue. A 95 percent confidence interval for the proportion of the population of voters who are
in favor of the bond issue is
(A) 0.5(0.5)
100 1.96432
(B) 0.5(0.5)
100 1.645432
(C) 0.231(0.769)
100 1.96432
(D) 0.231(0.769)
0.231 1.96432
(E) 0.231(0.769)
0.231 1.645432
MC VII-3.) [APSTATSMC2014-13] The manager of a car company will select a random sample of its
customers to create a 90 percent confidence interval to estimate the proportion of its customers who have
children. What is the smallest sample size that will result in a margin of error of no more than 6 percentage
points?
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MC VII-4.) [APSTATSMC2012-17] A large-sample 98 percent confidence interval for the proportion of
hotel reservation that is canceled on the intended arrival day is (0.048, 0.112). What is the point estimate
for the proportion of hotel reservations that are canceled on the intended arrival day from which this
interval was constructed?
(A) 0.032
(B) 0.064
(C) 0.080
(D) 0.160
(E) It cannot be determined form the information given.
MC VII-5.) [APSTATSMC2012-26] In 2009 a survey of Internet usage found that 79 percent of adults
age 18 years and older in the United States use the Internet. A broadband company believes that the percent
is greater now than it was in 2009 and will conduct a survey. The company plans to construct a 98 percent
confidence interval to estimate the current percent and wants to the margin of error to be no more than 2.5
percentage points. Assuming that at least 79 percent of adults use the Internet, which of the following
should be used to find the sample size (n) needed?
(A) 0.5
1.96 0.025n
(B) 0.5(0.5)
1.96 0.025n
(C) 0.5(0.5)
2.33 0.05n
(D) 0.79(0.21)
2.33 0.025n
(E) 0.79(0.21)
2.33 0.05n
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MC VII-6.) [APSTATSMC2007-21] A city is interested in building a waste management facility in a
certain area. One hundred randomly selected residents from this area were asked, “Do you support the
city’s decision to build a waste management facility in your area?” Of the 100 residents interviewed, 54
said no, 4 said yes, and 42 had no opinion. A large sample z-confidence interval, ˆ ˆ(1 )
ˆ *p p
p zn
,
was constructed from these data to estimate the proportion of this area’s residents who support building a
waste management facility in their area. Which of the following statements is correct for this confidence
interval?
(A) The confidence interval is valid because a sample size of more than 30 was used.
(B) The confidence interval is valid because each area resident was asked the same question.
(C) The confidence interval is valid because no conditions are required for constructing a large sample
confidence interval for a proportion.
(D) The confidence interval is not valid because the quantity ˆnp is too small.
(E) The confidence interval is not valid because “no opinion” was included as a response category for the
question.
MC VII-7.) [1997APSTATSMC1997-33] A 95 percent confidence interval of the form p̂ E will be
used to obtain an estimate for an unknown population proportion p . If p̂ is the sample proportion and
E is the margin of error, which of the following the smallest size that will guarantee a margin of error of at
most 0.08?
(A) 25
(B) 100
(C) 175
(D) 250
(E) 625
MC VII-8.) [2007APSTATSMC2007-34] A planning board in Elm County is interested in estimating the
proportion of its residents that are in favor of offering incentives to high-tech industries to build plants in
that county. A random sample of Elm County residents was selected. All of the selected residents were
asked, “Are you in favor of offering incentives to high-tech industries to build plants in your county?” A 95
percent confidence interval for the proportion of residents in favor of offering incentives was calculated to
be 0.54 0.05 . Which of the following statements is correct?
(A) At 95% confidence level, the estimate of 0.54 is within 0.05 of the true proportion of county residents
in favor of offering incentives to high-tech industries to build plants in the county.
(B) At 95% confidence level, the majority of area residents are in favor of offering incentives to high-tech
industries to build plants in the county.
(C) In repeated sampling, 95% of sample proportions will fall in the interval (0.49, 0.59)
(D) In repeated sampling, the true proportion of county residents in favor of offering incentives to high-tech
industries to build plants in the county will fall in the interval (0.49, 0.59).
(E) In repeated sampling, 95% of the time the true proportion of county residents in favor of offering
incentives to high-tech industries to build plants in the county will be equal to 0.54.
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MC VII-9.) [APSTATSMC1997-35] A survey was conducted to determine what percentage of college
seniors would have chosen to attend a different college if they had known then what they know now. In a
random sample of 100 seniors, 34 percent indicated that they would have attended a different college. A 90
percent confidence interval for the percentage of all seniors who would have attended a different college is
(A) 24.7% 43.3%to
(B) 25.8% 42.2%to
(C) 26.2% 41.8%to
(D) 30.6% 37.4%to
(E) 31.2% 36.8%to
MC VII-10.) [APSTATSMC1997-01]
MC VII-11.) [APSTATSMC2012-09] Based on a survey of a random sample of 900 adults in the United
States, a journalist reports that 60 percent of adults in the United States are in favor of increasing the
minimum hourly wage. If the reported percent has a margin error of 2.7 percentage points, what is the level
of confidence?
MC VII-12.) [CBAPStatsPracticeProblem] A large company is considering opening a franchise in St.
Louis and wants to estimate the mean household income for the area using a simple random sample of the
households. Based on information from a pilot study, the company assumes that the standard deviation of
household incomes is $7,200 . What is the least number of households that should be surveyed to
obtain an estimate that is within $200 of the true mean houshold income with 95 percent confidence?
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MC VII-13.) [APSTATSMC2002-08] A test engineer wants to estimate the mean gas mileage (in miles
per gallon) for a particular model of automobile. Eleven of these cars are subjected to a road test, and the
gas mileage is computed for each car.
A dotplot of the 11 gas-mileage values is roughly symmetrical and has no outliers. The mean and standard
deviation of these values are 25.5 and 3.01, respectively. Assuming that these 11 automobiles can be
considered a simple random sample of cars of this model, which of the following is a correct statement?
(A) A 95% confidence interval for is3.01
25.5 2.22811
.
(B) A 95% confidence interval for is3.01
25.5 2.20111
.
(C) A 95% confidence interval for is3.01
25.5 2.22810
.
(D) A 95% confidence interval for is3.01
25.5 2.20110
.
(E) The results cannot be trusted; the sample is too small.
MC VII-14.) [APSTATSMC2002-26] A quality control inspector must verify whether a machine that
packages snack foods is working correctly. The inspector will randomly select a sample of packages and
weigh the amount of snack food in each. Assume that the weights of food in packages filled by the
machine have a standard deviation of 0.30 ounce. An estimate of the mean amount of snack food in each
package must be reported with 99.6 percent confidence and a margin of error of no more than 0.12 ounce.
What would be the minimum sample size for the number of packages the inspector must select?
(A) 8
(B) 15
(C) 25
(D) 52
(E) 60
MC VII-15.) [APSTATSMC1997-24] A random sample of costs of repair jobs at a large muffler repair
shop produces a mean of $127.95 and a standard deviation of $24.03. If the size of this sample is 40, which
of the following is an approximate 90 percent confidence interval for the average cost of a repair at this
repair shop?
(A) $127.95 $4.87
(B) $127.95 $6.25
(C) $127.95 $7.45
(D) $127.95 $30.81
(E) $127.95 $39.53
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MC VII-16.) [APSTATSMC2002-37] A simple random sample procedure produces a sample mean, x , of
15. A 95 percent confidence interval for the corresponding population mean is 15 3. Which of the
following statements must be true?
(A) Ninety-five percent of the population measurements fall between 12 and 18.
(B) Ninety-five percent of the sample measurements fall between 12 and 18.
(C) If 100 samples were taken, 95 of the sample means would fall between 12 and 18.
(D) P( 12 x 18 ) = 0.95
(E) If = 19, this x of 15 would be unlikely to occur.
MC VII-17.) [APSTATS2002-33] An engineer for the Allied Steel Company has the responsibility of
estimating the mean carbon content of a particular day's steel output, using a random sample of 15 rods
from that day's output. The actual population distribution of carbon content is not known to be normal, but
graphic displays of the engineer's sample results indicate that the assumption of normality is not
unreasonable. The process is newly developed, and there are no historical data on the variability of the
process. In estimating this day's mean carbon content, the primary reason the engineer should use a t-
confidence interval rather than a z-confidence interval is because the engineer
(A) is estimating the population mean using the sample mean.
(B) is using the sample variance as an estimate of the population variance.
(C) is using data, rather than theory, to judge that the carbon content is normal.
(D) is using data from a specific day only.
(E) has a small sample, and a z-confidence interval should never be used with a small sample.
MC VII-18.) [APSTATSMC2002-13] A random sample has been taken from a population. A statistician,
using this sample, needs to decide whether to construct a 90 percent confidence interval for the population
mean or a 95 percent confidence interval for the population mean. How will these intervals differ?
(A) The 90% confidence interval will not be as wide as the 95 percent conf. Interval.
(B) The 90% confidence interval will be wider than the 95 percent conf. interval.
(C) Which interval is wider will depend on how large the sample is.
(D) Which interval is wider will depend on whether the sample is unbiased.
(E) Which interval is wider will depend on whether a z-statistic or a t-statistic is used.
MC VII-19.) [APSTATSMC2002-30] The population {2, 3, 5, 7} has mean = 4.25 and standard
deviation = 1.92. When sampling with replacement, there are 16 different possible ordered samples of
size 2 that can be selected from this population. The mean of each of these 16 samples is computed. For
example, 1 of the 16 samples is (2, 5), which has a mean of 3.5. The distribution of the 16 sample means
has its own mean x and its own standard deviation x . Which of the following statements is true?
(A) x = 4.25 and x = 1.92
(B) x = 4.25 and x > 1.92
(C) x = 4.25 and x < 1.92
(D) x > 4.25
(E) x < 4.25
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MC VII-20.) [APSTATSMC2002-40] A student working on a history project decided to find a 95 percent
confidence interval for the difference in mean age at the time of election to office for former American
Presidents versus former British Prime Ministers. The student found the ages at the time of election to
office for the members of both groups, which included all of the American Presidents and all of the British
Prime Ministers, and used a calculator to find the 95 percent confidence interval based on the t-distribution.
This procedure is not appropriate in this context because
(A) the sample sizes for the two groups are not equal.
(B) the entire population was measured in both cases, so the actual difference in means can be computed
and a confidence interval should not be used.
(C) election to office take place at different intervals in the two countries, so the distribution of ages
cannot be the same.
(D) ages at the time of election to office are likely to be skewed rather than bell-shaped, so the
assumptions for using this confidence interval formula are not valid.
(E) ages at the time of election to office are likely to have a few large outliers, so the assumptions for
using this confidence interval formula are not valid.
MC VII-21.) [APSTATSMC2012-22] A random sample of 50 students at a large high school resulted in a
95 percent confidence interval for the mean number of hours of sleep per day of (6.73, 7.67). Which of the
following statements best summarizes the meaning of this confidence interval?
(A) About 95% of all random samples of 50 students from this population would result in a 95%
confidence interval (6.73, 7.67).
(B) About 95% of all random samples of 50 students from this population would result in a 95%
confidence interval that covered the population mean number of hours of sleep per day.
(C) 95% of the students in the survey reported sleeping between 6.73 and 7.67 hours per day.
(D) 95% of the students in this school sleep between 6.73 and 7.67 hours per day.
(E) A student selected at random from this population sleeps between 6.73 and 7.67 hours per day for 95%
of the time.
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FRQ VII-1.) [APSTATSFRQ2016-05] A polling agency showed the following two statements to a
random sample of 1048 adults in the United States.
The order in which the statements were shown was randomly selected for each person in the sample. After
reading the statements, each person was asked to choose the statement that was most consistent with his or
her opinion. The results are shown in the table.
(a) Assume the conditions for inference have been met. Construct and interpret a 95 percent confidence
interval for the proportion of all adults in the United States who would have chosen in the economy
statement.
(b) One of the conditions for inference that was met is that the number who chose the economy statement
and the number who did not choose the economy statement are both greater than 10. Explain why it is
necessary to satisfy that condition.
(c) A suggestion was made to use a two-sample z-interval for a difference between proportions to
investigate whether the difference in proportions between adults in the United States who would have
chosen the environment statement and the adults in the United States who would have chosen the economy
statement is statistically significant.
Is the two-sample z-interval for a difference between proportions an appropriate procedure to investigate
the difference? Justify your answer.
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FRQ VII-2.) [AP STATSFRQ2011-06] Every year, each student in a nationally representative sample is
given tests in various subjects. Recently, a random sample of 9,600 12th
-grade students from US were
administered a multiple-choice US history exam. One of the multiple-choice questions is below. (The
correct answer is C.)
Of the 9,600 students, 28 percent answered the multiple-choice question correctly.
a.). Let p be the proportion of all United States twelfth-grade students who would answer the question
correctly. Construct and interpret a 99 percent confidence interval for p.
Assume that students who actually know the correct answer have a 100 percent chance of answering the
question correctly, and students who do not know the correct answer to the question guess completely at
random from among the four options. Let k represent the proportion of all United States twelfth-grade
students who actually know the correct answer to the question.
b.) A tree diagram of the possible outcomes for a randomly selected twelfth-grade student is provided
below. Write the correct probability in each of the five empty boxes. Some of the probabilities may be
expressions in terms of k.
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c.) Based on the completed tree diagram, express the probability, in terms of k, that a randomly selected
twelfth-grade student would correctly answer the history question.
d.) Using your interval from part (a) and your answer to part (c), calculate and interpret a 99 percent
confidence interval for k, the proportion of all United States twelfth-grade students who actually know the
answer to the history question. You may assume that the conditions for inference for the confidence
interval have been checked and verified.
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FRQ VII-3.) [APSTATSFRQ2015-02] To increase business, the owner of a restaurant is running a
promotion in which a customer’s bill can be randomly selected to receive a discount. When a customer’s
bill is printed, a program in the cash register randomly determines whether the customer will receive a
discount on the bill. The program was written to generate a discount with a probability of 0.2, that is, giving
20 percent of the bills a discount in the long run. However, the owner is concerned that the program has a
mistake that results in the program not generating the intended long-run proportion of 0.2.
The owner selected a random sample of bills and found that only 15 percent of them received discounts. A
confidence interval for p, the proportion of bills that will receive a discount in the long run, is
0.15 0.06 . All conditions for inference were met.
a.). Consider the confidence interval 0.15 0.06 .
i. Does the confidence interval provide convincing statistical evidence that the program is not
working as intended? Justify your answer.
ii. Does the confidence interval provide convincing statistical evidence that the program generates
the discount with a probability of 0.2? Justify your answer.
A second random sample of bills was taken that was four times the size of the original sample. In the
second sample, 15 percent of the bills received the discount.
b.) Determine the value of the margin of error based on the second sample of bills that would be used to
compute an interval for p with the same confidence level as that of the original interval.
c) Based on the margin of error in part (b) that was obtained from the second sample, what do you conclude
about whether the program is working as intended? Justify your answer.
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FRQ VII-4.) [2013APSTATSFRQ2013, #1] An environmental group conducted a study to determine
whether crows in a certain region were ingesting food containing unhealthy levels of lead. A biologist
classified lead levels greater than 6.0 parts per million (ppm) as unhealthy. The lead levels of a random
sample of 23 crows in the region were measured and recorded. The data are shown in the stemplot below.
a.) What proportion of crows in the sample had lead levels that are classified by the biologist as unhealthy?
b.) The mean lead level of the 23 crows in the sample was 4.90 ppm and the standard deviation was 1.12
ppm. Construct and interpret a 95 percent confidence interval for the mean lead level of crows in the region.
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VIII. HYPOTHESIS TESTS
MC VIII-1.) [APSTATSMC2012-27] A manufacturer claims its Brand A battery lasts longer than its
competitor’s Brand B battery. Nine batteries of each brand are tested independently, and the hours of
battery life are shown in the table below
Brand A 88 85 80 81 72 90 85 85 84
Brand B 80 79 77 82 75 81 77 73 78
Provided that the assumptions for inference are met, which of the following tests should be conducted to
determine if Brand A batteries do, in fact, last longer than Brand B batteries?
(A) A one-sided, paired t-test
(B) A one-sided, two-sample t-test
(C) A two-sided, two-sample t-test
(D) A one-sided, two-sample z-test
(E) A two-sided, two-sample z-test
MC VIII-2.) [APSTATSMC2012-29] A randomized experiment was performed to determine whether two
fertilizers, A and B, give different yields of tomatoes. A total of 33 tomato plants were grown; 16 using
fertilizer A, and 17 using fertilizer B. The distributions of the data did not show marked skewness and there
were no outliers in either data set. The results of the experiment are shown below.
Which of the following statements best describes the conclusion that can be drawn from this experiment?
(A) There is no statistical evidence of difference in the yields between fertilizer A and
fertilizer B (p > 0.15).
(B) There is a borderline statistically significant difference in the yields between fertilizer A and
fertilizer B (0.10 < p < 0.15).
(C) There is evidence of a statistically significant difference in the yields between fertilizer A and
fertilizer B (0.05 < p < 0.10).
(D) There is evidence of a statistically significant difference in the yields between fertilizer A and
fertilizer B (0.01 < p < 0.05).
(E) There is evidence of a statistically significant difference in the yields between fertilizer A and
fertilizer B (p < 0.01).
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MC VIII-3.) [APSTATSMC2012-35] Perchlorate is a chemical used in rocket fuel. People who live near a
former rocket-testing site are concerned that perchlorate is present in unsafe amounts in their drinking
water. Drinking water is considered safe when the average level of perchlorate is 24.5 parts per billion
(ppb) or less. A random sample of 28 water sources in this area produces a mean perchlorate measure of
25.3 ppb. Which of the following is an appropriate alternative hypothesis that addresses their concern?
MC VIII-4.) [APSTATSMC2012-38] A physician believes that the exercise habits of East Coast adults are
different from the exercise habits of West Coast adults. To study this, she gathers information on the
number of hours of exercise per week from a random sample of East Coast adults and a random sample of
West Coast adults. Which of the following might be an appropriate null hypothesis for this study?
(A) The average number of hours of exercise per week for East Coast adults is different from the average
number of hours of exercise per week for West Coast adults.
(B) The average number of hours of exercise per week for East Coast adults is the same as the average
number of hours of exercise per week for West Coast adults.
(C) The average number of hours of exercise per week for East Coast adults is greater than the average
number of hours of exercise per week for West Coast adults.
(D) The average number of hours of exercise per week for East Coast adults is less than the average
number of hours of exercise per week for West Coast adults.
(E) The probability is 0.5 that an East Coast adult and a West Coast adult exercise an equal number of
hours per week.
MC VIII-5.) [APSTATSMC2012-28] An experimenter conducted a two-tailed hypothesis test on a set of
data and obtained a p-value of 0.44. If the experimenter had conducted a one-tailed test on the same set of
data, which of the following is true about the possible p-value(s) that the experimenter could have
obtained?
(A) The only possible p-value is 0.22.
(B) The only possible p-value is 0.44.
(C) The only possible p-value is 0.88.
(D) The possible p-values are 0.22 and 0.78.
(E) The possible p-values are 0.22 and 0.88.
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MC VIII-6.) [APSTATSMC2007-5] A safety group claims that the mean speed of drivers on a highway
exceeds the posted speed limit of 65 miles per hour (mph). To investigate the safety group’s claim, which
of the following statements is appropriate?
(A) The null hypothesis is that the mean speed of drivers on this highway is less than 65 mph.
(B) The null hypothesis is that the mean speed of drivers on this highway is greater than 65 mph.
(C) The alternative hypothesis is that the mean speed of drivers on this highway is greater than 65 mph.
(D) The alternative hypothesis is that the mean speed of drivers on this highway is less than 65 mph.
(E) The alternative hypothesis is that the mean speed of drivers on this highway is greater than or equal to
65 mph.
MC VIII-7.) [APSTATSMC2007-13]
MC VIII-8.) [APSTATSMC2007-25] When performing a test of significance about a population mean, a t-
distribution, instead of a normal distribution, is often utilized. Which of the following is the most
approximate explanation for this?
(A) The sample size is not large enough to assume that the population distribution is normal.
(B) The sample does not follow a normal distribution
(C) There is an increase in variability of the test statistic due to estimation of the population standard
deviation.
(D) The sample standard deviation is unknown.
(E) The population standard deviation is too large.
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MC VIII-9.) [APSTATSMC2007-27]
MC VIII-10.) [APSTATSMC2007-37]
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MC VIII-11.) [APSTATSMC1997-6] Which of the following is a criterion for choosing a t-test rather than
a z-test when making an inference about the mean of a population?
(A) The standard deviation of the population is unknown.
(B) The mean of the population is unknown.
(C) The sample may not have been a simple random sample.
(D) The population is not normally distributed.
(E) The sample size is less than 100.
MC VIII-12.) [APSTATSMC1997-2]
MC VIII-13.) [APSTATSMC1997-29]
MC VIII-14.) [APSTATSMC1997-34]
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MC VIII-15.) [APSTATSMC2002-2] A manufacturer of balloons claims that p, the proportion of its
balloons that burst when inflated to a diameter of up to 12 inches, is no more than 0.05. Some customers
have complained that the balloons are bursting more frequently. If the customers want to conduct an
experiment to test the manufacturer's claim, which of the following hypotheses would be appropriate?
(A) Ho: p 0.05, Ha: p = 0.05
(B) Ho: p = 0.05, Ha: p > 0.05
(C) Ho: p = 0.05, Ha: p 0.05
(D) Ho: p = 0.05, Ha: p < 0.05
(E) Ho: p < 0.05, Ha: p = 0.05
MC VIII-16.) [APSTATSMC2002-12] The manager of a factory wants to compare the mean number of
units assembled per employee in a week for two new assembly techniques. Two hundred employees from
the factory are randomly selected and each is randomly assigned to one of the two techniques. After
teaching 100 employees one technique and 100 employees the other technique, the manager records the
number of units each of the employees assembles in one week. Which of the following would be the most
appropriate inferential statistical test in this situation?
(A) One-sample z-test
(B) Two-sample t-test
(C) Paired t-test
(D) Chi-square goodness-of-fit test
(E) One-sample t-test
MC VIII-17.) [APSTATSMC2002-24] A consulting statistician reported the results from a learning
experiment to a psychologist. The report stated that on one particular phase of the experiment a statistical
test result yielded a p-value of 0.24. Based on this p-value, which of the following conclusions should the
psychologist make?
(A) The test was statistically significant because a p-value of 0.24 is greater than a significance level of
0.05.
(B) The test was statistically significant because p = 1 – 0.24 = 0.76 and this is greater than a significance
level of 0.05.
(C) The test was not statistically significant because 2 times 0.24 = 0.48 and that is less than 0.5.
(D) The test was not statistically significant because, if the null hypothesis is true, one could expect to get
a test statistic at least as extreme as that observed 24% of the time.
(E) The test was not statistically significant because, if the null hypothesis is true, one could expect to get
a test statistic at least as extreme as that observed 76% of the time.
MC VIII-18.) [APSTATSMC2002-39] As lab partners, Sally and Betty collected data for a significance test.
Both calculated the same z-test statistic, but Sally found the results were significant at the = 0.05 level
while Betty found that the results were not. When checking their results, the women found that the only
difference in their work was that Sally used a two-sided test, while Betty used a one-sided test. Which of
the following could have been their test statistic?
(A) 1.980
(B) 1.690
(C) 1.340
(D) 1.690
(E) 1.780
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MC VIII-19.) [CollegeBoardAPStatsProblem] In a test of 0H : μ = 8 versus aH : μ ≠ 8, a sample of size
220 leads to a p-value of 0.034. Which of the following must be true?
(A) A 95% confidence interval for μ calculated from these data will not include μ = 8.
(B) At the 5% level if 0H is rejected, the probability of a Type II error is 0.034.
(C) The 95% confidence interval for μ calculated from these data will be centered at μ = 8.
(D) The null hypothesis should not be rejected at the 5% level.
(E) The sample size is insufficient to draw a conclusion with 95% confidence.
MC VIII-20.) [APSTATSMC2013-37] A university will add fruit juice vending machine to its classroom
building if the student body president is convinced that more than 20 percent of the students will use them.
A random sample of n students will be selected and asked whether or not they would use the wending
machines. A large-sample test for proportions at the significance level of 05.0 will be performed. The
null hypothesis that the proportion of all students who would use the wending machines is 20 percent will
be tested against the alternative that more than 20 percent of all students would use them. For which of the
following situations would the power of the test be highest?
(A) The sample size is n = 750, and 20 percent of all students use the wending machines.
(B) The sample size is n = 750, and 25 percent of all students use the wending machines.
(C) The sample size is n = 1000, and 25 percent of all students use the wending machines.
(D) The sample size is n = 500, and 50 percent of all students use the wending machines.
(E) The sample size is n = 1000, and 50 percent of all students use the wending machines.
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Answers
Part 1 – Review
VI. DISTRIBUTION
MC VI-1.) 42.7799.221
7.10452.1
x
x
MC VI-2.) D. The critical z value for 33% of the area is 44.0 . ( 44.0%)331(* invNormz )
MC VI-3.) 324 pounds
MC VI-4.) 0.16. 15
3035
z , 16.0)1( Z .
MC VI-5.) A. Let X be random variable of individual weight. Then, 190x , 10x . If XY 4 ,
then 760)190(4 y , 204 xy ,
220
760800
yz , 023.0)2()( ZzZp y .
FRQ VI-1.)
a.)
x $2 $1 -$8
( )p x 6 /10 3 /10 1/10
b.)
6 3 1 7[ ] 2 1 8
10 10 10 10E X
c.) Let 1 2 ... nY X X X , where n is the number of gamed played. Then,
1 2
7[ ] [ ... ] [ ] 500 714.286 715
10nE Y E X X X nE X n n
d.)
700, 92.79, 500
500 7002.155
92.79
( 2.155) 98.44%
x
z
P Z
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VII. CONFIDENCE INTERVAL
MC VII-1.) D. The proportion is asumed to be 0.5p . The error term (width) of the confidence interval
is calculated by
2
(1 )p pz
n
. So, when the sample size is increasing, the error term will be
decreasing. That is, the CI is narrowing when sample size is increasing.
MC VII-2.) D. 100
ˆ 0.231432
p
MC VII-3.) Since the proportion is normalized to 100%, so the question indicates that
(1 )1.64 6%
p p
n
without given what p is. The quadratic function ( ) (1 )f p p p reaches
maximum when 0.5p , therefore, for all 0 1p , (0.5) 0.5(1 0.5) 0.25f is maximum
value. That is 0.25
1.64 6% 187nn
.
MC VII-4.) C. 0.048 0.112
ˆ 0.082
p
MC VII-5.) D. ˆ 0.79p , 0.02 ,
2
0.022.33
2z invnorm
MC VII-6.) D. 4
100 4 10100
np
is too small.
MC VII-7.) C. Assume 0.5p , 0.5(1 0.5)
1.96 0.08 150.06nn
.
MC VII-8.) A.
MC VII-9.) C.
2
0.34(1 0.34)ˆ 0.34, 1.64, 100 0.34 1.64 0.34 0.078
100p z n
.
MC VII-10.) E
MC VII-11.)
2 2
2
(1 ) 0.6(1 0.6)2.7% 2.7%
900
1.6534 0.049 0.12
p pz z
n
z
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The CI is 90.0%. It is assumed that all the conditions are satisfied.
MC VII-12.) The variance is known.
2
7,200, 0.05, 1.96,z
2
200zn
2 2
2
1.96 72004979
200 200n z
MC VII-13.) A. 0.025
10 3.0125.5, 11, 10, 2.28, 25.5 2.28
11x n k df t
MC VII-14.) D. Assume that the conditions to use normal distribution are satisfied.
0.0020.3, 0.004, 2.878
0.302.878 0.12 51.7
z
nn
MC VII-15.) B. 24.03
1.64 6.2540
MC VII-16.) E. The population mean is out of CI.
MC VII-17.) B. The variance is unknown.
MC VII-18.) A. The lesser confidence, the narrower the CI.
MC VII-19.) C. The sample size is larger, so the sample variance is smaller.
MC VII-20.) B. The means calculated were the population means -- no need to estimate the mean.
MC VII-21.) B.
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FRQ VII-1.)
a.) ˆ
2
ˆ ˆ 0.37(1 0.37)ˆ1048, 0.37, 0.05, 1.96, 0.0149
1048p
pqn p z
n
0.37 1.96(0.0149) 0.37 0.03
(0.34, 0.40)
CI
CI
95% of chance that the interval contains the population proportion of selecting Economy Statement.
b.) One of the conditions is 10, 10np nq . Since p or q are less than one, the sample size should be
at least 10.
c.) No, the two samplings should be independent.
FRQ VII-2.)
a.) ˆ 0.280p , ˆ 9600(0.280) 2688 10np , ˆ(1 ) 9600(1 0.280) 6912 10n p ,
0.01 , 0.005
2
2.576z z , ˆ ˆ(1 )p p
n
.
2
ˆ ˆ(1 )ˆ
0.280(1 0.280)0.280 2.576
9600
0.28 0.012
(0.268, 0.292)
p pp z
n
The CI indicates that 99% confidence that the population proportion is between 0.268 and 0.292. That is,
we are 99 percent confident that the interval from 0.268 to 0.292 contains the population proportion of all
United States twelfth-grade students who would answer this question correctly.
b.)
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c.)
( _ )
( _ _ ) ( _ _ )
0.25(1 )
0.25 0.75
P Answer Correctly
P Know the Answers P Guess the Answers
k k
k
d.) Since 0.25 0.75p k , then
0.268 0.292
0.268 0.25 0.75 0.292
0.024 0.056
p
k
k
We are 99 percent confident that the interval from 0.024 to 0.056 contains the proportion of all United
States twelfth-grade students who actually know the answer to the history question.
FRQ VII-3.)
a.)
i. No. The assumed proportion is 0.2, and it is within the CI. So, there is no statistical evidence to
claim that the program is not working.
ii. No. Any number within CI could be the probability.
b.) 0.06
0.034
.
c.) Now the CI is 0.15 0.03 , so 0.2 is not within the CI. So, there is convincing evidence that the
program is not working.
FRQ VII-4.)
a.) 4
ˆ 0.17423
p
b.) 4.90x , 1.12xs , 23n 23 1 22df , 22
0.025
2
2.069dft t
1
2
1.124.90 2.069 4.90 0.483
23
(4.417, 5.383)
n xsx t
n
CI
We can be 95% confident that the population mean lead level among all crows in this region is between
4.416 and 5.384 parts per million.
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VIII. HYPOTHESIS TESTS
MC VIII-1.) B. Sample size is small, standard deviation is unknown and it is to compare greatness of two
means (one-sided).
MC VIII-2.) D. Assume that the standard deviations are equal. For the sample size and unknown standard
deviation, use the two-sample mean t test. Use Ti-84 Stats->TESTS->2-SampTTest. Or
1 1 116, 19.54, 3.68n x s , 2 2 217, 23.39, 4.93n x s
The degrees of freedom 16 17 2 31df .
2 2 2 2
1 1 2 2
1 2
( 1) ( 1) (16 1)(3.68) (17 1)(4.93)4.37
2 16 17 2pooled
n s n ss
n n
1 2
1 1 1 14.37 1.522
16 17x y pooleds
n n
1 2 19.54 23.392.53
1.522x y
x xt
, 2Pr | 2.53 |, 31 0.0167valuep T df
MC VIII-3.) D. 28, 25.3n x . Assume that 0 : 24.5H , : 24.5aH for unsafe level.
MC VIII-4.) B. The null hypothesis should be that there is no difference between the averages numbers of
exercise habits per week from both East Coast adults and West Coast adults.
MC VIII-5.) D. The probability on each outside the CI is 0.22. So depending on aH testing greater or less
than, the valuep could be 0.22 or 1 0.22 0.78 .
MC VIII-6.) C.
MC VIII-7.) D. The degrees of freedom are 1 2 2 638df n n ,
2Pr 3.27, 638 0.0011valuep T df
MC VIII-8.) C. xsn
is too large when n is small. Choice A is incorrect because the distribution of
population does not need to be normally distributed, the sample mean x will approach to normally
distributed when the sample size is large.
MC VIII-9.) B. To not include zero, or to reject the null hypothesis, 0.056valueP , regardless if it is
a two-sided, the confidence level should be 1 1 0.056 0.944 , and hence 94% is the answer.
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MC VIII-10.) A. 0.076 0.05valuep .
MC VIII-11.) A. The reasons are either the population standard deviation is unknown or the sample size is
too small.
MC VIII-12.) D.
MC VIII-13.) A. 0.017 0.02valuep . With at least 98% confidence, the test is significant. Choice
B is incorrect because only 1.7% of time, sample produces a mean larger than 10. Choice C is incorrect
because 1.7% of time, rejecting the null hypothesis is in error.
MC VIII-14.) A.
194.3, 200, 21, 70
194.3 200* 2.27
21/ 70
Pr( *) Pr( 2.27, 69) 0.013value
x s n
xt
s
n
p T t T df
MC VIII-15.) B.
MC VIII-16.) B. The two-sample mean test when the variance is unknown.
MC VIII-17.) D.
MC VIII-18.) A. Sally used 2-sided test, so 2Pr( 1.98) 0.047valuep Z .
MC VIII-19.) The answer is A. With 95% CI which is centered at the sample mean or sample proportion,
034.0%5 , the p-value, implies that μ = 8 should not be included in the CI. Choice B is
incorrect because is not given, only 0.034valueP is given. Choice C is incorrect the confidence
interval will be centered at the sample mean or sample proportion. Choice D is incorrect since the p-value
is less than 5%, the null hypothesis will be rejected. E is incorrect because the sample size is greater than
30, so it large enough.
MC VIII-20.) The answer is E. Increasing sample size reduces sample standard deviation to make the
curves narrower. Further move the aH away from 0H will reduce the overlapping of the two distributions.
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Part 2 -- Quarterly Exam Questions
VI. DISTRIBUTION (Cont. in ZSCORES)
MC VI-01.) [APSTATSMC1997-12] The heights of adult women are approximately normally distributed
about a mean of 65 inches with a standard deviation of 2 inches. If Rachael is at the 99th
percentile in height
for adult women, then her height, in inches, is closest to
(A) 60 inches
(B) 62 inches
(C) 68 inches
(D) 70 inches
(E) 74 inches
MC VI-02.) [APSTATSMC1997-17] Gina’s doctor told her that the standardized score (z-score) for her
systolic blood pressure, as compared to the blood pressure of other women her age, is 1.50. Which of the
following is the best interpretation of this standardized score?
(A) Gina’s systolic blood pressure is 150.
(B) Gina’s systolic blood pressure is 1.50 standard deviation about the average systolic blood pressure of
women her age.
(C) Gina’s systolic blood pressure is 1.50 above the average systolic blood pressure of women her age.
(D) Gina’s systolic blood pressure is 1.50 times the average systolic blood pressure for women her age.
(E) Only 1.5% of women Gina’s age have a higher systolic blood pressure than she does.
MC VI-03.) [APSTATSMC2002-10] The lengths of individual shellfish in a population of 10,000 shellfish
are approximately normally distributed with mean 10 centimeters and standard deviation 0.2 centimeter.
Which of the following is the shortest interval that contains approximately 4,000 shellfish lengths?
MC VI-04.) [APSTATSMC1997-25] At a college the scores on the chemistry final exam are approximately
distributed, with a mean of 75 and standard deviation of 12. The scores on the calculus final are also
approximately normally distributed, with a mean of 80 and a standard deviation of 8. A student scored 81
on the chemistry final and 84 on the calculus final. Relative to the students in each respective class, in
which subject did this student do better?
(A) The student did better in chemistry.
(B) The student did better in calculus.
(C) The student did equally well in each course.
(D) There is no basis for comparison, since the subjects are different form each other and are in different
departments.
(E) There is not enough information for comparison, because the number of students in each class is not
known.
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MC VI-05.) [APSTATSMC2002-7] Suppose that the distribution of a set of scores has a mean of 47 and a
standard deviation of 14.
If 4 is added to each score, what will be the mean and the standard deviation of the distribution of
new scores?
Mean Standard Deviation
(A) 51 14
(B) 51 18
(C) 47 14
(D) 47 16
(E) 47 18
MC VI-06.) [APSTATSMC2007-3] The distribution of the diameters of a particular variety of oranges is
approximately normal with a standard deviation of 0.3 inch. How does the diameter of an orange at the
67th percentile compare with the mean diameter?
(A) 0.201 inch below the mean
(B) 0.132 inch below the mean
(C) 0.132 inch above the mean
(D) 0.201 inch above the mean
(E) 0.440 inch above the mean
MC VI-07.) [APSTATSMC2007-22] The weights of a population of adult male gray whales are
approximately normally distributed with a mean weight of 18,000 kilograms and a standard deviation of
4,000 kilograms. The weights of a population of adult male humpback whales are approximately normally
distributed with a mean weight of 30,000 kilograms and a standard deviation of 6,000 kilograms. A certain
adult male gray whale weighs 24,000 kilograms. This whale would have the same standardized weight (z-
score) as an adult male humpback whale whose weight, in kilograms, is equal to which of the following?
(A) 21,000
(B) 24,000
(C) 30,000
(D) 36,000
(E) 39,000
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MC VI-08.) [APSTATSMC2015-9]
MC VI-09.) [APSTATSMC2015-31]
MC VI-10.) [APSTATSMC1997-20M] If a customer rolls the dice and rents a second movies every
Thursday for 30 consecutive weeks, what is the approximate probability that the total amount paid for these
second movies will exceed $15.00? Assume that X, amount paid for each of the second movie, is normally
distributed, and the mean (expected value) amount paid is $0.47 and standard deviation is $0.15. (The sum
of normally distributed random variables is also normally distributed.)
(A) 0
(B) 0.09
(C) 0.14
(D) 0.86
(E) 0.91
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MC VI-11.) [APSTATSMC2002-38] Suppose that public opinion in a large city is 35 percent against
increasing taxes to support the public school system. If random samples of 500 people from this city are
interviewed, what is the approximate probability that more than 200 of these people will be against
increasing taxes? Which of the following set-ups would answer the question?
(A) 200 300500
0.65 0.35200
(B) 200 300500
0.35 0.65200
(C)
0.40 0.65
0.65 0.35
500
P z
(D)
0.40 0.35
0.40 0.60
500
P z
(E)
0.40 0.35
0.35 0.65
500
P z
MC VI-12.) 10.) [APSTATSMC2007-30]
MC VI-13.) [APSTATSMC2007-39]
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MC VI-14.) [APSTATSMC1997-16]
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Answers
MC VI-01.) D
MC VI-02.) B
MC VI-03.)
10, 0.2x . The normal curve to cover middle 4000/10000 should set the critical z-scores to be
0.524 ( (0.3) 0.524invNorm ), and hence
* 10 0.524(0.2)
(9.895,10.105)
CI x z
CI
.
MC VI-04.) C.
MC VI-05.) A.
MC VI-06.) C.
MC VI-07.) E.
MC VI-08.) C.
MC VI-09.) D.
MC VI-10.) C. Let Y be the amount paid for 30 consecutive weeks, then
30n , [ ] 30 [ ] $0.47(39) $14.10E Y E X , 30 0.82158Y X ,
Pr( 15) (15,1000,14.10,0.82158) 0.136659X normalcdf
MC VI-11.) E.
200ˆ 0.4, 0.35, 500
500
0.40 0.35* ( *)
0.35 0.65
500
p p n
z and p z z
MC VI-12.) B. The variability is reduced by one-half : 1 1
2 2200 50new old
.
Hence, the CI is also reduced by one-half: 1 1
, ,2 2
new new old oldz z z z
.
MC VI-13.) A.
MC VI-14.) E.
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MC VIII-01. [APSTATSMC2002-35] In a test of the hypothesis Ho: = 100 versus Ha:
> 100, the power of the test when = 101 would be greatest for which of the following
choices of sample size n and significance level ?
(A) n = 10, = 0.05
(B) n = 10, = 0.01
(C) n = 20, = 0.05
(D) n = 20, = 0.01
(E) It cannot be determined from the information given.
Solution: C. Increase sample size and increase
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